Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 16 Inheritance Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 16 Inheritance

11th Computer Science Guide Inheritance Text Book Questions and Answers

Part I

Choose The Correct Answer:

Question 1.
Which of the following is the process of creating new classes from an existing class?
a) Polymorphism
b) Inheritance
c) Encapsulation
d) superclass
Answer:
b) Inheritance

Question 2.
Which of the following derives a dass student from the base class school?
a) school: student
b) class student: public school
c) student: public school
d) class school: public student
Answer:
b) class student: public school

Question 3.
The type of inheritance that reflects the transitive nature is
a) Single Inheritance
b) Multiple Inheritance
c) Multilevel Inheritance
d) Hybrid Inheritance
Answer:
c) Multilevel Inheritance

Question 4.
Which visibility mode should be used when you want the features of the base class to be available to the derived class but not to the classes that are derived from the derived dass?
a) Private
b) Public
c) Protected
d) All of these
Answer:
a) Private

Question 5.
Inheritance is the process of creating new class from
a) Base class
b) abstract
c) derived class
d) Function
Answer:
a) Base class

Question 6.
A class is derived from a class which is a derived class itself, then this is referred to as
a) multiple inheritances
b) multilevel inheritance
c) single inheritance
d) double inheritance
Answer:
b) multilevel inheritance

Question 7.
Which amongst the following is executed in the order of inheritance?
a) Destructor
b) Member function
c) Constructor
d) Object
Answer:
b) Member function

Question 8.
Which of the following is true with respect to inheritance?
a) Private members of base class are inherited to the derived class with private
b) Private members of base class are not inherited to the derived class with private accessibility
c) Public members of base class are inherited but not visible to the derived class
d) Protected members of base class are inherited but not visible to the outside class
Answer:
b) Private members of base class are not inherited to the derived class with private accessibility

Question 9.
Based on the following dass decoration answer the questions (from 9.1 o 9.5 )

class vehicle
{
int wheels;
public:
void input_data(float,float);
void output_data();
protected:
int passenger;
};
class heavy_vehicle: protected vehicle
{
int diesel_petrol;
protected:
int load;
protected:
int load;
public:
void read_data(float,float)
void write_data(); };
class bus: private heavy_vehicle
{
char Ticket[20];
public:
void fetch_data(char);
void display_data(); >;
};

Question 9.1.
Which is the base class of the class heavy, vehicle?
a) Bus
b) heavy_vehicle
c) vehicle
d) both (a) and (c)
Answer:
c) vehicle

Question 9.2.
The data member that can be accessed from the function displaydata()
a) passenger
b) load
c) Ticket
d) All of these
Answer:
d) All of these

Question 9.3.
The member function that can be accessed by an objects of bus Class is
a) input_data()
b) read_data() ,output_data()write_data()
c) fetch_data(),display_data()
d) All of these
Answer:
c) fetch_data(),display_data()

Question 9.4.
The member function that is inherited as public by Class Bus
a) input_data()
b) read_data(),output_data(),write_data()
c) fetch_data(), display_data()
d) None of these
Answer:
d) None of these

Question 10.
class x
{
int a;
public :
x()
{}
};
class y
{
x x1;
public:
y()
{}
};
class z : public y,x
{
int b;
public:
z()
{}
}z1;

What is the order of constructor for object z to be invoked?
a) z,y,x,x
b) x,y,z,x
c) y,x,x,z
d) x,y,z
e) x,y,x,z
Answer:
e) x,y,x,z

Part – II

Very Short Answers

Question 1.
What is inheritance?
Answer:
Inheritance is one of the most important features of Object-Oriented Programming. In object-oriented programming, inheritance enables a new class and its objects to take on the properties of the existing classes.

Question 2.
What is a base class?
Answer:
The class to be inherited is called a base class or parent class.

Question 3.
Why derived class is called a power-packed class?
Answer:

• Multilevel Inheritance: In multilevel inheritance, the constructors will be executed in the order of inheritance.
• Multiple Inheritance: If there are multiple base classes, then it starts executing from the leftmost base class.

Question 4.
In what multilevel and multiple inheritances differ though both contains many base class?
Answer:
In case of multiple inheritance derived class have more than one base classes (More than one parent). But in multilevel inheritance derived class have only one base class (Only one parent).

Question 5.
What is the difference between public and private visibility mode?
Answer:
Private visibility mode:
When a base class is inherited with private visibility mode the public and protected members of the base class become ‘private’ members of the derived class.

Public visibility mode:
When a base class is inherited with public visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.

Part – III

Short Answers

Question 1.
What are the points to be noted while deriving a new class?
Answer:

The following points should be observed for defining the derived class:

1. The keyword class has to be used.
2. The name of the derived class is to be given after the keyword class.
3. A single colon.
4. The type of derivation (the visibility mode), namely private, public or protected. If no visibility mode is specified, then by default the visibility mode is considered private.
5. The names of all base classes (parent classes) separated by a comma.

class derivedclass_name :visibility_mode
base_class_name
{
// members of derived class
};

Question 2.
What is differences between the members present in the private visibility mode and the members present in the public visibility mode?
Answer:
Private visibility mode:
When a base class is inherited with private visibility mode the public and protected members of the base class become ‘private’ members of the derived class.

Private visibility members can not be inherited further. So, it can not be directly accessed by its derived classes.
Public visibility mode:
When a base class is inherited with public visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.

Public visibility members can be inherited by its child and can access in it.

Question 3.
What is the difference between polymorphism and inheritance though are used for the reusability of code?
Answer:
Polymorphism:

• Reusability of code is implemented through functions (or) methods.
• Polymorphism is the ability of a function to respond differently to different messages.
• Polymorphism is achieved through overloading.

Inheritance:

• Reusability of code is implemented through classes.
• Inheritance is the process of creating derived classes from the base class or classes.
• Inheritance is achieved by various types of inheritances namely single, multiple, multilevel, hybrid and hierarchical inheritances.

Question 4.
What do you mean by overriding?
Answer:
When a derived class member function has the same name as that of its base class member function, the derived class member function shadows/hides the base class’s inherited function. This situation is called function overriding and this can be resolved by giving the base class name followed by :: and the member function name.

Question 5.
Write some facts about the execution of constructors and destructors in inheritance.
Answer:

1. Base class constructors are executed first, before the derived class constructors execution.
2. Derived class cannot inherit the base class constructor but it can call the base class constructor by using Base_class name: :base_class_constructor() in the derived class definition
3. If there are multiple base classes, then it starts executing from the leftmost base class
4. In multilevel inheritance, the constructors will be executed in the order of inheritance The destructors are executed in the reverse order of inheritance.

IV. Explain In Brief (Five Marks)

Question 1.
Explain the different types of inheritance.
Answer:
Types of Inheritance;
There are different types of inheritance viz., Single Inheritance, Multiple inheritance, Multilevel inheritance, hybrid inheritance and hierarchical inheritance.

1. Single Inheritance:
When a derived class inherits only from one base class, it is known as single inheritance.

2. Multiple Inheritance;
When a derived class inherits from multiple base classes it is known as multiple inheritance.

3. Hierarchical inheritance:
When more than one derived classes are created from & single base class known as Hierarchical inheritance.

4. Multilevel Inheritance
The transitive nature of inheritance is itself reflected by this form of inheritance. When a class is derived from a class which is a derived class – then it is referred to as multilevel inheritance.

5. Hybrid inheritance:
When there is a combination of more than one type of inheritance, it is known as hybrid inheritance. Hence, it may be a combination of Multilevel and Multiple inheritances or Hierarchical and Multilevel inheritance or Hierarchical, Multilevel and Multiple inheritances.

Question 2.
Explain the different visibility modes through pictorial representation.
Answer:
Private visibility mode:
When a base class is inherited with private visibility mode the public and protected members of the base class become ‘private’ members of the derived class.

Protected visibility mode:
When a base class is inherited with protected visibility mode the protected and public members of the base class become ‘protected members’ of the derived class.

Public visibility mode:
When a base class is inherited with public visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.

Question 3.
#include<iostream>
#include<string.h>
#include<stdio.h>
using name spacestd;
class publisher
{
char pname[15];
char hoffice[15];
char address[25];
double turnover;
protected:
char phone[3][10];
void register();
public:
publisher();
publisher();
void enter data();
void disp data();
};
class branch
{
char bcity[15];
char baddress[25];
protected:
int no_of_emp;
public:
char bphone[2][10];
branch();
~branch();
void havedata();
void givedata();
};
class author: public branch, publisher
{
int aut_code;
charaname[20];
float income;
public:
author();
~author();
void getdata();
void putdata();
};

Answer The Following Questions Based On The Above Given Program:

3.1. Which type of Inheritance is shown in the program?
3.2. Specify the visibility mode of base classes.
3.3 Give the sequence of Constructor/Destructor Invocation when object of class author is created.
3.4. Name the base class(/es) and derived class (/es).
3.5 Give number of bytes to be occupied by the object of the following class:
(a) publisher
(b) branch
(c) author
3.6. Write the names of data members accessible from the object of class author.
3.7. Write the names of all member functions accessible from the object of class author.
3.8 Write the names of all members accessible from member functions of class author.
Answer:
3.1 Multiple Inheritance
3.2 public
3.3 Constructors branch, publisher and author are executed.
Destructors author, publisher and branch will be executed.
3.4 Base classes : branch and publisher Derived class : author
3.5 a) publisher class object requires 93 bytes
b) branch class object requires 64 bytes
c) author class object requires 181 bytes

Question 4.
Consider the following C++ code and answer the questions.
class Personal
{
int Class, Rno;
char Section;
protected:
char Name[20];
public:
personal();
void pentry();
void Pdisplay();
};
class Marks:private Personal
{
float M{5};
protected:
char Grade[5];
public:
Marks();
void Mentry();
void Mdisplay();
};
class Result:public Marks
{
float Total,Agg;
public:
char FinalGrade, Commence[20];
Result();
void Rcalculate();
void Rdisplay();
};

4.1. Which type of Inheritance is shown in the program?
4.2. Specify the visibility mode of base classes.
4.3 Give the sequence of Constructor/Destructor Invocation when object of class Result is created.
4.4. Name the base class(/es) and derived class (/es).
4.5 Give number of bytes to be occupied by the object of the following class:
(a) Personal
(b) Marks
(c) Result
4.6. Write the names of data members accessible from die object of class Result.
4.7. Write the names of all member functions accessible from the object of class Result.
4.8 Write the names of all members accessible from member functions of class Result.
Answer:
4.1 Multilevel Inheritance
4.2 For Marks class – private visibility
For Result class – public visibility
4.3 Constructors Personal, Marks and Result be executed.
Destructors Result, Marks and Personal will be executed.
4.4 Base classes : Personal and Marks
Derived classes : Marks and Result

4.5 a) Personal class object requires 28 bytes (using Dev C++)
b) Marks class object requires 53 bytes (using Dev C++)
c) Result class requires 82 bytes (using Dev C++)

4.6 Data members FinalGrade, Commence(Own class members) alone can be accessed.
No members inherited under public, so base class members can not be accessed.

4.7 Member functions
Rcalculate( ), Rdisplay (own class member functions)
Mentry, Mdisplay (derived from Marks class) alone can be accessed.
Personal class public member functions can not be accessed because Marks class inherited under private visibility mode.

4.8 1) Data members

• Total, Agg, Final Grade and Commence of its own class
• M, Grade from Marks class can be accessed.

Personal class data members can not be accessed because Marks class inherited under private visibility mode.

2) Member functions
Mentry and Mdisplay from Marks class can be invoked from Result class member
Personal class member-functions can not be accessed because Marks class inherited under private visibility mode.

Question 5.
Write the output of the following program.
#include<iostream>
using namespace std;
class A
{
protected:
int x;
public:
void show()
{
cout<<“x = “<<x<<endl;
}
A()
{
cout<<endl<<” I am class A “<<endl;
}
~A()
{
cout<<endl<<” Bye”;
}
};
class B : public A
{
protected:
int y;
public:
B(int x, int y)
{
//this -> is used to denote the objects datamember this->x = x;
//this -> is used to denote the objects datamember this->y = y;
}
B()
{
cout<<endl<<“I am class B”<<endl;
}
~B()
{
cout<<endl<<” Bye”;
}
void show()
{
cout<<“x = “<<x<<endl;
cout<<“y = “<<y<<endl;
}
};
int main()
{
A objA;
B objB(30, 20);
objB.show();
return 0;
}
Output:

Question 6.
Debug the following program.
Output:
—————
15
14
13

Program :
%include(iostream.h)
#include<conio.h>
Class A
{
public;
int al,a2:a3;
void getdata[]
{
a1=15;
a2=13;a3=13;
}
}
Class B:: public A()
{
PUBLIC
voidfunc()
{
int b1:b2:b3;
A::getdata[];
b1=a1;
b2=a2;
a3=a3;
cout<<b1<<‘\t'<<b2<<‘t\'<<b3;
}
void main()
{
clrscr()
B der;
derl:func();
getch();
}
Answer:
Modified Error Free Program :
using namespace std;
#include<iostream>
#include<conio.h>
class A
{
public:
int a1,a2,a3;
void getdata()
{
a1=15;
a2=14;
a3=13;
}
};
class B : public A
{
public:
void func()
{
int b1,b2,b3;
A::getdata();
b1=a1;
b2=a2;
b3=a3;
cout<<b1<<‘\n'<<b2<<‘\n'<<b3;
}
};
int main()
{
B der;
der.func();
getch();
return 0;
}

11th Computer Science Guide Inheritance Additional Questions and Answers

Choose The Correct Answer:

Question 1.
When a derived class inherits only from one base class, it is known as ………………
(a) multiple inheritances
(b) multilevel inheritance
(c) hierarchical inheritance
(d) single inheritance
Answer:
(d) single inheritance

Question 2.
_____ enables new class and its objects to take on the properties of the existing classes.
a) Inheritance
b) Encapsulation
c) Overriding
d) None of these
Answer:
a) Inheritance

Question 3.
When more than one derived classes are created from a single base class, it is called ………………
(a) inheritance
(b) hybrid inheritance
(c) hierarchical inheritance
(d) multiple inheritances
Answer:
(c) hierarchical inheritance

Question 4.
A class that inherits from a superclass is called a _______ class.
a) Sub
b) Base class
c) Derived
d) Sub or Derived
Answer:
d) Sub or Derived

Question 5.
The ……………… are invoked in reverse order.
(a) constructor
(b) destructor
(c) pointer
(d) operator
Answer:
(b) destructor

Question 6.
There are ________ types of inheritance,
a) Two
b) Three
c) Four
d) Five
Answer:
d) Five

Question 7.
_________ inheritance is a type of inheritance.
a) Single or Hybrid
b) Multilevel / Hierarchical
c) Multiple
d) All the above
Answer:
d) All the above

Question 8.
When a derived class inherits only from one base class, it is known as ________ inheritance.
a) Single
b) Multilevel / Hierarchical
c) Multiple
d) Hybrid
Answer:
a) Single

Question 9.
When a derived class inherits from multiple base classes it is known as _________ inheritance.
a) Single
b) Multilevel / Hierarchical
c) Multiple
d) Hybrid
Answer:
c) Multiple

Question 10.
When more than one derived classes are created from a single base class, it is known
as_________inheritance.
a) Single
b) Hierarchical
c) Multiple
d) Hybrid
Answer:
b) Hierarchical

Question 11.
The transitive nature of inheritance is itself reflected by _____ form of inheritance.
a) Single
b) Multilevel
c) Multiple
d) Hybrid
Answer:
b) Multilevel

Question 12.
When a class is derived from a class which is a derived class – then it is referred to as ______ inheritance.
a) Single
b) Multilevel
c) Multiple
d) Hybrid
Answer:
b) Multilevel

Question 13.
When there is a combination of more than one type of inheritance, it is known as ________ inheritance.
a) Single
b) Multilevel
c) Multiple
d) Hybrid
Answer:
d) Hybrid

Question 14.
Hybrid inheritance may be a combination of ________inheritance.
a) Multilevel and Multiple
b) Hierarchical and Multilevel
c) Hierarchical, Multilevel and Multiple
d) All the above
Answer:
d) All the above

Question 15.
The order of inheritance by derived class, to inherit the base class is ________
a) Left to Right
b) Right to Left
c) Top to Bottom
d) None of these
Answer:
a) Left to Right

Question 16.
In ________ inheritance the base classes dc not have any relationship between them,
a) Single
b) Multilevel
c) Hybrid
d) Multiple
Answer:
d) Multiple

Question 17.
In ________inheritance a derived class itself acts as a base class to derive another class.
a) Single
b) Multilevel
c) Multiple
d) Multiple
Answer:
b) Multilevel

Question 18.
_________ inheritance is similar to relation between grandfather, father and child,
a) Single
b) Multilevel
c) Multiple
d) Multiple
Answer:
b) Multilevel

Question 19.
A class without any declaration will have ________byte size.
a) 1
b) 0
b) 2
d) 10
Answer:
a) 1

Question 20.
class x{}; x occupies
a) 1
b) 0
b) 2
d) 10
Answer:
a) 1

Question 21.
In inheritance, which member of the base class will be acquired by the derived class is done by using ________ .
a) Visibility modes
b) Data members
c) Member functions
d) None of these
Answer:
a) Visibility modes

Question 22.
The accessibility of base class by the derived class is controlled by ________
a) Visibility modes
b) Data members
c) Member functions
d) None of these
Answer:
a) Visibility modes

Question 23.
_______ is a visibility modes.
a) private
b) public
c) protected
d) All the above
Answer:
d) All the above

Question 24.
The default visibility mode is ________
a) private
b) public
c) protected
d) All the above
Answer:
a) private

Question 25.
When a base class is inherited with ________ visibility mode the public and protected members of the base class become ‘private’ members of the derived class.
a) private
b) public
c) protected
d) All the above
Answer:
a) private

Question 26.
When a base class is inherited with ________ visibility mode the protected and public members of the base class become ‘protected members’ of the derived class,
a) private
b) public
c) protected
d) All the above
Answer:
c) protected

Question 27.
When a base class is inherited with ________ visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.
a) private
b) public
c) protected
d) All the above
Answer:
b) public

Question 28.
When classes are inherited with ________the private members of the base class are not inherited they are only visible.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
d) Either A or B or C

Question 29.
When classes are inherited with ________ the private members of the base class are continue to exist in derived classes, and cannot be accessed.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
d) Either A or B or C

Question 30.
________inheritance should be used when you want the features of the base class to be available to the derived class but not to the classes that are derived from the derived class.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
a) private

Question 31.
________ inheritance should be used when features of base class to be available only to the derived class members but not to the outside world.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
c) protected

Question 32.
_____ inheritance can be used when features of base class to be available the derived class members and also to the outside world.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
b) public

Question 33.
When an object of the derived class is created, the compiler first call the ________class constructor.
a) Base
b) Derived
c) Either Base or Derived
d) None of these
Answer:
a) Base

Question 34.
When the object of a derived class expires first the ________class destructor is invoked.
a) Base
b) Derived
c) Either Base or Derived
d) None of these
Answer:
b) Derived

Question 35.
The ________ are executed in the order of inherited class.
a) Constructors
b) Destructors
c) Either A or B
d) None of these
Answer:
a) Constructors

Question 36.
The ________ are executed in the reverse order.
a) Constructors
b) Destructors
c) Either A or B
d) None of these
Answer:
b) Destructors

Question 37.
If there are multiple base classes, then it starts executing from the ________base class.
a) Leftmost
b) Rightmost
c) Compiler decided
d) None of these
Answer:
a) Leftmost

Question 38.
________ members of the base class can be indirectly accessed by the derived class using the public or protected member function of the base class.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
a) private

Question 39.
________ member function has the access privilege for the private members of the base class.
a) public
b) protected
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 40.
________ functions can access the private members.
a) Member
b) Non-member
c) Destructor
d) None of these
Answer:
a) Member

Question 41.
In case of inheritance there are situations where the member function of the base class and derived classes have the same name. The ________operator resolves this problem.
a) Conditional
b) Membership
c) Scope resolution
d) None of these
Answer:
c) Scope resolution

Question 42.
When a derived class member function has the same name as that of its base class member function, the derived class member function ________the base class’s inherited function.
a) Shadows
b) Hides
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 43.
When a derived class member function has the same name as that of its base class member function, the derived class member function shadows/hides the base class’s inherited function is called function ________
a) Overriding
b) Overloading
c) Shadowing
d) Either A or C
Answer:
d) Either A or C

Question 44.
________ pointer is a constant pointer that holds the memory address of the current object.
a) this
b) void
c) new
d) None of these
Answer:
a) this

Question 45.
________pointer is useful when the argument variable name in the member function and the data member name are same.
a) this
b) void
c) new
d) None of these
Answer:
a) this

Very Short Answers (2 Marks)

Question 1.
Write a short note on hierarchical inheritance.
Answer:
When more than one derived class is created from a single base class, it is known as Hierarchical inheritance.

Question 2.
What are the types of inheritance?
Answer:
There are different types of inheritance viz., Single Inheritance, Multiple inheritance, Multilevel inheritance, hybrid inheritance and hierarchical inheritance.

Question 3.
Give the syntax of deriving a class.
Answer:
Syntax:
class derived_dass_name :visibility_mode base_dass_name
{
// members of derivedclass
};

Question 4.
Write note on this pointer.
Answer:
‘this’ pointer is a constant pointer that holds the memory address of the current object. It identifies the currently calling object. It is useful when the argument variable name in the member function and the data member name are same. To identify the data member it will be given as this->data member name.

Short Answers (3 Marks)

Question 1.
What are inheritance and access control?
Answer:
When you declare a derived class, a visibility mode can precede each base class in the base list of the derived class. This does not alter the access attributes of the individual members of a base class but allows the derived class to access the members of a base class with restriction.

Classes can be derived using any of the three visibility modes:

1. In a public base class, public and protected members of the base class remain public and protected members of the derived class.
2. In a protected base class, public and protected members of the base class are protected members of the derived class.
3. In a private base class, public and protected members of the base class become private members of the derived class.
4. In all these cases, private members of the base class remain private and cannot be used by the derived class.
5. However, it can be indirectly accessed by the derived class using the public or protected member function of the base class since they have the access privilege for the private members of the base class.

Question 2.
Write a program for the working of constructors and destructors under inheritance.
Program
#include<iostream>
using namespace std;
class base
{
public:
base()
{
cout<<“\nConstructor of base class…”;
}
~base()
{
cout<<“\nDestructor of base class….”;
}
};
class derived:public base
{
public :
derived()
{
cout << “\nConstructor of derived …”;
}
~derived()
{
cout << “\nDestructor of derived…”;
}
};
class derived 1 :public derived
{
public :
derived 1()
{
cout << “\nConstructor of derived! //, …”;
}
~derived1()
{
cout << “\nDestructor of derived …”;
}
};
int main()
{
derivedl x;
return 0;
}

Output:
Constructor of base class…
Constructor of derived …
Constructor of derived …
Destructor of derived …
Destructor of derived …
Destructor of base class….

Question 3.
What about access control in a publicly derived class?
Answer:
From a publicly derived class, public and protected members of the base class remain public and protected members of the derived class. The public members can be accessed by the object of the derived class similar to its own members in public.

Question 4.
What about access control in the privately derived class?
Answer:
From a privately derived class, public and protected members of the base class become private members of the derived class. Hence it is not possible to access the derived members using the object of the derived class. The Derived members are invoked by calling it from the publicly defined members.

Explain in Detail

Question 1.
Write a program to implement single Inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno;
public:
void acceptnameO
{
cout<<“\n Enter roll no and name ..”;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name :-“<<name<<endl;
}
};
class exam : public student
//derived class with single base class
{
public:
int markl, mark2 ,mark3,mark4,marks, mark6, total;
void acceptmark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks..
cin>>mark1>>mark2>>mark3>>
mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t, Marks Obtained
cout<<“\n Language.. “<<mark1;
cout<<“\n English .. “<<mark2;
cout<<“\n Physics .. “<<mark3;
cout<<“\n Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout«”\n Maths .. “<<mark6;
}
};
int main()
{
exam e1;
el.acceptname();
//calling base class function using derived
class object
e1.acceptmark();
e1.displayname();
//calling base class function using derived
class object
e1l.displaymark();
return 0;
}
Output
Enter roll no and name . . 1201
KANNAN
Enter lang,eng,phy,che,esc,mat
marks.. 100 100 100 100 100 100
Roll no:-1201
Name:-KANNAN
Marks Obtained
Language.. 100
English .. 100
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100

Question 2.
Write a program to implement multiple inheritance.
Program :
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno;
public:
void acceptname()
{
cout<<“\n Enter roll no and name.. “;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name:-” << name << endl;
}
};
class detail //Base class
{
int dd,mm,yy;
char cl[4];
public:
void acceptdob()
{
cout<<“\n Enter date,month,year in digits and class ..”;
cin>>dd>>mm>>yy>>cl;
}
void displaydob()
{
cout<<“\n class:-“<<cl;
cout<<“\t\t DOB : “<<dd«” –
“<<mm<<“-” <<yy<<endl;
}
};
//derived class with multiple base class
class exam: public student, public detail
{
public:
int mark1, mark2 ,mark3,mark4,mark5, mark6, total;
void acceptmark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks..”;
cin>>mark1>>mark2>>mark3>> mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained
cout<<“\n Language.. “<<markl;
cout<<“\n English .. “<<mark2;
cout<<“\n Physics .. “<<mark3;
cout<<“\n Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout<<“\n Maths .. “<<mark6;
}
};
int main()
{
exam e1;
//calling base class function using derived
class object
e1.acceptname();
//calling base class function using derived
class object
e1.acceptdob();
e1.acceptmark();
//calling base class function using derived
class object
e1.displayname();
//calling base class function using derived
class object
e1.displaydob();
e1.displaymark();
return 0;
}
Output:
Enter roll no and name . . 1201 MEENA
Enter date, month, year in digits and
class .. 7 12 2001 XII
Enter lang, eng, phy, che, esc, mat
marks.. 96 98 100 100 100 100
Roll no:-1201
Name MEENA
class:-XII
DOB : 7 – 12 -2001
Marks Obtained
Language..96
English .. 98
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100

Question 3.
Write a program to implement multilevel inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno; public:
void acceptname()
{
cout<<“\n Enter roll no and name.. “;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name << name <<
endl;
}
};
//derived class with single base class class
exam: public student
{
public:
int mark1, mark2, mark3, mark4, mark5, mark6;
void accept mark()
{
cout<<“\n Enter
lang,eng,phy,che,esc,mat marks..’; cin>>mark1>>mark2>>mark3>> mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained
cout<<“\n Language… “<<markl;
cout<<“\n English… “<<mark2;
cout<<“\n Physics… “<<mark3;
cout<<“\n Chemistry… “<<mark4;
cout<<“\n Comp.sci… “<<mark5;
cout<<“\n Maths… “<<mark6;
}
};
class result: public exam
{
int total;
public:
void showresult()
{
total=markl+mark2+mark3+mark 4+mark5+mark6;
cout<<“\nTOTAL MARK SCORED : “<<total;
}
};
int main()
{
result r1;
//calling base class function using derived
class object
r1.acceptname();
//calling base class function which itself is a derived
r1.acceptmark();
// class function using its derived class object
r1.displayname();
//calling base class function
using derived class //object
//calling base class function which itself is a derived
r1.displaymark();
//class function using its derived class object
r1.showresult();
//calling the child class
function
return 0;
}

Output :
Enter roll no and name .. SARATHI
Enter lang,eng,phy,che,csc,mat
marks.. 96 98 100 100 100 100
Roll no:-1201
Name:-SARATHI
Marks Obtained
Language… 96
English… 98
Physics… 100
Chemistry… 100
Comp.sci… 100
Maths… 100
TOTAL MARK SCORED: 594

Question 4.
Write a program to implement hierarchical inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno;
public:
void acceptname()
{
cout<<“\n Enter roll no and name..”;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name :-” <<name <<
endl;
}
};
//derived class with single base class
class qexam: public student
{
public:
int mark1, mark2, mark3, mark4, marks, mark6;
void accent mark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks for quarterly exam”;
cin>>markl>>mark2>>mark3>>mark4>>mark5>>mark6;
}
void displaymark()
cout< <“\n\t\t Marks Obtained in quarterly”;
cout< <“\n Language.. “<<marki;
cout<<”\n English .. “<<mark2;
cout< <“\n Physics .. “<<mark3;
cout< <“\fl Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout< <“\n Maths .. “<<mark6;
}
};
//derived class with single base class
class hexam : public student
{
public:
int mark1, mark2, mark3, mark4, marks, mark6;
void acceptmark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks for half/early exam..”;
cin>>markl>>mark2>>mark3>>mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained in Halfyearly”;
cout<<“\n Language., “<< mark1;
coutcc”\n English .. “<< mark2;
coutcc”\n Physics .. “<< mark3;
coutcc”\n Chemistry.. “<< mark4;
coutcc”\n Comp.sci.. “<< mark5;
coutcc”\n Maths .. “<< mark6;
}
};
int main()
{
qexam q1;
hexam hi;
//calling base class function using derived class object
q1.acceptname();
//calling base class function
q1. acceptmark();
//calling base class function using derived
class object
h1.acceptname();
//calling base class function using derived class object
h1.displayname(); .
h1,acceptmark();
//calling base class- function using its // derived class object
h1.displaymark();
return 0
}
Output:
Enter roll no and name . . 1201
KANNAN
Enter lang,eng,phy,che,esc,mat
marks for quarterly exam. .
95 96 100 98 100 99
Roll no :-1201
Name :-KANNAN
Marks Obtained in quarterly
Language.. 95
English .. 96
Physics .. 100
Chemistry.. 98
Comp.sci.. 100
Maths .. 99
Enter roll no and name . . 1201
KANNAN
Enter lang,eng,phy,che,esc, mat marks for the half-yearly exam.
96 98 100 100 100 100
Roll no:-1201
Name:-KANNAN
Marks Obtained in Halfyearly
Language.. 96
English .. 98
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100

Question 5.
Write a program to implement hybrid inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno; public:
void acceptname()
{
cout<<“\n Enter roll no and name.. cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cou<<“\n Name name <<
endl;
}
};
//derived class with the single base class
class exam: public student
{
public:
int mark1, mark2, mark3, mark4, marks, mark6;
void accept mark()
{
cout<<“\n Enter larig, eng, phy, che, esc,mat marks..”;
cin>>markl>>mark2>>mark3>> mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained “;
cout<<“\n Language.. “<<markl;
cout<<“\n English .. “<<mark2;
cout<<“\n Physics .. “<<mark3;
cout<<“\n Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout<< “\n Maths .. “<<mark6;
}
};
class detail //base classs 2
{
int dd,mm,yy;
char cl[4];
public:
void acceptdob()
{
cout<<“\n Enter date,month,year in digits and class ..”;
cin>>dd>>mm>>yy>>cl;
}
void displaydob()
{
cou<<“\n class :-“<<cl;
cou<<“\t\t DOB : “<<dd<<” – ”
<<mm<<“-” <<yy<<endl;
}
};
//inherits from the exam, which itself is a //derived
class and also from class detail
class result: public exam, public detail
{
int total;
public:
void showresuit()
{
total=markl+mark2+mark3+ mark4+mark5+mark6;
cout<<“\nTOTAL MARK SCORED: ” <<total;
}
};
class detail //base classs 2
{
int dd,mm,yy;
char cl[4];
public:
void acceptdob()
{
cout<<“\n Enter date,month,year in digits and class ..
cin>>dd>>mm>>yy>>cl;
}
void displaydob()
{
cout<<“\n class :-“<<cl;
cout<<“\t\t DOB : “<<dd<<” – ”
<<mm<<“-” <<yy<<endl;
}
};
//inherits from the exam, which itself is a //derived class and also from class detail class result: public exam, public detail
{
int total;
public:
void showresuit()
{
total = markl + mark2-i-mark3 + mark4+mark5+mark6;
cout<<“\nTOTAL MARK SCORED : ” <<total;
}
};
int main()
{
result r1;
//calling base class function using derived class object
r1.acceptname();
//calling base class which itsel is a derived class function using its derived class object
r1.acceptmark();
r1.acceptdob();
cout<<“\n\n\t\t MARKS STATEMENT”; //calling base class function using derived class object
r1.displayname();
r1.displaydob();
//calling base class which itsel is a derived class function using its derived class object
r1.displaymark();
//calling the child class function
r1.showresuit();
return 0;
>
Output:
Enter roll no and name .. 1201 RAGU
Enter lang,eng,phy,che,esc,mat
marks.. 96 98 100 100 100 100
Enter date,month,year in digits and class .. 7 12 2001 XII
MARKS STATEMENT
Roll no :-1201
Name :-RAGU
class : -XII
DOB : 7 – 12 -2001
Marks Obtained
Language..96
English .. 98
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100
TOTAL MARK SCORED: 594

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 6 Gaseous State Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

11th Chemistry Guide Gaseous State Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the intermolecular interactions become significant
Answer:
(d) at high pressure the intermolecular interactions become significant

Question 2.
Rate of diffusion of a gas is
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) [P + \(\frac{a}{n^{2} V^{2}}\)](V – nb) = nRT

(b) [P + \(\frac{n a}{n^{2} V^{2}}\)](V – nb) = nRT

(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

(d) [P + \(\frac{n^{2} a^{2}}{V^{2}}\)(V – nb) = nRT]
Answer:
(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) collide without loss of energy
Answer:
(b) exert no attractive forces on each other

Question 5.
Equal weights of methane and oxygen are mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{3}\) × 273 × 298
Answer:
(a) \(\frac{1}{3}\)

Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature

Question 7.
In a closed room of 1000 m³ a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle

Question 9.
The value of universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Question 10.
The value of the gas constant R is
(a) 0.082 dm³atm
(b) 0.987 cal mol^-1K^-1
(c) 8.3J mol^-1K^-1
(d) 8erg mol^-1K^-1
Answer:
(c) 8.3J mol^-1K^-1

Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Question 12.
The table indicates the value of vanderWaals constant ‘a’ in (dm³)² atm. mol^-2. The gas which can be most easily liquefied is

(a) O₂
(b) N₂
(c) NH₃
(d) CH₄
Answer:
(c) NH₃

Question 13.
Consider the following statements
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises
Select the correct statement
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Answer:
(b) II and III

Question 14.
Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. The molar volume of CO₂ under these conditions is
(a) 22.04 dm³
(b) 2.24 dm³
(c) 0.41 dm³
(d) 19.5 dm³
Answer:
(c) 0.41 dm³

Question 15.
If temperature and volume of an ideal gas is increased to twice its valuesthe initial pressure P becomes
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(b) 2P

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 times that of a hydrocarbon having molecular formula C_nH_{2n – 2}. What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(d) 1

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape.
(a) \(\frac{3}{8}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{4}\)
Answer:
(c) \(\frac{1}{8}\)

Question 18.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion i.e., α = 1\(\left[\frac{\partial V}{\delta T}\right]\)Vp. For an ideal gas α is equal to
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(a) T

Question 19.
Four gases P, Q, R, and S have almost same values of ‘b’ but their a’ values (a. h are Vander Waals Constants) are in the order Q < R < S < p. At a particular temperature, a’nong the four gases the most easily liquelìable one is
(a) P
(b) Q
(c) R
(d) S
Answer:
(c) R

Question 20.
Maximum deviation from ideal gas is expected
(a) CH₄(g)
(b) NH₃(g)
(c) H₂ (g)
(d) N₂ (g)
Answer:
(b) NH₃(g)

Question 21.
The units of Vander Waals constants ‘b’ and ‘a’ respectively
(a) mol L^-1 and L atm² mol^-1
(b) mol L and L atm mol²
(c) mol^-1 L and L² atm mol^-2
(d) none of these
Answer:
(c) mol^-1 L and L² atm mol^-2

Question 22.
Assertion:
Critical temperature of CO₂ is 304 K it can be liquefied above 304 K.
Reason:
For a given mass of gas, volume is to directly proportional to pressure at constant temperature
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 23.
What is the density of N₂ gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^-1 mol^-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured in K)

Answer:

Question 25.
25g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI

II. Answer these questions briefly:

Question 26.
State Boyle’s law.
Answer:
At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, the Boyle’s law can be written as
V ∝ \(\frac{1}{P}\)     …………..(1)
(T and n are fixed, T-temperature, n-number of moles)
V = k × \(\frac{1}{P}\) …………(2)
k – proportionality constant
PV = k (at constant temperature and mass)

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law.
Answer:
Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. Volume of balloon decrease when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.

Question 28.
Name two items that can serve as a model for ‘Gay Lusaac’ law and explain.
Answer:
Firing a bullet:
When gunpowder burns, it creates; a significant amount of superheated gas. The high pressure of the hot gas behind the bullet forces it out, of the barrel of the gun.
Heating food in an oven:
When you keep food in an oven for heating, the air inside the oven is heated, thus pressurized.

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what j the mathematical expression means.
Answer:
The mathematical expression that relates gas volume and moles is Avogadro’s hypothesis. It may be expressed as
V ∝ n,
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = constant
where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂ are a different set of values of volume and number of moles of the same gas at same temperature and pressure.

Question 30.
What are ideal gases? In what way real gases differ from ideal gases?
Answer:
The kinetic theory of gases which is the basis for the gas equation (PV = nRT), assumes that the individual gas molecules occupy negligible volume when j compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases.

But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases f tend to approach the ideal behaviour under certain conditions.

Question 31.
Can a Vander Waals gas with a = 0 be liquefied? Explain.
Answer:
If the vander Waals constant (a) = 0 for a gas, then it behaves ideally, (i.e.,) there is no intermolecular forces of attraction. So it cannot be liquefied. Moreover,
P_c = \(\frac{a}{27 b^{2}}\)
If a = 0, then P_c = 0; therefore it cannot he liquefied.

Question 32.
Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:
Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The tyre of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude.
Answer:
(a) Aerated water bottles contains excess dissolved oxygen and minerals which dissolved under certain pressure and if this pressure suddenly decrease due to change in atmospheric pressure, the bottle will certainly burst with decrease the amount of dissolved oxygen in water , this tends to change the aerated water into normal-water.

(b) The vapour pressure of ammonia at room temperature is very high and hence the ammonia will evaporate unless the vapour pressure is decreased. On cooling the vapour pressure decreases so that the liquid remains in the same state. Hence, the bottle is cooled ’ before opening.

(c) In Summer due to hot weather condition, the air inside the tyre expands to large volume due to heat as compared to winter therefore inflated to lesser pressure in summer.

(d) As we move to higher altitude, the atmospheric pressure decreases, therefore balloon can J easily expand to large volume.

Question 34.
Give suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container.
Answer:
According to kinetic theory, gas molecules are moving continuous at random. Hence, they do not settle at the bottom of a container.

(b) Gases diffuse through all the space available to them.
Answer:
Gases have a tendency to occupy all the available space. The gas molecules migrate from region of higher concentration to a region of lowrer concentration. Hence, gases diffuse through all the space available to them.

Question 35.
Suggest why there is no hydrogen in our atmosphere. Why does the moon have no atmosphere?
Answer:
Hydrogen has tendency to combine with oxygen to from water vapour. Hence, the presence of hydrogen is negligible in the atmosphere. The gravitational pull in the moon is very less and hence, there is no atmosphere in the moon.

Question 36.
Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if
(a) it is compressed to a smaller volume at f constant pressure.
Answer:
When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior

(b) the temperature is raised at while keeping the volume constant.
Answer:
When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally.

(c) More gas is introduced into the same volume and at the same temperature.
Answer:
When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.

Question 37.
Which of the following gases would you expect to deviate from ideal behavior under conditions of low temperature F₂, Cl₂ or Br₂? Explain.
Answer:
Bromine has greater tendency to deviate from ideal behavior at low temperature. The compressibility factor tends to deviate from unity for bromine.

Question 38.
Distinguish between diffusion and effusion.
Answer:

Question 39.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
If aerosol cans are heated, then they will produce more vapour inside the can, which will make the pressure rise very quickly. The rises of temperature can double the pressure inside. Even though the cans are tested, they will burst if the pressure goes up too far. A bursting can could be dangerous.

Question 40.
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
Helium floats because it is buoyant; its molecules are lighter than the nitrogen and oxygen molecules of our atmosphere and so they rise above it. In the car, it’s the air molecules that are actually getting pulled and pushed around by gravity as the result of the accelerating frame.

That means it moves in a direction opposite to the force on the surrounding air. Normally, air is pulled downwards due to gravity, which pushes the balloon upwards. In this case, the surrounding air in the car is pulled forward by the deceleration of the car, which pushes the helium balloon backwards.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and Volume.
Answer:
Vander Waals equation for a real gas is given by
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Where n is the number of moles of gas and V is the volume of the container
⇒ P ∝ \(\frac{n^{2}}{V^{2}}\)
⇒ P = \(\frac{a n^{2}}{V^{2}}\)
Where a is proportionality constant and depends on the nature of gas
Therefore,
P = P + \(\frac{a n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\) π(2r)³ = 8V_m
where V_m is a volume of a single molecule.
Excluded volume for single molecule = \(\frac{8 V_{m}}{2}\) = 4V_m
Excluded volume for n molecule
= n(4V_m) = nb
Where b is van der waals constant which is equal to 4V_m
⇒ V’ = nb
V_ideal = V – nb
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas.

Question 43.
Derive the values of van der Vaals equation constants in terms of critical constants.
Answer:
The Van der walls equation for n moles is
(p + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT ……… (1)
For 1 mole
(P + \(\frac{n^{2}}{V^{2}}\)) (V – b) = RT ……………(2)

From the equation we can derive the values of critical constants P_c, V_c, and T_c in terms of a and b, the van der Waals constants, On expanding the above equation,
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0 ………….(3)

Multiply equation (3) by \(\frac{V^{2}}{P}\)
\(\frac{V^{2}}{P}\) (PV +\(\frac{a}{V}\) – pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V³ + \(\frac{a}{P}\)V – bV² – \(\frac{a b}{P}\) – \(\frac{R T V^{2}}{P}\) = 0 ………..(4)
When the above equation is rearranged in powers of V

V³ – [\(\frac{R T}{P}\) + b]V² + \(\frac{a}{P}\)V – \(\frac{a b}{P}\) = 0 ……………..(5)

The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_c. The pressure and temperature becomes P_c and T_c respectively
V = V_c
V – V_c = 0
(V – V_c)³ = 0
V³ – 3V_cV² + 3V_c²V – V_c³ = 0 …………..(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V₂, V and constant terms in (5) and (6).
-3V_cV² = –[\(\frac{R T_{c}}{P_{c}}\) + b] V²

3V_c = \(\frac{R T_{c}}{P_{c}}\) + b ………………(7)
3V_c² = \(\frac{a}{P_{c}}\) ……………(8)
V_c³ = \(\frac{ab}{P_{c}}\) …………….(9)

The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.
a = 3V_c² P_c and
b = \(\frac{V_{c}}{3}\)

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
Astronauts must wear spacesuits whenever they leave a spacecraft and are exposed to the environment of moon. In moon, there is no air to breath and no air pressure. Moon is extremely cold and filled with dangerous radiation. Without protection, an astronaut would quickly die in space. Spacesuits are specially designed to protect astronauts from the cold, radiation and low pressure in space. They also provide air to breathe. Wearing a spacesuit allows an astronaut to survive and work in moon.

Question 45.
When ammonia combines with HCl, NH₄Cl is formed as white dense fumes. Why do more fumes appear near HCl?
Answer:
HCl and NH₄Cl molecules diffuse through the air towards each other. When they meet, they reactto
form a white powder called ammonium chloride, NH₄Cl.
HCl_(g) + NH_3(g) ⇌ NHC1

Hydrogen chloride + ammonia ⇌ ammonium chloride.
The ring of white powder is closer to the HCl than
the NH₃. This is because the NH₃ molecules are lighter (smaller) and have diffùsed more quickly through the air in the tube.

Question 46.
A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm³. When the temperature is raised to 38°C at 1 atm does the volume of the gas Increase? If so, calculate the final volume.
Answer:
T₁ = 15°C + 273;
T₂ = 38 + 273
T₁ = 228;
T₂ = 311K
V₁ = 2.58dm³;
V₂ = ?
(P = 1 atom constant)
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

V₂ = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T₂

= \(\frac{2.58 d m^{3}}{288 K}\) × 311K
V₂ = 2.78 dm³ i.e, volume increased from 2.58 dm³ to 2.78 dm³

Question 47.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen In a vessel of volume of 37.6 dm³ at 298K, and the sample B Is in a vessel of volume 16.5 dm³ at 298K. Calculate the number of moles in sample B.
Answer:
n_A = 1.5mol; n_B = ?
V_A = 37.6 dm³; V_B = 16.5 dm³
(T = 298K constant)
\(\frac{V_{A}}{n_{A}}=\frac{V_{B}}{n_{B}}\)

n_A = \(\left[\frac{n_{A}}{n_{B}}\right] V_{B}\)

Question 48.
Sulphur hexafluoride Is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas In a steel vessel of volume 5.43 dm3 at 69.5°C, assumIng Ideal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P =?
PV = nRT
P = nRT/V
P =
P = 94.25 atm

Question 49.
Argon is an Inert gas used In light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C Is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P₁ = 1.2 atm
T₁ = 180°C + 273 = 291 K
T₂ = 850°C + 273 = 358 K
P₂ =?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P₂ = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\)

P₂ = \(\frac{1.2 a t m}{291 K}\) × 358 K
P₂ = 1.48 atm

Question 50.
A small bubble rises from the bottom of a lake where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure Is I arm. Calculate the final volume in (mL) of the bubble, If its initial volume 1.5 mL.
Answer:
T₁ = 6°C + 273 = 279 K
P₁ = 4 atm; V₂ = 1.5 ml
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm; V₂ =?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

= \(\frac{4 a t m \times 1.5 m l \times 298 K}{279 K \times 1 a t m}\) = 6.41 mol

Question 51.
Hydrochloric acid Is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 × 10^-3 dm³ of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 × 10^-3 dm³
P = 742 mm of Hg
T = 298K;
m =?
m = \(\frac{P V}{R T}\)

= \(\frac{742 m m H g \times 154.4 \times 10^{-3} L}{62 m m H g L K^{-1} m o l^{-1} \times 298 K}\)

n = \(\frac{\text { Mass }}{\text { Molar Mass }}\)

Mass = n × Molar Mass
= 0.0006 × 2.016
= 0.0121 g = 12.1 mg

Question 52.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknowa gas?
Answer:

Question 53.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure In the tanks Is 9.21 atm. calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:

Question 54.
A combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a maximum pressure of 12 atm after which It will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal 1100 K).
Answer:
Pressure of the gas in the tank at its melting point
T = 298 K;
P₁ = 2.98 atom;
T₂ = 1100 K;
P₂ =?
\(\frac{P_{1} P_{2}}{T_{1} T_{2}}\) =????
⇒ P₂ = \(\frac{P}{T_{1}}\) × T₂
= \(\frac{2.98 \text { atm }}{298 K}\) × 1100 K = 11 atm

At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

11th Chemistry Guide Gaseous State Additional Questions and Answers

I. Choose the best Answer:

Question 1.
The approximate volume percentage of nitrogen and oxygen in the atmosphere of air are _______ and ______ respectively.
(a) 21, 68
(b) 21, 78
(c) 78, 2
(d) 80, 21
Answer:
(c) 78, 2

Question 2.
The SI unit of pressure is
(a) pascal
(b) atmosphere
(c) bar
(d) torr
Answer:
(a) pascal

Question 3.
The compound widely used in the refrigerator as coolant is
(a) Freon-2
(b) Freon-12
(c) Freon-13
(d) Freon-14
Answer:
(b) Freon-12

Question 4.
“For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature”. This statement is
(a) Boyle’s law
(b) Gay-Lussac law
(c) Avogadro’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 5.
Hydrogen is placed in the ______ of the periodic table.
(a) Group -1
(b) group-17
(c) group-18
(d) group-2
Answer:
(a) Group -1

Question 6.
The plot of the volume of the gas against its temperature at a given pressure is called
(a) isotone
(b) isobar
(c) isomer
(d) isotactic
Answer:
(b) isobar

Question 7.
At constant temperature for a given mass, for each degree rise in temperature, all gases expand by of their volume at 0°C.
(a) 273
(b) 298
(c) \(\frac{1}{273}\)

(d) \(\frac{1}{298}\)
Answer:
(c) \(\frac{1}{273}\)

Question 8.
The precise value of temperature at which the volume of the gas becomes zero is
(a) -273.15 °C
(b) -273 °C
(c) -298.15°C
(d) -298 °C
Answer:
(a) -273.15 °C

Question 9.
‘At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature”. This statement is
(a) Charle’s law
(b) Boyle’s law
(c) Gay-Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay-Lussac’s law

Question 10.
The value of gas constant, R, in terms of JK^-1 mol^-1 is
(a) 8.314
(b) 4.184
(c) 0.0821
(d) 1.987
Answer:
(a) 8.314

Question 11.
A mixture of gases containing 4 mole of hydrogen and 6 mole of oxygen. The partial pressure of hydrogen, if the total pressure is 5 atm, is
(a) 10 atm
(b) 0.4 atm
(c) 20 atm
(d) 2 atm
Answer:
(d) 2 atm

Question 12.
A mixture of gases containing 1 mole of He, 4 mole of Ne and 5 mole of Xe. The correct order of partial pressure of the gases, if the total pressure is 10 atm is
(a) Xe < Ne < He
(b) He < Ne < Xe
(c) Xe < Ne < He
(d) He < Xe < Ne
Answer:
(b) He < Ne < Xe

Question 13.
The property of a gas which involves the movement of the gas molecules through another gases is called
(a) effusion
(b) dissolution
(c) difusion
(d) expansion
Answer:
(c) difusion

Question 14.
The process in which agas escapes from a container through a very small hole is called
(a) diffusion
(b) effusion
(c) occlusion
(d) dilution
Answer:
(a) diffusion

Question 15.
The rate of diffusion of a gas is inversely proportional to the
(a) square of molar mass
(b) square root of density
(c) square root of molar mass
(d) square of density
Answer:
(c) square root of molar mass

Question 16.
An unknown gas X diffuses at a rate of 2 times of oxygen at the same temperature and pressure. The molar mass (in g mol^-1) of the gas ‘X is (Molar mass of oxygen is 32 g mol^-1)
(a) 8
(b) 16
(c) 20
(d)12
Answer:
(a) 8

Question 17.
The deviation of real gases from ideal behavior is measured in terms of
(a) expansivity factor
(b) molar mass
(c) pressure
(d) compressibility factor
Answer:
(d) compressibility factor

Question 18.
The gases which deviate from ideal behavior at
(a) low temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) high temperature and high pressure
Answer:
(b) high temperature and low pressure

Question 19.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called temperature
(a) inversion
(b) ideal
(c) Boyle
(d) reversible
Answer:
(c) Boyle

Question 20.
Ideal gas equation for ‘n’ moles is
(a) \(\frac{P}{R}=\frac{n T}{V}\)

(b) PV = \(\frac{n R}{T}\)

(c) \(\frac{V}{T}=\frac{n P}{R}\)

(d) \(\frac{P}{T}=\frac{n R}{V}\)
Answer:
(d) \(\frac{P}{T}=\frac{n R}{V}\)

Question 21.
The pressure correction introduced by Van der Waals is,
P_ideal =
(a) P + \(\frac{a V^{2}}{n^{2}}\)

(b) P + \(\frac{a n}{V^{2}}\)

(c) P + \(\frac{a n^{2}}{V^{2}}\)

(d) P + \(\frac{a V}{n^{2}}\)
Answer:
(c) P + \(\frac{a n^{2}}{V^{2}}\)

Question 22.
The volume correction introduced by Van der Waals is, V_ideal =
(a) V – nb
(b) V + nb
(c) \(\frac{V}{n b}\)
(d) b – nV
Answer:
(b) V + nb

Question 23.
Van der waals equation for one mole of a gas is
(a) (P + \(\frac{a}{V}\))(V-b) = RT
(b) (P – \(\frac{a}{V^{2}}\))(V + -b) = RT
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT
(d) (P + \(\frac{a}{V}\))(V + b) = RT
Answer:
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT

Question 24.
The unit of Van der Waals constant V is
(a) atm lit mol^-1
(b) atm lit² mol^-2
(c) atm lit^-2 mol²
(d) atm lit^-1 mol²
Answer:
(b) atm lit² mol^-2

Question 25.
The unit of Van der waals constant ‘b’ is
(a) lit mol^-1
(b) lit mol
(c) atm lit mol^-1
(d) atm lif^-1 mol^-2
Answer:
(a) lit mol^-1

Question 26.
The temperature above which a gas cannot be liquefied even at high pressure is called is ________ temperature.
(a) ideal
(b) inversion
(c) critical
(d) real
Answer:
(c) critical

Question 27.
The gas used in pressure-volume isotherm study of Andrew’s experiment is
(a) N₂
(b) H₂S
(c) NH₃
(d) CO₂
Answer:
(d) CO₂

Question 28.
Which of the following gas has highest critical temperature?
(a) NH₃
(b) CO₂
(c) N₂
(d) CH₄
Answer:
(a) NH₃

Question 29.
The relationship between critical volume and Vander waals constant is
(a) V_c =\(\frac{a}{R b}\)
(b) V_c = 3b
(c) V_c = \(\frac{8 a}{27 R b}\)
(d) V_c = 8b
Answer:
(b) V_c = 3b

Question 30.
The temperature below which a gas obeys Joule-Thomson effect is called ______ temperature.
(a) critical
(b) Inversion
(c) ideal
(d) real
Answer:
(b) Inversion

Question 31.
In the equation PV = nRT, which one cannot be numerically equal to R?
(a) 8.31 × 10⁷ ergs K^-1 mol^-1
(b) 8.31 × 10⁷ dynes cm K^-1 mol^-1
(c) 8.317J K^-1 mol^-1
(d) 8.317L atm K^-1 mol^-1
Answer:
(d) 8.317L atm K^-1 mol^-1

Question 32.
A sample of a given mass of gas at constant temperature occupies a volume of 95 cm³ under a pressure of 10.13 × 10⁴ Nm^-2. At the same temperature, its volume at pressure of 10.13 × 10⁴ Vm^-2 is
(a) 190 cm³
(b) 93 cm³
(c) 46.5 cm³
(d) 4.75 cm³
Answer:
(b) 93 cm³

Question 33.
The number of moles of H₂ in 0.224 L of hydrogen gas at STP (273 K, 1 atm) assuming ideal gas behavior is
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01

Question 34.
Use of hot air balloons in sports and meteorological observations is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 35.
To what temperature must a neon gas sample be heated to double its pressure, if the initial volume of gas at 75°C is decreased by 15.0% by cooling the gas?
(a) 319°C
(b) 592°C
(c) 128°C
(d) 60°C
Answer:
(a) 319°C

Question 36.
7.0 g of a gas at 300K and 1 atm occupies a volume of 4.1 litre. What is the molecular mass of the gas?
(a) 42
(b) 38.24
(c) 14.5
(d) 46.5
Answer:
(a) 42

Question 37.
If most probable velocity is represented by a and fraction possessing it by / then with increase in temperature which one of the following is correct?
(a) α increases, f decreases
(b) α decreases, f increases
(c) Both α and f decrease
(d) Both α and f increase
Answer:
(a) α increases, f decreases

Question 38.
The rate of diffusion of gases A and B of molecular weight 36 and 64 are in the ratio
(a) 9 : 16
(b) 4 : 3
(c) 3 : 4
(d) 16 : 9
Answer:
(b) 4 : 3

Question 39.
To which of the following gaseous mixtures Dalton’s law is not applicable?
(a) Ne + He + SO₂
(b) NH₃ + HCl + HBr
(c) O₂ + N₂ +CO₂
(d) N₂ + H₂ + O₂
Answer:
(b) NH₃ + HCl + HBr

Question 40.
Which of the following is true about gaseous state?
(a) Thermal energy = molecular interaction
(b) Thermal energy >> molecular interaction
(c) Thermal energy << molecular interaction
(d) molecular forces >> those in liquids
Answer:
(b) Thermal energy >> molecular interaction

Question 41.
50 mL of gas A effuses through a pinhole in 146 seconds. The same volume of CO₂ under identical condition effuses in 115 seconds. The molar mass of A is
(a) 44
(b) 35.5
(c) 71
(d) None of these
Answer:
(c) 71

Question 42.
Gas deviates from ideal gas nature because molecules
(a) are colourless
(b) attract each other
(c) contain covalent bond
(d) show Brownian movement
Answer:
(b) attract each other

Question 43.
Which of the following gases is expected to have largest value of van der Waals constant ‘a’?
(a) He
(b) H₂
(c) NH₃
(d) O₂
Answer:
(c) NH₃

Question 44.
In van der Waals equation of state for a real gas, the term that accounts for intermolecular forces is
(a) V_m – b
(b) P + \(\frac{a}{V m^{2}}\)
(c) RT
(d) \(\frac{1}{R T}\)
Answer:
(b) P + \(\frac{a}{V m^{2}}\)

Question 45.
What are the most favourable conditions to liquefy a gas?
(a) High temperature and low pressure
(b) Low temperature and high pressure
(c) High temperature and high pressure
(d) Low temperature and low pressure
Answer:
(b) Low temperature and high pressure

Question 46.
Which of the following has a non-linear relationship?
(a) P vs V
(b) P vs \(\frac{1}{V}\)
(c) both (a) and (b)
(d) none of these
Answer:
(a) P vs V

Question 47.
Why is that the gases show ideal behavior when the volume occupied is large?
(a) So that the volume of the molecules can be neglected in comparison to it.
(b) So that the pressure is very high
(c) So that the Boyle temperature of the gas is constant
(d) all of these
Answer:
(a) So that the volume of the molecules can be neglected in comparison to it.

Question 48.
At a given temperature, pressure of a gas obeying Van der Waals equation is
(a) less than that of an ideal gas
(b) more than that of an ideal gas
(c) more or less depending on the nature of gas
(d) equal to that of an ideal gas
Answer:
(a) less than that of an ideal gas

Question 49.
NH₃ gas is liquefied more easily than N₂. Hence
(a) Van der Waals constant a and b of NH₃ > that of N₂
(b) van der Waals constants a and b of NH₃ < that of N₂
(c) a(NH₃) > a(N₂) but b(NH3) < b(N₂)
(d) a(NH₃) < a(N₂) but b(NH₃) > b(N₂)
Answer:
(c) a(NH₃) > a(N₂) but b(NH3) < b(N₂)

Question 50.
Maximum deviation from ideal gas is expected from
(a) CH_4(g)
(b) NH_3(g)
(c) H_2(g)
(d) N_2(g)
Answer:
(b) NH_3(g)

II. Very short question and answers (2 Marks):

Question 1.
Define Pressure? Give its unit?
Answer:
Pressure is defined as force divided by the area to which the force is applied. The SI unit of pressure is pascal which is defined as 1 Newton per square meter (Nm^-2).
Pressure = \(\frac{\text { Force }\left(\mathrm{N}(\text { or }) \mathrm{Kg} \mathrm{ms}^{-2}\right)}{\text {Area }\left(\mathrm{m}^{2}\right)}\)

Question 2.
State Gay Lussac ‘s law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T or \(\frac{P}{T}\) = constant K
If P₁ and P₂ are the pressures at temperatures T₁ and T₂, respectively, then from Gay Lussac’s law
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 3.
State Dalton’s law of partial pressure.
Answer:
The total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton s law of partial pressures.
For a mixture containing three gases 1, 2, and 3 with partial pressures p₁, p_2, and p₃ in a container with volume V, the total pressure Ptotal will be give by
P_total = p₁ + p₂ + p₃

Question 4.
What is Compressibility factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compressibility factor. Mathematically,
Z = \(\frac{P V}{n R T}\)

Question 5.
Define Critical temperature (T_c) of a gas?
Answer:
Critical temperature (T_c) of a gas is defined as the temperature above which it cannot be liquefied even at high pressure.

Question 6.
How does cooling is produced in Adiabatic Process?
Answer:
In Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10^-4 K i.e., as low as 0 K can be achieved.

Question 7.
How do you understand PV relationship?
Answer:
The PV relationship can be understood as follows. The pressure is due to the force of the gas particles on the walls of the container. If a given amount of gas is compressed to half of its volume, the density is doubled and the number of particles hitting the unit area of the container will be doubled. Hence, the pressure would increase two fold.

Question 8.
Write a note on Consequence of Boyle’s law.
Answer:
The pressure-density relationship can be derived from the Boyle’s law as shown below.
P₁V₁ = P₂V₂
P₁ \(\frac{m}{d_{1}}\) = P₂ \(\frac{m}{d_{2}}\)
where “m” is the mass, d₁ and d₂ are the densities of gases at pressure P₁ and P₂.
\(\frac{P_{1}}{d_{1}}=\frac{P_{2}}{d_{2}}\)
In other words, the density of a gas is directly proportional to pressure.

Question 9.
A gas cylinder can withstand a pressure of 15 aim. The pressure of cylinder is measured 12 atm at 27°C. Upto which temperature limit the cylinder will not burst?
Answer:
Cylinder will burst at that temperature when it attains the pressure of 15 atm
P₁ = 12 atm
T = 27° C = (27 + 273)K = 300K
P₂ = 15 atm; T = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

T₂ = \(\frac{15 \times 300}{12}\) = 375 K
= (375 – 273)°C = 102°C

Question 10.
Write a note on application of Dalton’s law.
Answer:
In a reaction involving the collection of gas by downward displacement of water, the pressure of dry vapor collected can be calculated using Dalton’s law.
P_{dry gas collected} = P_total – P_watervapour
P_watervapour has generally referred as aqueous tension and its values are available for air at various temperatures.

Question 11.
State Charles law.
Answer:
For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature.
\(\frac{V}{T}\) = constant (at constant pressure)

Question 12.
What happens when a balloon is moved from an ice cold water bath to a boiling water bath?
Answer:
If a balloon is moved from an ice cold water bath to a boiling water bath, the temperature of the gas increases. As a result, the gas molecules inside the balloon move faster and gas expands. Hence, the volume increases.

Question 13.
Write notes on coefficient of expansion(α).
Answer:
The relative increase in volume per °C (α) is equal to \(\frac{V}{V_{0} T}\)
Therefore, \(\frac{V}{V_{0} T}\)
TV = V₀(αT + 1)
Charles found that the coefficient of expansion is approximately equal to 1/273. It means that at constant temperature for a given mass, for each degree rise in temperature, all gases expand by 1/273 of their volume at 0°C.

Question 14.
State Gay-Lussac law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T (at constant volume)

Question 15.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 16.
Distinguish between diffusion and effusion.
Answer:
The property of gas that involves the movement of the gas molecules through another gases is called diffusion. Effusion is another process in which a gas escapes from a container through a very small hole.

Question 17.
State Graham’s law of diffusion.
Answer:
The rate of diffusion or effusion is inversely proportional to the square root of molar mass. This statement is called Graham’s law of diffusion/effusion.
Mathematically, rate of diffusion ∝ \(\frac{1}{M}\)

Question 18.
What is Boyle temperature?
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.

III. Short Question and Answers (3 Marks):

Question 1.
Write notes on Boyle’s point.
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point. The Boyle point varies with the nature of the gas. Above the Boyle point, for real gases, Z > 1, i.e., the real gases show positive deviation. Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.

Question 2.
What are the different methods of liquefaction of gases?
Answer:
There are different methods used for liquefaction of gases:

1. In Linde’s method, Joule-Thomson effect is used to get liquid air or any other gas.
2. In Claude’s process, the gas is allowed to perform mechanical work in addition to Joule-Thomson effect so that more cooling is produced.
3. In Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10^-4 K i.e., as low as 0 K can be achieved.

Question 3.
48 litre of dry N₂ is passed through 36g of H₂O at 27°C and this results In a loss of 1.20 g of water. Find the vapour pressure of water?
Answer:
Water loss is observed because of escape of water molecules with N₂ gas. These water vapour occupy the volume of N₂ gas i.e., 48 litres.
Using,
PV = \(\frac{m}{M}\)RT;
V = 48L
m = 1.2g;
M = 18u;
R = 0.082 L atm K^-1mol^-1
T = 27 + 273 = 300 K
P = \(\frac{1.2}{18} \times \frac{0.0821 \times 300}{48}\) = 0.034 atm

Question 4.
A flask of capacity one litre is heated from 25°C to 35°C. What volume of air will escape from the flask?
Answer:
Applying, \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V₁ = 1L;
T₁ = 25 + 273 = 298 K
V₂ = ?
T₂ = 35 + 273 = 308 K
V₂ = \(\frac{1}{298}\) × 308 = 1.033 L
Capacity of Flask = 1 L
So, volume of air escaped = 1.033 – 1 = 0.033 L = 33 mL

Question 5.
Discuss the graphical representation of Boyle’s law.
Answer:
Boyle’s law is applicable to all gases regardless of their chemical identity (provided the pressure is low). Therefore, for a given mass of a gas under two different sets of conditions at constant temperature we can write,

Question 6.
A certain gas takes three times as long to effuse out as helium. Find its molecular mass.
Answer:

Question 7.
The vapour pressure of water at 80°C is 355.5 mm of Hg. A 100 mL vessel contains water saturated with O₂ at 80°C, the total pressure being 760 mm of Hg. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What is the partial pressure of O₂?
Answer:
P_total = P_H2O + P_O2
P_O2 = 760 – 355.5 = 404.5
When the contents were pumped into 50 mL vessel
P₁ = 404.5
V₁ = 100 mL
P₂ =?
V₂ = 50 mL
P₁V₁ = P₂V₂
P₂ = \(\frac{404.5 \times 100}{50}\) = 809 mm
Thus, PO₂ in 50 mL vessel = 809 mm

Question 8.
Write notes on compressibility factor for real gases.
Answer:
The compressibility factor Z for real gases can be rewritten,
Z = PV_real / nRT …………..(1)
V_ideal = \(\frac{n R T}{P}\) ……………(2)
substituting (2) in (1)
where V_real is the molar volume of the real gas and V_ideal is the molar volume of it when it behaves ideally.

Question 9.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V, and n and their relationships were governed by the gas laws studied so far.
Boyle’s law V ∝ P¹
Charles law V ∝ T
Avogadro’s law V ∝ n
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
V ∝ \(\frac{n T}{P}\)

V = \(\frac{n R T}{P}\)
where, R is the proportionality constant called universal gas constant.
The above equation can be rearranged to give the ideal gas equation
PV = nRT

Question 10.
State and explain Dalton’s law of partial pressure.
Answer:
John Dalton stated that “the total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton’s law of partial pressures, i.e., for a mixture containing three gases 1, 2, and 3 with partial pressures p₁, p_2, and p₃ in a container with volume V, the total pressure P_total will be given by
P = p₁ + p₂ + p₃

IV. Long Question and Answers (5 Marks):

Question 1.
Derive Van der waals equation of state.
Answer:
J.D.Van der Waals made the first mathematical analysis of real gases. His treatment provides us an interpretation of real gas behaviour at the molecular level. He modified the ideal gas equation PV = nRT by introducing two correction factors, namely, pressure correction and volume correction.

Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.

Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
P’ ∝ p²
p = \(\frac{n}{v}\)

where n is the number of moles of gas and V is the volume of the container
⇒ P’ = \(\frac{n^{2}}{V^{2}}\)
⇒ P’ = a\(\frac{n^{2}}{V^{2}}\)

where a is the proportionality constant and depends on the nature of gas.
Therefore, ideal P = P + a\(\frac{n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V’ to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\)π(2r)³
= \(\left|\frac{-8}{(3 \pi)}\right|\) = 8V_m

where V_m is a volume of a single molecule
Excluded volume for single-molecule = \(\frac{8 V_{m}}{2}\) = 4V_m

Excluded volume for n molecule = n(4V_m) = nb
Where b is van der Waals constant which is equal to 4 V_m
=> V’ = nb
V_ideal = V nb

Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

Question 2.
Derive the relationship between Van der Waals constants and critical constants.
Answer:
The van der Waals equation for n moles is
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
For 1 mole
(p + \(\frac{a}{V^{2}}\))(V – b) = RT
From the equation we can derive the values of critical constants P_c, V_c, and T_c in terms of a and b, the van der Waals constants, On expanding the above equation
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0

Multiply above equation by
\(\frac{V_{2}}{P}\)(PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V³ + \(\frac{a V}{P}\) + -bV² – \(\frac{a b}{p}\) – \(\frac{R T V^{2}}{P}\) = 0

when the above equation is rearranged in powers of V

V³ \(\frac{R T}{P}\) + bV² \(\frac{a}{P}\) V – \(\frac{a b}{P}\) = 0.

The above equation is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point, all these three solutions of V_c are equal to the critical volume. The pressure and temperature becomes P_c and T_c respectively
i.e., V = V_c
V – V_c = 0
V – (V_c)³ = 0
V³ – 3V_cV² + 3VV_c² – V_c³ = 0.
As equation identical with equation above, we can equate the cocfficients of V_c, V and constant terms.

-3V_cV² = –\(\frac{R T_{C}}{P}\) + bV²

3V_c = \(\frac{R T_{C}}{P_{C}}\) + b ……….(1)

3V_c² = \(\frac{a}{P_{C}}\) …………..(2)

V_c³ = \(\frac{a b}{P_{C}}\) …………..(3)

Divide equation (3) by equation (2)

\(\frac{V_{C}}{3}\) = b
i.e., V_c = 3b ……………(4)

when equation (4) is substituted in (2)

The critical constants can be calculated using the values of vander walls constant of a gas and vice versa.

a = 3V_c² P_c and b = \(\frac{V_{C}}{3}\)

Question 3.
Explain Andrew’s isotherm for Carbon dioxide.
Answer:
Thomas Andrew gave the first complete data on pressure-volume-temperature of a substance in the gaseous and liquid states. He plotted isotherms of carbon dioxide at different temperatures which is shown in Figure. From the plots we can infer the following.

At low temperature isotherms, for example, at 130°C as the pressure increases, the volume decreases along AB and is a gas until the point B is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist and the pressure remains constant. At C, the gas is completely converted into liquid. If the pressure is higher than at C, only the liquid is compressed so, there is no significant change in the volume. The successive isotherms shows similar trend with the shorter flat region. i.e., The volume range in which the liquid and gas coexist becomes shorter.

At the temperature of 31.1°C the length of the shorter portion is reduced to zero at point P. In other words, the CO₂ gas is liquefied completely at this point. This temperature is known as the liquefaction temperature or critical temperature of CO₃. At this point the pressure is 73 atm. Above this temperature, CO₃ remains as a gas at all pressure values. It is then proved that many real gases behave in a similar manner to carbon dioxide.

Question 4.
Explain Boyle’s law experiment:
Answer:
Robert Boyle performed a series of experiments to j study the relation between the pressure and volume of gases. The schematic diagram of the apparatus j used Boyle is shown in figure.

Mercury was added through the open end of the apparatus such that the mercury level on both ends are equal as shown in the figure (a). Add more amount of mercury until the volume of the trapped air is reduced to half of its original volume as shown in figure (b). The pressure exerted on the gas by the addition of excess mercury is given by the difference in mercury levels of the tube.

Initially the pressure exerted by the gas is equal to 1 atm as the difference in height of the mercury levels is zero. When the volume is reduced to half, the difference in mercury levels increases to 760 mm. Now the pressure exerted by the gas is equal to 2 atm. It led him to conclude that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, the Boyle’s law can be written as
V ∝ \(\frac{1}{P}\) ……….(1)
(T and n are fixed, T-temperature, n- number of moles)
V = k × \(\frac{1}{P}\) ……….(2)
k – proportionality constant When we rearrange equation (2)
PV = k at constant temperature and mass.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 12 Mineral Nutrition Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition

11th Bio Botany Guide Mineral Nutrition Text Book Back Questions and Answers

Part -I

Question 1.
Identify correct match.
1. Die back disease of citrus -(i) Mo
2. Whip tail disease – (ii) Zn
3. Brown heart of turnip -(iii) Cu
4. Little leaf -(iv) B

Answer:
b) 1 (iii) 2 (i) 3 (iv) 4 (ii)

Question 2.
If a plant is provided with all mineral nutrients but, Mn concentration is increased, what will be the deficiency?
(a) Mn prevent the uptake of Fe, Mg but not Ca
(b) Mn increase the uptake of Fe, Mg and Ca
(c) Only increase the uptake of Ca
(d) Prevent the uptake Fe, Mg, and Ca
Answer:
(a) Mn prevent the uptake of Fe, Mg but not Ca

Question 3.
The element which is not remobilized?
a) Phosphorous
b) Potassium
c) Calcium
d) Sulphur
Answer:
c) Calcium

Question 4.
Match the correct combination.

a) A-1 B-3 C-4 D-2
b) A-2 B-1 C-3 D-4
c) A-4 B-3 C-1 D-2
d) A-4 B-2 C-1 D-3
Answer:
c) A-4 B-2 C-1 D-3

Question 5.
Identify the correct statement:
(i) Sulphur is essential for amino acids Cystine and Methionine
(ii) Low level of N, K, S and Mo affect the cell division
(iii) Non – leguminous plant Alnus which contain bacterium Frankia
(iv) Denitrification carried out by nitrosomonas and nitrobacter.

(a) (i), (ii) are correct
(b) (i), (ii), (iii) are correct
(c) I only correct
(d) all are correct
Answer:
(b) (i), (ii), (iii) are correct

Question 6.
Nitrogen is present in the atmosphere in huge amounts but higher plants fail to utilize it. Why?
Answer:
1. Plants absorb minerals from the soil along with water with the help of Roots. Minerals are absorbed as salts.

2. Nitrogen is present in large quantities in the atmosphere in a gaseous form, the gaseous nitrogen must be fixed in the form of Nitrate salts in the soil to facilitate absorption by plant roots.

3. Nitrogen fixation can occur 2 ways by

• Non – Biological means (Industrial process or by lighting)
• Biological means (Bacteria / Cyanobacteria Fungi)
• Therefore higher plants con not utilize the atmospheric Nitrogen.

Question 7.
Why is that in certain plants, deficiency symptoms appear first in younger parts of the plants while in others, they do so in mature organs?
Answer:
When deficiency symptoms appear first, we can notice the differences in old and younger leaves. It is mainly due to mobility’ of minerals. Based on this, they are classified into
1. Actively mobile minerals and
2. Relatively immobile minerals

a) Actively mobile minerals: Nitrogen, Phosphorus, Potassium, Magnesium, Chlorine, Sodium, Zinc and Molybdenum. Deficiency symptoms first appear on old and senescent leaves due to active movement of minerals to younger leaves, than the older leaves.

b) Relatively immobile minerals: Calcium, Sulphur, Iron, Boron and Copper. Here, deficiency symptoms first appear on young leaves due to the immobile nature of minerals.

Question 8.
Plant A in a nutrient medium shows whiptail disease plant B in a nutrient medium shows a little leaf disease. Identify mineral deficiency of plant A and B?
Answer:
Mineral deficiency of plant A and B:

1. Plant A is deficient in the mineral molybdenum (Mo).
2. Plant B is deficient in the mineral zinc (Zn).

Question 9.
Write the role of nitrogenase enzyme in nitrogen fixation?
Answer:
Nitrogen fixation is the first step in Nitrogen cycle, during which gaseous nitrogen from the atmosphere is fixed. It required nitrogenase enzyme complex nitrogenase is active only in anaerobic condition. To create this anaerobic condition, a pigment known as leghaemoglobin is synthesized in the nodules which acts as oxygen scavenger and removes oxygen.

Question 10.
Explain the insectivorous mode of nutrition in angiosperms?
Answer:
Plants which are growing in nitrogen deficient areas develop insectivorous habit to resolve nitrogen deficiency.

1. Nepenthes (Pitcher plant): Pitcher is a modified leaft and contains digestive enzymes. Rim of the pitcher is provided with nectar glands and acts as an attractive lid. When insect is trapped, proteolytic enzymes will digest the insect.
2. Drosera (Sundew): It consists of long club shaped tentacles which secrete sticky digestive fluid which looks like a sundew.
3. Utricularia (Bladder wort): Submerged plant in which leaf is modified into a bladder to collect insect in water.
4. Dionaea (Venus fly trap): Leaf of this plant modified into a colourful trap. Two folds of lamina consist of sensitive trigger hairs and when insects touch the hairs it will close.

Insectivorous Plants

1. Nepenthes (Pitcher Plant)

2. Drosera (Sundew)

3. Dlonaca (Venus Fly tray)

Part – II

11th Bio Botany Guide Mineral Nutrition Additional Important Questions and Answers

I. Choose the Correct Answers

Question 1.
Plants naturally obtain nutrients from:
(a) atmosphere
(b) water
(c) soil
(d) all of these
Answer:
(d) all of these

Question 2.
The minerals placed under the list of unclassified minerals are
a) Carbon. Hydrogen, & Oxygen
b) Sodium. Silicon. Cobalt and selenium
c) Copper, Iron, Cadmium, and selenium
d) Magnesium, Sulphur, & Manganese
Answer:
b) Sodium, Silicon, Cobalt, and Selenium

Question 3.
Who coined the term ‘Hydroponics’:
(a) Julius Von Sachs
(b) William Frederick Goerick
(c) Liebig
(d) Wood word
Answer:
(b) William Frederick Goerick

Question 4.
Skeletal elements are
a) Carbon, Hydrogen, and Oxygen
b) Nitrogen, Phosphorus, and Calcium
c) Potassium, Magnesium, and Sulphur
d) Nitrogen, Sulphur and Phosphorus
Answer:
a) Carbon, Hydrogen, and Oxygen

Question 5.
Actively mobile minerals are:
(a) nitrogen and phosphorus
(b) iron and manganese
(c) sodium and cobalt
(d) silicon and selenium
Answer:
(a) nitrogen and phosphorus

Question 6.
Which chelating agent found in soil are produced by bacteria?
a) Siderophores
b) EDTA
c) Auxin
d) Gibberellin
Answer:
a) Siderophores

Question 7.
Molybdenum is essential for the reaction of:
(a) hydrolase enzyme
(b) nitrogenase enzyme
(c) carboxylase enzyme
(d) dehydrogenase enzyme
Answer:
(b) nitrogenase enzyme

Question 8.
Minerals that play important role for activation of enzymes involved in Respiration are
a) Molybdenum and Boron
b) Boron and Silicon
c) Calcium and Magnesium
d) Magnesium and Manganese
Answer:
d) Magnesium and Manganese

Question 9.
Essential component of aminoacids like Cystine, Cysteine and Melhionine is
a) Potassium
b) Magnesium
c) Sulphur
d) Calcium
Answer:
c) Sulphur

Question 10.
Which of the element is involved in the synthesis of DNA and RNA:
(a) calcium
(b) magnesium
(c) sulphuric
(d) potassium
Answer:
(b) magnesium

Question 11.
Delay in flowering is due to the deficiency of
a) N, S, Mo
b) Ca, Mg, Mn
c) C, H, O
d) N,P,K
Answer:
a) N,S,Mo

Question 12.
Kheria disease of Rice and Internal cork of Apple are caused by the deficiency of
a) Calcium and Maganese
b) Zinc and Boron
c) Copper and Manganese
d) Boron and Nickel
Answer:
b) Zinc and Boron

Question 13.
Indicate the correct statements:
(i) Iron is the essential element for the synthesis of chlorophyll and carotenoid
(ii) Iron is the activator of carboxylene enzyme
(iii) Iton is the component of cytochrome
(iv) lvon is the component of plastocyanin

(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Question 14.
The enzyme that is a constituent of urease and dehydrogenase are
a) Molybdenum
b) Boron
c) Nickel
d) Zinc
Answer:
c) Nickel

Question 15.
A membrane bound bacterium formed inside the nodule is called
a) Bacteriod
b) Plasmid
c) Nucleoid
d) Noduloid
Answer:
a) Bacteriod

Question 16.
The increased concentration of manganese in plants will prevent the uptake of:
(a) calcium and potassium
(b) sodium and potassium
(c) boron and silicon
(d) iron and magnesium
Answer:
(d) iron and magnesium

Question 17.
Plants need one of the following minerals for ATP and meristematic tissue formation
a) K, N
b) N, Cu
c) N, Ca
d) P, N
Answer:
d) P, N

Question 18.
The techniques of Aeroponics was developed by:
(a) Goerick
(b) Amon and Hoagland
(c) Soifer Hillel and David Durger
(d) Von Sachs
Answer:
(c) Soifer Hillel and David Durger

Question 19.
Mo is a part of enzyme ……………..
a) Reverse transcriptase
b) Restriction endonuclease
c) Hexokinase
d) Nitrogenase
Answer:
d) Nitrogenase

Question 20.
Which of the bacterium causes denitrification?
a) Azotobacter
b) Nitrobacter
c) Nitrosomonas
d) Pseudomonas
Answer:
d) Pseudomonas

Question 21.
Beside paddy fields, cyanobacteria are also found inside the vegetative parts of
a) Psiloturn
b) Pinus
c) Cycas
d) Equiseturn
Answer:
c) Cycas

Question 22.
The legume plants secrete phenolics to attract:
(a) Azolla
(b) Rhizobium
(c) Nitrosomonas
(d) Streptococcus
Answer:
(b) Rhizobium

Question 23.
Element involved in Nitrogen fixation is
a) Zinc
b) Copper
c) iron
d) Chlorine
Answer:
c) Iron

Question 24.
The nitrogenase enzyme is active:
(a) only in aerobic condition
(b) only in anaerobic condition
(c) both in aerobic and anaerobic condition
(d) only in toxic condition
Answer:
(b) only in anaerobic condition

Question 25.
Plants that can grow in marshy places where there is scarcity of Nitrogen are
a) Halophytes
b) Psammophytes
c) Bryophytes
d) insectivorous plants
Answer:
d) Insectivorous plants

Question 26.
Decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called:
(a) nitrification
(b) ammonification
(c) nitrogen fixation
(d) denitrification
Answer:
(b) ammonification

Question 27.
Internal cork of apple and Exanthema in citrus and whiptail disease of cauliflower are produced by the deficiency of
1. Copper,
2. Zinc,
3. Boron
4. Molybdenum
a) 2,3, 1
b) 2, 3, 4
c) 4, 3, 1
d) 3, 1,4
Answer:
d) 3,1,4

Question 28.
Necrosis means
a) Discolouration of leaf
b) Stunted growth
c) Death of the tissue
d) Death of the root
Answer:
c) Death of the tissue

Question 29.
The transfer of amino group (NH₂) from glutamic acid to keto group of keto acid is termed as:
(a) Transamination
(b) Hydrogenation
(c) Nitrification
(d) Denitrification
Answer:
(a) Transamination

Question 30.
Denitrification process deplete important nutrients from soil. It also cause ………………………
a) Acidification of soil
b) Alkalification of soil
c) Neutralization of soil
d) Ammoniafication of soil
Answer:
a) Acidification of soil

Question 31.
Availability of Nitrogenase enzyme depend on
a) Non avoulability of ATP
b) Availability of Nitric acid
c) Availability of ATP
d) Non availability of Nitric acid
Answer:
c) Availability of ATP

Question 32.
Obligate or Total parasites are
a) Santalum albumn and orabanche
b) Vanda and Venilla
c) Cuscuta and Rafflesia
d) Viscum and Loranthus
Answer:
c) Cuscuta and Rafflesia

Question 33.
The association of mycorrhizae with higher plants is termed as:
(a) Parasitism
(b) Mutualism
(c) Symbiosis
(d) Saprophytic
Answer:
(c) Symbiosis

Question 34.
Major role of minor elements inside living organism is to act as
a) Binder of cell structure
b) Constituent of hormone
c) Building blocks of important amino acids
d) Co factors of enzymes
Answer:
d) Co factors of enzymes

Question 35.
Lichens are the indicators of:
(a) carbon monoxide
(b) nitrogen oxide
(c) sulphur di oxide
(d) hydrogen sulphide
Answer:
(c) sulphur di oxide

Question 36.
Free living aerobic nitrogen fixing bacterium is
a) Azotobacter, Beijemeckia and Derxia
b) Nostoc, Anabaena, and Oscullatoria
c) Saccharomyces, Pullularia, Pseudomonas
d) Chlorobium and Rhodospirillum
Answer:
a) Azotobacter, Beijerneckia and Derxia

Question 37.
Leguminous plants does not include
a) Black gram
b) Bengal gram
c) Pongamia
d) Casuarina
Answer:
d) Casuarina

Question 38.
Cyanobacteria does not include
a) Nostoc
b) Anabaena
c) Clostridium
d) Oscillatoria
Answer:
c) Clostridium

II. Match The Following & Find Out The Correct Option

Question 39.
Cuscuta – A) Giant flower
Dianaea – B) Pitcher plant
Rafflesia – C) Dodder
Utricularia – D) Venus fly trap
Nepenthus – E) Bladder wort

Answer:
b) C-D-A-E-B

Question 40.

Answer:
b) D-A-B-C

Question 41.

Answer:
a) D-A-B-C

Question 42.
I) Criteria required for essential minerals was given by – A) Julius von Sachs
II) Word – Hydroponics Was coined by – B)SoiferHillel& David Durger
III) Hydroponics was developed by – C)Amon& Stout
IV) Aeroponics was developed by – D) William Frederick Goerick

Answer:
c) C -D-A-B

III. Find Out The Incorrect Statement With Reference To Potassium

Question 43.
a. It is essential for opening & closing of stomata
b. It is an essential component of vitamins, hormones, alkaloids and chlorophyll
c. It maintains osmotic potential of the cell
d. It maintain anion, cation balance by ion exchange.
Answer:
b. It is an essential component of vitamins, hormones, alkaloids and chlorophyll

Question 44.
a. Magnesium is a constituent of chlorophyll
b. Iron is essential for the formation of chlorophyll
c. Phosphorus is a component of ATP
d. Copper is essential for the synthesis of IAA
Answer:
d. Copper is essential for the synthesis of IAA

Question 45.
Find out wrong choice with reference to symbiotic mode of Nutrition
a. Lichens
b. Mycorrhizae
c. Coralloid roots of cycas
d. Viscum
Answer:
d. Viscum

Question 46.
The deficiency of which two exhibit competitive behaviour and the deficiencey of the two showing same symptoms.
(I) Iron
(II) Magnesium
(III) Calcium
(IV) Manganese
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
d. I & IV

Question 47.
Statement
(I) Alnus and Casuarina are nonlegume nitrogen fixers containing bacterium Frankia
(II) Nostoc and Anabaena are present in the corolloid roots of cycas.
a) Both (I) & (IT) are correct
b) (I) is correct (II) is wrong
c) (I) is wrong (II) is correct
d) Both (I) & (II) are wrong
Answer:
a) Both (I) & (II) are correct

Question 48.
Statement
(I) Dionaea is a submerged hydrophyte in which leaf is modified into a bladder to trap insects
(II) Loranthus is a partial stem parasite, absorb water and minerals from the xylem of the host
a) Both (I) & (II) are correct
b) (I) is correct (II) is wrong
c) (I) is wrong (II) is correct
d) Both (I) & (II) are wrong
Answer:
c) (I) is wrong (II) is correct

Assertion ‘A’ & Reason ‘R’
a) Both ‘A’ and ‘R’ are True and ‘R’ is the correct explanation of A
b) Both A and R are True, but R is not the correct explanation of A
c) A is True but ‘R’ is False
d) A& Rare False

Question 49.
Assertion: A Manganese is a Micro element
Reason: R Micro elements are required in traces only, less than 1 mg/gm of dry matter
Answer:
a) Both A and R are True and R is the correct explanation of A

Question 50.
Assertion: Calcium is a constituent of cell wall
Reason: R Calcium is required in mitotic division.
Answer:
b) A and R are True but ‘R’ is not the correct explanation of A

Question 51.
Assertion: A Deficiency of sulphur causes chlorosis in plants
Reason: R Sulphur is a constituent of chlorophyll
Answer:
c) A is True but ‘R’ is false

Question 52.
Assertion: A Plants absorb Nitrogen in the form of Nitrate only
Reason: R Nitrogen is the most critical element
Answer:
d) Both A and R are false

Question 53.
Assertion: A Mineral salt absorption is an active process.
Reason: R Metabolic energy is not used in active absorption.
Answer:
c) A is true but ‘R’ is false

IV. 2 Mark Questions

Question 1.
Define micronutrients of plants.
Answer:
Essential minerals which are required in less concentration called Micronutrients.

Question 2.
Is there any mne monic for remembering essential minerals?
Answer:
CHOPKNs Cafe Mg B Mn Cu Zn Mo Cl (C) HOPKINS (name) Cafe managed by Mine CUZINS, Mo tnd Claude”.

Question 3.
What is the role of molybdenum in the conversion of nitrogen into ammonia?
Answer:
Molybdenum (Mo) is essential for nitrogenase enzyme during the reduction of atmospheric nitrogen into ammonia.

Question 4.
What are the minerls classifed as unclassified minerals and why?
Answer:
5ome minerals Such as Sodium, Silicon, Cobalt and Selenium some minerals are not included in the list ol essential nuitrients by they play some specific roles.
Eg. Silicon

• essential for pest resistance
• prevent water lodging
• aids in cell wall formation in Equisetaceae, Cyperaceae & Gramineae

Question 5.
What are the deficiency symptoms of nitrogen?
Answer:
Chlorosis, stunted growth, anthocyanin formation.

Question 6.
Distinguish between Hydroponics & Aeroponics
Answer:
Hydroponics: Growing plants in nutrient solution with roots immerse in it and air is supplied with the help of tube.
Hydroponics: It Is a system where roots suspended in air and nuitrients solution in a tank is sprayed over the roots by motor driven rotor – in the form of mist.

Question 7.
Define the term Siderophores.
Answer:
Siderophores (iron carriers) are iron-chelating agents produced by bacteria. They are used to chelate ferric iron (Fe³⁺) from environment and host.

Question 8.
What are called critical elements & complete fertilizers?
Answer:

• Macro elements which commonly remain deficient in the soil are called Critical elements, (ie) N.P.K.
• The fertilizer which contain critical elements are called complete fertilizer. They are expressed in the ratio 15: 15: 15(N:P: K)

Question 9.
Why is Iron kept between Macro and Micro nuitrients?
Answer:
Iron is required lesser than macro nuitrients and larger than the micronuitrient so it can be placed in any one of the two groups.

Question 10.
Write down the deficiency symptoms of molybdenum in plants.
Answer:
Chlorosis, necrosis, delayed flowering, retarded growth and whip tail disease of cauliflower.

Question 11.
List two purpose for which you think Magnesium is required essentially to the plants.
Answer:
(I) Synthesis of Chlorophyll
(II) Formation of nodules in legumes

Question 12.
Define Aeroponics.
Answer:
It is a system where roots are suspended in air and nutrients are sprayed over the roots by a motor driven rotor.

Question 13.
What is meant by Toxicity of Minerals
Answer:
If mineral nuitrients lesser than critical concentration cause deficiency, where as when there is increase in
mineral nuitrients more than normal concentration cause Toxicity Toxicity ¡s that particular concentration at which 10% of the dry weight of tissue is reduced.

Question 14.
Give examples for Nitrogen Fixation with out nodulation.
Answer:

Question 15.
Give examples for Non – symbiotic Nitrogen fixation by bacteria and Fungi.
Answer:

Question 16.
Define the term Nitrate assimilation.
Answer:
The process by which nitrate is reduced to ammonia is called nitrate assimilation and occurs during the nitrogen cycle.

Question 17.
What are the negative effects of denitrification.
Answer:

• Nitrate in the soil are converted back to atmospheric nitrogen.
• Denitrification process deplete important nuitrients from the soil.
• It also causes acidification of the soil.

Question 18.
Name 2 hormones involved in Nodule formation.
Answer:
During nodule formation in leguminous plants cytokinin from bacteria and Auxin from host (leguminous) plant promotes cell division and leads to nodule formation.

Question 19.
Give two examples of symbiotic mode of nutrition.
Answer:
Two examples of symbiotic mode of nutrition:

1. Lichens: It is a mutual association of Algae and Fungi. Algae prepares food and fungi absorb water and provides thallus structure.
2. Mycorrhizae: Fungi associated with roots of higher plants including Gymriosperms. eg: Pinus.

Question 20.
Decreased availability of the element results in early fall of fruits and flowers. Identify the element.
Answer:
Phosphorus, Magnesium and Copper (Any one of these three elements) may cause the above symptoms.

Question 21.
Name any 3 diseases caused by copper deficiency.
Answer:

1. Die back of Citrus.
2. Reclamation disease of cereals & legumes.
3. Exanthema in Citrus.

Question 22.
Notes on unclassified minerals.
Answer:
Required by some plants – for some specific functions, in trace amounts.
Example: Sodium, Silicon, Selenium & Cobalt.

Question 23.
Explain Nitrate Assimilation.
Answer:
Definaition: The process by which nitrate is reduced to ammonia is called Nitrate assimilation and it occurs during Nitrogen cycle.

Question 24.
Explain Aluminium Toxicity.
Answer:
Aluminium toxicity causes,

• Precipitation ofNucleic acid
• Inhibition of ATP ase
• Inhibition of cell division and binding of Plasma membrane with Calmodulin.

Question 25.
Differentiate between Nitrification & Denitrification
Answer:

Question 26.
Organisms like Pseudomonas and Thiobacillus are of great significance in nitrogen cycle. How?
Answer:
These microorganisms carry out denitrification they help to maintain the constant level of nitrogen in the atmosphere.

Question 27.
What is meant by Symbiotic association give examples?
Answer:
Close relationship between two organism, both being benefitted out of it is known as symbiosis.
Eg. 1. Nitrogen fixing bacteria Nitrosomonas living in the root nodules of leguminous plants.
2. Fungi associated with roots of higher plants is a symbiotic association known as Mycorrhiza

Question 28.
What is the use of FTWS.
Answer:

• FTWS – means floating treatment wet lands.
• It works on the principle of hydroponics recently FTWS work on the principle of hydroponics, helping to solve pollution that come up due to Eutrophication.

Question 29.
Notes on Lichens.
Answer:

• Lichens are pioneer species in xeric succession.
• Lichens are nothing but symbiotic association of Algae and Fungi partners.
• Lichens are also indicators of S0₂ pollution.

Question 30.
Notes on Haustoria.
Answer:
Total parasitic or partial parasites they have some special structures to absorb food or water from the host plant phloem and xylem. These special absorbing structures are known as Haustoria.

Question 31.
Identify the diagram A.
Answer:
Cycas corolloid roots – have symbiotic association with Nostoc helping to fix nitrogen.

Question 32.
Identify the diagram.
Answer:
Root nodules of leguminous plant inhabiting Rhizobium fixing nitrogen

3 Mark Questions.

V. Identify And Complete The Equations

Question 1.

Answer:

Question 2.
Explain the unclassified minerals required for plants.
Answer:
Minerals like Sodium,Silicon, Cobalt and Selenium are not included in the list of essential nutrients but are required by some plants, these minerals are placed in the list of unclassified minerals. These minerals play specific roles for example, Silicon is essential for pest resistance, prevent water lodging and aids cell wall formation in Equisetaceae (Equisetum), Cyperaceae and Gramineae.

Question 3.
Draw a model of Hydroponics.
Answer:

Question 4.
Explain briefly the functions and deficiency symptoms of potassium.
Answer:
Functions: Maintains turgidity and osmotic potential of the cell, opening and closure of stomata, phloem translocation, stimulate activity of enzymes, anion and cation balance by ion – exchange. It is absorbed as K⁺ ions. Deficiency symptoms: Marginal chlorosis, necrosis, low cambial activity, loss of apical dominance, lodging in cereals and curled leaf margin.

Question 5.
Draw the schematic representation of Nitrogenase enzyme function.
Answer:

Question 6.
Explain the term critical concentration of minerals.
Answer:
To increase productivity and also to avoid mineral toxicity knowledge of critical concentration is essential. Mineral nutrients lesser than critical concentration cause deficiency symptoms. Increase of mineral nutrients more than the normal concentration causes toxicity. A concentration, at which 10% of the dry weight of tissue is reduced, is considered a toxic critical concentration.

Question 7.
Nitrogen fixation is shown by Prokaryotes and not by Eukaryotes comment.
Answer:
Nitrogen fixation is the phenomenon that occurs in Prokaryotes but not in Eukaryotes, because the enzymes nitrogenase, which is capable of nitrogen reduction is present exclusively in prokaryotes and such microbes are often called fixers.

Question 8.
Who are people responsible for developing hydroponics?
Answer:
Hydroponics or Soil less culture: Von Sachs developed a method of growing plants in nutrient solution. The commonly used nutrient solutions are Knop solution (1865) and Amon and Hoagland Solution (1940). Later the term Hydroponics was coined by Goerick (1940) and he also introduced commercial techniques for hydroponics. In hydroponics roots are immersed in the solution containing nutrients and air is supplied with help of tube.

VI. 5 Mark Questions

Question 1.
Classify minerals on the basis of on their function.
a) Structural component – C, H, O & N
b) Enzyme function – Mo, Zn, Mg, &Ni

C) Osmotic potential – K:
Potassium -(K) – maintain osmotic Potential by 2 steps.

1. Absorption of water
2. Movement of stomata & turgidity

d) Energy components:
Mg – in chlorophyll
P- in ATP

Question 2.
Tabulate the mode of absorption, function and deficiency symptoms of any 5 microelements.
Answer:

Question 3.
Give the details of minerals and their deficiency symptoms.
Answer:
Name of the deficiency disease and symptoms:

1. Chlorosis (Overall)
   • Interveinal chlorosis
   • Marginal chlorosis
2. Necrosis (Death of the tissue)
3. Stunted growth
4. Anthocyanin formation
5. Delayed flowering
6. Die back of shoot, Reclamation disease, Exanthema in citrus (gums on bark)
7. Hooked leaf tip
8. Little Leaf
9. Brown heart of turnip and Internal cork of apple
10. Whiptail of cauliflower and cabbage
11. Curled leaf margin

Deficiency minerals:

1. Nitrogen, Potassium, Magnesium, Sulphur, Iron, Manganese, Zinc and Molybdenum. Magnesium, Iron, Manganese and Zinc Potassium
2. Magnesium, Potassium, Calcium, Zinc, Molybdenum and Copper.
3. Nitrogen, Phosphorus, Calcium, Potassium and Sulphur.
4. Nitrogen, Phosphorus, Magnesium and Sulphur
5. Nitrogen, Sulphur and Molybdenum
6. Copper
7. Calcium
8. Zinc
9. Boron
10. Molybdenum
11. Potassium

Question 4.
Why are NPK fertilizers important to plants?
Answer:
Nitrogen: It helps in plant growth and development.

• It required in large amount
• It is essential component of Proteins, Amino acids, Nucleic acids, Vitamins, Hormones, Chlorophyll etc.

Phosphorus:
It is an important constituent of Cell membrane, Proteins, Nucleic acids, ATP, NADP etc.

Potassium:

• It is essential to maintain turgidity and osmotic potential of the cell.
• Opening and closure of stomata.
• Phloem translocation.
• Ion exchange etc.
• So overall all the three in right proportion is used by farmers for various plants to enhance yield.

Question 5.
Tabulate the major Essential elements their function & Deficiency symptoms.
Answer:

Question 6.
What are the stages of Root nodule formation.
Answer:
1. Attraction:
Legume roots secretes Phenolics to attract Rhizobium.

2. Infection:

• Rhizobium – reaches rhizosphere
• Rhizosphere – to root hair.
• Curling of root hairs.

3. Spreading & multiplication:
Infection thread grows inwards and infected area is separated from normal tissue.

4. Bacteriod formation:
A membrane bound bacterium is formed inside the nodule ……………. called Bacterioid.

5. Nodule formation:

• Cytokinin from Bacteria.
• Auxin from legume roots together promote cell division and nodules are formed.

Question 7.
Explain the fate of Ammonia or Assimilation of Ammonia.
Answer:

• Ammonia ions are quite toxic to plants, and hence cannot accumulate in the plants.
• It should be converted into Amino acids.

There are 3 methods by which it is done.

I) Reductive amination:
In this ammonia reacts with Ketoglutaric acid and form glutamic acid.

II) Transamination:

• It involves the transfer of amino group from one amino acid to the ketogroup of another keto acid.
• Glutamic acid is the main amino acid from which the transfer of NH2 (amino group) takes place and other amino acids are fonned through transamination.
• The enzyme Transaminase + Pyridoxus phosphate (COenz) reactions.

Example:

III) Catalytic Amination (GS/GOGAT path way)

Question 8.
Explain parasitic mode of Nuitrition.
Answer:
Definition:
Organism deriving their nuitrients from another organism (host and causing damage/disease to the host is known as parasite. Stem parasite Root parasite Stem parasite Root parasite.

I) Obligate or Total parasite :

1. Completely depends on host for their survival produce haustoria.
   Total stem parasite:
2. Leafless plant twine around the host. Eg. Cuscuta on Zizipus, citrus etc.
   Total root parasite:
3. Plants do not have stem axis – so grow in the roots of host plants produce haustoria.
   Eg. Rafflesia, Orobanche and Balanophora.

II) Partial parasite:
Plant have chlorophyll on their leaves dependent on water and mineral requirements.

• Partial stem parasite: The plant grow an fig and mango and absorb water and minerals from xylem of host through haustoria.
  Eg. Loranthus.
• Partial root parasite: This plant in its juvenile stages produces haustoria which grow on roots of many forest trees.
  Eg. Sandal wood tree (santalum album)

Question 9.
Describe Saprophytic mode of nuitrition in Angiosperms?
Answer:
Definition:
Derving nuitrients from dead and decaying organic matter is known as saprophytic – nuitrition.
Eg. Bacteria, Fungi

Saprophytic Angiosperms:

• Neottia: (Bird’s nest orchid) Roots of Neottia get associated with the mycorrhizae and absorb nuitrients from the litter in the soil.
• The plant leaves lack chlorophyll so dependon mycorrhiza to absorb nuitrients from the decomposed litter in the soil.
• Monotropa: (Indian pipe) It also lack leaves, so absorb nuitrients from the soil through the mycorrhizal association.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 5 Alkali and Alkaline Earth Metals Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

11th Chemistry Guide Alkali and Alkaline Earth Metals Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
For alkali metals, which one of the following trends is incorrect?
(a) Hydration energy: Li > Na > K> Rb
(b) Ionisationenergy: Li> Na> K> Rb
(c) Density: Li < Na < K < Rb
(d) Atomic size: Li < Na < K < Rb
Answer:
(c) Density: Li < Na < K < Rb

Question 2.
Which of the following statements is incorrect?
(a) Li⁺ has minimum degree of hydration among alkali metal cations
(b) The oxidation state of K in KO₂ is +1
(c) Sodium is used to make Na / Pb alloy
(d) MgSO₄ is readily soluble in water
Answer:
(a) Li⁺ has minimum degree of hydration among alkali metal cations

Question 3.
Which of the following compounds will not evolve H₂ gas on reaction with alkali metals ?
(a) ethanoic acid
(b) ethanol
(c) phenol
(d) none of these
Answer:
(d) none of these

Question 4.
Which of the following has the highest tendency to give the reaction, M⁺_(g) M⁺_(aq)
(a) Na
(b) Li
(c) Rb
(d) K
Answer:
(b) Li

Question 5.
sodium is stored in
(a) alcohol
(b) water
(c) kerosene
(d) none of these
Answer:
(c) kerosene

Question 6.
RbO₂ is
(a) superoxide and paramagnetic
(b) peroxide and diamagnetic
(c) superoxide and diamagnetic
(d) peroxide and paramagnetic
Answer:
(a) superoxide and paramagnetic

Question 7.
Find the wrong statement
(a) sodium metal is used in organic qualitative analysis
(b) sodium carbonate is soluble in water and it is used in inorganic qualitative analysis
(c) potassium carbonate can be prepared by solvay process
(d) potassium bicarbonate is acidic salt
Answer:
(c) potassium carbonate can be prepared by solvay process

Question 8.
Lithium shows diagonal relationship with
(a) sodium
(b) magnesium
(c) calcium
(d) aluminium
Answer:
(b) magnesium

Question 9.
Incase of alkali metal halides, the ionic character increases in the order
(a) MF < MCl < MBr < MI
(b) MI < MBr < MCl < MF
(c) MI < MBr < MF < MCl
(d) none of these
Answer:
(b) MI < MBr < MCl < MF

Question 10.
In which process, fused sodium hydroxide is electrolysed for extraction of sodium?
(a) Castner’s process
(b) Cyanide process
(c) Down process
(d) All of these
Answer:
(a) Castner’s process

Question 11.
The product obtained as a result of a reaction of nitrogen with CaC2 is
(a) Ca(CN)₃
(b) CaN₂
(c) Ca(CN)₂
(d) Ca₃N₂
Answer:
(c) Ca(CN)₂

Question 12.
Which of the following has highest hydration energy
(a) MgCl₂
(b) CaCl₂
(c) BaCl₂
(d) SrCl₂
Answer:
(a) MgCl₂

Question 13.
Match the flame colours of the alkali and alkaline earth metal salts in the bunsen burner

(a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3
(b) p – 1, q – 2, r – 4, s – 5, t – 6, u – 3
(c) p – 4, q – 1, r – 2, s – 3, t – 5, u – 6
(d) p – 6, q – 5, r – 4, s – 3, t – 1, u – 2
Answer:
(a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3

Question 14.
Assertion:
Generally alkali and alkaline earth metals form superoxides
Reason:
There is a single bond between O and O in superoxides.
(a) both assertion and reason are true and reason is the correct explanation of assertion .
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 15.
Assertion:
BeSO₄ is soluble in water while BaSO₄ is not
Reason:
Hydration energy decreases down the group from Be to Ba and lattice energy remains almost constant.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Question 16.
Which is the correct sequence of solubility of carbonates of alkaline earth metals ?
(a) BaCO₃ > SrCO₃ > CaCO₃ > MgCO₃
(b) MgCO₃ > CaCO₃ > SrCO₃ > BaCO₃
(c) CaCO₃ > BaCO₃ > SrCO₃ > MgCO₃
(d) BaCO₃ > CaCO₃ > SrCO₃ > MgCO₃
Answer:
(b) MgCO₃ > CaCO₃ > SrCO₃ > BaCO₃

Question 17.
In context with beryllium, which one of the following statements is incorrect?
(a) It is rendered passive by nitric acid
(b) It forms Be₂C
(c) Its salts are rarely hydrolysed
(d) Its hydride is electron deficient and polymeric
Answer:
(c) Its salts are rarely hydrolysed

Question 18.
The suspension of slaked lime in water is known as
(a) lime water
(b) quick lime
(c) milk of lime
(d) aqueous solution of slaked lime
Answer:
(c) milk of lime

Question 19.
A colourless solid substance (A) on heating evolved CO₂ and also gave a white residue, soluble in water. Residue also gave CO₂ when treated with dilute HCl.
(a) Na₂CO₃
(b) NaHCO₃
(c) CaCO₃
(d) Ca(HCO₃)₂
Answer:
(b) NaHCO₃

Question 20.
The compound (X) on heating gives a colourless gas and a residue that is dissolved in water to obtain (5). Excess of CO₂ is bubbled through aqueous solution of B, C is formed. Solid (C) on heating gives back X. (B) is
(a) CaCO₃
(b) Ca(OH)₂
(c) Na₂CO₃
(d) NaHCO₃
Answer:
(b) Ca(OH)₂

Question 21.
Which of the following statement is false?
(a) Ca²⁺ ions are not important in maintaining the regular beating of the heart
(b) Mg²⁺ ions are important in the green parts of the plants
(c) Mg²⁺ ions form a complex with ATP
(d) Ca²⁺ ions are important in blood clotting
Answer:
(a) Ca²⁺ ions are not important in maintaining the regular beating of the heart

Question 22.
The name ‘Blue John’ is given to which of the following compounds?
(a) CaH₂
(b) CaF₂
(c) Ca₃(PO₄)₂
(d) CaO
Answer:
(b) CaF₂

Question 23.
Formula of Gypsum is
(a) CaSO₄ .2H₂O
(b) CaSO₄ .\(\frac{1}{2}\)H₂O
(c) 3CaSO₄ .H₂O
(d) 2CaSO₄ .2H₂O
Answer:
(a) CaSO₄ .2H₂O

Question 24.
When CaC2 is heated in atmospheric nitrogen in an electric furnace the compound formed is
(a) Ca(CN)₂
(b) CaNCN
(c) CaC₂N₂
(d) CaNC₂
Answer:
(b) CaNCN

Question 25.
Among the following the least thermally stable is
(a) K₂CO₃
(b) Na₂CO₃
(c) BaCO₃
(d) Li₂CO₃
Answer:
(d) Li₂CO₃

II. Write brief answer to the following questions:

Question 26.
Why sodium hydroxide is much more water-soluble than chloride?
Answer:

1. Sodium hydroxide is stronger base whereas sodium chloride is a salt.
2. Sodium hydroxide dissolve freely in water with evolution of much heat on account of intense hydration.
3. In other words when the Na⁺ and OH^– ions break up, the OH^– ions are much smaller than Cl^– ions and are able to form a hydrogen bond with water.
4. Thus sodium hydroxide dissolves easily in water.

Question 27.
Explain what to mean by efflorescence.
Answer:

1. Efflorescence is a process of losing water of hydration from hydrate.
2. Sodium carbonate crystallises as decahydrate which is white in colour.
3. Upon heating, it loses the water of crystallization to form monohydrate.
4. Monohydrate (Na₂CO₃.H₂O) is formed as a result of efflorescence.
   Na₂CO₃ .10H₂O → Na₂CO₃.H₂O + 9H₂O

Question 28.
Write the chemical equations for the reactions involved in solvay process of preparation of sodium carbonate.
Answer:
2NH₃ + H₂O + CO₂ → (NH₄)₂CO₃
(NH₄)₂CO₃ + H₂O + CO₂ → 2NH₄HCO₃
2NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O

Question 29.
An alkali metal (x) forms a hydrated sulphate, X₂SO₄ .10H₂O. Is the metal more likely to be sodium (or) potassium?
Answer:
Sodium: Because hydration is favoured by high charge density cations and of the two mono positive ions, sodium is smaller and will have higher charge density.
Thus, Na₂SO₄.10H₂O is more readily formed.

Question 30.
Write a balanced chemical equation for each of the following chemical reactions.
(i) Lithium metal with nitrogen gas
(ii) heating solid sodium bicarbonate
(iii) Rubidum with oxgen gas
(iv) solid potassium hydroxide with CO₂
(v) heating calcium carbonate
(vi) heating calcium with oxygen
Answer:
(i) 6Li(s) + N₂(g) → 2Li₃N(s)
(ii) 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
(iii) Rb + O₂ → RbO₂
(iv) 2KOH + CO₂ → K₂CO₃ + H₂O
(v) CaCO₃ → CaO + CO₂
(vi) 2Ca + O₂ → 2 CaO

Question 31.
Discuss briefly the similarities between beryllium and aluminium.
Answer:

1. Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms a polymeric chain structure in addition to dimer. Both are soluble in organic solvents and are strong Lewis acids.
2. Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)₄]^2- and hydrogen as aluminium hydroxide which gives aluminate ion, [Al(OH)₄]^2-.
3.  Beryllium and aluminium ions have strong tendency to form complexes, BeF₄^2-, AlF₆^3-.
4. Both beryllium and aluminium hydroxides are amphoteric in nature.
5. Carbides of beryllium (Be₂C) like aluminium carbide (Al₄C₃) give methane on hydrolysis.
6. Both beryllium and aluminium are rendered passive by nitric acid.

Question 32.
Give the systematic names for the following
(i) milk of magnesia
(ii) lye
(iii) lime
(iv) Caustic potash
(v) washing soda
(vi) soda ash
(vii) trona
Answer:
(i) Magnesium hydroxide
(ii) caustic soda(Sodium Hydroxide)
(iii) calcium oxide
(iv) Potassium Hydroxide
(v) sodium carbonate
(vi) sodium carbonate
(vii) Sodium Sesquicarbonate

Question 33.
Substantiate Lithium fluoride has the lowest solubility among group one metal fluorides.
Answer:
Lithium fluoride has the lowest solubility among alkali metal fluoride due to its small size of Li⁺ and F^– ions, lattice enthalpy is much higher than that of hydration enthalpy.

Question 34.
Mention the uses of plaster of paris.
Answer:
The largest use of Plaster of Paris is in the building industry as well as plasters. It is used for immobilising the affected part of organ where there is a bone fracture or sprain. It is also employed in dentistry, in ornamental work and for making casts of statues and busts.

Question 35.
Beryllium halides are Covalent whereas magnesium halides are ionic why?
Answer:
Halogens are non-metals and beryllium is also a non-metal. Since non-metals always form covalent bonds with each other due to almost similar ionization potential and electronegativity. And Beryllium is smaller in size and has high polarizing power therefore, beryllium halides are covalent.

Magnesium is a metal and metals mostly form ionic bonds with non-metals due to vast difference in their ionization potential and electronegativity, therefore magnesium halides are always ionic.

Question 36.
Alkaline earth metal (A), belongs to 3^rd period reacts with oxygen and nitrogen to form compound (B) and (C) respectively. It undergoes metal displacement reaction with AgNO₃ solution to form compound (D).
Answer:
Alkaline earth metal, 3^rd → Magnesium(Mg) ……….(A)
2 Mg + O₂ → 2MgO …………(B)
3 Mg + N₂ → Mg₃N₂ ……….(C)
Mg + 2 AgNO₃ → 2 Ag + Mg(NO₃)₂ ………….(D)
A – Magnesium
B – Magnesium oxide
C – Magnesium nitride
D – Magnesium nitrate

Question 37.
Write balanced chemical equation for the following processes
(a) heating calcium in oxygen
(b) heating calcium carbonate
(c) evaporating a solution of calcium hydrogen carbonate
(d) heating calcium oxide with carbon
Answer:
(a) 2 Ca + O₂ → 2CaO
(b) CaCO₃ → CaO + CO₂
(c) Ca(HCO₃)₂ → CO₂ + H₂O + CaCO₃.
(d) CaO + 3 C → CaC₂ + CO

Question 38.
Explain the important common features of Group 2 elements.
Answer:

1. Group 2 is known as alkaline earth metals. It contains soft, silver metals that are less metallic in character than the Group 1 elements. Although many characteristics are common throughout the group, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the Group 1 Alkali Metals.
2. General electronic configuration can be represented as [Noble gas] ns² where ‘n’ represents the valence shell.
3. All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2. This enables the metals to easily lose electrons, which increases their stability and allows them to form compounds via ionic bonds.
4. The atomic and ionic radii of alkaline earth metals are smaller than the corresponding members of the alkali metals.
5. On moving down the group, the radii increases due to gradual increase in the number of the shells and the screening effect.
6. Down the group the ionisation enthalpy decreases as atomic size increases. They are less electropositive than alkali metals.
7. Compounds of alkaline earth metals are more extensively hydrated than those of alkali metals, because the hydration enthalpies of alkaline earth metal ions are larger than those of alkali metal ions.

Question 39.
Discus the similarities between beryllium and aluminium.
Answer:
Similarities between Beryllium and Aluminium

1. Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms polymeric chain structure in addition to dimer. Both are soluble in organic solvents and are strong Lewis acids.
2. Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)₄]^2-and hydrogen as aluminium hydroxide which gives aluminate ion, [Al(OH)₄]^–.
3. Beryllium and aluminium ions have strong tendency to form complexes, BeF₄^2-, AlF₆^3-.
4. Both beryllium and aluminium hydroxides are amphoteric in nature.
5. Carbides of beryllium (Be₂C) like aluminium carbide (Al₄C₃) give methane on hydrolysis.
6. Both beryllium and aluminium are rendered passive by nitric acid.

Question 40.
Why alkaline earth metals are harder than alkali metals?
Answer:
The strength of metallic bond in alkaline earth metals is higher than the alkali metals due to presence of 2 electrons in its outermost shell as compared to alkali metal which have only 1 electron in valence shell. Therefore alkaline earth metals are harder than the alkali metals.

Question 41.
How is plaster of paris prepared?
Answer:
It is a hemihydrate of calcium sulphate. It is obtained when gypsum,
CaSO₄.2H₂O is heated to 393 K.
2CaSO₄.2H₂O(s) → 2CaSO₄.H₂O + 3H₂O
Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO₄ is formed. This is known as ‘dead burnt plaster’.

Question 42.
Give the uses of gypsum.
Answer:

1. Gypsum is used in making drywalls or plasterboards. Plasterboards are used as the finish for walls and ceilings, and for partitions.
2. Another important use of gypsum is the production of plaster of Paris. Gypsum is heated to about 300 degree Fahrenheit to produce plaster of Paris, which is also known as gypsum plaster. It is mainly used as a sculpting material.
3. Gypsum is used in making surgical and orthopaedic casts, such as surgical splints and casting moulds.
4. Gypsum plays an important role in agriculture as a soil additive, conditioner, and fertilizer. It helps loosen up compact or clay soil, and provides calcium and sulphur, which are essential for the healthy growth of a plant. It can also be used for removing sodium from soils having excess salinity.
5. Gypsum is used in toothpaste, shampoos, and hair products, mainly due to its binding and thickening properties.
6. Gypsum is a component of Portland cement, where it acts as a hardening retarder to control the speed at which concrete sets.

Question 43.
Describe briefly the biological importance of Calcium and magnesium.
Answer:
Magnesium:

1. A typical adult human body contains about 25 g of magnesium and 1200 g of calcium.
2. Magnesium plays an important role in many biochemical reactions catalyzed by enzymes.
3. It is the co-factor of all enzymes that utilize ATP in phosphate transfer and energy release.
4. It also essential for DNA synthesis and is responsible for the stability and proper functioning of DNA.
5. It is also used for balancing electrolytes in our body.
6. Deficiency of magnesium results into convulsion and neuromuscular irritation.
7. The main pigment that is responsible for photosynthesis, chlorophyll, contains magnesium which plays an important role in photosynthesis.

Calcium:

1. Calcium is a major component of bones and teeth.
2. It is also present in in blood and its concentration is maintained by hormones (calcitonin and parathyroid hormone).
3. Deficiency of calcium in blood causes it to take longer time to clot. It is also important for muscle contraction.

Question 44.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

1. Magnesium fluoride – 1263°C
2. Magnesium oxide – 2852°C
3. The strength of ionic bonds usually depends on two factors – ionic radius and charge. Mg²⁺ and O^2- have charges of +2 and -2 respectively. This is larger than the charge of other ions like.
4. Magnesium ions and oxygen ions also have small ionic radius.
5. Oxygen ion is smaller than fluoride
6. The smaller the ionic radii, the smaller the bond length and the stronger the bond. Therefore the ionic bond between magnesium and oxygen is very strong.

11th Chemistry Guide Alkali and Alkaline Earth Metals Additional Questions and Answers

I. Choose the best Answer:

Question 1.
The reducing property of alkali metals follows the order
(a) Na < K < Rb < Cs < Li
(b) K < Na < Rb < Cs < Li
(c) Li < Cs < Rb < K < Na
(d) Rb < Cs < K < Na < Li
Answer:
(a) Na < K < Rb < Cs < Li

Question 2.
Arrange the following in increasing order of hydration enthalpy.
(a) Rb⁺ > Li⁺ > Na⁺ > K⁺ > Cs⁺
(b) Cs⁺ > Rb⁺ > K⁺ > Na⁺ > Li⁺
(c) Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺
(d) K⁺ > Na⁺ > Li⁺ > Rb⁺ > Cs⁺
Answer:
(c) Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺

Question 3.
Li does not resemble other alkali metals in which of the following property?
(a) Li₂CO₃ decomposes into oxides while other alkali carbonates re thermally stable
(b) LiCl is predominantly covalent
(c) Li₃N stable
(d) All of the above
Answer:
(d) All of the above

Question 4.
1 mol of a substance (X) was treated with an excess of water, 2 mol of readily combustible gas were . produced along with solution which when reacted with CO₂ gas produced a white turbidity. The substance (X) could be
(a) Ca
(b) CaH₂
(c) Ca(OH)₂
(d) Ca(NO₃)₂
Answer:
(b) CaH₂

Question 5.
The alkali metal used in photoelectric cells is
(a) Na
(b) Cs
(c) Rb
(d) Fr
Answer:
(b) Cs

Question 6.
Na₂O₂ has light yellow colour. This is due to
(a) presence of unpaired electron in the molecule
(b) presence of trace of NaO₂
(c) presence of KO₂ ass an impurity
(d) none of the above
Answer:
(b) presence of trace of NaO₂

Question 7.
Be₂C + 4H₂O → 2X + CH₄
X + 2HCl + 2H₂O → Y
X and Y formed in the above two reactions is
(a) BeCO₃ and Be(OH)₂ respectively
(b) Be(OH)₂ and BeCl₂ respectively
(c) Be(OH)₂ and [Be(OH)₄]Cl₂ respectively
(d) [Be(OH)₄]^2- and BeCl₂ respectively
Answer:
(c) BQ(OH)₂ and [Be(OH)₄]Cl₂ respectively

Question 8.
When sodium reacts with excess of oxygen, oxidation number of oxygen changes from
(a) 0 to – 1
(b) 0 to 2
(c) – 1 to – 2
(d) +1 to -1
Answer:
(a) 0 to – 1

Question 9.
Magnesium burns in air to give
(a) MgO
(b) MgCO₃
(c) MgCO₃
(d) MgO and Mg₃N₂
Answer:
(d) MgO and Mg₃N₂

Question 10.
The composition of common baking powder is
(a) starch, sodium bicarbonate, citric acid
(b) Sodium bicarbonate, tartaric acid
(c) starch, sodium bicarbonate, citric acid
(d) Starch, sodium bicarbonate, calcium hydrogen phosphate.
Answer:
(d) Starch, sodium bicarbonate, calcium hydrogen phosphate.

Question 11.
The word ‘alkali’ used for alkali metals indicates
(a) ashes of plants
(b) metallic luster
(c) soft metals
(d) reactive metals
Answer:
(a) ashes of plants

Question 12.
Which salt can be used to identify coloured cation
(a) borax
(b) microcosmic salt
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)

Question 13.
Select the correct statement
LiOH > NaOH > KOH > RbOH
Li₂CO₃ > Na₂CO₃ > K₂CO₃ > Rb₂CO₃
(a) Solubility of alkali hydroxides is in order
(b) Solubility of alkali carbonates is in order
(c) both are correct
(d) None is correct
Answer:
(b) Solubility of alkali carbonates is in order

Question 14.
Which fumes in air?
(a) BeCl₂
(b) MgCl₂
(c) CaCl₂
(d) BaCl₂
Answer:
(a) BeCl₂

Question 15.
Which of the following ions form a hydroxide highly soluble in water?
(a) Ni²⁺
(b) K²⁺
(c) Zn²⁺
(d) Al³⁺
Answer:
(b) K²⁺

Question 16.
Which causes nerve signals in animals?
(a) Electrical potential gradient due to transfer of K⁺ ions
(b) Electrical potential gradient due to transfer of Na⁺ ions in (Na⁺ – K⁺) pumps
(c) Electrical potential gradient set up due to transfer of Ca²⁺ ions
(d) No nerve signal exists in animals.
Answer:
(a) Electrical potential gradient due to transfer of K⁺ ions

Question 17.
The carbide of which of the following metals on hydrolysis gives allylene or propyne?
(a) Be
(b) Ca
(c) Al
(d) Mg
Answer:
(d) Mg

Question 18.
Which of the following reaction produces hydrogen?
(a) Mg + H₂O
(b) H₂S₄O₈ + H₂O
(c) BaO₂ + HCl
(d) Na₂O₂ + 2HCl
Answer:
(a) Mg + H₂O

Question 19.
A major constituent of Portland cement (except lime) is
(a) Silica
(b) Alumina
(c) Iron oxide
(d) Magnesia
Answer:
(a) Silica

Question 20.
Which of the following is known as a variety of gypsum ?
(a) CaCO₃
(b) CaSO₄
(c) plaster of paris
(d) gypsum
Answer:
(d) gypsum

Question 21.
The elements belongs to group-1 are called as
(a) Alkali metals
(b) Alkaline earth metals
(c) halogens
(d) chalcogens
Answer:
(a) Alkali metals

Question 22.
Which one of the following is a radioactive element of alkali metal?
(a) Cesium
(b) Francium
(c) Potassium
(d) Sodium
Answer:
(b) Francium

Question 23.
Match the correct pair Element:

(a) A – ii, B – iii, C – i
(b) A – i, B – ii, C – iii
(c) A – ii, B – i, C – iii
(d) A – i, B – iii, C – ii
Answer:
(a) A – ii, B – iii, C – i

Question 24.
The colour of potassium salt in flame is
(a) Crimson red
(b) Lilac
(c) Blue
(d) Yellow
Answer:
(b) Lilac

Question 25.
Lithium reacts directly with carbon to form
(a) Li₂C₂
(b) Li₂C
(c) LiC₂
(d) LiC
Answer:
(a) Li₂C₂

Question 26.
Lithium shows diagonal relationship with
(a) Beryllium
(b) Carbon
(c) Magnesium
(d) Calcium
Answer:
(c) Magnesium

Question 27.
Which of the following element forms monoxide and peroxide?
(a) Lithium
(b) Potassium
(c) Rubedium
(d) Sodium
Answer:
(d) Sodium

Question 28.
Choose the correct pair:

(a) A – iii, B – i, C – iv, D – ii
(b) A – ii, B – iii, C – iv, D – i
(c) A – iv, B – ii, C – iii, D – i
(d) A – ii, B – iv, C – i, D – iii
Answer:
(a) A – iii, B – i, C – iv, D – ii

Question 29.
Sodium reacts with acetylene to give
(a) Sodium ethoxide
(b) Sodium acetylide
(c) Sodium hydroxide
(d) Sodamide
Answer:
(b) Sodium acetylide

Question 30.
The products obtained on reaction of Na₂O₂ with water are
(a) NaOH and H₂O
(b) NaOH and H₂O₂
(c) Na₂O and H₂O₂
(d) NaOH, Na₂O
Answer:
(c) Na₂O and H₂O₂

Question 31.
Choose the correct statement/s is are correct about alkali metals.
1. The oxides and peroxides are colourless when pure but the superoxides are yellow or orange in colour.
2. The peroxides are diamagnetic while the superoxides are paramagnetic.
3. Sodium peroxide is widely used as an oxidizing agent.
4. The alkali metal hydroxides are weak bases.
(a) 1, 2 and 4
(b) 2, 3 and 4
(c) 1, 2 and 3
(d) 1, 3 and 4
Answer:
(c) 1, 2 and 3

Question 32.
Statement – 1:
LiF has low solubility in water.
Statement – 2:
LiF has low lattice enthalpy.
In the above statements
(a) 1 alone is correct.
(b) Both 1 and 2 are correct
(c) 2 alone is correct
(d) Both 1 and 2 are incorrect
Answer:
(b) Both 1 and 2 are correct

Question 33.
Alkali metals except ______ form solid bicarbom
(a) Sodium
(b) Potassium
(c) Cesium
(d) Lithium
Answer:
(d) Lithium

Question 34.
The ammonia used in the Solvay proves- recovered by using
(a) calcium chloride
(b) Calcium hydroxide
(c) calcium carbonate
(d) calcium oxide
Answer:
(b) Calcium hydroxide

Question 35.
The by-product formed in the Solvay process
(a) calcium chloride
(b) calcium hydroxide
(c) calcium carbonate
(d) ammonium chloride
Answer:
(a) calcium chloride

Question 36.
Sodium carbonate decahydrate on heating ab 373K gives
(a) Na₂CO₃ .3H₂O
(b) N₂CO₃ .5H₂O
(c) Na₂CO₃
(d) Na₂CO₃ .H₂O
Answer:
(c) Na₂CO₃

Question 37.
_______ is used water treatment to convert the hard water to soft water.
(a) Sodium chloride
(b) Sodium bicarbonate
(c) Sodium hydroxide
(d) Sodium carbonate
Answer:
(d) Sodium carbonate

Question 38.
The product obtained on saturating a solution o sodium carbonate with carbon dioxide i,-.
(a) sodium bicarbonate
(b) sodium hydroxide
(c) sodium chloride
(d) sodium peroxide.
Answer:
(a) sodium bicarbonate

Question 39.
Choose the correct pair:

(a) A – ii, B – iv, C – iii, D – i
(b) A – ii, B – iii, C – iv, D – i
(c) A – iv, B – ii, C – i, D – iii
(d) A – iv, B – i, C – ii, D – iii
Answer:
(b) A – ii, B – iii, C – iv, D – i

Question 40.
The used in baking cakes, pastries etc., is
(a) sodium chloride
(b) sodium carbonate
(c) sodium bicarbonate
(d) sodium hydroxide
Answer:
(c) sodium bicarbonate

Question 41.
______ pump play an important role in transmitting nerve signals.
(a) sodium-Magnesium
(b) sodium-potassium
(c) sodium-calcium
(d) sodium-lithium
Answer:
(b) sodium-potassium

Question 42.
Fluoraptite is the ore of
(a) magnesium
(b) beryllium
(c) potassium
(d) calcium
Answer:
(d) calcium

Question 43.
The fifth most abundant element in the earth’s crust is
(a) magnesium
(b) beryllium
(c) calcium
(d) strontium
Answer:
(c) calcium

Question 44.
Strontium nitrate give _______ colour in fire works.
(a) violet
(b) bright red
(c) green
(d) orange
Answer:
(b) bright red

Question 45.
Which of the alkaline earth metal has highest hydration enthalpy?
(a) Be
(b) Mg
(c) Ca
(d) Sr
Answer:
(a) Be

Question 46.
The anomalous properties of beryllium is mainly due to its
1. small size
2. low electronegativity
3. high ionization energy
4. low polarizing power.
(a) 1,2, 3 and 4
(b) 2 and 4
(c) 1 and 3
(d) 1 and 3
Answer:
(d) 1 and 3

Question 47.
______ and ______ ions have strong tendency to form complexes.
(a) Be and Al
(b) Be and Mg
(c) Ca and Sr
(d) Be and B
Answer:
(a) Be and Al

Question 48.
______ is used as radiation windows for X-ray tubes.
(a) Mg
(b) Be
(c) Na
(d) Ca
Answer:
(b) Be

Question 49.
A compound of calcium used in makig surgical and orthopedic casts is
(a) Dolomite
(b) Gypsum
(c) Feldspar
(d) plaster of paris
Answer:
(b) Gypsum

Question 50.
______ is a major component of bones and teeth.
(a) Na
(b) Be
(c) Ca
(d) Mg
Answer:
(c) Ca

II. Very short question and answers (2 Marks):

Question 1.
What are s-block elements?
Answer:
The elements belonging to the group 1 and 2 in the modem periodic table are called s-block elements. The elements belonging to these two groups are commonly known as alkali and alkaline earth metals respectively.

Question 2.
What are alkali metals?
Answer:
Alkali metals consists of the elements: lithium, sodium, potassium, rubidium, caesium and francium. They are all metals, generally soft and highly reactive. They form oxides and hydroxides and these compounds are basic in nature.

Question 3.
Write the mineral source of lithium, sodium and potassium.
Answer:

Question 4.
Why does the ionization enthalpy of alkali metals decreases in a group?
Answer:
Alkali metals have the lowest ionisation enthalpy compared to other elements present in the respective period. As we go down the group, the ionisation enthalpy decreases due to the increase in atomic size. In addition, the number of inner shells also increases, which in turn increases the magnitude of screening effect and consequently, the ionisation enthalpy decreases down the group.

Question 5.
The second ionization enthalpies of alkali metals are very high. Give reason.
Answer:
The second ionisation enthalpies of alkali metals are very high. The removal of an electron from the alkali metals gives monovalent cations having stable electronic configurations similar to the noble gas. Therefore, it becomes very difficult to remove the second electron from the stable Configurations, already attained.

Question 6.
Why is the lithium salts are more soluble than the salts of other metals of group-1.
Answer:
Lithium salts are more soluble than the salts of other metals of group 1. eg., LiClO₄ is up to 12 times more soluble than NaClO₄. KClO₄, RbClO₄, and CsClO₄ have solubilities only 10-3 times of that of LiClO₄. The high solubility of Li salts is due to strong solvation of small size of Li⁺ ion.

Question 7.
What is diagonaol relatiobship?
Answer:
Similarity between the first member of group 1 (Li) and the diagonally placed second element of group 2 (Mg) is called diagonal relationship. It is due to similar size (rLi⁺ = 0.766 Å and Mg²⁺ = 0.12 Å) and comparable electronegativity values (Li = 1.0; Mg = 1.2).

Question 8.
Why does the solubility of carbonates and bicarbonates decreases in a group?
Answer:
All the carbonates and bicarbonates are soluble in water and their solubilities increase rapidly on descending the group. This is due to the reason that lattice energies decrease more rapidly than their hydration energies on moving down the group.

Question 9.
Lithium carbonate is considerably less stable and decompose readily. Give reason.
Answer:
Li₂CO₃ is considerably less stable and decomposes readily.
Li₂CO₃ → Li₂O + CO₂
This is presumably due to large size difference between Li⁺ and CO₂^3- which makes the crystal lattice unstable.

Question 10.
What is washing soda?
Answer:
Sodium carbonate, commonly known as washing soda, crystallises as decahydrate which is white in colour.

Question 11.
What is action of heating on sodium carbonate?
Answer:
Upon heating, it looses the water of crystallisation to form monohydrate. Above 313 K, the monohydrate becomes completely anhydrous and changes to a white powder called soda ash.
Na₂CO₃ .10H₂O → Na₂CO₃ .H₂O + 9H₂O
Na₂CO₃ .H₂O → Na₂CO₃ + H₂O

Question 12.
How does Lithium shows similar properties with magnesium in its chemical behavior?
Answer:

1. Both react with nitrogen to form nitrides.
2. Both react with oxygen to give monoxides.
3. Both the element have tendency to form covalent compounds.
4. Both can form complex compounds.

Question 13.
Why are potassium and caesium, rather than lithium used in photoelectric cells?
Answer:
Potassium and Caesium have much lower ionization enthalpy than that of Lithium. As a result, these metals easily emit electrons on exposure to light. Due to this, K and Cs are used in photoelectric cells.

Question 14.
Write the chemical formula of the following compounds.
(a) Chile salt petre; (b) marble; (c) Brine
Answer:
(a) Chile salt petre – NaNO₃
(b) marble – CaCO₃
(c) Brine – NaCl

Question 15.
The order ofionic mobility of the ions in aqueous solution is Cs⁺ > Rb⁺ > K⁺ > Na⁺. Account it.
Answer:
Smaller the size of cation, higher will be the hydration and its effective size will increase and hence mobility in aqueous solution will decrease. Larger size ions have more ionic mobility due to less hydration. Thus the degree of hydration of M⁺ ions decreases from Li⁺ to Cs⁺.

Cosequently the radii of the hydrated ion decreses from Li⁺ to Cs⁺. Hence, the ionic conductance of these hydrated ions increases from Li⁺ to Cs⁺.

Question 16.
(a) Lithium Iodide is more covalent than Lithium fluoride, (b) Lattice enthalpy of LiF is maximum among all the alkali metals halides. Explain.
Answer:
(a) According to Fajan’s rule, Li⁺ ion can polarise I^– ion more than the F^– ion due to bigger size of the anion. Thus, LiI has more covalent character than LiF.

(b) Smaller the size (intemuclear distance), more is the value of Lattice enthalpy since intemuclear distance is expected to be least in the LiF.

Question 17.
Write notes on flame test for alkali metals.
Answer:
When the alkali metal salts moistened with concentrated hydrochloric acid are heated on a platinum wire in a flame, they show characteristic coloured flame.
Eg:- Lithium – Crimson red; Potassium – Lilac

Question 18.
Give two uses of alkali metals.
Answer:

• Lithium metal is used to make useful alloys. For example with lead it is used to make ‘white metal’ bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. It is used in thermonuclear reactions.
• Lithium is also used to make electrochemical cells.

Question 19.
Write the uses of sodium bicarbonate.
Answer:
The uses of sodium bicarbonate are

• Primarily used as an ingredient in baking.
• Sodium hydrogen carbonate is a mild antiseptic for skin infections.
• It is also used in fire extinguishers.

Question 20.
What are alkaline earth metals?
Answer:
Group 2 in the modem periodic table contains the elements beryllium, magnesium, calcium, strontium, barium and radium are called alkaline earth metals.

Question 21.
Write notes on the physical state of alkaline earth metals.
Answer:
Beryllium is rare and radium is the rarest of all comprising only 10% of igneous rocks. Magnesium and calcium are very common in the earth’s crust, with calcium the fifth-most-abundant element, and magnesium the eighth. Magnesium and calcium are found in many rocks and minerals: magnesium in carnallite, magnesite, dolomite and calcium in chalk, limestone, gypsum.

Strontium is found in the minerals celestite and strontianite. Barium is slightly less common, much of it in the mineral barite. Radium, being a decay product of uranium, is found in all uranium-bearing ores.

Question 22.
Write the uses of Beryllium.
Answer:

1. Because of its low atomic number and very low absorption for X-rays, it is used as radiation windows for X-ray tubes and X-ray detectors.
2. The sample holder in X-ray emission studies usually made of beryllium
3. Since beryllium is transparent to energetic particles it is used to build the ‘beam pipe’ in accelerators.
4. Because of its low density and diamagnetic nature, it is used in various detectors.

Question 23.
What are the uses of Strontium.
Answer:

1. 90Sr is used in cancer therapy.
2. ⁸⁶Sr/ ₈₆Sr ratios are commonly used in marine investigations as well as in teeth, tracking animal migrations or in criminal forensics.
3. Dating of rocks.
4. As a radioactive tracer in determining the source of ancient archaeological materials such as timbers and coins.

Question 24.
Write the uses of Barium.
Answer:

1. Used in metallurgy, its compounds are used in pyrotechnics, petroleum mining and radiology.
2. Deoxidiser in copper refining.
3. Its alloys with nickel readily emits electrons hence used in electron tubes and in spark plug electrodes.
4. As a scavenger to remove last traces of oxygen and other gases in television and other electronic tabes,
5. An isotope of barium 133Ba, used as a source in the calibration of gamma-ray detectors in nuclear chemistry.

Question 25.
How does beryllium hydride can be prepared?
Answer:
All the elements except beryllium, combine with hydrogen on heating to fdrm their hydrides with general formula MH₂. BeH₂ can be prepared by the reaction of BeCl₂ with LiAlH₄.
2BeCl₂ + LiAlH₄ → 2BeH₂ + LiCl + AlCl₃

Question 26.
Why does solubility of sulphates of alkaline earth metal decreases? Explain.
Answer:
The sulphates of the alkaline earth metals are all white solids and stable to heat. BeSO₄, and MgSO₄ are readily soluble in water; the solubility decreases from CaSO₄ to BaSO₄. The greater hydration enthalpies of Be²⁺ and Mg²⁺ ions overcome the lattice enthalpy factor and therefore their sulphates are soluble in water.

Question 27.
Write notes on plaster of paris.
Answer:
It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO₄.2H₂O, is heated to 393 K.
2CaSO₄ 2H₂O{s) → 2CaSO₄ .H₂)O + 3H₂O
Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO₄ is formed. This is known as ‘dead burnt plaster’. It has a remarkable property of setting with water. On mixing with an adequate quantity of water it forms a plastic mass that gets into a hard solid in 5 to 15 minutes.
Uses:
The largest use of Plaster of Paris is in the building industry as well as plasters. It is used for immobilising the affected part of organ where there is a bone fracture or sprain. It is also employed in dentistry, in ornamental work and for making casts of statues and busts.

III. Short question and answers (3 Marks):

Question 1.
Write notes characteristic flame colouration of alkali metal salts.
Answer:
When the alkali metal salts moistened with concentrated hydrochloric acid are heated on a platinum wire in a flame, they show characteristic coloured flame as shown below.

The heat in the flame excites the valence electron to a higher energy level. When it drops back to its actual energy level, the excess energy is emitted as light, whose wavelength is in the visible region.

Question 2.
Discuss the reaction of alkali metals with liquid ammonia.
Answer:
Alkali metals dissolve in liquid ammonia to give deep blue solutions that are conducting in nature. The conductivity is similar to that of pure metals. This happens because the alkali metal atom readily loses its valence electron in ammonia solution. Both the cation and the electron are ammoniated to give ammoniated cation and ammoniated electron.
M + (x + y)NH₃ → [M(NH₃)_x]⁺ [e(NH₃)_y)]^–

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of an amide.
M⁺ + e^– + NH₃ → MNH₂ + \(\frac{1}{2}\)H₂
In concentrated solution, the blue colour changes to bronze colour and become diamagnetic.

Question 3.
Write the properties of oxides and peroxides of alkali metals.
Answer:
The oxides and the peroxides are colourless when pure, but the superoxides are yellow or orange in colour. The peroxides are diamagnetic while the superoxides are paramagnetic. Sodium peroxide is widely used as an oxidising agent.

The hydroxides which are obtained by the reaction of the oxides with water are all white crystalline solids. The alkali metal hydroxides are strong bases. They dissolve in water with evolution of heat on account of intense hydration.

Question 4.
Write the uses of sodium carbonate.
Answer:

1. Sodium carbonate known as washing soda is used heavily for laundering
2. It is an important laboratory reagent used in the qualitative analysis and in volumetric analysis.
3. It is also used in water treatment to convert the hard water to soft water
4. It is used in the manufacturing of glass, paper, paint etc…

Question 5.
Write the uses of sodium hydroxide.
Answer:

1. Sodium hydroxide is used as a laboratory reagent
2. It is also used in the purification of bauxite and petroleum refining
3. It is used in the textile industries for mercerizing cotton fabrics.
4. It is used in the manufacture of soap, paper, artificial silk and a number of chemicals.

Question 6.
What are the mineral sources of alkali metals?
Answer:

Question 7.
Comapare the ionization energy of alkali metals with alkaline earth metals.
Answer:
Members of group 2 have higher ionization enthalpy values than group 1 because of their smaller size, with electrons being more attracted towards the nucleus of the atoms. Correspondingly they are less electropositive than alkali metals.

Although IE₁ values of alkaline earth metals are higher than that of alkali metals, the IE₂ values of alkaline earth metals are much smaller than those of alkali metals. This occurs because in alkali metals the second electron is to be removed from a cation, which has already acquired a noble gas configuration. In the case of alkaline earth metals, the second electron is to be removed from a monovalent cation, which still has one electron in the outermost shell. Thus, the second electron can be removed more easily in the case of group 2 elements than in group 1 elements.

Question 8.
Write the uses of magnesium.
Answer:

1. Removal of sulphur from iron and steel
2. Refining of titanium in the ‘Kroll” process.
3. Used as photoengrave plates in printing industry.
4. Magnesium alloys are used in aeroplane and missile construction.
5. Mg ribbon is used in synthesis of Grignard reagent in organic synthesis.
6. It alloys with aluminium to improve its mechanical, fabrication and welding property.
7. As a desiccant.
8. As sacrificial anode in controlling galvanic corrosion.

Question 9.
Write the uses of calcium.
Answer:

1. As a reducing agent in the metallurgy of uranium, zirconium and thorium.
2. As a deoxidiser, desulphuriser or decarboniser for various ferrous and non-ferrous alloys.
3. In making cement and mortar to be used in construction.
4. As a getter in vacuum tubes.
5. In dehydrating oils
6. In fertilisers, concrete and plaster of paris.

Question 10.
Give uses of magnesium.
Answer:

1. Removal of sulphur from iron and steel
2. Refining of titanium in the “Kroll” process.
3. Used as photoengrave plates in printing industry.
4. Magnesium alloys are used in aeroplane and missile construction.
5. Mg ribbon is used in synthesis of Grignard reagent in organic synthesis.
6. It alloys with aluminium to improve its mechanical, fabrication and welding property.
7. As a desiccant.
8. As sacrificial anode in controlling galvanic corrosion.

Question 11.
Give the structure of BeCl₂ in the solid phase and Vapor phase.
Answer:
Beryllium chloride has a chain structure in the solid-state as shown below, (structure-a). In the vapour phase, BeCl₂ tends to form a chloro-bridged dimer (structure-c) which dissociates into the linear monomer at high temperatures of the order of 1200 K. (structure-b).

IV. Long Answer Questions(5 Marks):

Question 1.
Compare the properties of Lithium with other elements of the group.
Answer:
The distinctive behaviour of Li⁺ ion is due to its exceptionally small size, high polarising power, high hydration energy and non-availability of d-orbitals.

Question 2.
Discuss the diagonal relationship between Lithium and Magnesium.
Answer:
Similarity between the first member of group 1 (Li) and the diagonally placed second clement of group 2 (Mg) is called diagonal relationship. It is due to similar size (rLi₊ = 0.766 Å and Mg₂₊ = 0.72 Å) and comparable electronegativity values (Li = 1.0; Mg = 1.2). Similarities between Lithium and Magnesium are

1. Both lithium and magnesium are harder than other elements in the respective groups
2. Lithium and magnesium react slowly with water. Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating.
3. Both form a nitride, Li₃N and Mg₃N₂, by direct combination with nitrogen
4. They do not give any superoxides and form only oxides, Li₂O and MgO
5. The carbonates of lithium and magnesium decompose upon heating to form their respective oxides and CO₂.
6. Lithium and magnesium do not form bicarbonates.
7. Both LiCl and MgCl₂ are soluble in ethanol and are deliquescent. They crystallise from aqueous solution as hydrates, LiCl.2H₂O and MgCl₂8H₂O

Question 3.
Explain the reaction of alkali metals with (i) Oxygen (ii) hydrogen (iii) halogens.
Answer:
(i) Reaction with Oxygen:
All the alkali metals on exposure to air or oxygen bum vigorously, forming oxides on their surface. Lithium forms only monoxide, sodium forms the monoxide and peroxide and the other elements form monoxide, peroxide, and superoxides. These oxides are basic in nature.
4Li + O₂ → 2Li₂O (simple oxide)
2Na + O₂ → Na₂O₂ (peroxide)
M + O₂ → MO₂
(M = K, Rb, Cs; MO₂ – superoxide)

(ii) Reaction with hydrogen:
All alkali metals react with hydrogen at about 673 K (lithium at 1073 K) to form the corresponding ionic hydrides. Reactivity of alkali metals with hydrogen decreases from Li to Cs.
2M + H₂ → 2M⁺H^–
(M = Li, Na, K, Rb, Cs)
The ionic character of the hydrides increases from Li to Cs and their stability decreases. The hydrides behave as strong reducing agents and their reducing nature increases down the group.

(iii) Reaction with halogen:
Alkali metals combine readily with halogens to form ionic halides MX. Reactivity of alkali metals with halogens increases down the group because of corresponding decrease in ionisation enthalpy.
2 M + X₂ → 2 MX
(M = Li, Na,K, Rb, Cs) (X = F, Cl, Br, I)
All metal halides are ionic crystals. However Lithium iodide shows covalent character, as it is the smallest cation that exerts high, polarising power on the iodide anion. Additionally, the iodide ion being the largest can be polarised to a greater extent by Li⁺ ion.

Question 4.
Write the uses of alkali metals.
Answer:

1. Lithium metal is used to make useful alloys. For example with lead it is used to make ‘white metal’ bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. It is used in thermonuclear reactions.
2. Lithium is also used to make electrochemical cells.
3. Lithium carbonate is used in medicines
4. Sodium is used to make Na/Pb alloy needed to make Pb(Et)₄ and Pb(Me)₄. These organolead compounds were earlier used as anti-knock additives to petrol, but nowadays lead-free petrol in use.
5. Liquid sodium metal is used as a coolant in fast breeder nuclear reactors. Potassium has a vital role in biological systems.
6. Potassium chloride is used as a fertilizer. Potassium hydroxide is used in the manufacture of soft soap. It is also used as an excellent absorbent of carbon dioxide.
7. Caesium is used in devising photoelectric cells.

Question 5.
How is washing soda prepared? Discuss its properties.
Answer:
Sodium carbonate is one of the important inorganic compounds used in industries. It is prepared by Solvay process. In this process, ammonia is converted into ammonium carbonate which then converted to ammonium bicarbonate by passing excess carbon dioxide in a sodium chloride solution saturated with ammonia.

The ammonium bicarbonate thus formed reacts with the sodium chloride to give sodium bicarbonate and ammonium chloride. As sodium bicarbonate has poor solubility, it gets precipitated. The sodium bicarbonate is isolated and is heated to give sodium carbonate. The equations involved in this process are,
2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
(NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
2NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
2NaHCO → Na₂CO₃ + CO₂ + H₂O

The ammonia used in this process can be recovered by treating the resultant ammonium chloride solution with calcium hydroxide. Calcium chloride is formed as a by-product.

Properties:
Sodium carbonate, commonly known as washing soda, crystallises as decahydrate which is white in colour. It is soluble in water and forms an alkaline solution. Upon heating, it looses the water of crystallisation to form monohydrate. Above 373 K, the monohydrate becomes completely anhydrous and changes to a white powder called soda ash.
Na₂CO₃ .10H₂O → Na₂CO₃.H₂O + 9H₂O
Na₂CO₃.H₂O → Na₂CO₃ + H₂O

Question 6.
Compare the properties of Beryllium with other elements of the group.
Answer:

Question 7.
Explain the diagonal relationship of Beryllium with Aluminium.
Answer:
Beryllium (the first member of group 2) shows a diagonal relationship with aluminium. In this case, the size of these ions (r_Be2+ = 0.45 Å and r_Al3+ = 0.54 Å) is not as close. However, their charge per unit area is closer. (Be₂₊ = 2.36 and Al₃₊ = 2.50) They also have same electronegativity values (Be = 1.5; Al = 1.5).
Properties:

1. Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms polymeric chain structure in addition to dimer. Both are soluble in organic solvents and are strong Lewis acids.
2. Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)₄]^2- and hydrogen as aluminium hydroxide which gives aluminate ion. [Al( OH₄)^–
3. Beryllium and aluminum ions have strong tendency to form complexes, BeF₄^2-, AlF₆^3-.
4. Both beryllium and aluminium hydroxides are amphoteric in nature.
5. Carbides of beryllium (Be₂C) like aluminum carbide (Al₄C₃) give methane on hydrolysis
6. Both beryllium and aluminium are rendered passive by nitric acid.

Question 8.
Explain the properties and uses of Gypsum.
Answer:
Properties of Gypsum:

1. Gypsum is a soft mineral, which is moderately soluble in water. The solubility of this mineral in water is affected by temperature. Unlike other salts, gypsum becomes less soluble in water as the temperature increases. This is known as retrograde solubility, which is a distinguishing characteristic of gypsum.
2. Gypsum is usually white, colorless, or gray in color. But sometimes, it can also be found in the shades of pink, yellow, brown, and light green, mainly due to the presence of impurities.
3. Gypsum crystals are sometimes found to occur in a form that resembles the petals of a flower. This type of formation is referred to as ‘desert rose’, as they mostly occur in arid areas or desert terrains.
4. Gypsum is known to have low thermal conductivity, which is the reason why it is used in making drywalls or wallboards. Gypsum is also known as a natural insulator.
5. Alabaster is a variety of gypsum, that is highly valued as an ornamental stone. It has been used by the sculptors for centuries. Alabaster is granular and opaque.
6. Gypsum has hardness between 1.5 to 2 on Moh’s Hardness Scale. Its specific gravity is 2.3 to 2.4.

Uses of Gypsum:

1. The alabaster variety of gypsum was used in ancient Egypt and Mesopotamia by the sculptors. The ancient Egyptians knew how to turn gypsum into plaster of Paris about 5,000 years ago. Today, gypsum has found a wide range of uses and applications in human society, some of which are enlisted below.
2. Gypsum is used in making drywalls or plaster boards. Plaster boards are used as the finish for walls and ceilings, and for partitions.
3. Another important use of gypsum is the production of plaster of Paris. Gypsum is heated to about 300 degree Fahrenheit to produce plaster of Paris, which is also known as gypsum plaster. It is mainly used as a sculpting material.
4. Gypsum is used in making surgical and orthopedic casts, such as surgical splints and casting moulds.
5. Gypsum plays an important role in agriculture as a soil additive, conditioner, and fertilizer, It helps loosen up compact or clay soil, and provides calcium and sulphur, which are essential for the healthy growth of a plant. It can also be used for removing sodium from soils having excess salinity.

Question 9.
Mention the biological importance of sodium and potassium.
Answer:

1. Monovalent sodium and potassium ions are found in large proportions in biological fluids. These ions perform important biological functions such as maintenance of ion balance | and nerve impulse conduction.
2. A typical 70 kg man contains about 90 g of sodium and 170 g of potassium compared with only 5 g of iron and 0.06 g of copper.
3. Sodium ions are found primarily on the outside of cells, being located in blood plasma and in the interstitial fluid which surrounds the cells. These ions participate in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells.
4. Sodium and potassium, although so similar chemically, differ quantitatively in their ability to penetrate cell membranes, in their transport mechanisms and in their efficiency to activate enzymes.
5. Thus, potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and, with sodium, are responsible for the transmission of nerve signals.
   Sodium-potassium pump play an important role in transmitting nerve signals.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 10 Secondary Growth Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

11th Bio Botany Guide Secondary Growth Text Book Back Questions and Answers

Part -I

I. Consider the following statements.

Question 1.
In spring vascular cambium
i) is less active
ii) Produces a large number of xylary elements
iii) forms vessels with wide cavities of these,
a) (i) is correct but (ii) and (iii) are not correct
b) (i) is not correct but (ii) and (iii) are correct
c) (i) and (ii) are correct but (iii) is not correct
d) (i) and (ii) are not correct but (iii) is correct
Answer:
b) (i) is not correct but (ii) and (iii) are correct

Question 2.
Usually, the monocotyledons do not increase their girth, because
a) They possess actively dividing cambium
b) They do not possess actively dividing cambium
c) Ceases activity of cambium
d) All are correct
Answer:
b) They do not possess actively dividing cambium

Question 3.
In the diagram of lenticel identify the parts marked as A, B, C, D

a) A. phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.
b) A. Complementary tissue, B.Phellem, C. Phellogcn,
c) A. Phellogen, B.Phellem, C. Phelloderm, D. Complementary tissue
d) A. Phelloderm, B. Phellem, C. Complementary tissue, D. Phellogen
Answer:
a) A. phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.

Question 4.
The common bottle cork is a product of
a) Dermatogen
b) Phellogen
c) Xylem
d) Vascular cambium
Answer:
b) Phellogen

Question 5.
What is the fate of primary xylem in a dicot root showing extensive secondary growth?
a) It is retained in the center of the axis
b) It gets crushed
c) May or may not get crushed
d) It gets surrounded by primary phloem
Answer:
b) It gets crushed

Question 6.
In a forest, if the bark of a tree is damaged by the horn of a deer, How will the plant overcome the damage?
Answer:
When the bark is damaged, the phellogen forms a complete cylinder around the stem and it gives rise to ring barks.

Question 7.
In which season the vessels of angiosperms are larger in size, why?
Ans:
In spring season the vessels are larger in size, because the cambium cells are very active during spring season.

Question 8.
Continuous state of dividing tissue is called meristem. In connection to this, what is the role is lateral meristem?
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

1. Vascular cambium and
2. Cork cambium

1. Vascular cambium:
The vascular cambium is the lateral meristem that produces the secondary vascular tissues, i.e.. secondary xylem and secondary phloem.

Origin and Formation of Vascular Cambium:

• A strip of vascular cambium originate from the procambium is present between xylem and phloem of the vascular bundle. This cambial strip is known as intrafascicular or fascicular cambium.
• In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of vascular cambium. It is called interfascicular cambium.

A. Organization of Vascular cambium:

• The active vascular cambium possesses cells with large central vacuole (or vacuoles) surrounded by a thin, layers of dense cytoplasm.
• The most important character of the vascular cambium is the presence of two kinds of initials, namely fusiform initials and ray initials.

Fusiform Initials:

• These are vertically elongated cells. They give rise to the longitudinal or axial system of the secondary xylem (tracheary elements, fibres, and Axia? parenchyma) and pholem (sieve, elements, fibres, and axial parenchyma).
• Based on the arrangement of the fusiform initials two types of vascular cambium are recognized.

Stoned (Stratified cambium) and Non – storied (Non – stratified cambium)

• If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in tangential longitudinal section (TLS) it is called storied (stratified) cambium. It is the characteristic of the plants with short fùsiform initials.
• In plants with long fusiform initials, they strongly overlap at the ends, and this type of cambium is called non – storied (non stratified) cambium.

Ray Initials:
These are horizontally elongated cells. They give rise to the ray cells and form the elements of the radial system of secondary xylem and pholem.

Activity of Vascular Cambium:

• The vascular cambial ring, when active, cuts off new cells both towards the inner and outer side. The cells which are produced outward form secondary phloem and inward secondary xylem.
• Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.

B. Phellogen (Cork Cambium)

• It is a secondary lateral meristem. It comprises homogenous meristematic cells unlike vascular cambium. It arises from epidermis, cortex, pholem or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells.
• The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phelloderm (secondary cortex).

Question 9.
A timer merchant bought 2 logs of wood from, a forest & named them A & B, The log A was 50 year old & B was 20 years old. Which log of wood will last longer for the merchant? Why?
Answer:

• In wood, the older it is, the stronger it becomes.
• Log A – Which was 50 years old is stronger and it will last longer.
• In a tree the central part of the wood will be darker in colour, dead in nature known as Heartwood or Duramen, and the outer sad wood is lighter in colour, living and conducting water.
• In the central Heartwood the conduction is blocked by the formation of tyloses from the nearby parenchyma cells, and dead.
• In the fully developed tyloses, starch crystals, resins, gums, oils tannins and coloured substances are found and it becomes very hard and durable.
• It is more resistant to the attack of microbes and insects like termites.
• Older woods have more heartwood than sapwood.
• Here log ‘A’ is older, has more heartwood and it is stronger and will last longer.

Question 10.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What are the significance of these rings?
Answer:
Growth (or) Annual Rings:
1. In the spring season cambium is very active and produces large number of xylary elements called Earlywood or Springwood. In the Winter season – cambium is less active and form few xylary elements – Latewood or Autumn Wood.

2. The springwood is lighter in color and has a lower density whereas the autumn wood is darker and has a higher density. The annual ring denotes the combination of earlywood and latewood and the ring becomes evident to our eye due to the high density of latewood. Sometimes annual rings are called growth rings

3. Pseudo – Annual Rings:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost defoliation, flood, mechanical, injury and biotic factors. Such rings arc called pseudo – or false – annual rings

4. Dendrochronology:
Each annual ring corresponds to one year’s growth and on the basis of these rings, the age of a particular plant can easily be calculated. The determination of the age of a tree by counting the annual rings is called dendrochronology.

Part – II.

11th Bio Botany Guide Secondary Growth Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
The roots and stems grow in length with the help of:
(a) cambium
(b) secondary growth
(c) apical meristem
(d) vascular parenchyma
Answer:
(c) apical meristem

Question 2.
The Gymnosperm in which vessel is present
a) Pinus
b) Cýcas
c) Ginkgo
d) Gnetum
Answer:
d. Gnetum

Question 3.
The secondary vascular tissues include:
(a) secondary xylem and secondary phloem
(b) secondary xylem, cambium strip and secondary phloem
(c) secondary phloem and fascicular cambium
(d) secondary xylem and primary phloem
Answer:
(a) secondary xylem and secondary phloem

Question 4.
In a dicotyledonous stem, the sequence of tissues from the outside to the inside of
a) Phellem, Pericycle, Endodermis, phloem
b) Phellem phloem, Endodermis, Pericycle
c) Phellem, Endodermis, Pericycle, phloem
d) Pericycle, Phellem, Endodermis, Phloem
Answer:
c. Phellem, Endodermis, Pericycle, Phloem

Question 5.
For a critical study of secondary growth in plants which one of the following pairs is suitable?
a) Sugarcane and sunflower
b) Teak and pine
c) Bamboo and Fem
d) Wheat and Fem
Answer:
b. Teak and pine

Question 6.
The axial system of the secondary xylem includes:
(a) treachery elements, sieve elements, fibers and axial parenchyma
(b) treachery elements, fibers and axial parenchyma
(c) treachery elements and fibers
(d) sieve elements and axial parenchyma
Answer:
(b) treachery elements, fibers and axial parenchyma

Question 7.
Tissues considered in an annual ring is/are
a) Secondary xylem and phloem
b) Primary xylem and phloem
c) Secondary xylem only
d) Primary phloem and secondary xylem
Answer:
c. Secondary xylem only

Question 8.
Ray cells are present between:
(a) primary xylem and phloem
(b) primary xylem and secondary xylem
(c) secondary xylem and phloem
(d) secondary phloem and cambium
Answer:
(c) secondary xylem and phloem

Question 9.
Interfascicular cambium is a
a) Primary meristematic tissue
b) Primordial meristem
c) Type of protoderm
d) Secondary meristematic tissue
Answer:
d. Secondary meristematic tissue

Question 10.
Interfascicular cambium develops from the cells of
a) Xylem parenchyma
b) endodermis
c) Pericycle
d) Medullary rays
Answer:
d. Medullary rays

Question 11.
Which of the statement is not correct?
(a) In temperate regions, the cambium is very active in the winter season.
(b) In temperate regions, the cambium is very active in the spring season.
(c) In temperate regions, cambium is less active in the winter season.
(d) In temperate regions earlywood is formed in the spring season.
Answer:
(a) In temperate regions, the cambium is very active in the winter season.

Question 12.
At maturity, the sieve plates become impregnated with
a) Cellulose
b) Pectin
c) Suberin
d) Callose
Answer:
d. Callose

Question 13.
You are given a fairly old piece of dicot stem and a dicot root, which of the following anatomical structures will you distinguish between the two?
a) Secondary xylem
b) Secondary phloem
c) Protoxylem
d) Cortical cells
Answer:
c. Protoxylem

Question 14.
determination of the age of a tree by counting the annual rings is called:
(a) chronology
(b) dendrochronology
(c) palaeology
(d) histology
Answer:
(c) palaeology

Question 15.
Which one of the following is dead and works efficiently?
a) Sieve tube
b) Companian cells
c) Vessels
d) Both (b) and (c)
Answer:
c. Vessels

Question 16.
Which one of the following pairs is an example for meristematic tissue
a) Phellogen and phelloderm
b) Phellogen and Fascicular cambium
c) Procambium and phelloderm
d) Interfascicular cambium and phellem
Answer:
b. Phellogen and Fascicular cambium

Question 17.
In fully developed tyloses:
(a) only starchy crystals are present
(b) resin and gums only are present
(c) oil and tannins are present
(d) starchy crystals, resins, gums, oils, tannins, or colored substances are present
Answer:
(d) starchy crystals, resins, gums, oils, tannins, or colored substances are present

Question 18.
Which wood is also known as Non-porous
a) Softwood
b) Heartwood
c) Hardwood
d) Sapwood
Answer:
a. Softwood

Question 19.
Which of the statement is not correct?
(a) Sapwood and heartwood can be distinguished in the secondary xylem
(b) Sapwood is paler in colour
(c) Heartwood is darker in colour
(d) The sapwood conducts minerals, while the heartwood conduct water
Answer:
(d) The sapwood conducts minerals, while the heartwood conduct water

Question 20.
Removal of a ring of wood tissue outside the vascular cambium from the tree trunk kills it because
a) Water cannot move up
b) Food does not travel down and root become starved
c) Shoot apex become starved
d) Annual rings are not produced
Answer:
food does not travel down root become starved

Question 21.
The trees growing in the desert will
a) Show alternate rings of xylem and sclerenchyma
b) Have only conjunctive tissue and phloem formed by the activity of cambium
c) do show distinct annual rings
d) do not show distinct annual rings
Answer:
d. do not show distinct annual rings.

Question 22.
Canada balsam is produced from:
(a) Pisum sativum
(b) the resin of Arjuna plant
(c) Abies balsamea
(d) the root of Vinca rosea
Answer:
(c) Abies balsamea

Question 23.
Choose the living cells from the given
I) Phellem
II) Phloem
III) Phellogen
IV) Xylem parenchyma
a) (I) (II) & (III)
b) (II) (III) & (IV)
c) (I) (III) & (IV)
d) (I) (II) & (IV)
Answer:
b) (II) (III) & (IV)

Question 24.
When we peel the skin of a potato tuber we remove
a) Periderm
b) Epidermis
c) Cuticle
d) Sapwood
Answer:
a) Periderm

Question 25.
Phelloderm is otherwise called as:
(a) primary cortex
(b) corkwood
(c) secondary cortex
(d) rhytidome
Answer:
(c) secondary cortex

Question 26.
The waxy substances associated with cell walls of cork cells are impervious to water because of the presence of ………………. which gets deposited on the cork cells
a) Cutin
b) Suberin
c) Eignin
d) Hemicellulose
Answer:
b) Suberin

Question 27.
The antimalarial compound quinine is, extracted from:
(a) seeds of cinchona
(b) bark of cinchona
(c) leaves of cinchona
(d) flowers of cinchona
Answer:
(b) bark of cinchona

Question 28.
Annual rings are distinct in plants growing in
a) Tropical regions
b) Arctic regions
c) Grasslands
d) Temperate region
Answer:
d) Temperate region

Question 29.
The bark does include three except one from the options.
I) Cortex
II) Periderm
III) Pith
IV) Secondary phloem
a) (I) (II) & (Ill)
b) (II) (III) & (IV)
c) (I) (III) & (IV)
d) (I) (II) & (IV)
Answer:
d) (I) (II) & (IV)

Question 30.
Rubber is obtained from:
(a) Bombax mori
(b) Hevea brasiliensis
(c) Quercus suber
(d) Morus rubra
Answer:
(b) Hevea brasiliensis

Question 31.
Wood actually means
a) Primary xylem
b) Secondary xylem
c) Primary phloem
d) Secondary phloem
Answer:
b) Secondary xylem

Question 32.
When a plant is wounded, the wound is healed by the formation of new cells, by the cavity of
a) Primary meristem
b) Apical meristem
c) Secondary meristem
d) Intercalary meristem
Answer:
c) Secondary meristem

Question 33.
Commercial cork is obtained from
a) Oak
b) Silver oak
c) Pine
d) Ficus
Answer:
a) Oak

II. Match Correctly And Choose The Right Answer

Question 1.
I) Spicy bark – A Turpentine
II) Ornamental Antique – B Quinine
III) Active drug – C Cinnamon
IV) Thinner and solvent – D Amber

Answer:
d) C-D- B-A

Question 2.
I) Phellogen – A. Cork
II) Phelloderm – B. Cork cambium
III) Phellem – C. Lack suberin
IV) Phelloids – D. Secondary cortex

Answer:
a) B-D- A-C

Question 3.
I) Sapwood – A. Softwood
II) Heartwood – B. Hardwood
III) Porous wood – C. Albumum
IV) Nonporous wood – D. Duramen

Answer:
c) C-D-B-A

III Identify True Or False And From The Given Option Choose The Right Answer:

Question 1.
A) In pinus wood is Non-porous
B) In Morns wood is porous
C) In Quercus the wood is Diffuse porous
D) In Acer the wood is Ring porous
a) A&DTrue B&C False
b) A & B True C & D False
c) A & C True B & D False
d) A&BFalseC &DTrue
Answer:
b) A & B True C & D False

Question 2.
Arrange the given plants in order from more distinct Annual rings to least distinct Annual rings.
a) Seashore plants, Desert plants, Tropical plants & Temperature plants
b) Temperature plants, Tropical plants, Desert plants, Seashore plants
c) Tropical plants, Desert plants, Temperature plants, Seashore plants
d) Temperature plants, Seashore plants, Tropical plants, Desert plants
Answer:
b) Temperature plants, Tropical plants, Desert plants, Seashore plants.

Question 3.
In wood, the Annual rings become clearly evident to our eyes due to
a) The high density and dark coloured latewood or Autumn wood
b) The low density and light coloured early wood or springwood
c) The high density and dark coloured early wood or springwood
d) The low density and light coloured latewood or Autumn wood
Answer:
a) The high density and dark coloured of latewood or Autumn wood

Question 4.
Column I and Column II – Match them correctly and Find out the right option.
Answer:

Which of the following combination is correct?
a) A – 2, 4, 7, 8 B- 1,3, 5,6
b) A- 1,2, 7,8 B-3,4, 5, 6
c) A – 1,3, 5,6 B – 2,4, 7, 8
d) A- 1,3,7, 8 B – 2, 4, 5, 6
Answer:
b) A – 1, 2, 7,8 B-3,4,5, 6

IV. Assertion And Reason

Question 1.
ASSERTION: – A All tissues lying inside the Vascular Cambium are called as Bark
REASON -R: Bark is made up of Phellogen. Phellem and Phelloderm Cortex, primary and secondary phloem
a) Both A and R are true and R is the correct explanation of A
b) Both A and R are true, but R is not the correct explanation of A
c) A is true but ‘R’ is false
d) A is false and ‘R’ is true
e) Both A and ‘R’ are false
Answer:
d) ‘A’ is false and ‘R’ is true

Question 2.
ASSERTION: -A In angiosperms, the conduction of water is more efficient because their xylem has vessels
REASON – R: Conduction of water by vessel element is an active process in which energy is supplied by xylem parenchyma with a large number of Mitochondria
Answer:
a) Both A and R – are true and ‘R’ is the correct explanation of A

Question 3.
ASSERTION: -A. All the endodermal cells of the root do not contain Casparian thickenings on their radial and transverse walls.
REASON-R: Passage cells are found in the root endodermis, which conducts water in to the xylem
Answer:
a) Both A and R are true and ‘R’ is the correct explanation of A

Question 4.
ASSERTION:-A Cambium is a lateral meristem and causes growth in width
REASON-R Cambium is made up of fusiform and ray initials in stem
Answer:
b) Both A and R are true, but R is not the correct explanation of A

Question 5.
ASSERTION: – A The lenticel is meant for gaseous exchange.
REASON-R Lenticel checks excessive evaporation of water.
Answer:
b) Both A and R are true but R is not the correct explanation of A

Question 6.
ASSERTION:-A Heartwood is more durable
REASON – R Heartwood contains organic compounds like tannins, resins, oil, gums, aromatic substances and essential oils help to resist microbial and termites attack.
Answer:
a) Both A and R are true and R is the correct explanation of A

V. 2 Marks Questions

Question 1.
Distinguish between Primary and Secondary growth.
Answer:
Primary growth

1. The roots and stem grow in length with the help of Apical meristem
2. It is known as longitudinal growth
3. Eg. Angiosperms & Gymnosperms

Secondary growth

1. The roots and stem show an increase in thickness or width with the help of Lateral meristem
2. It is also known as latitudinal growth or growth in girth
3. Eg. Most Angiosperms, including some Monocots and Gymnosperms

Question 2.
Mention the two Lateral meristems responsible for secondary growth.
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

1. Vascular Cambium and
2. Cork Cambium

Question 3.
Define interfascicular cambium?
Answer:
In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of the vascular cambium. It is called interfascicular cambium.

Question 4.
Fill in the blanks

Answer:

1. The resin used as a mounting medium for microscopic slide preparation.
2. Gum Arabic
3. Quercus suber
4. Haematoxylon campechianum

Question 5.
Distinguish between the stratified cambium and Non-stratified cambium.
Answer:
Stratified cambium: Plants with short fusiform initials, produced, storied, cambium in horizontal tiers known as
stratified cambium

Non-stratified cambium: Plants with long fusiform initials produced non-storied cambium, strongly overlap at the ends, known as Non-stratified cambium

Question 6.
Why does porous wood be harder than non-porous wood?
Answer:

• Porous wood is wood with xylem vessels which appear as a pore in cross-section.
• When a tree stem become old, most of its vessels are blocked by tyloses with deposition of gum, resin, tannin, oils, etc. (Heartwood)
• So porous wood is harder and commercially important.

Question 7.
What is the source of turpentine? and What is its use?
Answer:

• Turpentine is a resin obtained from the bark of conifers Eg. pinus
• It is also used as a thinner for oil-based paints.
• It is also used as an organic solvent.
• It is also used as a balm to relieve muscular pain.

Question 8.
Distinguish between Periderm and Polydor
Answer:
Periderm:

1. The secondary growth replaces the epidermis and primary cortex and forms the Periderm
2. It consists of
   • Phellem
   • Phellogen
   • Phelloderm
3. Eg. Dicot stem & roots

Polyderm:

1. It is a special type of protective tissue consisting of a miserable suberized layer, alternating with multiseriate nonsuberized cells in periderm
2. Eg. Roots and underground stems of Rosaceae plants

Question 9.
Define Rhytidome?
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues ? formed during successive secondary growth, eg: Quercus.

Question 10.
What is the use of Canada balsam?
Answer:
From the resin ducts, the Abies balsamea plant produces an organic gum-like substance, used as a permanent mounting medium for microscopic slide preparation.
Eg. A slide of 60 years old holotype specimen of a flatworm is permanently mounted in Canada balsam.

VI. 3 Mark Questions

Question 1.
Distinguish between primary and secondary growth.
Answer:
1. Primary growth: The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is tailed primary growth or longitudinal growth.

2. Secondary growth: The gymnosperms and most angiosperms, including some monocots, show an increase in the thickness of stems and roots by means of secondary growth or latitudinal growth.

Question 2.
Notes on Lenticels.
Answer:

• The aerating pores are seen as raised opening on the surface of bark as scars on old stems and roots.
• It is fonned during secondary growth in stems.
• In this portion phellogen activity is more than elsewhere, a filling tissue known as complementary tissue (loosely arranged parenchyma) is formed.
• Lenticel is helpful in the exchange of gases and also facilitate the little amount of transpiration

Question 3.
Explain briefly about false annual rings.
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo – or false – annual rings.

Question 4.
Differences between Diffuse porous wood and Ring porous wood
Answer:
Diffuse porous wood:

1. This type of wood is formed where the climatic conditions are uniform
2. The vessels are more or less equal in diameter in any annual ring
3. The vessels are uniformly distributed throughout the wood

Ring porous wood:

1. This type of wood is formed where the climatic conditions are not uniform
2. The vessels are wide and narrow within an annual ring
3. The vessels are not uniformly distributed throughout the wood

Question 5.
Differences between Porous Wood and Non – porous wood
Answer:
Porous wood or Hardwood, Ex: Morus:

1. Common in Angiosperms
2. Porous because it contains vessels

Non-porous wood or softwood Ex: Pinus

1. Common in Gymnosperms
2. Non – porous because it does not contain vessels

Question 6.
Differences between Sapwood (alburnum) and Heart Wood (duramen)
Answer:

Question 7.
Differences Between Phellem and Phelloderm
Answer:

Question 8.
Explain the term lenticel.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems. When phellogen is more active in the region of lenticels, a mass of loosely arranged thin-walled parenchyma cells is formed. It is called complementary tissue or filling tissue. Lenticel is helpful in the exchange of gases and transpiration called lenticular transpiration.

Question 9.
Mention the benefits of bark in a tree.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation, and guards against variations of external temperature. It is insect repellent, decay proof, fireproof, and is used in obtaining drugs or spices. The phloem cells of the bark are involved in the conduction of food while secondary cortical cells involved in storage.

Question 10.
Annual rings are not clear and distinct in desert trees and seashore plants – Justify.
Answer:

• In the desert, as well seashore regions the climatic condition remain the same throughout the year.
• Secondary growth in plants is influenced by seasonal changes since in these areas seasonal changes are not significant enough to bring in distinct Annual rings with early and latewood formation alternatively.

Question 11.
A plant dies if the sapwood is damaged, but not with that of Heartwood – Give factual justification.
Answer:

• Sapwood is a living part of the wood, perform water conduction, that’s why it is known as Sapwood.
• Heartwood is a dead part of the wood, do not perform water conduction so if destroyed, no vital function of the plant is affected.
• If sapwood is damaged, or exposed conduction of water will be blocked, water loss is rapid leading to decay and decomposition of tissues and leads to death of the plant.

Question 12.
What is Dendrochronology? Add a note on the significance of studying the growth rings.
Answer:

• The annual ring of a tree corresponds to one year’s growth.
• If we count the rings we can determine the very age of the plant.
• This method of calculating the age of a tree by counting the annual ring is known as dendrochronology.

Significance of studying growth rings:

• Age of wood – calculated
• Age verified by Radioactive carbon dating
• Provides evidence in Forensic investigation.

Question 13.
If  ‘A’ is vascular cambium, then label other parts with reference to cambial activity.

Answer:
A – Vascular cambium
B – First formed phloem – (Primary phloem)
C – First formed xylem – (Primary xylem)
D – Second formed phloem – (Secondary phloem)
E – Second formed xylem – (Secondary xylem)

Question 14.

Answer:
Cross-section of wood showing Annual rings.
A-Bark
B-Sapwood
C – heartwood
D -Annual rings

Question 15.
Identify the diagram & Label the parts.

Answer:
Structure of Tyloses A – Parenchyma cell
B – Tyloses
C – Vessel wall
D – Vessel Lumen

Question 16.
In the given diagram, the parts labelled are A, B, C, D identify the part correctly with respect to its function.

a) A – Periderm for gaseous exchange
b) C – Secondary cortex for protection
c) B – Complementary tissues for gaseous exchange
d) D – Phellogen for xylem and phloem formation
Answer:
c) B – Complementary tissues for gaseous exchange

Question 17.

Answer:
Different stages of secondary growth in Dicot root
A – Cambial ring
B – Primary’ xylem
C – Secondary xylem
D – Primary phloem
E – Secondary phloem

VII. 5 Mark Questions

Question 1.
Distinguish between Phellem and Phelloderm.
Answer:
Phellem (Cork):

1. It is formed on the outer side of phellogen.
2. Cells are compactly arranged in regular tires and rows without intercellular spaces.
3. Protective in function.
4. Consists of nonliving cells with suberized walls.
5. Lenticels are present.

Phelloderm (Secondary cortex):

1. It is formed on the inner side of phellogen.
2. Cells are loosely arranged with intercellular spaces.
3. As it contains chloroplast, it synthesises and stores food.
4. Consists of living cells, parenchymatous in nature and does not have suberin.
5. Lenticels are absent.

Question 2.
Distinguish the significance of Cork and Bark
Answer:
Cork:

1. It includes only phellem layer of bark
2. It is composed of suberin a hydrophobic substance
3. It has impermeable buoyant, elastic and fibre retartant properties
4. Used in making bottle stoppers Eg. Bark of Quercus suber

Bark:

1. It includes all tissues outside vascular cambium (Periderm, Cortex, Primary and secondary phloem)
2. It has insect repellent, decay proof, fire proof properties.
3. Used as Drugs or spices.
4. Eg. Bark of Chichona – (AntimalariaJ drug), Bark of Cinnamomum (Used as spice)

Question 3.
Differentiate between, the vascular cambial components Fusiform initials and Ray initials.
Answer:
Fusiform initials:

1. Vertically elongated cells
2. Give rise to axial system of secondary tissues, xylem and phloem
3. Secondary xylem includes tracheary elements, fibres and axial parenchyma
4. Secondary phloem includes sieve elements
5. Based on arrangement of fusiform initials 2 types of vascular cambium recognised
   a – stratified cambium
   b – Nonstratified cambium

Ray initials

1. Horizontally elongated cells
2. Give rise to radial system of secondary xylem and phloem
3. Radial system consists of rows of parenchymatous cells oriented at right angles to the longitudinal axis of xylem elements
4. Secondary phloem include phloem rays fibres and axial parenchyma

Question 4.
Give an account of Primary & Secondary structure of Dicto Stern. (Flow chart)
Answer:

Question 5.
Give an account of Any 5 Commercial barks their, properties and uses
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 4 Hydrogen Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

11th Chemistry Guide Hydrogen Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
Which of the following statements about hydrogen is incorrect?
(a) Hydrogen ion, H₃O⁺ exists freely in solution.
(b) Dihydrogen acts as a reducing agent.
(c) Hydrogen has three isotopes of which tritium is the most common.
(d) Hydrogen never acts as cation in ionic salts.
Answer:
(c) Hydrogen has three isotopes of which tritium is the most common.

Question 2.
Water gas is
(a) H₂O(g)
(b) CO + H₂O
(c) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

Question 3.
Which one of the following statements is incorrect with regard to ortho and para dihydrogen ?
(a) They are nuclear spin isomers
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
(c) The para isomer is favoured at low temperatures
(d) The thermal conductivity of the para isomer is 50% greater than that of the ortho isomer.
Answer:
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin

Question 4.
Ionic hydrides are formed by
(a) halogens
(b) chalogens
(c) inert gases
(d) group one elements
Answer:
(d) group one elements

Question 5.
Tritium nucleus contains
(a) 1p + 0n
(b) 2p + 1n
(c) 1p + 2n
(d) none of these
Answer:
(c) 1p + 2n

Question 6.
Non-stoichiometric hydrides are formed by
(a) palladium, vanadium
(b) carbon, nickel
(c) manganese, lithium
(d) nitrogen, chlorine
Answer:
(b) carbon, nickel

Question 7.
Assertion:
Permanent hardness of water is removed by treatment with washing soda.
Reason:
Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 8.
If a body of a fish contains 1.2 g hydrogen in its total body mass, if all the hydrogen is replaced with deuterium then the increase in body weight of the fish will be
(a) 1.2 g
(b) 2.4 g
(c) 3.6 g
(d) \(\sqrt{4.8}\) g
Answer:
(a) 1.2 g

Question 9.
The hardness of water can be determined by volumetrically using the reagent
(a) sodium thio sulphate
(b) potassium permanganate
(c) hydrogen peroxide
(d) EDTA
Answer:
(d) EDTA

Question 10.
The cause of permanent hardness of water is due to
(a) Ca(HCO₃)²
(b) Mg(HCO₃)²
(c) CaCl₂
(d) MgCO₃
Answer:
(c) CaCl₂

Question 11.
Zeolite used to soften hardness of water is, hydrated
(a) Sodium aluminium silicate
(b) Calcium aluminium silicate
(c) Zinc aluminium borate
(d) Lithium aluminium hydride
Answer:
(a) Sodium aluminium silicate

Question 12.
A commercial sample of hydrogen peroxide marked as 100 volume H₂O₂, it means that
(a) 1 ml of H₂O₂ will give 100 ml O₂ at STP
(b) 1 L of H₂O₂ will give 100 ml O₂ at STP
(c) 1 L of H₂O₂ will give 22.4 L O₂
(d) 1 ml of H₂O2 will give 1 mole of O₂ at STP
Answer:
(a) 1 ml of H₂O₂ will give 100 ml O₂ at STP

Question 13.
When hydrogen peroxide is shaken with an acidified solution of potassium dichromate in presence of ether, the ethereal layer turns blue due to the formation of
(a) Cr₂O₃
(b) CrO₄^2-
(c) CrO(O₂)₂
(d) none of these
Answer:
(c) CrO(O₂)₂

Question 14.
For decolourisation of 1mole of acidified KMnO₄, the moles of H₂O₂ required is
(a) \(\frac{1}{2}\)

(b) \(\frac{3}{2}\)

(c) \(\frac{5}{2}\)

(d) \(\frac{7}{2}\)
Answer:
(c) \(\frac{5}{2}\)

Question 15.
Volume strength of 1.5 NH₂O₂ is
(a) 1.5
(b) 4.5
(c) 16.8
(d) 8.4
Answer:
(d) 8.4

Question 16.
The hybridisation of oxygen atom is H₂O and H₂O₂ are, respectively
(a) sp and sp³
(b) sp and sp
(c) sp and sp²
(d) sp³ and sp³
Answer:
(d) sp³ and sp³

Question 17.
The reaction H₃PO₂ + D₂O → H₂DPO₂ + HDO indicates that hypo-phosphorus acid is
(a) tribasic acid
(b) dibasic acid
(c) mono basic acid
(d) none of these
Answer:
(c) mono basic acid
Solution:
Hypophosphorus acid on reaction with D₂O, only one hydrogen is replaced by deuterium and hence it is mono basic.

Question 18.
In solid ice, oxygen atom is surrounded
(a) tetrahedrally by 4 hydrogen atoms
(b) octahedrally by 2 oxygen and 4 hydrogen atoms
(c) tetrahedrally by 2 hydrogen and 2 oxygen atoms
(d) octahedrally by 6 hydrogen atoms
Answer:
(a) tetrahedrally by 4 hydrogen atoms
Solution:

Question 19.
The type of H-bonding present in ortho nitro phenol and p-nitro phenol are respectively
(a) inter molecular H-bonding and intra molecular f H-bonding
(b) intra molecular H-bonding and inter molecular H-bonding
(c) intra molecular H – bonding and no H – bonding
(d) intra molecular H -bonding and intra molecular H-bonding
Answer:
(b) intra molecular H-bonding and inter molecular H-bonding
                         

Question 20.
Heavy water is used as
(a) modulator in nuclear reactions
(b) coolant in nuclear reactions
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
(c) Heavy water is used as moderator as well as coolant in nuclear reactions.

Question 21.
Water is a
(a) basic oxide
(b) acidic oxide
(c) amphoteric oxide
(d) none of these
Answer:
(c) amphoteric oxide
Solution:
Water is a amphoteric oxide.

II. Write brief answer to the following questions:

Question 22.
Explain why hydrogen is not placed with the halogen in the periodic table.
Answer:
The electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halide ions. In most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group -1 along with alkali metals and not placed with halogens.

Question 23.
A cube at the 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:
In ice, each atom is surrounded tetrahedrally by four water molecules through hydrogen bonds. That is, the presence of two hydrogen atoms and two lone pairs of electron on oxygen atoms in each water molecule allows formation of a three-dimensional structure.

This arrangement creates an open structure, which accounts for the lower density of ice compared with water at 0°C. While in liquid water, unlike ice where hydrogen bonding occurs over a long-range, the strong hydrogen bonding prevails only in a short range and therefore the denser packing. Hence, ice cube sinks in water.

Question 24.
Discuss the three types of Covalent hydrides.
Answer:
Covalent hydrides are the compound in which hydrogen is attached to another element by sharing of electrons. The most common examples of covalent hydrides of non-metals are methane, ammonia, water and hydrogen chloride. Covalent hydrides are further divided into three categories, viz., electron
precise (CH₄ C₂H₆), electron deficient (B₂H₆) and electron-rich hydrides (NH₃H₂ O). Since most of the covalent hydrides consists of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

Question 25.
Predict which of the following hydrides is a gas on a solid (a) HCl (b) NaH. Give your reason.
Answer:
Sodium hydride (NaH) is a gas on a soiid. NaH is prepared by direct reaction of hydrogen gas with liquid sodium and it has NaCl crystal structure. It is a tendency of deprotonating in many organic reactions. NaH is used in fuel cells.

Question 26.
Write the expected formulas for the hydrides of 4^th period elements. What is the trend in the formulas? In what way the first two numbers of the series different from the others?
Answer:
The expected formulas of the hydrides of 4^th period elements are MH or MH₂. However, except the first two members, many of the elements form non- stoichiometric interstitial hydrides with variable composition. The first two members of the period alkali metal (Potassium) and alkali earth metal (Calcium) forms ionic hydrides.

Question 27.
Write chemical equation for the following reactions.
Answer:
(i) reaction of hydrogen with tungsten (VI) oxide NO₃ on heating.
(ii) hydrogen gas and chlorine gas.
Answer:
(i) Hydrogen can be used to reduce metal oxide into metal. Hydrogen reduces tungsten(VI) oxide into tungsten.
WO₃ + 3H₂ → W + 3H₂O

(ii) Hydrogen reacts with chlorine at room temperature under light gives hydrogen chloride.
H_2(g) + Cl_2(g) → 2HCl_(g)

Question 28.
Complete the following chemical reactions and classify them into (a) hydrolysis (b) redox (c) hydration reactions.
(i) KMnO₄ + H₂O₂ →
(ii) CrCl₃ + H₂O →
(iii) CaO + H₂O →
Answer:
(i) KMnO₄ + H₂O₂ → 2KMnO₂ + 2KOH + 2H₂O + 3O₂
The reaction of potassium permanganate with hydrogen peroxide is a redox reaction.

(ii) CrCl₃ + H₂O → [Cr(H₂O)₆]Cl₃
It is a hydration reaction. Many salts crystallized from aqueous solutions form hydrated crystals. The water in the hydrated salt may form co-ordinate bonds.

(iii) CaO + H₂O → Ca(OH)₂
It is a hydrolysis reaction. Calcium oxide hydrolyses to calcium hydroxide.

Question 29.
Hydrogen peroxide can function as an oxidising agent as well as reducing agent. Substantiate this statement with suitable examples.
Answer:
Hydrogen peroxide can act both as an oxidizing agent and a reducing agent. Oxidation is usually performed in acidic medium while the reduction reactions are performed in basic medium.

In acidic conditions:
H₂O₂ + 2H⁺ + 2e^– → 2H₂O (E° = +1.77 V)
For example
2FeSO₄ + H₂SO₄ + H₂O₂ → Fe₂(SO₄)₃ + 2H₂O

In basic conditions:
HO₂^– + OH^– → O₂ + H₂O + 2e^– (E° = + 0.08V)

For Example,
2 KMnO_4(aq) + 3H₂O_2(aq) → 2MnO₂ + 2KOH + 2H₂O + 3O₂(g)

Question 30.
Do you think that heavy water can be used for drinking purposes?
Answer:
Heavy water cannot be used for drinking purposes because it does not form hydrogen bonding.

Question 31.
What is water-gas shift reaction?
Answer:
The carbon monoxide of the water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.
CO + H₂O → CO₂ + H₂
The CO₂ formed in the above process is absorbed in a solution of potassium carbonate.
CO₂ + K₂CO₃ + H₂O → 2KHCO₃

Question 32.
Justify the position of hydrogen in the periodic table?
Answer:
The hydrogen has the electronic configuration of 1s¹ which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H⁺) like alkali metals {Na⁺, K⁺, Cs⁺)
2. It forms halides (HX), oxides, (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, NaH₂OH₂, NaH₂S)
3. It also acts as a reducing agent.

However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1,314 kJ mol^-1 which is much higher than alkali metals.

Like the formation of halides (X^–) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion (H^–) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\(\frac{1}{2}\) H₂ + e^– → H^–                         ∆H = +36 kcalmol^-1
\(\frac{1}{2}\) Br₂ + e^– → Br^–                       ∆H = -55 kcalmol^-1

Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Question 33.
What are isotopes? Write the names of isotopes of hydrogen.
Answer:
Atoms of the same element having same atomic number and different mass number are called isotopes.
Hydrogen has three naturally occurring isotopes, viz., protium (₁H¹ or H), deuterium (₁H² or D) and tritium (₁H³ or T). Protium ₁H¹ is the predominant form (99.985 %) and it is the only isotope that does not contain a neutron. Deuterium, also known as heavy hydrogen, constitutes about 0.015 %. The third isotope, tritium is a radioactive isotope of hydrogen which occurs only in traces (~1 atom per 1018 hydrogen atoms). Due to the existence of these isotopes naturally occurring hydrogen exists as H₂, HD, D₂, HT, T₂, and DT.

Question 34.
Give the uses of heavy water.
Answer:
The uses of heavy water are as follows
(i) Heavy water is widely used as moderator in nuclear reactors as it can lower the energies of fast neutrons
(ii) It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
(iii) It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 35.
Explain the exchange reactions of deuterium.
Answer:
When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for deuterium. This reaction is known as exchange reaction of deuterium.
2NaOH + D₂O → 2NaOD + HOD

HCl + D₂O → DCl + HOD

NH₄Cl + 4D₂O → ND₄Cl + 4HOD

These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D₂O is treated with of hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
H₃PO₂ + D₂O → H₂DPO₂ + HDO
It is also used to prepare some deuterium compounds:

Al₄C₃ + 12D₂O → 4Al(OD)₃ + 3CD₄
CaC₂ + 2D₂O → Ca(OD)₂ + C₂D₂
Mg₃N₂ + 6D₂O → 3Mg(OD)₂ + 2ND₃
Ca₃P₂ + 6D₂O → 3Ca(OD)₂ + 2PD₃

Question 36.
How do you convert parahydrogen into ortho hydrogen?
Answer:
At room temperature, normal hydrogen consists of about 75% ortho-form and 25% para-form. As the ortho-form is more stable than para-form, the conversion of one isomer into the other is a slow process. However, the equilibrium shifts in favour of para hydrogen when the temperature is lowered.

The para-form can be catalytically transformed into I ortho-form using platinum or iron. Alternatively, it can also be converted by passing an electric j discharge, heating above 800°C and mixing with paramagnetic molecules such as O₂, NO, NO₂ or with nascent/atomic hydrogen.

Question 37.
Mention the uses of deuterium.
Answer:
(i) Deuterium is used to prepare heavy water which is used as moderator in nuclear reactors.
(ii) Deuterium exchange reactions are useful in determining the number of ionic hydrogens present in a given compound.
(iii) It is also used to prepare some deuterium compounds.

Question 38.
Explain preparation of hydrogen using electrolysis.
Answer:
High purity hydrogen (> 99.9 %) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
At anode:
2OH^– → H₂O + \(\frac{1}{2}\)O₂ + 2e^–
At cathode:
2H₂O + 2e^– → 2OH^– + H₂
Overall reaction:
H₂O → H₂ + \(\frac{1}{2}\)O₂

Question 39.
A group-1 metal (A) which is present ¡n common salt reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state. (B) on reaction with a gas to give universal solvent (D). The compound (D) on reacts with (A) to give (B), a strong base. Identify A, B, C, D and E. Explain the reactions.
Answer:
The metal belongs to Group-1 which is present in common salt is Sodium(A).
The metal reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state.
2Na + H₂ → 2NaH
Hence, the (B) is hydrogen and (C) is sodium hydride. (B) on reaction with a gas to give universal solvent (D).
H₂ + O₂ → 2NaOH
The universal solvent (D) is water. The compound (D) on reaction with (A) to give (E) which is a strong base.
H₂O + 2Na → 2NaOH
The Compound (E) is Sodium hydroxide.
A – Sodium (Na)
B – Hydrogen (H₂)
C – Sodium Hydride (NaH)
D – Water (H₂O)
E – Sodium hydroxide (NaOH)

Question 40.
An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderatorin nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C₃H₆ to give (D). Identify A, B, C and D.
Answer:
An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderator in nuclear reaction.
2D₂ + O₂ → 2D₂O
The isotope of hydrogen is deuterium(A), diatomic element is oxygen and the compound (B) is heavy water.
(A) adds on to a compound (C), which has the molecular formula C₃H₆ to give (D).
CH₃ – CH = CH₂ + D₂ → CH₃ – CHD – CH₂D
(C)                                     (D)
The compound (C) is propene and (D) is 1, 2 deutropropane.
A – Deuterium (D₂)
B – Heavy water (D₂O)
C – Propene (CH₃ – CH = CH₂)
D – 1, 2 deutero propene (CH₃ – CHD – CH₂D)

Question 41.
NH₃ has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15- Explain.
Answer:
When a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, the bond is polarized. Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity.

This interaction is called hydrogen bonding. Hence, NH₃ has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15 due to intermolecular hydrogen bonding.

Question 42.
Why interstitial hydrides have a lower density than the parent metal?
Answer:
In interstitial hydrides, hydrogen occupies the interstitial sites. These hydrides show properties similar to parent metals. Most of these hydrides are non-stoichiometric with variable composition. Hence, interstitial hydrides have lower density than the parent metal.

Question 43.
How do you expect the metallic hydrides to be useful for hydrogen storage?
Answer:
Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Most of the hydrides are non-stoichiometric with variable composition (TiH _{1.5 – 1.8} and PdH_{0.6 – 0.8}), some are relatively light, inexpensive and thermally unstable which make them useful for hydrogen storage applications.

Question 44.
Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.
Answer:
When a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, the bond is polarized. Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity. This interaction is called hydrogen bonding. The magnitude of hydrogen bonding increases with the increase in electronegativity of the atom. Hence, the increasing magnitude of hydrogen bonding of NH₃, H₂O and HF follows the order NH₃ < H₂O < HF.

Question 45.
Compare the structures of HO and HO.
Answer:
Both in gas-phase and liquid-phase, the molecule adopts a skew conformation due to repulsive interaction of the OH bonds with lone-pairs of electrons on each oxygen atom. Indeed, it is the smallest molecule known to show hindered rotation about a single bond.

H₂O₂ has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in figure. Structurally, H₂O₂ is represented by the dihydroxyl formula in which the two OH-groups do not lie in the same plane.

One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spine. In the solid phase of molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°. Water has bent structure and the H-O-H bond angle is 104.5°.

11th Chemistry Guide Hydrogen Additional Questions and Answers

I. Choose the best Answer:

Question 1.
A simplest atom which contains one electron and one proton is
(a) Helium
(b) Deuterium
(c) Hydrogen
(d) Tritium
Answer:
(c) Hydrogen

Question 2.
Hydrogen has similarities with
(a) Alkali metals and halogens
(b) Alkaline earth metals and halogens
(c) Alkalimetals and noble gases
(d) Halogens and noble gases.
Answer:
(a) Alkali metals and halogens

Question 3.
Which of the following properties of hydrogen similar to alkali metals?
1. It forms uni negative ion.
2. It forms halides, oxides and sulphides similar to alkali metals.
3. It also acts as a reducing agent.
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1,2 and 3
Answer:
(b) 2 and 3

Question 4.
Ionisation energy of hydrogen is (in kJ mol^-1)
(a) 377
(b) 520
(c) 1413
(d) 1314
Answer:
(d) 1314

Question 5.
Hydrogen is placed in the ________ of the periodic table.
(a) group – 1
(b) group – 17
(c) group – 18
(d) group – 2
Answer:
(a) group – 1

Question 6.
The predominant isotope of hydrogen is
(a) deuterium
(b) protium
(c) tritium
(d) heavy hydrogen
Answer:
(b) protium

Question 7.
The radioactive isotope of hydrogen is
(a) protium
(b) deuterium
(c) tritium
(d) heavy hydrogen
Answer:
(c) tritium

Question 8.
The number of naturally occurring hydrogens due to existence of isotopes is
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(a) 6

Question 9.
The half life period of tritium is ______(in years).
(a) 13.2
(b) 10.5
(c) 12.3
(d) 15.8
Answer:
(c) 12.3

Question 10.
The correct order of melting point of isotopes of hydrogen is
(a) H < D < T
(b) T < D < H
(c) H < T < D
(d) D < H < T
Answer:
(a) H < D < T

Question 11.
The percentage of ortho and para forms of normal hydrogen at room temperature is
(a) 25% and 75%
(b) 40% and 60%
(c) 60% and 40%
(d) 75% and 25%
Answer:
(d) 75% and 25%

Question 12.
Which of the following statement is correct about ortho-para hydrogen?
(a) The magnetic moment of para hydrogen is twice that of a proton.
(b) Ortho form is more stable than para form
(c) Ortho form can be catalytically converted into para form using platinum.
(d) At room temperature, normal hydrogen consists of 75% para form.
Answer:
(b) Ortho form is more stable than para form

Question 13.
The composition of syngas is
(a) CO + N₂
(b) CO + H₂O
(c) CO + H₂
(d) CO₂ + H₂
Answer:
(c) CO + H₂

Question 14.
During electrolysis of water containing traces of acid hydrogen is liberated at
(a) cathode
(b) anode
(c) both anode and cathode
(d) none of the above
Answer:
(a) cathode

Question 15.
The conversion of carbon monoxide of the water gas into carbon dioxide is called water gas _______ reaction.
(a) displacement
(b) decomposition
(c) shift
(d) conversion
Answer:
(c) shift

Question 16.
The catalyst used in the water gas shift reaction is
(a) Copper
(b) Nickel
(c) Platinum
(d) Palladium
Answer:
(a) Copper

Question 17.
The CO₂ formed in the water gas shift reaction is absorbed in a solution of
(a) Potassium bicarbonate
(b) Sodium chloride
(c) Potassium sulphate
(d) Potassium carbonate
Answer:
(d) Potassium carbonate

Question 18.
The percentage of heavy water in normal water is
(a) 1.6 × 10⁴
(b) 1.6 × 10^-4
(c) 1.6 × 10^-3
(d) 1.6 × 10²
Answer:
(b) 1.6 × 10^-4

Question 19.
When water is completely electrolysed, the gas liberated is/are
(a) H₂
(b) D₂
(c) H₂ and D₂
(d) H₂ and T₂
Answer:
(c) H₂ and D₂

Question 20.
Lithium Aluminium Hydride is
(a) [LiAlH₃]
(b) Li[Al₂H₄]
(c) Li[AlH₂]
(d) Li[AlH₄]
Answer:
(d) Li[AlH₄]

Question 21.
Deuterium oxide is called
(a) hard water
(b) soft water
(c) heavy water
(d) heavy hydrogen
Answer:
(c) heavy water

Question 22.
Ammonia is synthesized by ______ process.
(a) Haber’s
(b) Bergius
(c) Decon’s
(d) Solvay
Answer:
(b) Bergius