Answer the following questions.<\/span><\/p>\nQuestion 1.
\nBriefly account for the trend in atomic radius of elements in the nitrogen family.
\nAnswer:
\nAtomic radii increase on going down the group. The increase is due to addition of new energy level in each succeeding element. There is a considerable increase in covalent radius from N to P but from As to Bi only a small increase is observed.
\nThe increase in nitrogen to phosphorus is attributed to strong shielding effect of \u2018s\u2019 and \u2018p\u2019 electrons present in inner shells. Small increase in covalent radii from As to Bi is due to poor shielding effect of \u2018d\u2019 and \u2018f\u2019 electrons present in inner shells on valency electrons. The increased nuclear charge reduces the effect of the addition of new shell to same extent.<\/p>\n
Question 2.
\nBriefly explain the trend in melting and boiling point in the nitrogen family elements.
\nAnswer:
\nThe melting point first increase from nitrogen to arsenic and then decreases from antimony and bismuth. The boiling point gradually increases from nitrogen to antimony. Lower values of melting points for antimony and bismuth may be due to availability of only three electrons instead of five for metallic bonding due to inert pair effect.<\/p>\n
<\/p>\n
Question 3.
\nWhat happens when (i) Sodium azide is heated? (ii) Ammonia is treated with bromine. Give equations.
\nAnswer:
\nIn both cases, nitrogen is formed.
\n<\/p>\n
Question 4.
\nExplain why nitrogen is inert at room temperature.
\nAnswer:
\nThis is due to high bond enthalpy of N \u2261 N. The triple bond is short and hence large amount of heat is necessary to break the bond. Hence N2<\/sub> is unreactive at room temperature.<\/p>\nQuestion 5.
\nWhat are nitrides? Give the preparation of (i) Lithium nitride, (ii) Calcium nitride, (iii) Boron nitride by means of chemical equation.
\nAnswer:
\nNitrides are binary compounds of nitrogen which contains N-3<\/sup> ion. They are formed by direct combination of metals with nitrogen.
\n<\/p>\nQuestion 6.
\nMention the conditions under which maximum amount of ammonia is formed from nitrogen to hydrogen.
\nAnswer:
\nThe formation of NH3<\/sub> is represented by
\n
\nis an exothermic reaction. The reaction is favoured at high temperatures and at an optimum temperature in the presence of iron as catalyst.<\/p>\nQuestion 7.
\nGive equation for (i) hydrolysis of urea,
\n(ii) heating ammonium chloride with CaO,
\n(iii) Heating magnesium nitride with water.
\nAnswer:
\n<\/p>\n
<\/p>\n
Question 8.
\nHow is ammonia manufactured?
\nAnswer:
\nAmmonia is manufactured by passing nitrogen and hydrogen over iron catalyst (a small amount of K2<\/sub>O and Al2<\/sub>O3<\/sub> is also used to increase the rate of attainment of equilibrium) at 750 K at 200 atm pressure. In the actual process the hydrogen required is obtained from water gas and nitrogen from fractional distillation of liquid air.<\/p>\nQuestion 9.
\nCompare the properties of liquid ammonia and water.
\nAnswer:
\n(i) Both liquid ammonia and water are highly associated through strong hydrogen bonding.
\n(ii) Both ammonia and water are good ionising solvents.
\n(iii) The ionisation of ammonia and water is given as
\n
\n
\nBoth OH–<\/sup> and NH2<\/sub>–<\/sup> are strong bases,
\neg: NaOH and NaNH2<\/sub>.<\/p>\nQuestion 10.
\nGive equations for the following reactions.
\n(i) Ammonia is heated over 500\u00b0C
\n(ii) Ammonia is burnt in oxygen.
\n(iii) Ammonia is burnt in oxygen in the presence of a metal catalyst (pt)
\n(iv) Ammonia is treated with excess of chlorine.
\n(v) Excess of ammonia is treated with chlorine.
\nAnswer:
\n<\/p>\n
Question 11.
\nGive an example for a reducing property of ammonia. [OR] Ammonia is a reducing agent. Give an example to prove this statement.
\nAnswer:
\nIt reduces metal oxides to metals when passed over heated metallic oxide.
\n
\nIn this reaction, ammonia is oxidised to nitrogen.<\/p>\n
Question 12.
\nWhat are amides and nitrides? Give an example for each. How are they formed?
\nAnswer:
\nAmides are salts formed by ammonia and a strong electropositive metal like Na, eg: NaNH2<\/sub> (sodamide). It is formed by the action of ammonia on sodium.
\n2Na + 2NH3<\/sub> \u2192 2NaNH2<\/sub> + H2<\/sub>
\nNitrides are salts which contain N-3<\/sup> ion.
\neg: Mg3<\/sub>N2<\/sub> (magnesium nitride). It is formed by the action of magnesium with ammonia.
\n3Mg + 2NH3<\/sub> \u2192 Mg3<\/sub> N2<\/sub> + 3H2<\/sub><\/p>\n<\/p>\n
Question 13.
\nWhat happens when an aqueous solution of ammonia is treated with (i) aqueous solution of ferric chloride, (ii) an aqueous solution of cupric chloride, (iii) an aqueous solution of aluminium chloride.
\nAnswer:
\nAn aqueous solution of ammonia or ammonium hydroxide precipitates metal hydroxides from metallic salts solution.
\n
\nThis property is made use of in detecting group III metals in qualitative analysis.<\/p>\n
Question 14.
\nExplain with examples, that ammonia acts as a Lewis base.
\nAnswer:
\nDue to the presence of lone pair of electrons on the nitrogen atom, ammonia acts as a Lewis base. It can donate its electron pair to form coordinate covalent bond with electron deficient molecules eg: BF3<\/sub> or transition metal cation, having vacant \u2018d\u2019 orbitals) to form complexes. For example,
\n<\/p>\nQuestion 15.
\nExplain why (i) insoluble silver chloride dissolves in aqueous ammonia?
\nAnswer:
\nThis is due to the formation of a complex Ag (NH3<\/sub>)2<\/sub> Cl (diammine silver (I) chloride)
\n<\/p>\nQuestion 16.
\nWhen aqueous ammonia is treated with a copper sulphate solution, a blue precipitate is formed It dissolves an adding excess ammonia. Explain this observation.
\nAnswer:
\nThe precipitate formed is cupric hydroxide, which dissolves in excess ammonia to form a blue solution due to the formation of soluble [Cu (NH3<\/sub>)4<\/sub>] SO4<\/sub> complex, i.e., tetrammine copper(II) sulphate.
\n<\/p>\nQuestion 17.
\nExplain the structure of ammonia.
\nAnswer:
\nThe nitrogen in ammonia is sp3<\/sup> hybridised. Each of the sp3<\/sup> hybrid orbital containing one unpaired electron forms a covalent bond with the 1s orbital of hydrogen. The fourth sp3<\/sup> hybrid orbital contain a lone pair of electrons. Because of the lone pair – bond pair repulsion, the tetrahedral bond angle is reduced to 107\u00b0. Hence, it assumes a pyramidal shape.
\n<\/p>\n<\/p>\n
Question 18.
\nHow is nitric acid prepared?
\nAnswer:
\nNitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
\nKNO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 KHSO4<\/sub> + HNO3<\/sub>
\nThe temperature is kept as low as possible to avoid decomposition of nitric acid.<\/p>\nQuestion 19.
\nGive reason for the following:
\nAnswer:
\nThe nitric acid prepared by heating potassium nitrate and conc. H2<\/sub>SO4<\/sub> is brown coloured.
\nThe acid formed by heating KNO3<\/sub> and H2<\/sub>SO4<\/sub> dioxide formed by the decomposition of nitric acid.
\n4HNO3<\/sub> \u2192 4NO2<\/sub> + 2H2<\/sub>O + O2<\/sub><\/p>\nQuestion 20.
\nExplain the manufacture of Oswalld\u2019s process of nitric acid.
\nAnswer:
\nThe method is based as upon the catalystic oxidation of ammonia by atmospheric oxygen.
\n
\nNitric oxide thus formed readily combines with oxygen to form nitrogen dioxide (NO2<\/sub>).
\n2NO(g) + O2<\/sub>(g) \u2192 2NO2<\/sub>(g)
\nNitrogen dioxide, thus formed dissolves in water to form nitric acid.
\n3NO2<\/sub> (g) + H2<\/sub>O (l) \u2192 2HNO3<\/sub> (aq) + NO (s)
\nThe nitric oxide formed is recycled and the aqueous nitric acid is concentrated by distillation to give 68% nitric acid by mass. Further concentration to 98% can be achieved by dehydration with conc. H2<\/sub>SO4<\/sub>.<\/p>\nQuestion 21.
\nExplain with examples to show that nitric acid acts as an (i) acid, (ii) oxidising agent and (iii) as nitrating agent.
\nAnswer:
\n(i) It reacts with bases and basic oxides to form salts and water.
\n
\n(ii) It oxidises non metals to their highest oxy acids.
\n
\n(iii) As a nitrating agent, it forms nitronium ion (NO2<\/sub>+<\/sup>) in aromatic electrophillic substitution reactions.
\n<\/p>\nQuestion 22.
\nBriefly explain the action of nitric acid on metals.
\nAnswer:
\nNitric acid, both dilute and concentrated are powerful oxidising agents and metals are good reducing agents. The products of oxidation depending on the concentration of the acid and the nature of the metal.
\n(i) Gold, platinum, rhodium, iridium and tantalum do not react with nitric oxide.
\n(ii) Metals which are more electropositive than hydrogen (Na, K, Ca, Mg, Al, Mn, Zn, Cr, Cd, Fe, CO, Ni, Sn, Pb, etc.,) reduces HNO3<\/sub> to a variety of products such as NO2<\/sub>, NO, NH3<\/sub>, NH4<\/sub> NO3<\/sub> and N2<\/sub>O, depending upon the conditions.
\n
\n(iii) Magnesium and manganese are the only metals that produce H2<\/sub> with very dilute (1-2%) HNO3<\/sub>.
\n
\n(iv) More active metals like magnesium, zinc, tin and iron react with cold, dilute nitric acid to form ammonium nitrate.
\n
\nLead, under similar condition, give lead nitrate and nitric oxide.
\n
\n(v) With hot, dilute nitric acid, ammonium nitrate thus formed undergoes decomposition to form nitrous oxide (N2<\/sub>O)
\n
\n(vi) Metals like Zn, Mg, Bi, Pb, etc., form metallic nitrate and nitrogen dioxide when treated with conc. HNO3<\/sub>.
\n
\n(vii) Similarly,
\n
\nException: Tin forms metastannic acid H2<\/sub>SnO3<\/sub> and NO2<\/sub>
\n
\n(viii) Metals which are less electropositive than hydrogen, reduce nitric acid to NO2<\/sub> or NO.
\n
\neg:
\n
\n(ix) Metals like iron, chromium, nickel and aluminium because passive as treatment with cone. HNO3<\/sub>.
\n(x) Noble metals like gold, platinum do not react with conc. HNO3<\/sub>.<\/p>\nQuestion 23.
\nGive uses of nitric acid.
\nAnswer:<\/p>\n
\n- Nitric acid is used as a oxidising agent and in the preparation of aquaregia.<\/li>\n
- Salts of nitric acid are used in photography (AgNO3<\/sub>) and gunpowder for firearms. (NaNO3<\/sub>).<\/li>\n<\/ol>\n
<\/p>\n
Question 24.
\nComplete and balance the following equations.
\n
\nAnswer:
\n<\/p>\n
Question 25.
\nGive the oxidation states of nitrogen in the following oxides.
\n(i) N2<\/sub>O, (ii) NO, (iii) N2<\/sub>O3<\/sub>, (iv) NO2<\/sub>, (v) N2<\/sub>O4<\/sub>, (vi) N2<\/sub>O5<\/sub>.
\nAnswer:
\n<\/p>\nQuestion 26.
\nWrite the structure of the following oxides of nitrogen, (i) N2<\/sub>O, (ii) NO, (iii) N2<\/sub>O3<\/sub>, (iv) NO2<\/sub>, (v) N2<\/sub>O4<\/sub>, (vi) N2<\/sub>O5<\/sub>.
\nAnswer:
\n<\/p>\nQuestion 27.
\nGive equation for the preparation of the following acids.
\n(i) Hyponitrous acid,
\n(ii) Nitrous acid,
\n(iii) Pemitrous acid,
\n(iv) Nitric acid,
\n(v) Pemitric acid.
\nAnswer:
\n<\/p>\n
<\/p>\n
Question 28.
\nGive reason for the following:
\n(i) Freshly prepared phosphorus becomes yellow on standing.
\n(ii) Yellow phosphorus glows in dark.
\n(iii) Nitrogen is a gas while phosphorus is a solid.
\nAnswer:
\n(i) This is due to the formation of a red phosphorus upon standing.
\n(ii) This is due to oxidation which is known as phosphorus.
\n(iii) The nitrogen has N \u2261 N structure while in phosphorus four atoms in phosphorus have a polymeric structure with chains of P4<\/sub> linked tetrahedrally. P \u2261 P is less stable than P \u2014 P bonds. Hence nitrogen is a gas while phosphorus is a solid.<\/p>\nQuestion 29.
\nWrite the structure of the following:
\n(i) Hyponitrous acid, (ii) Hydronitrous acid, (iii) Nitrous acid, (iv) Pemitrous acid, (v) Nitric acid, (vi) Pemitric acid.
\nAnswer:
\n<\/p>\n
Question 30.
\nGive the formula and the oxidation state of nitrogen in the following acids.
\n(i) Hyponitrous acid, (ii) Nitrous acid, (iii) Pemitrous acid, (iv) Nitric acid, (v) Pemitric acid.
\nAnswer:
\n<\/p>\n
Question 31.
\nExplain the structure of white phosphorus.
\nAnswer:
\nIt exists as P4<\/sub> units. The four sp3<\/sup> hybridised phosphorus atoms lie at the corners of a regular tetrahedral with \u2220PPP=60\u00b0 as shown in figure.
\n
\nEach phosphorus atom is linked to other three phosphorus atoms by covalent bonds so that each P-atom completes its octet.<\/p>\n<\/p>\n
Question 32.
\nExplain why white phosphorus is kept under water.
\nAnswer:
\nThe P4<\/sub> units of white phosphorus are held together by weak vanderwaals forces of attraction. As a result, its ignition temperature (303K) is very low and easily catches fire. Hence, it is kept under water.<\/p>\nQuestion 33.
\nThe red phosphorus is less reactive than white phosphorus. Explain with examples.
\nAnswer:
\n(i) Yellow phosphorus readily catches fire in air forming phosphorus pentoxide whereas red phosphorus forms phosphorus trioxide or phosphorus pentoxide only on heating.
\n
\n(ii) With chlorine, phosphorus forms phosphorus tri and penta chlorides. White phosphorus reacts violently at room temperature while red phosphorus react on heating.
\n
\n(iii) Yellow phosphorus reacts with alkali an boiling in an inert atmosphere to form phosphine. Red phosphorus has no action an alkali.<\/p>\n
Question 34.
\nDistinguish between white and red phosphorus interms of their properties.
\nAnswer:
\n<\/p>\n
Question 35.
\nCompare the chemical properties of white and red phosphorus.
\nAnswer:
\n
\n<\/p>\n
Question 36.
\nHow phosphine is formed from (i) White phosphorus, (ii) Calcium phosphide, (iii) Phosphorus acid, (iv) Phosphonium iodide. Give equations.
\nAnswer:
\n(i) By heating white phosphorus and sodium hydroxide in an inert atmosphere of carbon dioxide or hydrogen.
\n
\n(ii) By the hydrolysis of calcium phosphide with water or dilute mineral acids.
\n
\n(iii) By heating phosphorus acid (H3<\/sub>PO3<\/sub>)
\n
\n(iv) By heating phosphorus iodide with sodium hydroxide.
\n<\/p>\n<\/p>\n
Question 37.
\nGive equations for the following:
\n(i) Phosphine is heated at 317 K in the absence of air.
\n(ii) Phosphine is heated with oxygen.
\n(iii) Phosphine is treated with chlorine.
\nAnswer:
\n
\n(iii) PH3<\/sub> + 4Cl3<\/sub> \u2192 PCl5<\/sub> + 3HCl<\/p>\nQuestion 38.
\nGive examples for (i) basic nature and (ii) reducing property of phosphine.
\nAnswer:
\n(i) Basic nature of PH3<\/sub>: Because it contains a lone pair of electrons as phosphorus, it accepts a proton and form phosphorium ion.
\n
\nIt is a weaker base than NH3<\/sub> because of larger size and lesser electronegativity of phosphorus than nitrogen.
\n
\n(ii) Reducing property of PH3<\/sub>: Phosphine reduces aqueous solutions of copper and mercury salts to their corresponding phosphides.
\n<\/p>\nQuestion 39.
\nBriefly explain the structure of phosphine.
\nAnswer:
\nThe phosphorus in phosphine is sp3 hybridised. The three sp3<\/sup> hybrid orbitals each containing an electron overlaps with 1s orbital of hydrogen, containing are electron form three P \u2014 H covalent bonds. The fourth sp3<\/sup> hybrid orbital contains a lone pair of electron. The lone pair-bond pair interaction, due to the larger size and lesser electro negativity of phosphorus, compared to nitrogen, reduces the tetrahedral bond angle to 94\u00b0. Hence phosphine has a pyramidal shape, like ammonia.<\/p>\nQuestion 40.
\nWhat are Holmes signals? Mention its use.
\nAnswer:
\nPhosphine (PH3<\/sub>) is used in Holme\u2019s signals in deep seas and oceans for signalling danger points to steamers. Containers containing a mixture of calcium phosphide and calcium carbide are pierced and thrown into the sea.
\nIn contact with water, a mixture of acetylene and phosphine is produced. Phosphine contains traces of highly inflammable P2<\/sub>H4<\/sub> which catches five spontaneously. This ignites acetylene with a luminous flame and thus serves as a signal to the approaching ship.<\/p>\nQuestion 41.
\nHow is phosphorus trichloride prepared? [OR] What happens when white phosphorus is treated with a stream of chlorine gas.
\nAnswer:
\nP4<\/sub> + 3Cl2<\/sub> \u2192 P4<\/sub>Cl6<\/sub><\/p>\n<\/p>\n
Question 42.
\nWhat happens when (i) White phosphorus is treated with thionyl chloride.
\n(i) White phosphorus is treated with thionyl chloride.
\n(ii) Phosphorus trichloride is hydrolysed by cold water.
\nAnswer:
\n(i) Phosphorus trichloride is formed.
\n4P4<\/sub> + 8SOCl2<\/sub> \u2192 4PCl3<\/sub> + 4SO2<\/sub> + 2S2<\/sub>Cl2<\/sub>
\n(ii) Hydrolysis of phosphorus trichloride gives phosphorus acid. (H3<\/sub>PO3<\/sub>)
\nPCl3<\/sub> + 3H2<\/sub>O \u2192 H3<\/sub>PO3<\/sub> + 3HCl<\/p>\nQuestion 43.
\nHow does phosphorus trichloride react with
\n(i) ethanol and (ii) propionic acid? Give equations.
\nAnswer:
\nThe \u2018OH\u2019 group in alcohols and acids are replaced by \u2018Cl\u2019.
\n<\/p>\n
Question 44.
\nExplain the structure of phosphorus trichloride.
\nAnswer:
\nIts structure is similar to ammonia. P atoms undergoes sp3<\/sup> hybridisation. In the tetrahedral configuration one of the positions is occupied by the lone pair. Thus, it is pyramidal in shape. The Cl \u2014 P \u2014 Cl bond angle is 100.4\u00b0 and P \u2014 Cl bond length is 204 pm.
\n<\/p>\nQuestion 45.
\nGive equations for (i) the formation of PCl5<\/sub> from PCl3<\/sub> (ii) H3<\/sub>PO4<\/sub> from PCl5<\/sub>.
\nAnswer:
\n<\/p>\nQuestion 46.
\nHow does phosphorus pentachloride react with metals? Give examples.
\nAnswer:
\nPhosphorus penta chloride (PCl5<\/sub>) reacts with metals to give metal chlorides.
\n<\/p>\nQuestion 47.
\nBriefly outline the structure of PCl5<\/sub>.
\nAnswer:
\nIt has a trigonal bipyramidal shape in which P undergoes sp3<\/sup>d hybridisation.
\n<\/p>\nQuestion 48.
\nMention the uses of phosphorus pentachloride.
\nAnswer:
\nPhosphorus pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.<\/p>\n
<\/p>\n
Question 49.
\nHypophosphorus acid is monobasic and a good reducing agent. Explain.
\nAnswer:
\nThe structure of hypophosphorus acid is
\n
\nIt has One P \u2014 OH bond, two P \u2014 H and P \u2192 O or P = O bonds. Of the three hydrogen atoms, only one hydrogen atom attached to the oxygen atom (OH group) is replaceable. Hence it is a monobasic acid. Because of the presence of P \u2014 H bond, it acts as a reducing agent.<\/p>\n
Question 50.
\nBriefly explain the structure of phosphorus trioxide and phosphorus pentoxide.
\nAnswer:
\n(i) Structure of phosphorus trioxide: The phosphorus atom lie at tetrahedral positions with respect to each other and six oxygen atoms are inserted between them. Each phosphorus atom is covalently bonded to three oxygen atoms and each oxygen is bonded to two phosphorus atom. The P \u2014 O bond length is 165.6 pm.
\n
\n(ii) Structure of phosphorus pentoxide: In P4<\/sub>O10<\/sub>, each P atom form three bonds with oxygen atom an also an additional coordinate bond with an oxygen atom. Terminal coordinate P \u2014 O bond length is 143 pm.
\n<\/p>\nQuestion 51.
\nWhat is the basicity of orthophosphorus acid? Explain with its structure. Will it act as a reducing agent or not?
\nAnswer:
\nThe structure of orthophosphorus acid is
\n
\nIt has two P \u2014 P \u2014 OH bonds, one P \u2192 O or P = O bond, and one P \u2014 H bond. There are two \u2018OH\u2019 groups attached to the phosphorus atom, and the hydrogen atoms are replaceable.
\nHence it is a dibasic acid. The acid and their salts are good reducing agent because of P \u2014H bond.<\/p>\n
Question 52.
\nWrite the formula and structure of hypo phosphoric acid. What information do you get from its structure regarding its basicity and its reducing action?
\nAnswer:
\nThe formula is H4<\/sub>P2<\/sub>O6<\/sub> and its structure is
\n
\nIt contains \u20184\u2019 OH group and 4 replaceable hydrogen atoms. Hence it is a tetrabasic acid or its basicity is four. It does not contain a P \u2014 H bond. Hence, it does not act as a reducing agent.<\/p>\nQuestion 53.
\nOrthophosphoric acid is tribasic and is not a reducing agent. Explain.
\nAnswer:
\nOrthophosphoric acid (H3<\/sub>PO4<\/sub>) is tribasic, because it contains three OH groups attached to the phosphorus atom where all the hydrogens are replaceable.
\nHence, it is a tribasic acid.
\n
\nIt forms three series salt, by successive replacement of H atom of the \u2018OH\u2019 group,
\n
\nThese salts are known as dihydrogen phosphate, hydrogen phosphate and phosphate respectively. It is not a reducing agent because it does not contain a P \u2014 H bond.<\/p>\n<\/p>\n
Question 54.
\nWrite the structure of pyrophosphoric acid and explain its basicity on the basis of the structure.
\nAnswer:
\nPhosphoric acid (H4<\/sub>P2<\/sub>O7<\/sub>) has the structure
\n
\nIt is a tetra basic acid.<\/p>\nQuestion 55.
\nWrite the formula and the oxidation state of phosphorus in the following oxyacids.
\n(i) Hypophosphorus acid
\n(ii) Orthophosphorus acid
\n(iii) Hypophosphoric acid
\n(iv) Orthophosphoric acid
\n(v) Pyrophosphoric acid
\nAnswer:
\n<\/p>\n
Question 56.
\nComplete and balance the following equations:
\n
\nAnswer:
\n<\/p>\n
Question 57.
\nBriefly explain the trend in physical properties of group 16 elements.
\nAnswer:
\nGroup 16 elements consist of oxygen (O), sulphur (S), selenium (Se), tellurium (Te) and polonium (PO). They are also called oxygen family or chalcogens, because many metals occur as oxides and sulphides.<\/p>\n
\n- The electronic configuration of these elements are:
\n<\/li>\n - The atomic and ionic radii of group 16 elements are smaller than those of the corresponding elements of group 15. Further, the atomic radii increases down the group. The increase in atomic radii of group 16 elements in primarily due to increase in the number of electron shells.<\/li>\n
- The melting, boiling points and densities increase regularly as we go down the group upto tellunium. However, the melting and boiling points of polonium are lower than those of tellurium. This is because, the atomic size increases down the group. As a result, vanderwaals forces of attraction among their atoms also increase and hence melting and boiling points regularly increase from oxygen to tellurium. However due to presence of inner \u2018d\u2019 and \u2018f\u2019 electrons, the inert pair effect is maximum in polonium. Consequently, the \u2018s\u2019 valence electrons in polonium are less available as compared to those in tellurium for bonding. As a result, vanderwaals forces of attraction will be weaker in Po than in Te and therefore melting point and boiling point of Po will be lower than that of Te.<\/li>\n
- Oxygen exists as a diatomic gas while other elements exist as octa atomic solids.<\/li>\n
- All the elements of this group show allotropy. Oxygen exists in two non metallic forms viz dioxygen (O2<\/sub>), tri oxygen O3<\/sub>. Sulphur exists in crystalline as well as amorphous allotropic forms. The crystalline form includes rhombic sulphur (\u03b1 sulphur) and mono clinic sulphur (\u03b2 sulphur). Amorphous allotropic forms include plastic sulphur (\u03b3 sulphur), milk of sulphur, and colloidal sulphur.<\/li>\n<\/ol>\n
<\/p>\n
Question 58.
\nGive two methods of preparation of oxygen in the laboratory.
\nAnswer:
\nIn the laboratory, oxygen is prepared by one of the following methods.
\nThe decomposition of hydrogen peroxide in the presence of a catalyst (MnO2<\/sub>) or by oxidation with potassium permanganate.
\n
\nThe thermal decomposition of certain metallic oxides or oxoanions gives oxygen.
\n<\/p>\nQuestion 59.
\nGive reasons for
\n(i) Oxygen exists as a diatomic gas.
\n(ii) Oxygen is paramagnetic.
\n(iii) Oxygen forms strong hydrogen bonds.
\nAnswer:
\n(i) Due to small size and high electronegativity, oxygen forms p\u03c0 \u2014 p\u03c0, double bond with another oxygen atom to form O = O molecule. The intermolecular forces of attraction between oxygen molecules are weak and hence oxygen exists as a diatomic gas.
\n(ii) The oxygen is paramagnetic because it contains two unpaired electrons in the antibonding molecular orbitals.
\n(iii) The high electronegativity of the oxygen atom is responsible to form hydrogen bonding eg: H2<\/sub>O.<\/p>\nQuestion 60.
\nHow is ozone prepared in the laboratory?
\nAnswer:
\nOzone is prepared in the laboratory by passing silent electrical discharge through oxygen. At a potential of 20,000 V about 10% of oxygen is converted into ozone it gives a mixture known as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.
\n<\/p>\n
Question 61.
\nWrite a note on the structure of ozone molecule.
\nAnswer:
\nThe central oxygen atom in ozone is sp2<\/sup> hybridised containing a lone pair of electrons. As a result, ozone has an angular structure.
\n
\nIt is actually a resonance hybrid of the following two resonating structural (I and II) as shown is figure.
\n<\/p>\nQuestion 62.
\nHow does oxygen react with metals and non metals? Give examples.
\nAnswer:
\nOxygen reacts with metals and non metals and form oxides.
\n(i) Oxygen combines with many metals and non metals to form oxides.
\n
\n(ii) Some less reactive metals react when powdered finely and made to react exothermically with oxygen at room temperature.
\n<\/p>\n
Question 63.
\nExplain why ozone is a more powerful
\noxidising agent than oxygen.
\nAnswer:
\nThis is due to the reason that ozone has higher energy content than oxygen and undergo decomposition.
\n2O3<\/sub> (g) \u2192 O2<\/sub> (g) + [O]
\nThe atomic oxygen, thus liberated brings about oxidation.<\/p>\n<\/p>\n
Question 64.
\nGive an example for the oxidising action of ozone.
\nAnswer:
\nOzone oxidises potassium iodide to iodine.
\n<\/p>\n
Question 65.
\nExplain how ozone can be estimated quantitatively.
\nAnswer:
\nOzone gas is passed through an aqueous solution of potassium iodide. The liberated iodine is titrated with a standard solution of theo sulphate. From the volume of the standard solution of sodium theo sulphate, the amount of ozone can be calculated.
\n<\/p>\n
Question 66.
\nMention the uses of oxygen.
\nAnswer:<\/p>\n
\n- Oxygen is one of the essential components for the survival of living organisms.<\/li>\n
- It is used in welding (oxy acetylene welding).<\/li>\n
- Liquid oxygen is used as fuel in rockets etc.<\/li>\n<\/ol>\n
Question 67.
\nName the stable allotrope of sulphur at ordinary temperature and pressure. What happens when it is heated to avoid 96\u00b0C?
\nAnswer:
\nRhombic sulphur is the most stable allotrope of sulphur at ordinary temperature and pressure. When heated slowly at 96\u00b0C. It is converted to monoclinic sulphur.<\/p>\n
Question 68.
\nGive the characteristics of rhombic and monoclinic sulphur.
\nAnswer:
\nRhombic sulphur consist of yellow coloured crystals and made up of S8<\/sub> molecules. Monoclinic sulphur also contains S8<\/sub> molecules in addition to S6<\/sub> molecules (small amount). It exists as a long needle like prism. It is stable between 96 – 119\u00b0C.<\/p>\nQuestion 69.
\nHow is sulphur dioxide prepared from
\n(i) sulphur, (ii) galena, (iii) iron pyrites.
\nAnswer:
\n(i) Burning sulphur in air.
\nS + O2<\/sub> \u2192 SO2<\/sub>
\n<\/p>\nQuestion 70.
\nGive equation for the preparation of sulphur dioxide in the laboratory.
\nAnswer:
\n<\/p>\n
<\/p>\n
Question 71.
\nWith an example, prove sulphur dioxide in an acidic oxide.
\nAnswer:
\nIt dissolves in water producing sulphurous acid. (H2<\/sub>SO3<\/sub>)
\n
\nIt reacts with sodium hydroxide and sodium carbonate to form sodium hydrogen sulphate and sodium sulphate respectively.
\n<\/p>\nQuestion 72.
\nGive two examples for the oxidising property of sulphur dioxide.
\nAnswer:
\nIt oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
\n<\/p>\n
Question 73.
\nGive two examples to show that sulphur dioxide acts as a reducing agent.
\nAnswer:
\n(i) It reduces chlorine water to hydrochloric acid.
\n
\n(ii) It reduces potassium permanganate and potassium dichromate to Mn+2<\/sup> and Cr+3<\/sup> respectively.
\n<\/p>\nQuestion 74.
\nGive equation for the reaction in which sulphur dioxide is used for the manufacture of sulphuric acid by contact process.
\nAnswer:
\n<\/p>\n
Question 75.
\nExplain the use of sulphur dioxide as a bleaching agent.
\nAnswer:
\nIn presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.
\n
\nHowever, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.<\/p>\n
<\/p>\n
Question 76.
\nDraw the structure of sulphur dioxide.
\nAnswer:
\nIn sulphur dioxide, sulphur atom undergoes sp2<\/sup> hybridisation. A double bond arises between S and O is due to p\u03c0- d\u03c0 overlapping.
\n
\nIt is a resonance hybrid of the canonical form I and II.<\/p>\nQuestion 77.
\nMention the uses of sulphur dioxide.
\nAnswer:<\/p>\n
\n- Sulphur dioxide is used in bleaching hair, silk, wool etc…<\/li>\n
- It can be used for disinfecting crops and plants in agriculture.<\/li>\n<\/ol>\n
Question 78.
\nDiscuss the various steps involved in the manufacture of sulphuric acid by contact process?
\nAnswer:
\nThe contact process involves the following steps.<\/p>\n
\n- Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen\/air.
\n<\/li>\n - Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as V2<\/sub>O5<\/sub> or platinised asbestos.<\/li>\n
- The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum (H2<\/sub>S2<\/sub>O7<\/sub>). The oleum is converted into sulphuric acid by diluting it with water.
\n
\nTo maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric acid obtained in this process is over 96 % pure.<\/li>\n<\/ol>\nQuestion 79.
\nHow do you prove that sulphuric acid is dibasic acid?
\nAnswer:
\nThe structure of sulphuric acid is
\n
\nIt has two replaceable hydrogen atoms. Hence, it forms two types of salts viz bi sulphates and sulphates. Hence H2<\/sub>SO4<\/sub> is a dibasic acid.
\n<\/p>\n<\/p>\n
Question 80.
\nGive two examples to show that cone, sulphuric acid is an oxidising agent.
\nAnswer:
\n(i) It oxidises carbon to carbon dioxide and gets reduced to sulphur dioxide
\n
\n(ii) It oxidises hydrogen sulphide to sulphur. It gets reduced to sulphur dioxide.
\n<\/p>\n
Question 81.
\nComplete and balance the following equations.
\n
\nAnswer:
\n<\/p>\n
Question 82.
\nGive a brief account of the action of sulphuric acid on metals.
\nAnswer:
\nSulphuric acid reacts with metals and gives different products depending on the reactants and reacting condition. Dilute sulphuric acid reacts with metals like tin, aluminium, zinc to give corresponding sulphates.
\n
\nHot concentrated sulphuric acid reacts with copper and lead to give the respective sulphates as shown below.
\n
\nSulphuric acid doesn\u2019t react with noble metals like gold, silver and platinum.<\/p>\n
Question 83.
\nWhat happens when
\n(i) Potassium chloride is heated with concentrated sulphuric acid.
\n(ii) Potassium nitrate is heated with concentrated sulphuric acid.
\n(iii) Sodium carbonate is heated with dilute sulphuric acid.
\n(iv) Sodium bromide is heated with cone, sulphuric acid?
\nAnswer:
\n(i) Hydrogen chloride is produced.
\nKCl + H2<\/sub>SO4<\/sub> \u2192 KHSO4<\/sub> + HCl
\n(ii) Sulphuric acid, being a strong acid displaces relatively weaker nitric acid from its salt.
\nKNO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 KHSO4<\/sub> + HNO3<\/sub>
\n(iii) It decomposes sodium carbonate to carbon dioxide and water.
\n
\n(iv) It oxidises Br–<\/sup> to Br2<\/sub>.
\n<\/p>\n<\/p>\n
Question 84.
\nHow will you detect sulphate radical in qualitative analysis.
\nAnswer:
\nDilute solution of sulphuric acid\/aqueous solution of sulphates gives white precipitate (barium sulphate) with barium chloride solution. It can also be detected using lead acetate solution. Here a white precipitate of lead sulphate is obtained.
\n<\/p>\n
Question 85.
\nWrite the formula and the oxidation state of sulphur in the following oxoacids.
\n(i) Sulphurous acid,
\n(ii) Sulphuric acid,
\n(iii) Thio sulphuric acid,
\n(iv) Peroxy mano sulphuric acid,
\n(v) Peroxydithionic acid,
\n(vi) Dithionic acid,
\n(vii) Pyrosulphuric acid.
\nAnswer:
\n<\/p>\n
Question 86.
\nWrite the structure of the following:<\/p>\n
\n- Sulphurous acid,<\/li>\n
- Sulphuric acid,<\/li>\n
- Thiosulphuric acid,<\/li>\n
- Dithionous acid,<\/li>\n
- Pyro sulphuric acid,<\/li>\n
- peroxy mono sulphuric acid,<\/li>\n
- Peroxy di sulphuric acid,<\/li>\n
- Dithionic acid,<\/li>\n
- Polythionic acid.<\/li>\n<\/ol>\n
Answer:
\n
\n<\/p>\n
<\/p>\n
Question 87.
\nBriefly outline the trend in the physical properties of halogens.
\nAnswer:<\/p>\n
\n- Fluorine, chlorine, bromine, iodine and astatine constitute group 17 elements or halogens.<\/li>\n
- The electronic configuration of these elements are
\n<\/li>\n - Halogens have the smallest atomic radii in their respective periods. Both atomic and ionic radii increase from fluorine to iodine as the number of shells increases.<\/li>\n
- All halogens exist as diatomic molecules in the elemental state. The different molecules of halogens are held together by vanderwaals forces of attraction. The strength of vanderwaals forces increases as. the size of the halogen atom increases from fluorine to iodine. As a result, F2<\/sub> and Cl2<\/sub> are gases, at room temperature, Br2<\/sub> is a liquid, whereas I2<\/sub> is a solid. Due to increase in vanderwaals force of attraction increases down the group, melting and boiling points increase with increase in atomic mass from fluorine to chlorine.<\/li>\n
- All halogens are highly electronegative and electronegativity decreases down the group.<\/li>\n<\/ol>\n
Question 88.
\nHow is chlorine prepared from (i) NaCl (ii) HCl (iii) bleaching powder?
\nAnswer:
\n(i) By the action of conc.H2<\/sub>SO4<\/sub> in the presence of manganese dioxide an sodium chloride.
\n
\n(ii) Oxidation of hydrochloric acid using oxidising agents such as MnO2<\/sub>, PbO2<\/sub>, KMnO4<\/sub> or K2<\/sub>Cr2<\/sub>O7<\/sub>.
\n
\n(iii) By treating to bleaching powder with a mineral acid.
\n<\/p>\nQuestion 89.
\nComplete and balance the following equations.
\n
\nAnswer:
\n<\/p>\n
<\/p>\n
Question 90.
\nBriefly outline the manufacture of chlorine by
\n(i) Electrolytic process, (ii) Deacon\u2019s process.
\nAnswer:
\n(i) Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na+<\/sup> and Cl