2<\/sub>O.<\/p>\n<\/p>\n
Question 4.
\nOxygen generally exhibits an oxidation state of – 2 only while the other members of the group show oxidation states of +2, + 4 and + 6 also – Explain.
\nAnswer:
\nThe multiple oxidation states in the case of elements other than oxygen are due to the availability of d-orbitals in the valence shell of the atoms.<\/p>\n
Oxygen exhibits -2 oxidation state only generally except with F, where it shows + I in F2<\/sub>O2<\/sub> and + II in F2<\/sub>O.<\/p>\nQuestion 5.
\nWrite any two compounds in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.
\nAnswer:<\/p>\n
\n- In peroxides oxidation state of oxygen is – 1. Ex: H2<\/sub>O2<\/sub><\/li>\n
- In super oxides oxidation state of oxygen is -1\/2. Ex : KO2<\/sub><\/li>\n
- In F2<\/sub>O, oxidation state of oxygen is + 2.<\/li>\n
- In F2<\/sub>O2<\/sub>, oxidation state of oxygen is + 1.<\/li>\n<\/ol>\n
Question 6.
\nOxygen molecule has the formula O2<\/sub> while sulphur has S8<\/sub> – explain.
\nAnswer:
\nDue to small atomic size and high electro-negativity in oxygen molecule, each oxygen atom is linked to other oxygen atom by a double bond. Hence its formula is O2<\/sub>.
\n\\(: \\ddot{\\mathrm{O}}=\\ddot{\\mathrm{O}}:\\)
\nDue to large atomic size and less electro-negativity in sulphur molecule, eight ‘S’ atoms are linked together by single covalent bonds forming.
\nPuckered S8<\/sub> rings with crown configuration. Hence formula of sulphur is S8<\/sub>.
\n<\/p>\nQuestion 7.
\nWhy is H2<\/sub>O a liquid while H2<\/sub>S is a gas? [IPE ’14]
\nAnswer:
\nThe O-H bond in H2<\/sub>O is highly polar. There are hydrogen bonds among the molecules of H2<\/sub>O. Hence it is present as a liquid.
\n
\nThere are no hydrogen bonds among H2<\/sub>S molecules. So it exists as a gas at room temperature.<\/p>\n<\/p>\n
Question 8.
\nH2<\/sub>O is neutral while H2<\/sub>S is acidic-explain.
\nAnswer:
\nBond dissociation enthalpy of H-S bond is less than H – O bond. Hence H2<\/sub>S is acidic.<\/p>\nQuestion 9.
\nName the most abundant element present in earth’s crust.
\nAnswer:
\nOxygen<\/p>\n
Question 10.
\nWhich element of group-16 shows highest catenation?
\nAnswer:
\nSulphur<\/p>\n
Question 11.
\nAmong the hydrides of chalcogens, which is most acidic and which is most stable?
\nAnswer:<\/p>\n
\n- Most acidic hydride of chalcogens is H2<\/sub>Te.<\/li>\n
- Most stable hydride of chalcogens is H2<\/sub>O.<\/li>\n<\/ol>\n
Question 12.
\nGive the hybridisation of sulphur in the following.
\na) SO2<\/sub>
\nb) SO3<\/sub>
\nc) SF4<\/sub>
\nd) SF6<\/sub>
\nAnswer:
\nHybridisation of S in SO2<\/sub> is sp2<\/sup>.
\nHybridisation of S in SO3<\/sub> is sp2<\/sup>.
\nHybridisation of S in SF4<\/sub> is sp3<\/sup>d.
\nHybridisation of S in SF2<\/sub> is sp3<\/sup>d2<\/sup>.<\/p>\n<\/p>\n
Question 13.
\nWrite the names and formulae of any two oxyacids of sulphur. Indicate the oxidation state of sulphur in them.
\nAnswer:<\/p>\n
\n- Sulphurous acid = H2<\/sub>SO3<\/sub>
\nOxidation state of sulphur is + 4.<\/li>\n- Sulphuric acid = H2<\/sub>SO4<\/sub>
\nOxidation state of sulphur is + 6.<\/li>\n- Pyrosulphuric acid = H2<\/sub>S2<\/sub>O7<\/sub>
\nOxidation state of sulphur is + 6.<\/li>\n<\/ol>\nQuestion 14.
\nExplain the structures of SF4<\/sub> and SF6<\/sub>.
\nAnswer:
\nStructure of SF4<\/sub> :
\nExcited state configuration of S is
\n
\nHybridisation of Sulphur is sp3<\/sup>d.
\nSF4<\/sub> has distorted trigonal bipyramidal structure with one orbital being occupied by a lone pair of electrons.
\n<\/p>\nStructure of SF6<\/sub>:
\nExcited state configuration of Sulphur is
\n
\nSF6<\/sub> has octahedral symmetry.<\/p>\nQuestion 15.
\nGive one example each for
\na) a neutral oxide
\nb) a peroxide
\nc) a super oxide
\nAnswer:
\na) Nitric oxide NO is a neutral oxide.
\nb) Na2<\/sub>O2<\/sub> is a peroxide. Peroxides contain O – O bond.
\nc) KO2<\/sub> is potassium super oxide.<\/p>\n<\/p>\n
Question 16.
\nWhat is tailing of mercury? How is it removed ? [AP & TS ’15]
\nAnswer:
\nOzone reacts with mercury to form Hg2<\/sub>O. Due to dissolution of Hg2<\/sub>O in Hg mercury loses its meniscus and sticks to the sides of glass. This is called tailing of mercury.
\nThe menisus can be regained by shaking with water which dissolves Hg2<\/sub>O.<\/p>\nQuestion 17.
\nWrite the principle involved in the quantitative estimation of ozone gas.
\nAnswer:
\nOzone liberates I2<\/sub> from KI solution which can be titrated against a standard solution of Hypo using starch as an indicator.
\n2KI + H2<\/sub>O + O3<\/sub> \u2192 2KOH + I2<\/sub> + O2<\/sub>
\n2Na2<\/sub>S2<\/sub>O3<\/sub> + I2<\/sub>\u2192 Na2<\/sub>S4<\/sub>O6<\/sub> + 2NaI<\/p>\nQuestion 18.
\nWrite the structure of Ozone.
\nAnswer:
\n
\nO-O bond lengths in ozone are identical (128pm) and molecule is angular with a bond angle of 117\u00b0.<\/p>\n
Question 19.
\nSO2<\/sub> can be used as anti-chlor. Explain.
\nAnswer:
\nSulphur dioxide reacts with chlorine in presence of charcoal to give sulphuryl chloride.
\nSO2<\/sub>(g) + Cl2<\/sub> (g) \u2192 SO2<\/sub>Cl2<\/sub> (l)
\nSo it can remove chlorine and can be used as anti-chlor.<\/p>\n<\/p>\n
Question 20.
\nHow is ozone detected?
\nAnswer:<\/p>\n
\n- Ozone turns startch iodide paper blue.<\/li>\n
- It tails mercury.<\/li>\n<\/ol>\n
Question 21.
\nHow does ozone react with ethylene? [Mar. 18 – A.P.]
\nAnswer:
\nWhen ozone is bubbled through the solution of ethylene in an inert solvent like CCl4<\/sub>, at 195K, ethylene ozonide is formed.
\n
\nNO + O3<\/sub> \u2192 NO2<\/sub> + O2<\/sub><\/p>\nQuestion 22.
\nOut of O2<\/sub> and O3<\/sub>, which is paramagnetic?
\nAnswer:
\nO2<\/sub> is paramagnetic. It contains two unpaired electrons in its molecular form O2<\/sub>.<\/p>\nQuestion 23.
\nBetween O3<\/sub> and O2<\/sub>, ozone is a better oxidising agent – why?
\nAnswer:
\nDue to the ease with which it liberates atoms of nascent oxygen O3<\/sub> \u2192 O2<\/sub> + (O). Ozone acts as a powerful oxidising agent.<\/p>\n<\/p>\n
Question 24.
\nWrite any two uses each for O3<\/sub> and H3<\/sub>SO4<\/sub>.
\nAnswer:
\nUses of Ozone :<\/p>\n\n- It is used as a germicide, disinfectant, and for sterilising water.<\/li>\n
- It is used for bleaching oils, ivory, flour, starch, etc.<\/li>\n<\/ol>\n
Uses of H2<\/sub>SO4<\/sub> :<\/p>\n\n- H2<\/sub>SO4<\/sub> is used in the manufacture of fertilisers e.g. ammonium sulphate, super phosphate.<\/li>\n
- Petroleum refining<\/li>\n<\/ol>\n
Question 25.
\nWhich form of sulphur shows paramagnetism?
\nAnswer:
\nIn vapour state sulphur partly exists as S2<\/sub> molecule which has two unpaired electrons in the antibonding \u03c0* orbitals. Hence exhibits paramagnetism.<\/p>\nQuestion 26.
\nHow is the presence of SO2<\/sub> detected?
\nAnswer:
\nSO2<\/sub> decolourises KMnO4<\/sub> solution in acid medium.
\n5SO2<\/sub> + 2\\(\\mathrm{MnO}_4^{-}\\) + 2H2<\/sub>O \u2192 5\\(\\mathrm{SO}_4^{2-}\\) + 4H+<\/sup> + 2Mn++<\/sup><\/p>\n<\/p>\n
Question 27.
\nWhy are group-16 elements called chal- cogens ?
\nAnswer:
\nThe elements Oxygen, Sulphur, Selenium, Tellurium of group 16- are collectively known as chalcogens meaning ‘ore forming’. Since many metals occur as oxides or sulphides in nature. The name is derived from the Greek word for brass. It indicates the association of sulphur and its congeners with copper.<\/p>\n
Question 28.
\nAmong chalcogens, which has highest electronegativity and which has highest electron gain enthalpy? ‘
\nAnswer:
\nOxygen has highest electronegativity (3.5 Pauling scale) among chalcogens. Sulphur has highest electron gain enthalpy among chalcogens.<\/p>\n
Question 29.
\nWhich hydride of group-16 has highest boiling point and weakest acidic character ?
\nAnswer:
\nH2<\/sub>O (water)<\/p>\nShort Answer Questions (4 Marks)<\/span><\/p>\nQuestion 30.
\nJustify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxida-tion states and hydride formation.
\nAnswer:
\nElectronic configuration: All the elements of group 16 have six electrons in the outermost shell and have ns2<\/sup>np4<\/sup> general electronic configuration.<\/p>\nOxidation states:
\nGeneral oxidation states of Group 16 elements are -2, +2, +4 and +6. Oxygen cannot show higher oxidation states +4, +6 because of non-availability of d-orbitals in the valence shell. Sulphur, Selenium, and Tellurium usually show +4 oxidation state in their compounds with oxygen and +6 with fluorine. The stability of +6 oxidation state decreases down the group and stability of +4 oxidation state increases. This is due to inert pair effect.<\/p>\n
Hydride formation :
\nAll the chalcogens form covalent hydrides of the formula H2<\/sub>E. (E = S, Se, Te, Po)<\/p>\nThermal stability of the hydrides decreases from H2<\/sub>O to H2<\/sub>Po. This is due to an increase in E – H bond length.<\/p>\nWater is a liquid while others are gases. Water exists as associated liquid due to hydrogen bonding.<\/p>\n
The acidic character in aqueous solution increases from H2<\/sub>O to H2<\/sub>Te. This is due to decrease in charge density on conjugate bases OH–<\/sup>, SH–<\/sup>, SeH–<\/sup>, TeH–<\/sup>.<\/p>\nAll the hydrides except water possess reducing property. The reducing property increases from H2<\/sub>O to H2<\/sub>Po. This trend can be attributed to decrease in thermal stability of hydrides from H2<\/sub>O to H2<\/sub>Po.<\/p>\nThus there is a regular gradation in properties of these elements. Hence the inclusion of these elements in the same group is justified.<\/p>\n
<\/p>\n
Question 31.
\nDescribe the manufacture of H2<\/sub>SO4<\/sub> by contact process. [TS ’16]
\nAnswer:
\nManufacture of H2<\/sub>SO4<\/sub> by Contact process involves three steps.
\ni) Burning of sulphur or sulphide ores in air to generate SO2<\/sub>.
\nS + O2<\/sub> \u2192 SO2<\/sub>
\n4FeS2<\/sub> + 11O2<\/sub> \u2192 2Fe2<\/sub>O3<\/sub> + 8SO2<\/sub><\/p>\nii) Conversion of SO2<\/sub> to SO3<\/sub> by the reaction with oxygen in the presence of a catalyst (V2<\/sub>O5<\/sub>).
\n<\/p>\niii) Absorption of SO3<\/sub> in H2<\/sub>SO4<\/sub> to give oleum.
\nSO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 H2<\/sub>S2<\/sub>O7<\/sub><\/p>\niv) Dilution of oleum with water gives sul-phuric acid.
\nH2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2H2<\/sub>SO4<\/sub><\/p>\nQuestion 32.
\nHow is ozone prepared? How does it react with the following ? [Mar. 19, 18, 17, AP]
\na) PbS
\nb)KI
\nc) Hg
\nd) Ag
\nAnswer:
\nPreparation of Ozone :
\nWhen a slow dry stream of oxygen is passed through a silent electrical discharge, oxygen is converted to ozone. This product is called ozonised oxygen.
\n3O2<\/sub> \u2192 2O3<\/sub>, \u2206H\u00b0 (298K) = +142 kJ\/mol.
\nIf high concentration of ozone (> 10%) is required, a battery of ozonisers can be used. Pure ozone can be condensed in a vessel surrounded by liquid oxygen.
\nProperties:
\na) Pbs is oxidised to PbSO4<\/sub>.
\nPbS (s) + 4O3<\/sub> (g) \u2192 PbSO4<\/sub> + 4O2<\/sub>
\nb) I2<\/sub> is liberated. I–<\/sup> is oxidised to I2<\/sub>.
\n2KI(aq)<\/sub> + H2<\/sub>O (l) + O3<\/sub>(g) \u2192 2KOH(aq) + I2<\/sub>(s) + O3<\/sub>(g)<\/p>\nc) Hg2<\/sub>O formed sticks to the glass surface and mercury loses its meniscus due to dissolution of Hg2<\/sub>O in Hg. This is called tailing of mercury.
\n2Hg + O3<\/sub> \u2192 Hg2<\/sub>O + O2<\/sub><\/p>\nd) Silver metal is blackened. This is due to oxidation of the metal to its oxide
\n2Ag + O3<\/sub> \u2192 Ag2<\/sub>O + O2<\/sub> (oxidation)<\/p>\n<\/p>\n
Question 33.
\nWrite a short note on the allotropy of sulphur.
\nAnswer:
\nSulphur forms numerous allotropes. The two common crystalline forms are \u03b1 (alpha) or rhombic sulphur and \u03b2 (beta) or monoclinic sulphur. The stable form at room temperature is rhombic sulphur. It transforms to monoclinic sulphur when heated above 369K.<\/p>\n
Rhombic sulphur:
\nThis is yellow in colour. It is formed by evaporating the solution of roll sulphur in CS2<\/sub>. It is soluble in CS2<\/sub>. m.p. 385.8K.<\/p>\nMonoclinic sulphur:
\nThis form of sulphur is prepared by melting rhombic sulphur in a dish and cooling the solution till crust is formed. Two holes are made in the crust and remaining liquid poured out.<\/p>\n
\u03b1 – sulphur transforms to \u03b2 – sulphur above 369K. At 369K both are stable. This temperature is called transition temperature.
\nBoth forms have S8<\/sub> puckered ring forms. It has a crown shape.
\n
\nIn cyclo – S6<\/sub>, the ring adopts chair form. At 1000K, S2<\/sub> is the dominant species and is paramagnetic like O2<\/sub>.