JAI GURUDEV CHEMISTRY CLASSES// BY: - AJEET KUMAR// M.N. 7209985022// f- ajeet.jha3 GROUP-16 ELMENTS 1. Why does O3 act as powerful oxidizing agent? Ans: - Ozone decomposes rapidly to give nascent oxygen; hence it is a good oxidizing agent. O 3 → O 2 + [O] 2. How is O3 estimated quantitatively? Ans: - O 3 is treated with excess of KI solution. I 2 is liberated which is titrated against a standard solution of Na 2 S 2 O 3 using starch as an indicator. 3. Comment on the nature of two S-O bonds formed in SO 2 molecule. Are the two S-O bonds in the molecule equal? Ans: - 4. How is the presence of SO 2 detected? Ans: - SO 2 is a pungent smelling gas. It reduces acidified orange solution of K 2 Cr 2 O 7 into green, Cr 2 (SO 4 ) 3 . K 2 Cr 2 O 7 + H 2 SO 4 + SO2 = Cr 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O 5. Why is Ka 2 less than Ka 1 for H 2 SO 4 in water? Ans: - Ka 2 less than Ka 1 for H 2 SO 4 in water because it is difficult to give H + ion from HSO 4 - . 6. Compare the parameter such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F 2 and Cl 2 . Ans:- The electrode potential of F2 (+ 2.87 V) is much higher than that of Cl 2 (+1.3eV), therefore, F2 is much stronger oxidizing agent than Cl2.

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JAI GURUDEV CHEMISTRY CLASSES// BY: - AJEET KUMAR// M.N. 7209985022// f- ajeet.jha3 GROUP-16 ELMENTS

1. Why does O3 act as powerful oxidizing agent? Ans: - Ozone decomposes rapidly to give nascent oxygen; hence it is a good oxidizing agent. O3 → O2 + [O] 2. How is O3 estimated quantitatively? Ans: - O3 is treated with excess of KI solution. I2 is liberated which is titrated against a standard solution of Na2S2O3 using starch as an indicator. 3. Comment on the nature of two S-O bonds formed in SO2 molecule. Are the two S-O bonds in the molecule equal? Ans: - 4. How is the presence of SO2 detected? Ans: - SO2 is a pungent smelling gas. It reduces acidified orange solution of K2Cr2O7 into green, Cr2(SO4)3. K2Cr2O7 + H2SO4 + SO2 = Cr2(SO4)3 + K2SO4 + H2O 5. Why is Ka2 less than Ka1 for H2SO4 in water? Ans: - Ka2 less than Ka1 for H2SO4 in water because it is difficult to give H+ ion from HSO4

-. 6. Compare the parameter such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F2 and Cl2. Ans:- The electrode potential of F2 (+ 2.87 V) is much higher than that of Cl2 (+1.3eV), therefore, F2 is much stronger oxidizing agent than Cl2.

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