Q4. What volume of chlorine, at 273.15 K and 1 bar, is needed to set free 25.38 g of iodine from potassium iodide solution? (I = 126.9 g mol−1; molar volume 22.7 L mol−1.)
M(I2) = 253.8 g mol−1, so 25.38 g is 0.1000 mol. From Cl2 + 2KI → 2KCl + I2, one Cl2 gives one I2, so 0.1000 mol Cl2 is needed: 0.1000 × 22.7 = 2.27 L (reasoned from NCERT Class 11 Chemistry, page 243).
Q5. Why are displacement reactions of fluorine not carried out in water?
Fluorine is so reactive that it attacks water itself, 2F2 + 2H2O → 4HF + O2, instead of only displacing the other halogen (NCERT Class 11 Chemistry, page 243).
Q6. Arrange HF, HCl, HBr and HI in order of increasing boiling point.
HCl < HBr < HI < HF. From HCl to HI the boiling point rises with the van der Waals forces; HF is highest because it alone forms hydrogen bonds (Wikipedia, “Hydrogen halide”).
Q7. Which of ClO−, ClO3− and ClO4− cannot disproportionate, and why?
ClO4−: chlorine in it is +7, its highest state, so it cannot be oxidised further, and a disproportionation needs both a higher and a lower state (NCERT Problem 7.5, Class 11 Chemistry, page 244).
Q8. Gaseous ClF3 has ΔfH⊖ = −163.2 kJ mol−1. Taking the bond enthalpies Cl–Cl = 243 and F–F = 155 kJ mol−1, find the mean Cl–F bond enthalpy in ClF3. What does it suggest about the reactivity of ClF3 compared with chlorine?
Making the gaseous atoms from the elements takes ½ × 243 + 1.5 × 155 = 121.5 + 232.5 = 354.0 kJ mol−1. Breaking ClF3(g) into atoms therefore takes 354.0 + 163.2 = 517.2 kJ mol−1, and the mean Cl–F bond is 517.2 ÷ 3 = 172 kJ mol−1, far weaker than Cl–Cl, 243 (bond enthalpies from NCERT Class 11 Chemistry, Table 5.3(a), page 154; ΔfH⊖ from Wikipedia, “Chlorine trifluoride”). Its weaker bonds can contribute to high reactivity, but this average alone cannot determine reaction rates or products; the full reaction pathway matters. Interhalogens are generally reactive fluorinating or halogenating agents (Wikipedia, “Halogen”).