Question

(a) Account for the following:

(i) Interhalogens are more reactive than pure halogens.

(ii) $N_2$ is less reactive at room temperature.

(iii) Reducing character increases from $N{H}_{3}toBi{H}_3$.

(b) Draw the structures of the following:

(i) $H_4{P}_2{O}_7$ (Pyrophosphoric acid)

(ii) $Xe{F}_4$

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.

(b) Which noble gas has the lowest boiling point?

(c) Fluorine is a stronger oxidising agent than chlorine. Why?

(d) What happens when $H_{3}P{O}_3$ is heated?

(e) Complete the equation:

$PbS+O_3\to$

Answer 1

(a) (i) This is because the covalent bond between dissimilar atoms (X–X′) in interhalogen compounds is weaker than that between similar atoms (X–X or X′–X′) in halogens except the F–F bond.

(ii) $N_2$ is less reactive at room temperature because its molecule is very small and the bond dissociation enthalpy of the molecule is high.

(iii) The reducing character depends on the stability of the hydride. The greater the instability of a hydride, the greater is its reducing character. Because the stability decreases from $N{H}_{3}toBi{H}_3$, the reducing character increases.

(b)

(i) $H_4{P}_2{O}_7$ (pyrosulphuric acid)

(ii) $Xe{F}_4$

OR

(a) On heating with caustic soda solution, white phosphorus forms poisonous gas phosphine.

$P_4+3NaOH+3H_{2}O\to P{H}_3+Na{H}_{2}P{O}_2$

(b) Helium has the lowest boiling point.

(c) The oxidising power of halogens decreases as we go down the group. The electrode potential of $F_2$ is +2.87 V, while that of $C{l}_2$ is +1.36 V. This means that fluorine can be reduced more easily than chlorine. Therefore, fluorine is the strongest oxidising agent.

(d) On heating, $H_{3}P{O}_3$ decomposes into phosphoric acid and phosphine.

$4H_{3}P{O}_3\stackrel{457K}{\to }3H_{3}P{O}_4+P{H}_3$

(e) $PbS+4O_3\to PbS{O}_4+4O_2$