Question

Which of the following statements are correct?
(I) Both melting and boiling points of $H_{2}O$ are higher than those of $H_{2}Te$ .
(II) In both $N_2{O}_5$ and $N_2{O}_4$ all $N-O$ bond lengths are equivalent.
(III) In both crystalline $NaHC{O}_3$ and $KHC{O}_3.HC{O}_3^{-}$ forms only dimeric anion through hydrogen bond.
(IV) $B_2,C_2,N_2$ on further ionization (removal of single electron) form thermodynamically less stable species.

• A. (I) and (II)
• B. (III) and (IV)
• C. (II) and (III)
• D. (I) and (IV)

Answer 1

The correct option is D (I) and (IV)

(I) On account of $H-bonding$ $H_{2}O$ has higher melting and boiling points.
$\begin{array}{ccc}& H_{2}O& H_{2}Te\\ m.p/K& 273& 222\\ b.p/K& 373& 269\end{array}$

(II) (III) (a) In $NaHC{O}_3,$ the $HC{O}_3^{-}$ ions are linked into an infinite chain and (b) in $KHC{O}_3,HC{O}_3^{-}$ forms a dimeric anion.

(IV) Each molecule through ionization (after removal of single electron) corresponding cation is formed.
$B_2-B.O=1and{B}_2^{+}-B.O=\frac{1}{2}{C}_2-B.O.=2and{C}_2^{+}-B.O.=1.5$
$N_2-B.O=3and{N}_2^{+}-B.O=2.5$
As , bond strength $\propto$ bond order.
So , with ionization bond strength decreases which eventually decreases thermal stability of the molecule.