Question

Out of given reaction which show change in hybridisation of central atom?

• A. $H_{3}B{O}_3$ dissolve in water
• B. $H_{2}S{O}_4$ dissolve in water
• C. $N_2{O}_{5\left(g\right)}\to N_2{O}_{5\left(s\right)}$
• D. $PB{r}_{5\left(g\right)}\to PB{r}_{5\left(s\right)}$
• E. $C_2{H}_6\underset{bondcleavageofC-Cbond}{\overset{Homolytic}{\to }}$

Answer 1

Answer: (A), (C), (D), (E)

The correct options are
A $H_{3}B{O}_3$ dissolve in water
C $N_2{O}_{5\left(g\right)}\to N_2{O}_{5\left(s\right)}$
D $C_2{H}_6\underset{bondcleavageofC-Cbond}{\overset{Homolytic}{\to }}$
E $PB{r}_{5\left(g\right)}\to PB{r}_{5\left(s\right)}$

(A) When $H_{2}B{O}_3^{-}$ dissolves in water, the hybridization changes.
$H_{2}B{O}_3^{-}+H_{2}O\rightleftharpoons HB{O}_3^{2-}+H_3{O}^{+}$
(C) In the gas phase, nitrogen pentoxide exists as $N_2{O}_5$.
in solid phase, it exists as $\left[N{O}_2{]}^{+}\right[N{O}_3{]}^{-}$
Hence, in the transformation $N_2{O}_{5\left(g\right)}\to N_2{O}_{5\left(s\right)}$ the oxidation state changes.
(D) In the gas phase, phosphorus pentabromide exists as $PB{r}_5$.
in solid phase, it exists as $[PB{r}_4{]}^{+}B{r}^{-}$
Hence, in the transformation $PB{r}_{5\left(g\right)}\to PB{r}_{5\left(s\right)}$ the oxidation state changes.
(E) In the transformation $C_2{H}_6\underset{bondcleavageofC-Cbond}{\overset{Homolytic}{\to }}$
the hybridization changes from $s{p}^3$ to $s{p}^2$