Question

Hybridisation of the underlined atom does/do not change when:

• A. $\underset{–}{S}{O}_3$ changes to $\underset{–}{S}{O}_4^{2-}$
• B. $\underset{–}{Al}{H}_3$ changes to ${\underset{–}{Al}{H}_4}^{-}$
• C. $\underset{–}{B}(OH{)}_3$ change to $\left[\underset{–}{B}\right(OH{)}_4{]}^{-}$
• D. $H_2\underset{–}{O}$ changes to $H_3{\underset{–}{O}}^{+}$

Answer 1

Answer: (D)

$H_2\underset{–}{O}$ changes to $H_3{\underset{–}{O}}^{+}$

Hybridisation of the underlined atom does change when $\underset{–}{S}{O}_3$ changes to $\underset{–}{S}{O}_4^{2-}$
In $S{O}_3$, $S$ atom is $s{p}^2$ hybridized whereas in $S{O}_4^{2-}$, S atom is $s{p}^3$ hybridized.
Hybridisation of the underlined atom does change when $\underset{–}{Al}{H}_3$ changes to $\underset{–}{Al}{H}_4^{-}$
In $Al{H}_3$, $Al$ atom is $s{p}^2$ hybridized whereas in $Al{H}_4^{-}$, Al atom is $s{p}^3$ hybridized.
Hybridisation of the underlined atom does change when $\underset{–}{B}(OH{)}_3$ change to $\left[\underset{–}{B}\right(OH{)}_4]$
In $B(OH{)}_3$, $B$ atom is $s{p}^2$ hybridized whereas in $\left[B\right(OH{)}_4{]}^{-}$, B atom is $s{p}^3$.
Hybridisation of the underlined atom does not change when $H_2\underset{–}{O}$ change to $H_3{\underset{–}{O}}^{+}$.
In both $H_2\underset{–}{O}$ and $H_3{\underset{–}{O}}^{+}$, the $O$ atom is $s{p}^3$ hybridized.