Question

Which of the following ions has tetrahedral geometry and $s{p}^3$ -hybridisation for its central atom?

• A. $B{H}_4^{-}$
• B. $N{H}_2^{-}$
• C. $C{O}_3^{2-}$
• D. $H_3{O}^{+}$

Answer 1

Answer: (A)

(a) Boron-hydride ion $\left(B{H}_4^{-}\right)$ has $s{p}^3$ -hybridisation. Therefore, it is tetrahedral geometry.



(b) In $N{H}_2^{-}$ molecule, the number of sigma bond is 2 and the number of lone pairs is 2 , i.e. $s{p}^3$ -hybridisation. Due to presence of 2 lone pairs, $N{H}_2^{-}$ molecule will be bent shape similar to water.



(c) The carbonate ion, $C{O}_3^{2-}$ has a trigonal planar geometry, i.e. carbon is $s{p}^2$ hybridised.



(d) In $H_3{O}^{+}$, hybridisation of oxygen is $s{p}^3$ and structure of $H_3{O}^{+}$ ion will be pyramidal.