Question

Predict the geometry of the following species and describe the hybridization on the central atom:
(i) $PbC{l}_4$
(ii) $Sb{F}_6^{-}$
(iii) $B{H}_4^{-}$
(iv) $PC{l}_3$

• A. Tetrahedral $s{p}^2$, octahedral $s{p}^3{d}^3$, tetrahedral $s{p}^3$, tetrahedral $s{p}^3$, respectively
• B. Tetrahedral $s{p}^3$, octahedral $s{p}^3{d}^2$, tetrahedral $s{p}^3$, tetrahedral $s{p}^3$, respectively
• C. Tetrahedral $s{p}^3$, octahedral $s{p}^3{d}^2$, tetrahedral $s{p}^3$, pyramidal $s{p}^3$, respectively
• D. Trigonal planar $s{p}^2$, octahedral $s{p}^3{d}^2$, tetrahedral sp, tetrahedral $s{p}^2$, respectively

Answer 1

The correct option is C Tetrahedral $s{p}^3$, octahedral $s{p}^3{d}^2$, tetrahedral $s{p}^3$, pyramidal $s{p}^3$, respectively

$PbC{l}_4$ is a covalent compound that has a tetrahedral geometry.

Like $P{F}_6^{-}$, $Sb{F}_6^{-}$ has an octahedral shape with the central
atom having an $s{p}^3{d}^2$ hybridization.

The borohydride anion is also $s{p}^3$ and hence tetrahedral.
$PC{l}_3$, like $N{H}_3$ is pyramidal with an $s{p}^3$ hybridization.