Chemistry - CHEMICAL BONDING Question with Solution | TestHub

For ${\text{PCl}}_5$, the central atom $\text{P}$ has $5$ valence electrons. It forms $5$ bond pairs with $\text{Cl}$ atoms and has $0$ lone pairs. The hybridization is $s{p}^{3}d$, and the geometry is trigonal bipyramidal. (A $\to$ III)

For ${\text{BrF}}_5$, the central atom $\text{Br}$ has $7$ valence electrons. It forms $5$ bond pairs with $\text{F}$ atoms and has $1$ lone pair. The hybridization is $s{p}^3{d}^2$, and the geometry is square pyramidal. (B $\to$ IV)

For ${\text{BF}}_4^{-}$, the central atom $\text{B}$ has $3$ valence electrons plus $1$ from the negative charge, giving $4$ electrons. It forms $4$ bond pairs with $\text{F}$ atoms and has $0$ lone pairs. The hybridization is $s{p}^3$, and the geometry is tetrahedral. (C $\to$ I)

For $[\text{Ni}(\text{CN}{)}_4{]}^{2-}$, the central metal ion ${\text{Ni}}^{2+}$ has a $3d^8$ electronic configuration. Since ${\text{CN}}^{-}$ is a strong field ligand, it causes the pairing of electrons in the $3d$ orbitals. The hybridization is $ds{p}^2$, and the geometry is square planar. (D $\to$ II)

The correct matching is A-III, B-IV, C-I, D-II.

Answer: A-III, B-IV, C-I, D-II