Question

Match the geometry (given in column A) with the complexes (given in column B)

Geometry : A	Complex : B
I: Octahedral	P: $\left[Ni\right(CN{)}_4{]}^{2-}$
II: Square planar	Q: $Ni(CO{)}_4$
III. Tetrahedral	R: $\left[Fe\right(CN{)}_6{]}^{4-}$

• A. I-P, II-Q, III-R
• B. I-R, II-P, III-Q
• C. I-R, II-Q, III-P
• D. I-Q, II-P, III-R

Answer 1

Answer: (B)

I-R, II-P, III-Q

$\left[Ni\right(CN{)}_4{]}^{2-}-N{i}^{2+}\left(\right[Ar\left]3d^8\right)$
The electrons are paired in 3d obitals because $C{N}^{-}$ is strong field ligand.Hence the hybridisation will be $ds{p}^2$(Square planar)
$Ni(CO{)}_4-Ni(\left[Ar\right]3d^{8}4s^2)$
$CO$ is strong field ligand. First the unpaired electrons will pair and then the electrons from 4s orbital will go in 3d orbital. Now 4s and 4p is empty hence it will be $s{p}^3$ hybridised.
$\left[Fe\right(CN{)}_6{]}^{4-}-F{e}^{2+}\left(\right[Ar\left]3d^6\right)$
The electrons are paired in 3d obitals because $C{N}^{-}$ is strong field ligand.Hence the hybridisation will be $d^{2}s{p}^3$(Octahedral)