Question

Consider following complexes :

I: $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
II :$\left[Fe\right(CN{)}_6{]}^{4-}$
III :$\left[Ni\right(CO{)}_4]$
IV :$\left[Ni\right(H_{2}O{)}_4{]}^{2+}$
V :$\left[Ni\right(CN{)}_4{]}^{2-}$

Out of this, select the complexes with tetrahedral geometry :

• A. $\left[Fe\right(CN{)}_6{]}^{4-}$
• B. $\left[Ni\right(CO{)}_4]$
• C. $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$

Answer 1

Answer: (B), (C)

The correct options are
B $\left[Ni\right(CO{)}_4]$
C $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$

The complexes $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ and $\left[Fe\right(CN{)}_6{]}^{4-}$ have octahedral geometry.

The central iron atom undergoes $s{p}^3{d}^2$ hybridization in $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$. It results in outer orbital complex or high spin complex.

The central iron atom undergoes $d^{2}s{p}^3$ hybridization in $\left[Fe\right(CN{)}_6{]}^{4-}$. It results in outer orbital complex or high spin complex.

$\left[Ni\right(CO{)}_4]$ and $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$ have tetrahedral geometry due to $s{p}^3$ hybridisation.

$\left[Ni\right(CN{)}_4{]}^{2-}$ has square planar geometry due to $ds{p}^2$ hybridization.

Therefore, option B and C both are correct.