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 octahedral geometry :

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

Answer 1

Answer: (A), (B)

$\left[Fe\right(CN{)}_6{]}^{4-}$
$\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

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.

Therefore, option A and B both are correct.