Question

Which of the following complexes is/are correctly matched with their hybridization and magnetic behavior?

• A. $\left[Fe\right(N{H}_3{)}_6{]}_2(S{O}_4{)}_3;d^{2}s{p}^3;$ paramagnetic
• B. $\left[Fe\right(N{H}_3{)}_6\left]\right(N{O}_3{)}_2;s{p}^3{d}^2;$ paramagnetic
• C. $\left[Fe\right(CO{)}_5];s{p}^{3}d;$ diamagnetic
• D. $\left[Ni\right(CN{)}_4{]}^{2-};ds{p}^2;$ diamagnetic

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\left[Fe\right(N{H}_3{)}_6{]}_2(S{O}_4{)}_3;d^{2}s{p}^3;$ paramagnetic
B $\left[Fe\right(N{H}_3{)}_6\left]\right(N{O}_3{)}_2;s{p}^3{d}^2;$ paramagnetic
D $\left[Ni\right(CN{)}_4{]}^{2-};ds{p}^2;$ diamagnetic

(a) $\left[Fe\right(N{H}_3{)}_6{]}_2(S{O}_4{)}_3$ has $Fe$ ($d^{5}system$) is in +3 oxidation state. Here $N{H}_3$ acts as strong field ligand and facilitates pairing. So hybridisation will be $d^{2}s{p}^3$.


(b) $\left[Fe\right(N{H}_3{)}_6\left]\right(N{O}_3{)}_2$ has $Fe$ ($d^{6}system$) is in +2 oxidation state. Here $N{H}_3$ acts as weak field ligand and can not facilitate pairing. So outer orbital complex will be formed with hybridisation $s{p}^3{d}^2$. Due to presence of unpaired electrons complex show paramagnetic behaviour.


(c) In $\left[Fe\right(CO{)}_5]$, $Fe$ is in ground state $(3d^{6}4s^2$).
$CO$ is strong field ligand which cause pairing to form low spin complex. So hybridisation will be $ds{p}^3$ and complex is diamagnetic.

(d) In $\left[Ni\right(CN{)}_4{]}^{2-}$ has $Ni$ ($d^{8}system$) is in +2 oxidation state. Here $C{N}^{-}$ acts as strong field ligand facilitates pairing. So inner orbital square planar complex will be formed with hybridisation $ds{p}^2$. Since all electrons are paired up hence3 complex will be diamagnetic.