Question

Which of the following complex compound(s) is/are paramagnetic and low spin?

• A. $K_3\left[Fe\right(C{N}_6\left)\right]$
• B. $\left[Ni\right(CO{)}_4{]}^0$
• C. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
• D. $\left[Mn\right(CN{)}_6{]}^{4-}$

Answer 1

Answer: (A), (C), (D)

$K_3\left[Fe\right(C{N}_6\left)\right]$
$\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
$\left[Mn\right(CN{)}_6{]}^{4-}$

$K_3\left[Fe\right(C{N}_6\left)\right],\left[Ni\right(CO{)}_4{]}^0,\left[Mn\right(CN{)}_6{]}^{4-},\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ contains 1, 0, 1 and 3 unpaired electrons respectively.

$K_3\left[Fe\right(C{N}_6\left)\right]and\left[Mn\right(CN{)}_6{]}^{4-}$ undergoes $d^{2}s{p}^3$ hybridization and are low spin complexes. They contains 1 unpaired electron each. Hence, they are paramagnetic in nature.

$\left[Ni\right(CO{)}_4{]}^0$ is diamagnetic in nature contains 0 unpaired electron.

$\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ undergoes $d^{2}s{p}^3$ hybridization and is a low spin complex. It contains 3 unpaired electrons. Hence, it is paramagnetic in nature.