Question

Which of the following is paramagnetic ?

• A. $[Co(N{H}_3{)}_6]3+$
• B. $[Ni(CO{)}_4]$
• C. $[Ni(N{H}_3{)}_4{]}^{2+}$
• D. $[Ni(CN{)}_4{]}^{2-}$

Answer 1

Answer: (C)

Ni(Z = 28) : $[Ar{]}^{18}4s^{2}3d^8$
$N{i}^{2+}:[Ar{]}^{18}3d^8$
$[Ni(N{H}_3{)}_4{]}^{2+}:N{i}^{2+}$ undergoes $s{p}^3$-hybridisation, the resulting ion is paramagnetic due to the presence of two unpaired electrons.
$Ni(CO{)}_4$ : Two $4s$-electrons in Ni pair up with $3d$-electrons followed by $s{p}^3$-hybridisation. The resulting complex has no unpaired electron and is diamagnetic.
$[Ni(CN{)}_4{]}^{2-}$ : CN' being a strong ligand, pairs up the two unpaired electrons in $d$-subshell of $N{i}^{2+}$.
This is followed by $ds{p}^2$-hybridisation. The complex ion has no unpaired electron and is diamagnetic.
$[Co(N{H}_3{)}_6{]}^{3+}:C{o}^{3+}$ with $3d^6$-configuration undergoes $d^{2}s{p}^3$-hybridisation, after pairing up of all the six $d$-electrons in three $d$-orbitals. As the ion has no unpaired electron, it is diamagnetic