Question

A correct statement is

• A. $[MnB{r}_4{]}^{-2}$ is tetrahedral
• B. $[Ni(N{H}_3{)}_6{]}^{+2}$ is an inner orbital complex
• C. $[Co(N{H}_3{)}_6{]}^{+2}$ is paramagnetic
• D. $[CoB{r}_2(en{)}_2{]}^{-}$ exhibits linkage isomerism

Answer 1

Answer: (C)

${\left[Co{\left(N{H}_3\right)}_6\right]}^{2+}$ In this ion, Co is present as $C{o}^{2+}.N{H}_3$, being strong field ligand, pair up the unpaired electrons.

Electronic configuration of $=C{o}^{2+}=3d^7,4s^{\circ },4p^0$



Electronic configuration of $C{O}^{2+}$ in $=$

${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$



Due to the presence of one unpaired electron, ${\left[Co{\left(N{H}_3\right)}_6\right]}^{+}$ is paramagnetic.

${\left[MnB{r}_4\right]}^{2-}$ In this ion, Mn is present as $M{n}^{2+}$. Br, being weak field ligand, leave unpaired electrons of the metal ion as such.

Electronic configuration of $M{n}^{2+}=3d^{5}4s^{0}4p^0$



Electronic configuration of $M{n}^{2+}$ in ${\left[MnB{r}_4\right]}^2$



Due to $s{p}^3$ hybridisation, shape of ${\left[MnB{r}_4\right]}^{2-}$ is tetrahedral.

(c) Linkage isomerism occurs when two or more atoms in a monodentate ligand may function as a donor.

Therefore, ${\left[Co{B}_2(en{)}_2\right]}^{-}$ does not exhibit linkage isomerism.

(d) ${Ni{\left(N{H}_3\right)}_6]}^{2+}$ In this ion, $Ni$ is present as $N{i}^{2+}$.

Electronic configuration of $N{i}^{2+}=[Ar]3d^{8}4s^{\circ }$



$N{H}_3$ is a strong field ligand, it can pair up electrons only if they are more than 4 and less than $8.$

Electronic configuration of nine ${\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}$



Since. $(n-1)d$, i.e., $3d$ orbital does not take part in hybridisation, it is not an inner orbital complex. It is an outer orbital complex.