Question

Hybridisation, shape and magnetic moment of $K_3[Co{\left(C{O}_3\right)}_3]$ is

• A. ${\text{d}}^2\text{s}{\text{p}}^3,\text{o}\text{c}\text{t}\text{a}\text{h}\text{e}\text{d}\text{r}\text{a}\text{l},4.9\text{B}\text{M}$
• B. $\text{s}{\text{p}}^3{\text{d}}^2,\text{o}\text{c}\text{t}\text{a}\text{h}\text{e}\text{d}\text{r}\text{a}\text{l},4.9\text{B}\text{M}$
• C. $\text{d}\text{s}{\text{p}}^2,\text{s}\text{q}\text{u}\text{a}\text{r}\text{e}\text{p}\text{l}\text{a}\text{nar},4.9\text{B}\text{M}$
• D. $\text{s}{\text{p}}^3,\text{t}\text{e}\text{t}\text{r}\text{a}\text{h}\text{e}\text{d}\text{r}\text{a}\text{l},4.9\text{B}\text{M}$

Answer 1

Answer: (B)

In $K_{3}CoC{O}_{33}$, cobalt shows the $+3$ oxidation state i.e., $\left(d^6\right)$ ion. Hence, $Co(+3)$ has four unpaired electrons so, it is paramagnetic.
The magnetic moment of $C{o}^{+3}$
In $K_{3}CoC{O}_{3_3}=\sqrt{nn+2}BM$ $=\sqrt{44+2}BM=4.9BM$
Where, $n=number$ of unpaired electrons
$C{O}_3^{2-}$ is a weak field bidentate ligand, so $3C{O}_3^{2-}$ ligands occupy six orbitals, thus it shows $s{p}^3{d}^2$ hybridisation and octahedral.