Question

Hybridisation, shape and magnetic moment of ${\mathrm{K}}_3\left[\mathrm{Co}{\left({\mathrm{CO}}_3\right)}_3\right]$ 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 ${\mathrm{K}}_3\left[\mathrm{Co}{\left({\mathrm{CO}}_3\right)}_3\right]$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 Co⁺³
In ${\mathrm{K}}_3\left[\mathrm{Co}{\left({\mathrm{CO}}_3\right)}_3\right]=\sqrt{\mathrm{n}(\mathrm{n}+2)}\mathrm{BM}$
$=\sqrt{4\left(4+2\right)}$ BM=4.9 BM
Where, n=number of unpaired electrons
$\text{C}{\text{O}}_3^{2-}$ is a weak field bidentate ligand, so $3\text{C}{\text{O}}_3^{2-}$ ligands occupy six orbitals, thus it shows $\text{s}{\text{p}}^3{\text{d}}^2$ hybridisation and octahedral.