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.