Question

Hybridisation, shape and magnetic moment of $K_{3}\left[\right.Co\left(C O_{3}\right)_{3}\left]\right.$ 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 $K _3 CoCO _{33}$, 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 $Co ^{+3}$
In $K _3 CoCO _{3_3}=\sqrt{ nn +2} BM$ $=\sqrt{44+2} BM =4.9 BM$
Where, $n=n u m b e r$ of unpaired electrons
$CO _3^{2-}$ is a weak field bidentate ligand, so $3 CO _3^{2-}$ ligands occupy six orbitals, thus it shows $sp ^3 d ^2$ hybridisation and octahedral.