JEE Chemistry 2023 Question with Solution

Given: The complexes are $\mathrm{K_3[Co(CN)_6]}$, $\mathrm{[Ni(NH_3)_6]Cl_2}$, $\mathrm{[Co(H_2O)_6]Cl_2}$ and $\mathrm{Na_3[CoCl_6]}$.

Find: Which complex is octahedral, diamagnetic and the most stable.

For $\mathrm{K_3[Co(CN)_6]}$, let the oxidation state of cobalt be $x$. Then

so,

Therefore, cobalt is $\mathrm{Co^{3+}}$.

The electronic configuration of $\mathrm{Co^{3+}}$ is $3d^6$.

$\mathrm{CN^-}$ is a strong field ligand. Strong field ligands produce large splitting of the $d$-orbitals and cause pairing of electrons in the lower energy orbitals.

Hence, in $\mathrm{[Co(CN)_6]^{3-}}$ all the electrons become paired, so the complex is diamagnetic.

Its coordination number is $6$, therefore the complex is octahedral.

Also, $\mathrm{CN^-}$ forms strong bonds with the metal centre due to its strong field nature, so $\mathrm{K_3[Co(CN)_6]}$ is the most stable among the given options.

Therefore, the complex which is octahedral, diamagnetic and the most stable is $\mathrm{K_3[Co(CN)_6]}$. The correct option is A.

Given: We need a complex that is simultaneously octahedral, diamagnetic and highly stable.

Find: The correct option using oxidation state and ligand strength.

In $\mathrm{K_3[Co(CN)_6]}$, the complex ion is $\mathrm{[Co(CN)_6]^{3-}}$. Since each $\mathrm{CN^-}$ carries charge $-1$,

So the metal ion is $\mathrm{Co^{3+}}$ with configuration $3d^6$.

Because $\mathrm{CN^-}$ is a strong field ligand, the octahedral splitting is large and the $3d^6$ configuration becomes low spin. Thus all electrons are paired.

A low-spin octahedral $d^6$ complex is diamagnetic and generally quite stable, especially with $\mathrm{CN^-}$ as ligand.

So the required complex is option A.