Consider the following metal complexes: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$ The spin-only magnetic moment value of the complex that absorbs light with shortest wavelength is _____ B.M. (Nearest integer)

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Coordination Compounds

Consider the following metal complexes:

[Co (NH_{3})_{6}]^{3 +}

[CoCl (NH_{3})_{5}]^{2 +}

[Co (CN)_{6}]^{3 -}

[Co (NH_{3})_{5} (H_{2} O)]^{3 +}

The spin-only magnetic moment value of the complex that absorbs light with shortest wavelength is _____ B.M. (Nearest integer)

Answer: 0

The crystal field splitting energy

Δ_{0}

is inversely proportional to the wavelength of absorbed light:

Δ_{0} \propto \frac{1}{λ}

. The shortest wavelength corresponds to the largest

Δ_{0}

.

Among the given ligands,

CN X^{-}

is the strongest field ligand, producing the maximum crystal field splitting. Therefore,

[Co (CN)_{6}]^{3 -}

absorbs light with the shortest wavelength.

For

[Co (CN)_{6}]^{3 -}

Co X^{3 +}

has

d^{6}

configuration. With strong field

CN X^{-}

ligands, all electrons pair up in the

t_{2 g}

orbitals, giving an inner orbital

d^{2} s p^{3}

complex with zero unpaired electrons.

Spin-only magnetic moment:

μ = n (n + 2)

where

n = 0

, so

μ = 0

B.M.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer