The homoleptic and octahedral complex of $\mathrm{Co}^{2+}$ and $\mathrm{H}_{2} \mathrm{O}$ has \_\_\_\_ unpaired electron(s) in the $\mathrm{t}_{2 \mathrm{~g}}$ set of orbitals.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Coordination Compounds

The homoleptic and octahedral complex of

Co^{2 +}

and

H_{2} O

has ____ unpaired electron(s) in the

t_{2 g}

set of orbitals.

Answer: 1

Solution

The complex is

[Co (H_{2} O)_{6}]^{2 +}

. Co²⁺ has electronic configuration 3d⁷. Water is a weak field ligand, so it forms a high-spin octahedral complex. The d⁷ configuration splits as

t_{2 g}^{5} e_{g}^{2}

in an octahedral field. In the

t_{2 g}

set, the electron distribution is:

t_{2 g}^{↑↓} t_{2 g}^{↑↓} t_{2 g}^{↑}

, giving 1 unpaired electron in the

t_{2 g}

set.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Integer answer

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