The total number of unpaired electrons present in the complex $\mathrm{K}_{3}\left[\mathrm{Cr}(\text { oxalate })_{3}\right]$ is \_\_\_\_ .

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Coordination Compounds

The total number of unpaired electrons present in the complex

K_{3} [Cr (oxalate)_{3}]

is ____ .

Answer: 3

Solution

Oxalate is a dinegative bidentate ligand, so charge balance

3 (+ 1) + x + 3 (- 2) = 0

puts chromium in the

+ 3

state.

C r^{3 +}

has the configuration

[Ar] 3 d^{3}

, and in an octahedral field these three electrons go one each into the three

t_{2 g}

orbitals, because pairing can only begin once every orbital of that lower set already holds an electron. Strong and weak field therefore give the same arrangement here, and the number of unpaired electrons is 3.

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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