The total number of unpaired electrons present in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is ___.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Coordination Compounds

The total number of unpaired electrons present in

[Co (NH_{3})_{6}] Cl_{2}

and

[Co (NH_{3})_{6}] Cl_{3}

is ___.

Answer: 3

Solution

[Co (N H_{3})_{6}] C l_{2}

cobalt is in the

+ 2

state with a

3 d^{7}

configuration. The larger, less highly charged

C o^{2 +}

ion is not split far enough by ammonia for the electrons to pair, so the complex is high spin,

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

, with three unpaired electrons. In

[Co (N H_{3})_{6}] C l_{3}

cobalt is in the

+ 3

state with

3 d^{6}

; the higher charge raises the splitting above the pairing energy, giving the low spin

t_{2 g}^{6}

arrangement with no unpaired electron. The total number of unpaired electrons in the two complexes is

3 + 0 = 3

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer

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