Considering that $\Delta_{0}>\mathrm{P}$, the magnetic moment (in BM ) of $\left[\mathrm{Ru}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ would be \_\_\_\_ .

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Coordination Compounds

Considering that

Δ_{0} > P

, the magnetic moment (in BM ) of

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

would be
____ .

Answer: 0

Solution

Ruthenium in

[Ru (H X_{2} O) X_{6}] X^{2 +}

is in the

+ 2

state, a

4 d^{6}

configuration. The stated condition

Δ_{0} > P

means the splitting outweighs the pairing energy, so the complex is low-spin and all six

d

electrons pair in the lower

t_{2 g}

set, leaving

t_{2 g}^{6} e_{g}^{0}

with no unpaired electrons. The spin-only moment is then

n (n + 2) = 0 = 0

BM.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Integer answer

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