Considering that Δ 0 > P , the magnetic moment (in BM) of Ru H 2 O 6 2 + would be

Considering that Δ0>P, the magnetic moment (in BM) of RuH2O62+ would be

Solution

Ru2*=4d6=t2g2,2.2eg0,0 since Δ0>P
No. of unpaired electrons = zero
Magnetic Moment =0

Asked in: JEE Main 2020 (05 Sep Shift 2)

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