JEE MAIN - Chemistry (2020 - 5th September Evening Slot - No. 4)
Considering that $$\Delta $$0
> P, the magnetic moment
(in BM) of [Ru(H2O)6]2+ would be _________.
(in BM) of [Ru(H2O)6]2+ would be _________.
Answer
0
Explanation
Ru(44) : [Kr] 4d75s1
Ru+2 = [Kr]4d6
As $$\Delta $$0 > P,
$$ \therefore $$ Pairing of eās will take place.
No. of unpaired eās = 0
$$ \therefore $$ Magnetic moment = 0 B.M
Ru+2 = [Kr]4d6
As $$\Delta $$0 > P,
$$ \therefore $$ Pairing of eās will take place.
No. of unpaired eās = 0
$$ \therefore $$ Magnetic moment = 0 B.M
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