Magnetic moment of [Cu(NH_3)_4]^{2+} ion is ( | vedclass

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(A) The magnetic moment is calculated using the spin-only formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.For the $

Vedclass · Accepted Answer

$1.73$

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(A) The magnetic moment is calculated using the spin-only formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.For the $[Cu(NH_3)_4]^{2+}$ ion,the central metal ion is $Cu^{2+}$,which has an electronic configuration of $[Ar] 3d^9$.In the $3d^9$ configuration,there is $n=1$ unpaired electron.Substituting $n=1$ into the formula: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$

Magnetic moment of $[Cu(NH_3)_4]^{2+}$ ion is (in $B.M.$)

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