[Cu(NH_3)_4]^{2+} has hybridisation and magne | vedclass

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Question

(D) In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. The $3d^9$ configuration has

Vedclass · Accepted Answer

$dsp^2, 1.73 \ B.M.$

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(D) In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. The $3d^9$ configuration has one unpaired electron. The magnetic moment $\mu$ is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. For $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$ Due to the square planar geometry of this complex,the hybridisation is $dsp^2$.

$[Cu(NH_3)_4]^{2+}$ has hybridisation and magnetic moment

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