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Q.
The shape of $\left[Cu\left(NH_{3}\right)_{4}\right]^{2+}$ is
Coordination Compounds
Solution:
Oxidation state of $Cu$ in $\left[Cu\left(NH_{3}\right)_{4}\right]^{2+}= +2$
$Cu^{2+}\left(Z =29\right) \to3d^{9}$
One electron is shifted from 3d to 4p-orbital. Thus, $\left[Cu\left(NH_{3}\right)_{4}\right]^{2+}$ is $dsp^{2}$ hybridised and has square planar geometry.
