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(D) The chemical formula for Cuproammonium sulphate is $[Cu(NH_{3})_{4}]SO_{4} \cdot H_{2}O$.In the complex ion $[Cu(NH_{3})_{4}]^{2+}$,the oxidation

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Paramagnetic and square planar

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(D) The chemical formula for Cuproammonium sulphate is $[Cu(NH_{3})_{4}]SO_{4} \cdot H_{2}O$.In the complex ion $[Cu(NH_{3})_{4}]^{2+}$,the oxidation state of $Cu$ is $+2$.The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^{9}$.Due to the presence of $NH_{3}$ (a strong field ligand),the complex undergoes $dsp^{2}$ hybridization,resulting in a square planar geometry.Since there is one unpaired electron in the $3d$ orbital,the complex is paramagnetic.

What is the type of magnetic behavior and geometry respectively in Cuproammonium sulphate (Atomic no. of $Cu=29$)?

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