If the bidentate ligand ethane-1, 2-diamine ( | vedclass

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(D) The reaction sequence is as follows:$[Ni(H_2O)_6]^{2+} + en \rightarrow [Ni(en)(H_2O)_4]^{2+} + 2H_2O$$[Ni(en)(H_2O)_4]^{2+} + en \rightarrow [Ni(

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$\sqrt{8}, \sqrt{8}, \sqrt{8}$

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(D) The reaction sequence is as follows:$[Ni(H_2O)_6]^{2+} + en \rightarrow [Ni(en)(H_2O)_4]^{2+} + 2H_2O$$[Ni(en)(H_2O)_4]^{2+} + en \rightarrow [Ni(en)_2(H_2O)_2]^{2+} + 2H_2O$$[Ni(en)_2(H_2O)_2]^{2+} + en \rightarrow [Ni(en)_3]^{2+} + 2H_2O$In all these complexes,$Ni^{2+}$ has a $d^8$ electronic configuration. The coordination number is $6$ in all cases,and the geometry is octahedral.For $Ni^{2+}$ $(d^8)$ in an octahedral field,the crystal field splitting does not cause pairing of electrons because the energy difference is not sufficient to overcome the pairing energy. Thus,the number of unpaired electrons remains $2$ for all three complexes.Using the spin-only formula $\mu_s = \sqrt{n(n+2)}$,where $n = 2$:$\mu_s = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$Therefore,the magnetic moments for all three complexes are $\sqrt{8}, \sqrt{8}, \sqrt{8}$.

If the bidentate ligand ethane-$1, 2$-diamine $(en)$ is progressively added to a green aqueous solution of $[Ni(H_2O)_6]Cl_2$ in the molar ratios $en : Ni$ as $1 : 1, 2 : 1, 3 : 1$,then their associated spin-only magnetic moments (in $B.M.$ unit) are respectively

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