An aqueous solution of [Ni(H_2O)_6]^{2+} cont | vedclass

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(A) The initial complex is $[Ni(H_2O)_6]^{2+}$. In this complex,$Ni^{2+}$ has a $d^8$ configuration. $H_2O$ is a weak field ligand,so the number of un

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It remains the same.

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(A) The initial complex is $[Ni(H_2O)_6]^{2+}$. In this complex,$Ni^{2+}$ has a $d^8$ configuration. $H_2O$ is a weak field ligand,so the number of unpaired electrons is $2$. The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$. When ammonia $(NH_3)$ is added,it acts as a stronger ligand than $H_2O$ and replaces it to form $[Ni(NH_3)_6]^{2+}$. However,$NH_3$ is not strong enough to cause pairing of electrons in the $d^8$ configuration of $Ni^{2+}$. Thus,the number of unpaired electrons remains $2$. Therefore,the magnetic moment remains $2.83 \ BM$.

An aqueous solution of $[Ni(H_2O)_6]^{2+}$ contains $Ni^{2+}$ ions and has a magnetic moment of $2.83 \ BM$. What is the effect on the magnetic moment of the solution when ammonia is added?

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