The magnetic moment of the complex [Ni(NH_3)_ | vedclass

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(B) $1$. In the complex $[Ni(NH_3)_6]Cl_2$,the oxidation state of $Ni$ is $+2$. $2$. The atomic number of $Ni$ is $28$,so its electronic configuration

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$\sqrt{8} \ BM$

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(B) $1$. In the complex $[Ni(NH_3)_6]Cl_2$,the oxidation state of $Ni$ is $+2$. $2$. The atomic number of $Ni$ is $28$,so its electronic configuration is $[Ar] 3d^8 4s^2$. $3$. For $Ni^{2+}$,the configuration is $[Ar] 3d^8$. $4$. $NH_3$ is a ligand,but in an octahedral field with $d^8$ configuration,the electrons are arranged as $t_{2g}^6 e_g^2$. $5$. There are $2$ unpaired electrons in the $3d$ orbitals. $6$. The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. $7$. Substituting $n = 2$,we get $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.

The magnetic moment of the complex $[Ni(NH_3)_6]Cl_2$ is

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