The 'Spin only' Magnetic moment for is ________ BM (given = Atomic number of Ni : 28)

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Solution

Q.
The 'Spin only' Magnetic moment for ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ is ________ $\times 10^{- 1}$ BM (given = Atomic number of Ni : 28) [2024]

Ans.
(28)

   ${[{Ni(NH}_{3})_{6}]}^{2 +}$

  Oxidation number of Ni = +2

   ${Ni}^{2 +} : {}_{18}{[Ar]} 3 d^{8}$

  Irrespective of strength of ligand, $3 d^{8}$ configuration after octahedral splitting becomes $t_{2 g}^{2, 2, 2} e_{g}^{1, 1}$ . It has two unpaired electrons $(n)$ ,

  Spin only magnetic moment,

   $μ = \sqrt{n (n + 2)} BM = \sqrt{2 (2 + 2)} BM = 2.83 BM$

$= 28.3 \times 10^{- 1} BM$

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