In the ground state of atomic Fe(z=26), The spin only magnetic moment is --------------------- ×10−1BM. (Round off to nearest Integer) (given 3=1.73, 2 =1.41)

Fe(z=26) Electronic configuration is [Ar]3d64s2↑↓↑↑↑↑ NO .of unpaired electrons =4μ=n(n+2) BM=4(4+2)=24=4.89 =48.9x10−1 BM=49x10−1BM

In the ground state of atomic Fe(z=26), The spin only magnetic moment is --------------------- ×10−1BM. (Round off to nearest Integer) (given 3=1.73, 2 =1.41)

Fe(z=26) Electronic configuration is [Ar]3d64s2↑↓↑↑↑↑ NO .of unpaired electrons =4μ=n(n+2) BM=4(4+2)=24=4.89 =48.9x10−1 BM=49x10−1BM

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In the ground state of atomic Fe(z=26), The spin only magnetic moment is --------------------- $\times 10^{- 1}$ BM. (Round off to nearest Integer) (given $\sqrt{3} = 1.73, \sqrt{2} = 1.41$ )

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Fe(z=26) Electronic configuration is $[A r] 3 d^{6} 4 s^{2}$

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NO .of unpaired electrons =4

$\begin{array}{l} μ = \sqrt{n (n + 2)} B M = \sqrt{4 (4 + 2)} = \sqrt{24} = 4.89 \\ = 48.9 x 10^{- 1} B M = 49 x 10^{- 1} B M \end{array}$

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In the ground state of atomic Fe(z=26), The spin only magnetic moment is --------------------- ×10−1BM. (Round off to nearest Integer) (given 3=1.73, 2 =1.41)

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In the ground state of atomic Fe(z=26), The spin only magnetic moment is --------------------- $\times 10^{- 1}$ BM. (Round off to nearest Integer) (given $\sqrt{3} = 1.73, \sqrt{2} = 1.41$ )