Question

In the ground state of atomic Fe (Z = 26), the spin-only magnetic moment is $x\times {10}^{-1}BM$. The value of x is : (Round off to the Nearest Integer)

[Given: √3 = 1.73, √2 = 1.41]

Answer 1

Answer: 49

$Fe:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$

In $3d^6$, no. of unpaired electrons are 4

$\text{Spin only magnetic moment}=\sqrt{n(n+2)}BM$

$=\sqrt{4(4+2)}=\sqrt{24}=4.89\approx 4.9=49\times {10}^{-1}$
The value of x is 49