Question

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

• A. 49.00
• B. 49
• C. 49.0

Answer 1

Answer: (A), (B), (C)

$Fe(Z=26$)
Electrons configuration :
$\left[Ar\right]4s^{2}3d^6$

Unpaired electrons:
$n=4$

$\therefore \mu =\sqrt{4(4+2)}=\sqrt{24}BM$

$\mu =4.89$

$\mu =49\times {10}^{-1}B.Mx=49$