In the ground state of atomic $\mathrm{Fe}(\mathrm{Z}=26)$, the spin-only magnetic moment is ______ $\times 10^{-1} \mathrm{BM}$. (Round off to the nearest integer). [Given $\sqrt{3}=1.73, \sqrt{2}=1.41$]

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · d and f Block Elements

In the ground state of atomic

Fe (Z = 26)

, the spin-only magnetic moment is ______

\times 1 0^{- 1} BM

.
(Round off to the nearest integer).
[Given

3 = 1.73, 2 = 1.41

]

Answer: 49

Iron,

Z = 26

, has the ground state configuration

[Ar] 3 d^{6} 4 s^{2}

. The six

3 d

electrons occupy the five orbitals singly before any pairing, so one orbital holds a pair and four electrons are left unpaired, while the filled

4 s

orbital contributes nothing. The spin-only moment is

μ = n (n + 2) = 4 \times 6 = 26 = 223 = 2 (1.41) (1.73) = 4.88 BM

which is

48.8 \times 1 0^{- 1} BM

, and to the nearest integer the spin-only magnetic moment is

49 \times 1 0^{- 1} BM

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer