In the ground state of atomic Fe(Z = 26), the spin-only magnetic moment is ____________ $\times$ 10$-$1 BM. (Round off to the Nearest Integer). [Given : $\sqrt 3$ = 1.73, $\sqrt 2$ = 1.41 ]
Answer (integer)
49
Solution
${}_{26}Fe = [Ar]3{d^6}4{s^2}$<br><br>No. of unpaired electrons = 4<br><br>$\mu = \sqrt {n(n + 2)} BM$<br><br>$= \sqrt {4(4 + 2)} = \sqrt {24}$<br><br>$= 2\sqrt {3 \times 2} = 2[1.73 \times 1.41]$<br><br>= 4.8786 BM<br><br>= $48.78 \times {10^{ - 1}}$ BM
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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