JEE MAIN - Chemistry (2021 - 17th March Evening Shift - No. 18)
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
49
Explanation
$${}_{26}Fe = [Ar]3{d^6}4{s^2}$$
No. of unpaired electrons = 4
$$\mu = \sqrt {n(n + 2)} BM$$
$$ = \sqrt {4(4 + 2)} = \sqrt {24} $$
$$ = 2\sqrt {3 \times 2} = 2[1.73 \times 1.41]$$
= 4.8786 BM
= $$48.78 \times {10^{ - 1}}$$ BM
No. of unpaired electrons = 4
$$\mu = \sqrt {n(n + 2)} BM$$
$$ = \sqrt {4(4 + 2)} = \sqrt {24} $$
$$ = 2\sqrt {3 \times 2} = 2[1.73 \times 1.41]$$
= 4.8786 BM
= $$48.78 \times {10^{ - 1}}$$ BM
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