Question

M^x+ ion has magnetic moment of 4.90 BM.Calculate the value of x?(Atomic number is 24)​

Answer 1

according to formula of magnetic moment, M = $\sqrt{n(n+2)}$ BM

where n is maximum number of unpaired electrons in given ion.

here, magnetic moment = 4.9 BM

or, √{n(n + 2)} = 4.9

or, n(n + 2) = (4.9)²

or, n² + 2n = 24.01 ≈ 24

or, n² + 2n - 24 = 0

or, n² + 6n - 4n - 24 = 0

or, n(n + 6) - 4(n + 6) = 0

or, (n - 4)(n + 6) = 0

or, n = 4 and -6 but n ≠ -6

so, maximum number of unpaired electrons , n = 4

now, write electron configuration of element (Z = 24) : 1s², 2s², $2p^6$, 3s², $3p^6$ , 4s¹, $3d^5$

now, when we remove two electrons, electronic configuration becomes : 1s², 2s², $2p^6$, 3s², $3p^6$ , $4s^0$, 3d⁴ .

here we see, 4 unpaired electrons are present in d subshell.

so, after 2 electrons removing from M, M²+ ion forms.

hence, value of x = 2