Spin-only magnetic moment is mu=sqrtn(n+2) BM. Find the sum of the values of mu^2 (that is, n(n+2) for each ion) for Cr^3+, Mn^2+, Fe^3+ and Ni^2+ in their ground states.

JEE Advanced levelNumerical answerDifficulty 3 of 3as214

Spin-only magnetic moment is $μ = n (n + 2)$ BM. Find the sum of the values of $μ^{2}$ (that is, $n (n + 2)$ for each ion) for $Cr X^{3 +}$ , $Mn X^{2 +}$ , $Fe X^{3 +}$ and $Ni X^{2 +}$ in their ground states.

Numerical answer type, answer in BM^2, tolerance 0.1.

Cr X^{3 +}

3 d^{3}

3 \times 5 = 15

Mn X^{2 +}

3 d^{5}

5 \times 7 = 35

Fe X^{3 +}

3 d^{5}

: 35.

Ni X^{2 +}

3 d^{8}

2 \times 4 = 8

. Sum

= 15 + 35 + 35 + 8 = 93