For the maximum possible value of 'x' the value of spin only magnetic moment will be.

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Identify high-spin configurations: High-spin and low-spin distinctions only exist for $d^{4}$ to $d^{7}$ electronic configurations in octahedral fields.

Find the maximum value of $x$ : For these high-spin cases, the $t_{2 g}^{x} e_{g}^{y}$ distributions are $d^{4} (t_{2 g}^{3} e_{g}^{1})$ , $d^{5} (t_{2 g}^{3} e_{g}^{2})$ , $d^{6} (t_{2 g}^{4} e_{g}^{2})$ , and $d^{7} (t_{2 g}^{5} e_{g}^{2})$ , making the maximum value of $x = 5$ .

Count unpaired electrons: For the $t_{2 g}^{5} e_{g}^{2}$ configuration, there is $1$ unpaired electron in the $t_{2 g}$ level and $2$ in the $e_{g}$ level, giving a total of $n = 3$ unpaired electrons.

Calculate magnetic moment: Using the spin-only formula $μ = n (n + 2)$ , we get $μ = 3 (3 + 2) = 15 B.M.$ (Option B).

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