A 3d metal M forms an aqueous $M X^{3 +}$ ion with spin-only magnetic moment 5.92 BM. The atomic number of M and the spin-only magnetic moment of the high-spin aqua $M X^{2 +}$ ion are respectively:

Answer: B

5.92 BM means

n = 5

, so

M X^{3 +}

3 d^{5}

. M has

3 d^{6}

after adding three electrons removed from

4 s^{2} 3 d^{6}

, so

Z = 26

(Fe).

Fe X^{2 +}

3 d^{6}

with 4 unpaired electrons (high spin), so

μ = 24 = 4.90 BM