A lanthanoid has atomic number 64 and the ground-state configuration $[Xe] 4 f^{7} 5 d^{1} 6 s^{2}$ . The spin-only magnetic moment (in BM) of its tripositive ion is closest to

Answer: D

The three electrons lost are

5 d^{1}

and

6 s^{2}

, so the ion is

4 f^{7}

with n = 7 unpaired electrons. Spin-only

μ = n (n + 2) = 63 = 7.94

BM. Values 6.93, 5.92 and 8.94 correspond to n = 6, 5 and 8.