A solid contains, per formula unit, one $Fe X^{3 +}$ (high spin) and two $Cu X^{2 +}$ ions that behave as independent paramagnetic centres. Treating the moments as adding in quadrature, the spin-only moment per formula unit is

Answer: A

μ^{2}

values add:

Fe X^{3 +}

(

n = 5

) gives

5 \times 7 = 35

and each

Cu X^{2 +}

(

n = 1

) gives

3

. Total

μ^{2} = 35 + 2 \times 3 = 41

, so

μ = 41 = 6.40 BM

. Adding moments directly or ignoring one

Cu X^{2 +}

gives wrong values.