Dissolving $CrCl X_{3} \cdot 6 H X_{2} O$ in water and adding excess $AgNO X_{3}$ to 0.1 mol of one isomer precipitates 0.2 mol of AgCl. Which option gives the correct formula of the isomer and its spin-only magnetic moment?

Answer: B

Two chlorides precipitate per formula unit, so two

Cl X^{-}

are free and one is bound:

[Cr (H X_{2} O) X_{5} Cl] Cl X_{2} \cdot H X_{2} O

(five

H X_{2} O

plus one

Cl X^{-}

fill the six coordination sites; the sixth water is outside).

Cr X^{3 +}

3 d^{3}

with

n = 3

, so

μ = 15 = 3.87

BM whichever ligands are bound.