Treat $CN X^{-}$ as a strong-field ligand (low spin) and use the spin-only formula. Which statements about $K X_{3} [Fe (CN) X_{6}]$ and $K X_{4} [Fe (CN) X_{6}]$ are correct?

A $K X_{3} [Fe (CN) X_{6}]$ gives 4 ions per formula unit and its anion has $μ \approx 1.73$ BM.
B $K X_{4} [Fe (CN) X_{6}]$ gives 5 ions per formula unit and its anion is diamagnetic.
CAt equal molar concentration, $K X_{4} [Fe (CN) X_{6}]$ has a higher molar conductivity than $K X_{3} [Fe (CN) X_{6}]$ (ideal dissociation).
DThe anion of $K X_{4} [Fe (CN) X_{6}]$ has two unpaired electrons.

Answer: A, B, C

K X_{3} [Fe (CN) X_{6}]

3 K X^{+}

plus one anion gives 4 ions, Fe(III)

d^{5}

low spin has

t_{2 g}^{5}

, 1 unpaired,

μ = 3 = 1.73

BM.

K X_{4} [Fe (CN) X_{6}]

: 5 ions, Fe(II)

d^{6}

low spin

t_{2 g}^{6}

has 0 unpaired electrons. More ions mean higher conductivity, so the first three are correct and the fourth is wrong.