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JEE Advanced levelSingle correct optionDifficulty 3 of 3qa163

Prussian blue, $Fe X_{4} [Fe (CN) X_{6}] X_{3}$ , is formed from $Fe X^{3 +}$ and $[Fe (CN) X_{6}] X^{4 -}$ . Assume the 4 outer iron ions are Fe(III), N-bonded and high spin, and the 3 inner iron ions are Fe(II), C-bonded and low spin. Ignoring magnetic coupling, the total number of unpaired electrons per formula unit is:

A $16$
B $15$
C $35$
D $20$

Answer: D

Charge balance:

4 \times 3 = 3 \times 4

, so four Fe(III) and three Fe(II). High-spin Fe(III) (

d^{5}

) has 5 unpaired electrons each, giving

4 \times 5 = 20

. Low-spin Fe(II) (

d^{6}

) in cyanide has none. The total is 20.

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