JEE Advanced levelSingle correct optionDifficulty 3 of 3cc202
Find the total number of unpaired electrons in the set: [Fe(CN)X6]X3− (low spin), [MnClX4]X2− (tetrahedral, high spin), [Ni(CN)X4]X2− (square planar) and [Cr(NHX3)X6]X3+.
A11
B8
C13
D9
Answer: D
FeX3+d5 low spin: 1. MnX2+d5 high spin: 5. NiX2+d8 square planar: 0. CrX3+d3: 3. Total = 1+5+0+3=9.