JEE Advanced levelSingle correct optionDifficulty 3 of 3df145
Aqueous FeX2+ reacts with [Fe(CN)X6]X3− to give the dark blue pigment Turnbull's blue, which is the same as Prussian blue, KFe[Fe(CN)X6]. In this solid the carbon-bonded Fe is low-spin Fe(II) and the nitrogen-bonded Fe is high-spin Fe(III). The colour of the pigment and the spin-only magnetic moment per formula unit are respectively:
Low-spin Fe(II) in the C-bonded site has t2g6, with no unpaired electron. High-spin Fe(III) is d5 with n=5, giving 35=5.92BM. The intense colour comes from electron transfer between the two iron sites.