Question

Berlin Green is a pigment related to Prussian Blue, except all the metal atoms are Fe(III). The formula is nominally Fe^III[Fe^III(CN)₆]. The Fe(III) inside the brackets is bound to carbon and low spin, while the Fe^III outside the brackets nitrogen-bound and high spin.

What is the expected μ_so for this compound?

Note that there are two different types of metals in the formula (both Fe^III, but one high spin and one low spin).

Answer 1

Both Fe are +3, so electronic configuration is [Ar] 3d⁵

For High spin Fe³⁺, unpaired electrons = 5

For low spin Fe³⁺, unpaired electrons = 1

So, total unpaired electrons = 6

Magnetic moment = [n(n+2)]^1/2

= [6(6+2)]^1/2

= (48)^1/2

= 6.93 Bohr Magneton.