Chemistry - A Molecular Science

Chapter 14 Inorganic Chemistry

as chloride ions, the higher energy set of d orb

itals is close in energy to the lower set (

< Δ

PE), and no pairing occurs. Each of the five electrons has a spin of +

1 /^2

, which gives the

ion a total

spin

of (5)(

1 /^2

) =

5 /^2

, and MnCl

4- 6

is said to be

high spin

. However, if the

ligands are strong-field ligands, such as cyan

ide ions, occupation of the higher energy set

of orbitals requires more energy than does pair

ing, so the electrons pair. In this case, there

are three electrons with spins of +

1 /^2

and two with spins of -

1 /^2

for a total spin of (3)(

1 /

)+ 2

(2)

-1(

/)^2

=

1 /^2

, and Mn(CN)

4- 6

is said to be a

low spin

ion.

Example 14.2 a) When FeSO

dissolves in water, it forms the Fe(H 4

O) 2

2+ 6

ion. Given that water is a

weak-field ligand, how many unpaired el

ectrons would be present in the Fe(H

O) 2

2+ 6

ion? Iron has a 4s

2 3d

6 valence electron configuration, and it

loses the two 4s electrons to form

iron(II). Consequently, Fe

2+ is 3d

6. Water is a weak-field ligand, so

Δ < PE, and the iron is

expected to be high spin. The corresponding diagram is shown below in Figure a. Because

Δ < PE, each of the d orbitals is occupied

before any pairing occurs. The result is

that there are four unpaired electrons.

Δ < PE high spin iron(II)

Δ > PE low spin iron(II)

a.

b.

b) When CN

1- ion is added to a solution of Fe(H

O) 2

2+ 6

, the deeply colored Fe(CN)

4- 6

ion

is produced. How many unpaired electrons does it contain? Cyanide is a strong-field ligand, so

Δ > PE

, which produces a low spin iron(II). The

situation is shown in Figure b. The result is

that the six electrons pair rather than occupy

the higher energy set of orbitals to produce a complex with no unpaired electrons.

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