SME
Inorganic Chemistry
Crystal field theory
Use this for “Fe²⁺ (d⁶) with water has Δₒ = 10,400 cm⁻¹ and P = 17,600 cm⁻¹. Is it high or low spin? Find the unpaired electrons, the CFSE and μ.”
Diagram
Δₒ = 10,400 cm⁻¹ < P = 17,600 cm⁻¹: high spin, every orbital takes one electron before any pairs; t2g⁴ eg².
4
unpaired;
μ
=
√
4 × 6
=
4.9
BM.
CFSE = (−0.4 × 4 + 0.6 × 2) × 10,400 + 0 × 17,600 = −4,160 cm⁻¹.
Each pair beyond the free ion's 1 costs P: 0 extra.
This is an example. Type your own values to start a new problem.
d
d electrons
example
Δₒ
Splitting
example · cm⁻¹
P
Pairing energy
example · cm⁻¹
n(t₂g)
Electrons in t₂g
calculated
n(eg)
Electrons in eg
calculated
n
Unpaired electrons
calculated
CFSE
Stabilization energy
calculated · cm⁻¹
μ
Spin-only moment
calculated · BM
Formulas
n
(t₂g)
=
t₂g
filling
of
d
electrons
with
Δₒ
against
P
4 = t₂g filling of 6 electrons with 10,400 against 17,600
n
(eg)
=
d
−
n
(t₂g)
2 = 6 − 4
n
=
unpaired
count
of
d
electrons
with
Δₒ
against
P
4 = unpaired count of 6 electrons with 10,400 against 17,600
CFSE
=
(−0.4
×
n
(t₂g)
+
0.6
×
n
(eg))
×
Δₒ
+
(
d
−
n
2
−
the
free
ion’s
pairs
among
d
electrons)
×
P
−4,160
=
(−0.4
×
4
+
0.6
×
2)
×
10,400
+
(
6 − 4
2
−
the
free
ion’s
pairs
among
6
electrons)
×
17,600
μ
=
√(
n
×
(
n
+
2))
4.899 = √(4 × (4 + 2))
Assumptions
•
Octahedral: t₂g sits 0.4Δₒ below the free ion’s d level and eg 0.6Δₒ above it.
•
Each electron goes where it costs least, so the complex is low spin when Δₒ > P; only d⁴–d⁷ have a choice.
•
Pairs beyond the free ion’s (d⁶ has one) cost P each in the CFSE; Δₒ, P and CFSE are in cm⁻¹.
•
μ is spin only, in Bohr magnetons. Tetrahedral Δₜ is about 4/9 Δₒ, so those complexes are high spin.
Step-by-step
WE KNOW
d = 6, Δₒ = 10,400 cm⁻¹, P = 17,600 cm⁻¹
FIND
electrons in t₂g (n(t₂g)), electrons in eg (n(eg)), unpaired electrons (n), stabilization energy (CFSE), spin-only moment (μ)
Step 1 · Find electrons in t₂g (n(t₂g))
Use n(t₂g) = t₂g filling of d electrons with Δₒ against P
Δₒ
<
P
:
one
electron
in
each
of
the
five
orbitals
first,
then
pairs
from
t₂g
up.
n
(t₂g)
=
d
−
2
n
(t₂g)
=
6
−
2
n
(t₂g)
=
4
→
Δₒ
<
P
:
high
spin,
the
most
unpaired
electrons.
Step 2 · Find electrons in eg (n(eg))
Use n(eg) = d − n(t₂g)
The
rest
of
the
d
electrons
sit
in
eg.
n
(eg)
=
6
−
4
n
(eg)
=
2
Step 3 · Find unpaired electrons (n)
Use n = unpaired count of d electrons with Δₒ against P
Count the boxes holding one arrow.
n
=
10
−
d
n
=
10
−
6
n
=
4
Step 4 · Find stabilization energy (CFSE)
Use CFSE = (−0.4 × n(t₂g) + 0.6 × n(eg)) × Δₒ + ((d − n) ÷ 2 − the free ion’s pairs among d electrons) × P
Each
t₂g
electron
is
lowered
0.4Δₒ
and
each
eg
one
raised
0.6Δₒ;
each
pair
beyond
the
free
ion’s
costs
P
.
CFSE
=
(−0.4
×
n
(t₂g)
+
0.6
×
n
(eg))
×
Δₒ
+
(
d
−
n
2
−
(
d
−
5))
×
P
CFSE
=
(−0.4
×
4
+
0.6
×
2)
×
10,400
+
(
6 − 4
2
−
(6
−
5))
×
17,600
CFSE
=
(−0.4
×
4
+
0.6
×
2)
×
10,400
+
(
2
2
−
1)
×
17,600
CFSE = (−1.6 + 1.2) × 10,400 + (1 − 1) × 17,600
CFSE = −0.4 × 10,400 + 0 × 17,600
CFSE = −4,160 cm⁻¹
Step 5 · Find spin-only moment (μ)
Use μ = √(n × (n + 2))
The spin-only moment grows with the unpaired electrons, in Bohr magnetons.
μ
=
√(4
×
(4
+
2))
μ
=
√
4 × 6
μ
=
√
24
μ
=
4.899
BM
Check
Put the numbers back in. Both sides should match.
4 = t₂g filling of 6 electrons with 10,400 against 17,600 ✓
2 = 6 − 4 ✓
4 = unpaired count of 6 electrons with 10,400 against 17,600 ✓
−4,160
=
(−0.4
×
4
+
0.6
×
2)
×
10,400
+
(
6 − 4
2
−
the
free
ion’s
pairs
among
6
electrons)
×
17,600
✓
4.899 = √(4 × (4 + 2)) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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