Explains Factors affecting Crystal Field Splitting

1. FACTORS AFFECTING
CRYSTAL FIELD
SPLITTING
Dr. Narinderjit Kaur
Assistant Professor
PG Dept. OfChemistry
Kanya Maha Vidyalaya, Jalandhar
Factors
affecting
Crystal field
Splitting
Dr. Narinderjit Kaur
Assistant professor
PG Dept. of Chemistry
Kanya Maha Vidyalaya,
Jalandhar

2. Dr. Narinderjit Kaur
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Why magnitude of
Crystal Field Splitting
is different for
different complexes?

3. The Magnitde of Δo
depend on following
factors:
• Nature of the ligand (Spectrochemical Series)
• Oxidation state of the transition metal
• Type of d-orbitals (transition metals)
• Geometry of the Complex
Dr. Narinderjit Kaur
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4. 1. Nature of Ligands
• Greater the ease with which ligands can approach the metal ion, greater will be
the CFS.
• Complexes with weak field ligands have smaller Δo than complexes with
strong field ligands.
How to decide which
ligand is weak field &
which one is strong field?
The ligands which cause small degree of CFS Δo are called weak field
ligands while which cause large splitting are called strong field ligands.
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Dr. Narinderjit Kaur

5. Spectrochemical series
The arrangement of ligands in an increasing or decreasing order of CFS
Δo is called spectrochemical series. This is an experimentally determined
series. The order remains almost same for different metal ions.
CO > CN
-
= C H > PR > NO
-
= phen > bipy > SO 2-
> en = py = NH3 > edta
2 4 3 2 3
- - -
> NCS-
> H2O > C2O4
2- > OSO3
2- > OH-
= ONO-
> F-
> Cl-
> SCN > Br > I
Decreasing Crystal Field
Strong
Weak
In fact, it is difficult to explain the order as it incorporates the effects of both σ & π
bonding. In general the variation in the splitting depends on number of other factors
Such as:
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Dr. Narinderjit Kaur

6. Other factors:
 Small ligands can cause greater CFS.
 Ligands containing polarizable electron pair will be drawn more closely to metal
ion.
 Ligands which can form multiple bonds such as CN-
& CO cause greater CFS.
CFS by CN-
ligand is about double that for weak field ligands like halide
ions, this is due to the π bonding in metal. Metal donates e’s filled t2g
orbitals into vacant orbitals onligand.
 Increasing σ donation occurs in the order:
Halide donors < O donors < N donors < C donors
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Dr. Narinderjit Kaur

7. Weak field ligands, such as Cl-
and SCN-
, give rise to small values for
Δo, while strong field ligands, such as CN-
and CO, give rise to large
values forΔo.
It is also possible to arrange the metals according to a
spectrochemical series as well. The approximate order is:
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Dr. Narinderjit Kaur
Mn2+ < Ni2+ < Fe2+ < V2+ < Fe3+ < Co3+ < Mn3+ < Mo3+ < Rh3+ < Ru3+ < Pd4+ <Ir3+
< Pt4+

8. Which of the following would
have the largest value ofΔo?
A. [Cr(H2O)6]3+
B. [CrF6]3-
C. [Cr(NH3)6]3+
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Dr. Narinderjit Kaur

9. 6 2 3For [Co(III)L ], Δ in cm-1: 13,100 (F-
); 20,760 (H O); 22,870 (NH )
For [Cr(III)L ], Δ in cm-1: 15,260 (F-
); 17,830 (H O); 26,280 (CN-
)
6 2
Complex ligand Donor atom Δ in cm-1
[CrCl6]3- Cl-
Cl 13640
[CrF6]3- F-
F 15260
[Cr(H2O)6]3+ H2O O 17830
[Cr(NH3)6]3+ NH3 N 21680
[Cr(en)3]3+ en N 21900
[Cr(CN-
)6]3- CN-
C 26280
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Dr. Narinderjit Kaur

10. Why CFS is different
even if
complexes have
same metal ion and
same ligands?
Dr. Narinderjit Kaur 10

11. 2. Oxidation State of Metal ion
 Δo =10,200 cm-1 for[Co(NH3)6]2+
 Δo =22,870 cm-1 for[Co(NH3)6]3+
 Δo =32,200 cm-1 for [Fe(CN)6]4-
 Δo =35,000 cm-1for[Fe(CN)6]3-
Let’s take an example……
Metal ion with higher oxidation state causes largerΔo
As the oxidation state of the transition metal (the charge on the metal) is increased, the
surrounding ligands are attracted more closely to the metal centre. The orbitals on the
ligands interact more strongly with the d orbitals andΔo increases.
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Dr. Narinderjit Kaur

12. Which of the following would
have the largest value ofΔo?
A. [Fe(H2O)6]3+
B. [Fe(H2O)6]2+
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Dr. Narinderjit Kaur

13. 13
Dr. Narinderjit Kaur

14. 3. Type of d-orbitals (transition metals)
-1
Δo = 22,870 cm (Co)
Δo = 34,100 cm-1
(Rh)
oΔ = 41,200 cm-1
(Ir)
In groups, heavier analogues have larger Δ.
For hexaammine complexes[MIII(NH3)6]3+
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Dr. Narinderjit Kaur

15. On going from the first to the second row of a transition
metal triad, there is approximately 30 to 50% increase in the
size of Δo, and another 30 to 50% increase on going from the
second row to the third row. This is due to the fact that as
you descend a transition metal triad the size of the d-orbitals
increases (3d < 4d < 5d). The larger d-orbitals interact more
with the orbitals on the ligands, hence Δo is larger. It should
also be noted that the pairing energies for the larger d-
orbitals are smaller, which means that low spin
configurations are favoured
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Dr. Narinderjit Kaur

16. 4. Geometry of Complex
CFS energy of tetrahedral complexes (Δt) is nearly half the
value for octahedral complexes (Δo)
𝟒
∆ 𝒕 =
𝟗
∆ 𝒐
Since magnitude of CFS in tetrahedral complex is quite small and
is always less than pairing energy.
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Dr. Narinderjit Kaur

17. Complex Oxidation
state of
metal
Geometry Δ in cm-1
[Co(NH3)4]2+ II Tetrahedral 59,00
[Co(NH3)6]2+ II Octahedral 10,200
VCl4 IV Tetrahedral 7,900
[VCl6]2- IV Octahedral 15,400
Thus, all tetrahedral complexes are high spin complexes.
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Dr. Narinderjit Kaur

18. Which of the following would have the smallest
value ofΔo?
A. [Os(OH2)6]3+
B. [Fe(OH2)6]3+
C. [Ru(OH2)6]3+
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Dr. Narinderjit Kaur

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Dr. Narinderjit Kaur