NEET 2025 Exam Date Announcement Delayed by NTA: Know Possible Reasons

Crystal Field Theory (CFT) MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:24 PM | #NEET

Quick Facts

  • Applications of CFT is considered one the most difficult concept.

  • Crystal Field Splitting in Octahedral Field is considered one of the most asked concept.

  • 75 Questions around this concept.

Solve by difficulty

Crystal field stabilization energy for high spin d4 octahedral complex is :

How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca^{2+} ion ?

 Nickel (Z=28)   combines with a uni-negative monodentate ligand to form a diamagnetic complex [NiL4]2-. The hybridization involved and the number of unpaired electrons present in the complex are respectively :            

 

NEET 2024: Cutoff (OBC, SC, ST & General Category)

NEET 2024 Admission Guidance: Personalised | Study Abroad

NEET 2025: SyllabusMost Scoring concepts NEET PYQ's (2015-24)

NEET PYQ's & Solutions: Physics | ChemistryBiology

Of the following complex ions, which is diamagnetic in nature?

Which one of the following complexes is an outer orbital complex?

\left [ Atomic\; nos.:Mn=25,Fe=26,Co=27,Ni=28 \right ]

 The pair having the same magnetic moment is :
[At. No. : Cr=24, Mn=25, Fe=26, Co=27]

 

Which of the following facts about the complex \left [ Cr(NH_{3})_{6} \right ]Cl_{3}  is wrong ?

Pearson | PTE

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally

Consider the given complexes

(i) $K_3\left[\mathrm{Cr}(\mathrm{CN})_6\right]$
(ii) $K_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
(iii) $K_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
(iv) $K_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$

Out of the following, select the set containing only paramagnetic complexes

Concepts Covered - 6

Main Postulates of Crystal Field Theory

The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field. 

Crystal Field Splitting in Octahedral Field

In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands. Such a repulsion is more when the metal d orbital is directed towards the ligand than when it is away from the ligand. Thus, the dx2-y2 and dzd orbitals which point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy; and the dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field. Thus, the degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy, t2g set and two orbitals of higher energy, eg set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by ∆o. Thus, the energy of the two eg orbitals will increase by (3/5) ∆o and that of the three t2g will decrease by (2/5)∆o.

The crystal field splitting, ∆o, depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields in which case, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below:

I < Br < SCN < Cl < S2– < F < OH < C2O42– < H2O < NCS < edta4– < NH3 < en < CN < CO

Such a series is termed as spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands. Let us assign electrons in the d orbitals of metal ion in octahedral coordination entities. Obviously, the single d electron occupies one of the lower energy t2g orbitals. In d2 and d3 coordination entities, the d electrons occupy the t2g orbitals singly in accordance with the Hund’s rule. For d4 ions, two possible patterns of electron distribution arise: (i) the fourth electron could either enter the t2g level and pair with an existing electron, or (ii) it could avoid paying the price of the pairing energy by occupying the eg level. Which of these possibilities occurs, depends on the relative magnitude of the crystal field splitting, ∆o and the pairing energy, P (P represents the energy required for electron pairing in a single orbital). The two options are:

  • If ∆o < P, the fourth electron enters one of the eg orbitals giving the configuration t32ge1g. Ligands for which ∆o < P are known as weak field ligands and form high spin complexes.
  • If ∆o > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g4eg0. Ligands which produce this effect are known as strong field ligands and form low spin complexes.

Calculations show that d4 to d7 coordination entities are more stable for strong field as compared to weak field cases.

Crystal Field Splitting in Tetrahedral Field

In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting. For the same metal, the same ligands and metal-ligand distances, it can be shown that ∆t = (4/9) ∆0. Consequently, the orbital splitting energies are not sufficiently large for forcing pairing and, therefore, low spin configurations are rarely observed. The ‘g’ subscript is used for the octahedral and square planar complexes which have centre of symmetry. Since tetrahedral complexes lack symmetry, ‘g’ subscript is not used with energy levels.

Factors Affecting CFSE
  1. Nature of central metal atom: As we move down the group, the CFSE increases. From 3d to 4d, there is a 30% increase in CFSE and from 4d to 5d, there is a 50% increase in CFSE.
  2. Oxidation state of central metal atom: Oxidation state is directly proportional to CFSE.
    The CFSE of [Fe(CN)6]3- is greater than [Fe(CN)6]4-.
  3. Nature of ligand: In case of strong field ligand, magnitude of CFSE is high. In case of weak field ligand, magnitude of CFSE is low.
  4. Nature of complex: In octahedral complex, 6 ligands approach the central metal atom and thus repulsion is higher, due to which the CFSE is higher. But, in tetrahedral complex, 4 ligands approach the central metal atom and thus repulsion is lower and thus the CFSE is lower.
Applications of CFT

These are the various applications of crystal field theory.

  • Configuration of metal ion: In weak field ligand, the crystal field splitting energy difference is very less. Thus, pairing of electrons is done by Hund's rule. But for strong field ligand, the splitting energy difference is high due to which the pairing of electrons is not done by Hund's rule. 
  • Magnetic nature of complex: On the basis of magnetism, all the complexes can be divided into two categories, paramagnetism and diamagnetism. Paramagnetic complexes are weakly attracted by magnetic field and have unpaired electron, while diamagnetic complexes are weakly repelled by magnetic field and these molecules have no unpaired electron.
  • Colour in coordination compounds: The colour of the complex is complementary to that which is absorbed. The complementary colour is the colour generated from the wavelength left over; if green light is absorbed by the complex, it appears red. The colour in the coordination compounds can be readily explained in terms of the crystal field theory. Consider, for example, the complex [Ti(H2O)6]3+, which is violet in colour. This is an octahedral complex where the single electron (Ti3+ is a 3d1 system) in the metal d orbital is in the t2g level in the ground state of the complex. The next higher state available for the electron is the empty eg level. If light corresponding to the energy of blue-green region is absorbed by the complex, it would excite the electron from t2g level to the eg level (t2g1eg0 → t2g0eg1). Consequently, the complex appears violet in colour. The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.
     
Limitations of CFT

The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds to a large extent. However, from the assumptions that the ligands are point charges, it follows that anionic ligands should exert the greatest splitting effect. The anionic ligands actually are found at the low end of the spectrochemical series. Further, it does not take into account the covalent character of bonding between the ligand and the central atom. These are some of the weaknesses of CFT, which are explained by ligand field theory (LFT) and molecular orbital theory which are beyond the scope of the present study.

Study it with Videos

Main Postulates of Crystal Field Theory
Crystal Field Splitting in Octahedral Field
Crystal Field Splitting in Tetrahedral Field

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Main Postulates of Crystal Field Theory

Chemistry Part I Textbook for Class XII

Page No. : 257

Line : 19

Crystal Field Splitting in Octahedral Field

Chemistry Part I Textbook for Class XII

Page No. : 257

Line : 32

E-books & Sample Papers

Get Answer to all your questions

Back to top