Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 350 off

Bonding in Co-ordination Compounds

  • Transition metals have greater positive charge which attract the negatively charged ligands to form stable complexes.

  • Transition metals also have vacant-d-orbitals which can accommodate the lone pair of electrons donated by the ligands in co-ordinate bond formation.

Effective Atomic Number

Co-ordinate bonds are formed between ligands and the central metal ion in a complex; that is, a ligand donates an electron pair to the metal ion. Co-ordination compounds are formed very readily by the transition metals since they have vacant-d-orbitals which can accommodate donated electron pairs.

The number of co-ordinate bonds which can be formed largely depends on the number aof vacant orbitals of suitable energy. In many cases ligands are added until the central metal in the complex possesses (or) shares the same number of electrons as the next inert gas. The total number of the electrons on the central metal in the complex including those gained by bonding, is called effective atomic number (EAN).

Thus by forming complexes many metals obtain an EAN of the next inert gas. However a significant number of exceptions are known where EAN is one (or) two units more than the corresponding inert gas.

For example,

Fe ⇒ atomic number is 26

and forming a complex, [Fe(CN)6]4–

the number of electrons lost = 2

the number of electrons gained = 12 ⇒ EAN = 36

The tendency to attain an inert gas configuration is a significant factor but not a necessary condition for complex formation, because it is also necessary to produce a symmetrical structure irrespective of the number of electrons involved.

The d-orbitals

Since d-orbitals are frequently used in co-ordination complexes it is important to study their shapes and distribution in space. There is no unique way of representing the five d-orbitals, but the most eminent representations are shown below

In fact there are six wave functions that can be written for orbitals having the typical four-lobed form. In as much as there can be only five d-orbitals having any physical reality, one of them  (dz2) is conventionally regarded as a linear combination of two others, the (dz2-y2) and (dz2-x2). Thus these latter two orbitals have no independent existence, but the dz2 can be thought of as having the average properties of the two.

Therefore since both have high electron density along the z axis, the dz2 orbital has a large fraction of its electron density concentrated along the same axis. Also, since one of the component wave functions (dz2-y2)  has lobes along the y-axis and the other (dz2-x2) along the x-axis, the resultant dz2 orbital has a torus of electron density in the xy plane. The xy component which is often referred to as a ‘doughnut’ (or) a ‘collar’, is frequently neglected in pictorial representations, especially when an attempt is being made to portray all five d-orbitals simultaneously.

The five d-orbitals in an isolated gaseous metal ion are degenerate. If a spherically symmetric field of negative charges is placed around the metal, the orbitals will remain degenerate, but all of them will be raised in energy as a result of repulsion between the negative field and the negative electrons in the orbitals. If the field results from the influence of real ligands the symmetry of the field will be less than spherical and the degeneracy of the d-orbitals will be removed.

Formation of an Octahedral Complexes

Let us consider the case of six ligands forming an octahedral complex. For convenience, we may regard the ligands as being symmetrically positioned along the axes of a Cartesian co-ordinate system with the metal ion at the origin. To simplify the situation, we can consider an octahedral complex as a cube, having the metal ion at the centre of the body and the 6 ligands at the face centres.

and if we take the metal ion as the origin of a Cartesian co-ordinate, the ligands will be along the axes. As in the case of a spherical field, all of the d-orbitals will be raised in energy relative to the free ion because of negative charge repulsions. However it should be pictorially obvious that not all of the orbitals will be affected to the same extent. The orbitals lying along the axes (dz2 and dz2-y2) will be more strongly repelled than the orbitals with lobes directed  between the axes (dxy, dxz, dyz). The d-orbitals are thus split into two sets with the dz2 and dz2-y2  at a higher energy than the other three.

Factors Affecting the Magnitude of \Delta _{0}

Oxidation state of the metal ion: The magnitude of D0 increases with increasing ionic charge on the central metal ion. As the ionic charge on the metal ion increases greater is the attraction for the ligands, greater the repulsion and hence greater the magnitude of D0.

Nature of the ligands: Based on experimental observation for a wide variety of complexes, it is possible to list ligands in order of increasing field strength in a spectrochemical series. Although it is not possible to form a complete series of all ligands with a single metal ion, it is possible to construct one from overlapping sequences, each constituting a portion of the series:

I < Br < S2– < SCN < Cl < N3–, F < urea, OH < ox, O2–, H2O < NCS < py, NH3 < en < bpy, phen < NO2 < C6H5 < CN < CO.

The spectrochemical series and other trends described allow one to rationalise differences in spectra and permit some predictabiltiy. If the splitting of the d-orbitals resulted simply from the effect of point charges (ions (or) dipoles), one should expect that anionic ligands would exert the greatest effect. To the contrary most anionic ligands lie at the low end of the spectrochemical series. Further more, OH– lies below the neutral H2O molecule and NH3 produces a greater splitting than H2O. Despite its imperfections, the basic theory can be used to interpret a number of effects in co-ordination chemistry.

Outer Orbital and Inner Orbital Complexes

Consider the complexes [CoF6]3– and [Co(NH3)6]3+

The electronic configuration of Co3+ ion is

In a weak ligand field such as [CoF6]3–, the approach of the ligand causes only a small split in the energy level.

Since the ligand is a weak field ligand, its repulsions with the electrons in dz2 and dx2-y2  orbitals are very less (or) in other words we can say that the electrons in dz2 and dx2-y2  cannot move away from the approaching ligands since they have insufficient energy to pair up with the electrons in dxy, dyz and dxz orbitals.

Thus there are no vacant orbitals in the 3d shell and the ligands occupy the first six vacant orbitals (one 4s, three 4p and two 4d). Since outer d orbitals are used, this is an outer orbit complex. The orbitals are hybridised and are written sp3d2 to denote this. Since none of the electrons has been forced to pair off, this is a high spin complex and will be strongly paramagnetic because it contains 4 unpaired 3d electrons.

Under the influence of a strong ligand field as in the complex [Co(NH3)6]3+, the approach of the ligand causes a greater split in the energy level.

Since, the split is very high, we can say that the energy difference between the two sets of orbitals is much greater and this energy difference is sufficient to allow the electrons in dz2 and dx2-y2  orbitals to move into the half filled dxz, dxy and dyz orbitals, even though this pairing requires energy. We can also view this like, the ligand repel the electrons in higher energy level to an extent such that they get paired up against Hund's rule

So, The dz2 and dx2-y2  orbitals become vacant. The six ligands each donate a lone pair to the first six vacant orbitals, which are: two 3d, one 4s and three 4p. Inner d-orbitals are used and so this is an inner orbital complex. The orbital are hybridised and written d2sp3 to denote the use of inner orbitals.

Since, the orginal unpaired electrons have been forced to pair off, theis is a low spin complex and is in fact diamagnetic.

The inner and outer orbital complexes may be distinguished by magnetic measurements. Since the outer orbital complexes use high energy levels, they tend to be more reactive. The inner orbital are sometimes called inert orbitals.

Formation of a Square Planar Complex

If the central metal ion has eight d-electrons, these will be arranged as







In a weak octahedral ligand field, a regular octahedral complex is thus formed by using outer d-orbitals.

However, under the influence of a strong ligand field, the electrons in the dz2 and dx2- y2  orbitals may pair up, leaving one vacant d-orbital, which can accept a lone pair from a ligand.

For example consider [Ni(CN)4]2–

The electronic configuration of Ni2+ ion is

Consider, a Ni2+ ion with one electron in the dx2- y2 orbital and one in the dz2 orbital. The approach of ligands along x, y and z axes will result in the energy of these orbitals increasing. Because the dx2-y2 orbital is attacked by four ligands and the dx2 by only two, the energy of dx2-y2 orbital will increase most. If the ligands have enough strong field, the electrons will be forced out of the dx2-y2 orbital into the dx2. Thus four ligands can approach along x, –x, y and –y axes. A ligand approaching in the z (or) – z direction attempting to donate a lone pair meets the very strong repulsive forces from a completely filled dz2 orbital. Thus only four ligands succeed in bonding to the metal.

Refer to the following video for octahedral complexes

Formation of Tetrahedral Complexes

The directions x,y and z point to the centre of the faces. The  dz2 and dx2-y2 orbitals point along x,y and z axis and dxy, dyz and dxz orbitals point in between x,y and z.

The directions of approach of the ligands does not coincide exactly with either the dz2 and dx2-y2 orbitals (or) the dxy, dyz and dxz orbitals. The approach of ligands raises the energy of both sets of orbitals, but since the dxy,dyz and dxz orbitals correspond more closely to the position of the ligands, their energy increases most and the dz2 and dx2-y2 orbitals are filled first. This is opposite to what happens in octahedral complexes.

Consider, the complex ion, [FeCl4]. The electronic configuration  of Fe3+ ion is

Since Cl ion is a weak field ligand it is unable to pair the unpaired electrons and hence, the Cl ion uses 4s and 4p orbitals to form a tetrahedral complex of sp3 hybridisation.


Some of the factors which favour complex formation have already been mentioned. Small highly charged ions with suitable vacant orbitals of right energy, the satisfaction of the effective atomic number rule and the attainment of a symmetrical shape, all favour complex formation.

In some complexes a group occupies more than one co-ordination position, that is more than one atom in the group is bonded to the central metal.

For example ethylene diamine (en) forms a complex with copper ions

In this complex the copper is surrounded by four –NH2 groups, each nitrogen atom donating a lone pair of electrons and forming a co-ordinate bond. Thus each ethylene diamine molecule is bonded to the copper in two places, hence it is called a bidentate group (or) ligand. A ring structure is thus formed and such ring structures are called chelates. Normally chelate complexes are more stable than similar non-chelated complexes.
The more rings that are formed, the more stable the complex is chelating agent with three –, four – and six – donor atoms are known and are termed tri –, tetra – and hexa– dentate ligands.

Chelate compounds are even more stable when they contain a system of alternate double and single bonds. The p electron density is delocalised and spread over the ring, which is said to be stabilized by resonance

Solved Problems

Question 1:
Why is hydrochloric acid not used to acidify a permangnate solution is volumetric estimation of Fe2+ and C2O42–.

This is because a part of the oxygen produced from KMnO4 + HCl is used is oxidizing HCl to Cl2

4HCl + 2(O) →  2H2O + Cl2


Question 2:
Why is copper sulphate pentahydrate coloured?

In the presence of H­2O as ligand d-orbitals of Cu(II) ions split into two sets, one with lower energy and the other with higher energy. From the white light falling on it, red wavelength is absorbed or excitation of electron from lower to higher energy level. The complimentary colour, viz blue is reflected.


Question 3:
Most of the transition metals do not displace hydrogen from dilute acids. Why?

This is because most of the transition metals have negative oxidation potentials.


Question 4:
K2 [Pt Cl6] is well known compound whereas corresponding Ni compound is not known. State a reason for it.

This is because Pt4+ is more stable than Ni4+ as the sum of four ionization energies of Pt is less than that of Ni.


Question 5:
Why have the transition elements high enthalpy of hydration?

This is due to their small size and large nuclear charge. This is so because when we move along any transition series the nuclear charge increases and size decreases.


Question 6:
Though copper, silver and gold have completely filled sets of a d-orbitals yet they are considered as transition metals. Why?

These metals in their common oxidation states have incompletely filled d-orbitals e.g. Cu2+ has 3d9 and Au3+ has 5d8 configuration

Question 1: The transition elements are so named because

(a)  they have partly filled d-orbitals    

(b)  their properties are similar to other elements

(c) their properties are different from other elements

(d)  they lie between the s- and p-blocks

Question 2: CrO3 dissolves in aqueous NaOH to give

(a)  Cr2O72–

(b)  CrO42–

(c) Cr(OH)3

(d)  Cr(OH)2

Question 3: The complex Cr2[CoCl4] has

(a)  square planar structure with all electrons paired

(b)  tetrahedral structure with three unpaired electrons

(c) square planar structure with one unpaired electron

(d)  tetrahedral structure with oine unnpaired electron

Question 4: The splitting of d-orbitals in tetrahedral complexes is

(a)  not comparable to octahedral complexes since the d-orbitals in tetrahedral complexes does not split.

(b)  reverse of the octahedral complexes

(c) same as in octahedral complexes

(d)  none of these









Related Resources

To read more, Buy study materials of Coordination Compounds comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 3,750 off