Bonding Of Co-ordination Compounds

Bonding in coordination compounds is explained by three models: Sidgwick's effective atomic number (EAN) rule, valence bond theory (VBT), which uses hybrid orbitals to predict shape and magnetism, and crystal field theory (CFT), which splits the d orbitals and explains high and low spin. This page draws the orbital boxes, shapes and splitting diagrams for every key complex and ends with the synergic bonding in metal carbonyls. Bonding in coordination compounds is tested every year in JEE Main, JEE Advanced and NEET.

1. Theories of Bonding: An Overview

Werner's theory (see Introduction and Nomenclature) explained which groups are bonded to the metal, but not why the bonds form, why they point in fixed directions, or why complexes are coloured and magnetic. Later theories answered these questions one by one:

More advanced models, ligand field theory and molecular orbital theory, combine the two and are beyond the NCERT syllabus.

2. Sidgwick's Effective Atomic Number (EAN) Rule

Here Z is the atomic number of the metal, x is its oxidation state (electrons lost) and CN is the coordination number (each ligand donates 2 electrons).

The rule works beautifully for metal carbonyls, where the metal is in the zero oxidation state. A metal with an even atomic number reaches 36 as a monomer: $[\mathrm{Cr}(\mathrm{CO}{)}_6]$ ($24+12$), $[\mathrm{Fe}(\mathrm{CO}{)}_5]$ ($26+10$) and $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ ($28+8$). A metal with an odd atomic number cannot reach 36 alone, so it forms a metal-metal bond that adds one electron to each metal: $[\mathrm{M}{\mathrm{n}}_2(\mathrm{CO}{)}_{10}]$ ($25+10+1=36$) and $[\mathrm{C}{\mathrm{o}}_2(\mathrm{CO}{)}_8]$ ($27+8+1=36$).

3. Valence Bond Theory (VBT)

3.1 Main Postulates

1. The central metal ion provides a number of empty orbitals equal to its coordination number, to accept the electron pairs donated by the ligands.
2. These empty orbitals (s, p and d) mix (hybridise) to give equivalent hybrid orbitals with definite directions. The type of hybridisation fixes the shape.
3. Each hybrid orbital overlaps with a filled orbital of a ligand to form a coordinate (dative) covalent bond.
4. The d orbitals used can be the inner (n-1)d orbitals, giving an inner orbital (low spin) complex, or the outer nd orbitals, giving an outer orbital (high spin) complex.
5. If there are not enough empty orbitals, unpaired metal electrons may be forced to pair up (against Hund's rule). Strong field ligands such as $\mathrm{C}{\mathrm{N}}^{-}$, $\mathrm{CO}$ and usually $\mathrm{N}{\mathrm{H}}_3$ do this; weak field ligands such as ${\mathrm{F}}^{-}$, $\mathrm{C}{\mathrm{l}}^{-}$ and ${\mathrm{H}}_2\mathrm{O}$ do not.
6. A complex with unpaired electrons is paramagnetic; with none it is diamagnetic. So magnetic measurements reveal the hybridisation.

3.2 Hybridisation and Geometry

Octahedral hybridisation uses only the two d orbitals that point along the axes, ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ and ${\mathrm{d}}_{{\mathrm{z}}^2}$, together with one s and three p orbitals. The other three d orbitals (${\mathrm{d}}_{\mathrm{xy}}$, ${\mathrm{d}}_{\mathrm{yz}}$, ${\mathrm{d}}_{\mathrm{xz}}$) point between the axes and take no part.

3.3 Octahedral Complexes: Inner and Outer Orbital

• Inner orbital (low spin) complex: the hybrid set uses the inner (n-1)d orbitals, ${\text{d}}^2{\text{sp}}^3$. Examples: $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$, $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$, $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$, $[\mathrm{Co}({\mathrm{C}}_2{\mathrm{O}}_4{)}_3{]}^{3-}$, $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$.
• Outer orbital (high spin) complex: the hybrid set uses the outer nd orbitals, ${\text{sp}}^3{\text{d}}^2$. Examples: $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$, $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$, $[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$, $[\mathrm{Ni}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{2+}$.

Figure 3 shows the classic pair. $\mathrm{C}{\mathrm{o}}^{3+}$ has six 3d electrons. Ammonia pairs them into three orbitals, leaving two inner 3d orbitals free for ${\text{d}}^2{\text{sp}}^3$ hybridisation. Fluoride cannot, so the outer 4d orbitals are used.

For iron, cyanide pairs up the electrons in both $\mathrm{F}{\mathrm{e}}^{2+}$ and $\mathrm{F}{\mathrm{e}}^{3+}$, while fluoride does not:

3.4 Four-Coordinate Complexes: Tetrahedral and Square Planar

With four ligands the metal uses either ${\text{sp}}^3$ (tetrahedral) or ${\text{dsp}}^2$ (square planar) hybrids. For a ${\mathrm{d}}^8$ ion such as $\mathrm{N}{\mathrm{i}}^{2+}$ the ligand decides:

• $[\mathrm{Ni}(\mathrm{CO}{)}_4]$: nickel is in the zero oxidation state ($3{\mathrm{d}}^8$ $4{\mathrm{s}}^2$). The strong field CO pushes the two 4s electrons into 3d, giving $3{\mathrm{d}}^{10}$. The empty 4s and 4p orbitals form ${\text{sp}}^3$ hybrids: tetrahedral, diamagnetic.
• $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$: weak field $\mathrm{C}{\mathrm{l}}^{-}$ leaves the two unpaired 3d electrons alone, so ${\text{sp}}^3$ is used: tetrahedral, paramagnetic (2 unpaired).
• $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$: strong field $\mathrm{C}{\mathrm{N}}^{-}$ pairs them, one 3d orbital becomes free and ${\text{dsp}}^2$ hybrids form: square planar, diamagnetic. The same holds for all ${\mathrm{d}}^8$ ions of the 4d and 5d series, such as $[\mathrm{Pt}(\mathrm{N}{\mathrm{H}}_3{)}_4{]}^{2+}$ and $[\mathrm{PdC}{\mathrm{l}}_4{]}^{2-}$, which are square planar even with weak ligands.
• $[\mathrm{Zn}(\mathrm{N}{\mathrm{H}}_3{)}_4{]}^{2+}$ ($3{\mathrm{d}}^{10}$) and $[\mathrm{FeC}{\mathrm{l}}_4{]}^{-}$ ($3{\mathrm{d}}^5$) have no empty 3d orbital to offer, so both are ${\text{sp}}^3$ tetrahedral; the first is diamagnetic and the second has 5 unpaired electrons.

3.5 Magnetic Criterion of Bond Type

The number of unpaired electrons (n) is found from the measured magnetic moment, using the spin-only formula $\mu =\sqrt{n(n+2)}$ BM (details in Colour, Magnetism and Stability). It tells which orbitals were used. For example, $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$ is diamagnetic ($\mu =0$), so its six 3d electrons are paired and the complex must be inner orbital ${\text{d}}^2{\text{sp}}^3$. $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$ has $\mu \approx 5.92$ BM, so it has 5 unpaired electrons and must be outer orbital ${\text{sp}}^3{\text{d}}^2$. This use of magnetic data to decide the bond type is called the magnetic criterion of bond type.

3.6 Limitations of Valence Bond Theory

• It explains a shape once it is known, but cannot predict it: it cannot say in advance whether a four-coordinate complex will be tetrahedral or square planar.
• It is qualitative: it gives no quantitative explanation of magnetic data or of thermodynamic and kinetic stability, or the relative stability of isomers.
• It cannot explain the colour and absorption spectra of complexes.
• It does not distinguish between weak and strong field ligands; it simply assumes which ones cause pairing.

4. Crystal Field Theory (CFT)

4.1 Basic Assumptions

1. The metal-ligand bond is treated as purely electrostatic: no covalent bond is formed.
2. Anionic ligands are treated as negative point charges, and neutral ligands as point dipoles with their negative end towards the metal.
3. The positive metal ion attracts the ligands, but the electrons in the metal d orbitals are repelled by the ligand electrons.
4. In a free ion the five d orbitals have the same energy (they are degenerate). The ligands repel some d orbitals more than others, so the degeneracy is lifted. This is crystal field splitting, and it explains colour and magnetic properties.

4.2 Why the d Orbitals Split

In an octahedral complex the six ligands approach along the x, y and z axes. The lobes of ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ and ${\mathrm{d}}_{{\mathrm{z}}^2}$ point straight at them, while the lobes of ${\mathrm{d}}_{\mathrm{xy}}$, ${\mathrm{d}}_{\mathrm{xz}}$ and ${\mathrm{d}}_{\mathrm{yz}}$ point between them:

4.3 Octahedral Crystal Field

The five d orbitals split into two sets: the higher ${\mathrm{e}}_{\mathrm{g}}$ set (${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$, ${\mathrm{d}}_{{\mathrm{z}}^2}$) and the lower ${\mathrm{t}}_{2g}$ set (${\mathrm{d}}_{\mathrm{xy}}$, ${\mathrm{d}}_{\mathrm{xz}}$, ${\mathrm{d}}_{\mathrm{yz}}$). The energy gap between them is the crystal field splitting energy, ${\Delta }_o$ (o for octahedral). The average energy (barycentre) does not change, so the three ${\mathrm{t}}_{2g}$ orbitals drop by $0.4{\Delta }_o$ and the two ${\mathrm{e}}_{\mathrm{g}}$ orbitals rise by $0.6{\Delta }_o$: $3(-0.4)+2(+0.6)=0$.

An electron can be promoted from ${\mathrm{t}}_{2g}$ to ${\mathrm{e}}_{\mathrm{g}}$ by absorbing light whose energy matches the gap, $E=h\nu =\frac{hc}{\lambda }={\Delta }_o$. This d-d transition is the origin of the colour of most complexes (see Colour, Magnetism and Stability).

4.4 Tetrahedral Crystal Field

In a tetrahedral complex the four ligands sit at alternate corners of a cube around the metal, between the axes. Now the ${\mathrm{t}}_2$ orbitals (${\mathrm{d}}_{\mathrm{xy}}$, ${\mathrm{d}}_{\mathrm{xz}}$, ${\mathrm{d}}_{\mathrm{yz}}$) point closer to the ligands than the $\mathrm{e}$ orbitals do, so the splitting is reversed: $\mathrm{e}$ is lower and ${\mathrm{t}}_2$ is higher (no "g" label, because a tetrahedron has no centre of symmetry). With only four ligands, none pointing straight at an orbital, the splitting is much smaller:

Because ${\Delta }_t$ is small, it is almost always less than the pairing energy, so tetrahedral complexes are practically always high spin.

4.5 Factors Affecting the Size of $\Delta$

(a) Nature of the ligand. Arranging ligands in order of increasing field strength (increasing $\Delta$) gives the spectrochemical series (NCERT order):

Ligands at the left are weak field (small $\Delta$, high spin); those at the right are strong field (large $\Delta$, low spin). Note that the order cannot be explained by charge alone: anions such as ${\mathrm{I}}^{-}$ are the weakest, $\mathrm{O}{\mathrm{H}}^{-}$ is weaker than neutral ${\mathrm{H}}_2\mathrm{O}$, and neutral CO is the strongest.