Valence Bond Theory (VBT) for Coordination Compounds

What VBT Says

According to Valence Bond Theory, the central metal atom or ion, under the influence of the ligands, makes available a set of empty orbitals and mixes them into an equivalent set of hybrid orbitals of a definite geometry. Each ligand donates a lone pair of electrons that overlaps with one empty hybrid orbital to form a coordinate (dative) $\sigma$ bond. The metal therefore behaves as a Lewis acid and the ligands as Lewis bases.

The three operating principles, as set out in the sources, are: hybridisation of the valence orbitals of the central metal ion; coordinate bond formation by ligand-to-metal lone-pair donation; and a definite relationship between the type of hybridisation adopted and the observed magnetic behaviour. The orbitals available for mixing are the inner $(n-1)d$, the $ns$ and the $np$ set, or alternatively the $ns$, $np$ and outer $nd$ set. Which set is used decides whether the complex is "inner orbital" or "outer orbital".

The power of VBT for NEET lies in this last principle. Because magnetic moment is dictated by the number of unpaired electrons, and the number of unpaired electrons is fixed by the choice of hybridisation, predicting one quantity predicts all three: hybridisation, geometry and magnetism are a single linked chain.

Historically, the theory answered three questions Werner's coordination theory could not. Werner could describe that a metal bound six groups in an octahedral arrangement, but he could not say why the bonds had directional character, why only certain elements form complexes readily, or why those complexes carry characteristic magnetic and optical signatures. By proposing that the metal supplies a definite set of hybrid orbitals into which ligand lone pairs are donated, VBT supplies the directional bonding picture and links it directly to the experimentally measurable magnetic moment. It treats the metal-to-ligand interaction as essentially covalent, in contrast to the electrostatic picture used by Crystal Field Theory.

Hybridisation and Geometry

Each coordination number is served by a characteristic hybridisation, and each hybridisation projects its hybrid orbitals into a fixed spatial arrangement. NCERT Table 5.2 lays out the mapping that you must recall on sight in an exam.

The two octahedral entries are the heart of the topic. Both give six coordinate bonds and an octahedral shape, but $d^2sp^3$ recruits the inner $(n-1)d$ orbitals while $sp^3d^2$ recruits the outer $nd$ orbitals. Likewise, coordination number 4 splits into two competing geometries — square planar via $dsp^2$ and tetrahedral via $sp^3$ — and VBT decides between them using the metal's electron count and the ligand's field strength.

Inner vs Outer Orbital Complexes

Whether a metal ion offers up its inner $d$ orbitals depends on whether the ligand is strong enough to force the metal's $d$ electrons to pair up. A strong field ligand pairs electrons, vacating inner $d$ orbitals for $d^2sp^3$ hybridisation; the result is an inner orbital, low spin, spin-paired complex. A weak field ligand cannot pair the electrons, so the metal must reach out to its higher-energy outer $nd$ orbitals for $sp^3d^2$ hybridisation; the result is an outer orbital, high spin, spin-free complex.

The dependence on ligand strength comes from the spectrochemical series, the experimental ordering of ligands by field strength. A useful working sequence runs $\ce{I^- < Br^- < SCN^- < Cl^- < F^- < OH^- < C2O4^{2-} < H2O < NH3 < en < CN^- < CO}$. Ligands to the right of water tend to drive inner-orbital, low-spin behaviour; ligands to the left tend to leave the complex high spin. NCERT itself notes a complication worth remembering: for $d^4$ and $d^5$ ions the outcome can swing either way, which is exactly why $\ce{[Fe(CN)6]^{3-}}$ (one unpaired electron) and $\ce{[FeF6]^{3-}}$ (five unpaired electrons) behave so differently despite both being $\ce{Fe^{3+}}$, $d^5$.

A word of caution about the reasoning direction is in order. VBT does not predict which ligand is strong and which is weak; it borrows that knowledge from experiment and then uses it to decide whether pairing happens. This is why, for ions with up to three $d$ electrons such as $\ce{Ti^{3+}}$ ($d^1$), $\ce{V^{3+}}$ ($d^2$) and $\ce{Cr^{3+}}$ ($d^3$), the magnetic behaviour of the free ion and of its octahedral complexes is essentially the same — two empty inner $d$ orbitals are always available for hybridisation, so no electron has to be moved. Genuine complications only begin once the metal carries four or more $d$ electrons, where a vacant pair of inner $d$ orbitals can be created only by pairing existing electrons. The terms "inner orbital", "low spin" and "spin paired" are therefore synonyms in this context, as are "outer orbital", "high spin" and "spin free".

Coordination Number 6: Octahedral

For six-coordinate complexes the VBT recipe is mechanical. Write the $d^n$ configuration of the metal ion; decide if the ligand is strong or weak; if strong, pair electrons to liberate two inner $(n-1)d$ orbitals and hybridise as $d^2sp^3$; if weak, leave the electrons alone and reach to the outer $nd$ orbitals for $sp^3d^2$. The six ligand lone pairs then fill the six hybrid orbitals.

The canonical contrast is the pair of $\ce{Co^{3+}}$ ($d^6$) complexes. In $\ce{[Co(NH3)6]^{3+}}$ the strong field ammine ligands compress the six $3d$ electrons into three orbitals, leaving two empty inner $3d$ orbitals for $d^2sp^3$; with no unpaired electrons it is diamagnetic — an inner orbital, low spin complex. In $\ce{[CoF6]^{3-}}$ the weak field fluoride ligands cannot pair the electrons, so the ion uses $sp^3d^2$ with its outer $4d$ orbitals; four unpaired electrons make it paramagnetic — an outer orbital, high spin complex.

The same machinery resolves the iron cyanide pair. $\ce{[Fe(CN)6]^{4-}}$ has $\ce{Fe^{2+}}$ ($d^6$); strong field $\ce{CN^-}$ pairs all electrons, giving $d^2sp^3$ with zero unpaired electrons, so it is diamagnetic. $\ce{[Fe(CN)6]^{3-}}$ has $\ce{Fe^{3+}}$ ($d^5$); $\ce{CN^-}$ pairs electrons down to a single unpaired electron, still $d^2sp^3$, so it is weakly paramagnetic. Both are inner orbital complexes; the difference of one electron between $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ is the difference between diamagnetic and one-unpaired-electron.

Coordination Number 4: Square Planar vs Tetrahedral

Four-coordinate complexes branch into two geometries. In square planar complexes the metal uses $dsp^2$ hybridisation, borrowing one inner $(n-1)d$ orbital; this requires the field to pair electrons and free that $d$ orbital. In tetrahedral complexes the metal uses $sp^3$ hybridisation with only $ns$ and $np$ orbitals, requiring no inner $d$ orbital and therefore no pairing.

The textbook pair is again a single metal in one oxidation state. $\ce{[Ni(CN)4]^{2-}}$ has $\ce{Ni^{2+}}$ ($d^8$); the strong field $\ce{CN^-}$ pairs the two unpaired $3d$ electrons, vacating one $3d$ orbital for $dsp^2$ — the complex is square planar and diamagnetic. $\ce{[NiCl4]^{2-}}$ also has $\ce{Ni^{2+}}$ ($d^8$), but weak field $\ce{Cl^-}$ cannot pair the electrons, so the ion uses $sp^3$ — the complex is tetrahedral and paramagnetic with two unpaired electrons.

A third instructive case is $\ce{Ni(CO)4}$. Here nickel is in the zero oxidation state with a $3d^8 4s^2$ ground configuration; the strong field $\ce{CO}$ ligand shifts the $4s$ electrons into the $3d$ shell and pairs all the $d$ electrons, leaving $sp^3$ hybridisation with no unpaired electrons. The complex is therefore tetrahedral but diamagnetic — geometrically identical to $\ce{[NiCl4]^{2-}}$ yet magnetically opposite, purely because of the difference in oxidation state and ligand field strength.

Predicting Magnetic Moment

Once VBT fixes the number of unpaired electrons $n$, the spin-only magnetic moment follows from a single formula:

$$\mu = \sqrt{n(n+2)} \;\text{ BM (Bohr magneton)}$$

A complex with $n=0$ is diamagnetic; any $n \geq 1$ is paramagnetic, with the moment growing as the unpaired count rises. The workflow is invariant: (1) find the oxidation state and $d^n$ count, (2) read the ligand's field strength, (3) place electrons, (4) count $n$, (5) substitute. The table below records the standard values that recur in NEET matching questions.

VBT can also be run in reverse: an experimentally measured magnetic moment pins down the geometry. NCERT Example 5.7 illustrates this for $\ce{[MnBr4]^{2-}}$, whose moment of 5.9 BM corresponds to five unpaired electrons. Since a square planar ($dsp^2$) arrangement would force pairing, the only structure consistent with five unpaired electrons is tetrahedral $sp^3$.

Worked Examples

Limitations of VBT

VBT explains the formation, geometry and broad magnetic behaviour of many complexes, but NCERT §5.5.3 lists six shortcomings that are themselves examinable. These gaps are precisely what motivate the next theory in the syllabus.

Points 3, 5 and 6 are the most heavily tested. VBT borrows the strong-versus-weak field distinction from experiment (the spectrochemical series) rather than deriving it; it offers no language for the colour of complexes, since it has no concept of $d$–$d$ electronic transitions; and for four-coordinate species it cannot predict a priori whether the geometry will be square planar or tetrahedral. Crystal Field Theory was developed to address exactly these failings, and the still more refined Ligand Field Theory and Molecular Orbital Theory go further, though both lie beyond the NEET syllabus.

For exam purposes, the practical takeaway is that VBT remains the fastest tool for the specific job it does well: given a named complex with a known geometry or magnetic moment, deduce the hybridisation and the number of unpaired electrons. When a question instead asks you to rank wavelengths of absorbed light, explain why a complex is coloured, or compare the relative magnitudes of splitting energy, the question has moved out of VBT's territory and into Crystal Field Theory. Recognising which theory a question belongs to is itself a frequently rewarded skill in the coordination compounds unit.