Valence bond theory

$[Ni(NH_3)_6]^{2+}$ is paramagnetic; $[Ni(CN)_4]^{2-}$ is diamagnetic — yet both start from the very same $Ni^{2+}$ ion. Why does the ligand change the magnetism so drastically? Valence Bond Theory (VBT) answers this by treating each ligand as donating an electron pair into an empty hybrid orbital on the metal, and it ties the type of hybridisation used directly to the complex's geometry and to how many electrons remain unpaired. This article expands the Pass-1 treatment of VBT for octahedral, tetrahedral and square planar coordination complexes — the bonding theory prescribed alongside Crystal Field Theory in the ISC Class 12 Coordination Compounds unit — with a fuller derivation, a second worked example, and exam-ready traps and quiz checks. The scope stays exactly where the syllabus draws the line: strong-field versus weak-field ligands, inner- versus outer-orbital hybridisation, and counting unpaired electrons to classify a complex as paramagnetic or diamagnetic.

The core insight

VBT explains a complex's geometry and magnetism from a single choice — which empty metal orbitals hybridise to accept the ligand lone pairs, a choice controlled entirely by whether the ligand is strong-field (forces pairing, inner-orbital) or weak-field (does not force pairing, outer-orbital).

The core idea

Every ligand in a coordination complex is a Lewis base: it offers a lone pair, and the central metal ion offers an empty hybrid orbital to receive it, forming a coordinate covalent (dative) bond. VBT's whole predictive power comes from matching the coordination number and the ligand's field strength to one specific hybridisation scheme.

Step 1 — fix the metal ion's configuration. Work out the metal's oxidation state from the complex's overall charge and the ligands' known charges, then write the resulting $d^n$ configuration (removing $s$ electrons before $d$ electrons, since the outer $ns$ shell is lost first on ionisation).

Step 2 — classify the ligand. Strong-field ligands such as $CN^-$ , $CO$ , and $NO_2^-$ force the metal's $d$ electrons to pair up, freeing an inner $(n-1)d$ orbital. Weak-field ligands such as $F^-$ and $Cl^-$ , and moderate ones like $H_2O$ and $NH_3$ , generally leave the free-ion arrangement of electrons undisturbed, so the metal must reach for its outer $nd$ orbital instead.

Step 3 — assign the hybridisation and geometry. The coordination number and the availability (or not) of an inner $d$ orbital together fix which hybrid set is used:

Coordination number	Hybridisation	Geometry
4	$sp^3$	Tetrahedral
4	$dsp^2$	Square planar
6	$sp^3d^2$	Octahedral, outer-orbital (high spin)
6	$d^2sp^3$	Octahedral, inner-orbital (low spin)

Step 4 — count unpaired electrons and read off magnetism. Any unpaired electron makes the complex paramagnetic; zero unpaired electrons makes it diamagnetic. Inner-orbital (low spin) complexes pack electrons into fewer orbitals and so typically end up with fewer unpaired electrons than outer-orbital (high spin) complexes of the same metal ion. The number of unpaired electrons $n$ connects to a measurable quantity, the spin-only magnetic moment:

\mu = \sqrt{n(n+2)}\ \text{BM}

where: $\mu$ is the magnetic moment in Bohr magnetons (BM), and $n$ is the number of unpaired electrons in the metal ion's $d$ orbitals after hybridisation. This formula is the bridge between the electron-counting VBT does and an experimentally measured number — a sample with a larger $\mu$ has more unpaired spins.

Figure

Two paths from the same free nickel two plus ion to a square planar diamagnetic complex with cyanide and a tetrahedral paramagnetic complex with chloride

Derivation APredicting hybridisation, geometry and magnetism from oxidation state and ligand strength

The goal is to show, starting only from a complex's formula and a short list of known strong-field and weak-field ligands, how a fixed four-step procedure — find the oxidation state, write the $d^n$ configuration, classify the ligand, then arrange and count electrons — determines hybridisation, geometry and magnetic character without any ambiguity, and to make explicit why the same metal ion can land on opposite outcomes for two different ligands.

Step 1. Determine the oxidation state of the metal. Let the metal's oxidation state be $x$ . Since the complex's overall charge equals the sum of the metal's oxidation state and the ligands' charges, solve $x + (\text{sum of ligand charges}) = (\text{overall charge})$ . This is pure bookkeeping, but it must be done first: every later step depends on having the correct $d^n$ configuration, and that depends on $x$ .

Step 2. Write the free-ion $d^n$ configuration. Take the metal's neutral-atom configuration, remove the $ns$ electrons first (since the outer $s$ subshell is always lost before the $(n-1)d$ subshell on ionisation), then remove any further electrons needed to reach oxidation state $x$ from the $(n-1)d$ subshell. Fill the resulting $d^n$ electrons into the five $d$ orbitals following Hund's rule (maximum unpaired spins first) to get the free-ion arrangement.

Step 3. Classify the ligand and decide whether pairing is forced. If the ligand is strong-field ( $CN^-$ , $CO$ , $NO_2^-$ , and similar), it forces the free-ion's unpaired $d$ electrons to pair up as far as possible, which — for a four- or six-coordinate complex — frees one or two of the $(n-1)d$ orbitals. Those freed inner orbitals combine with $s$ and $p$ orbitals to give inner-orbital hybridisation ( $dsp^2$ for four-coordinate, $d^2sp^3$ for six-coordinate). If the ligand is weak-field ( $F^-$ , $Cl^-$ , or moderate ligands like $H_2O$ , $NH_3$ ), the free-ion arrangement of $d$ electrons is left undisturbed, no inner $(n-1)d$ orbital is available, and the metal instead uses its outer $nd$ orbital together with $s$ and $p$ orbitals, giving outer-orbital hybridisation ( $sp^3$ for four-coordinate, $sp^3d^2$ for six-coordinate).

Step 4. Read off geometry and magnetism from the hybridisation chosen in Step 3, and count the unpaired electrons remaining in the final $d$ -orbital arrangement to classify the complex: zero unpaired electrons means diamagnetic, and one or more unpaired electrons means paramagnetic, with the magnitude checked against $\mu = \sqrt{n(n+2)}$ BM if a numerical moment is asked for. Verified against Worked example 1 and Worked example 2 below: both start from the identical $Ni^{2+}$ , $3d^8$ free-ion configuration from Step 1–2, and Step 3's ligand classification alone is what sends them to opposite hybridisations, geometries and magnetic classes.

Worked example

Problem. Predict the hybridisation, geometry and magnetic character of $[Ni(CN)_4]^{2-}$ and of $[NiCl_4]^{2-}$ , and state the magnetic moment of whichever one is paramagnetic.

Step 1 — oxidation state. In both ions each ligand carries charge $-1$ and there are four of them, with overall ion charge $-2$ : $x + 4(-1) = -2 \implies x = +2$ . So the metal is $Ni^{2+}$ in both complexes.

Step 2 — free-ion configuration. Nickel is $[Ar]3d^8 4s^2$ ; removing the two $4s$ electrons for the $+2$ state leaves $Ni^{2+}$ as $3d^8$ . Across five $3d$ orbitals by Hund's rule this is $\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow$ — three filled pairs and two unpaired electrons.

Step 3 — classify the ligands. $CN^-$ is strong-field: it forces the two unpaired electrons to pair up, leaving one $3d$ orbital completely empty: $\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \, (\text{empty})$ . $Cl^-$ is weak-field: it does not disturb the free-ion arrangement, so the two unpaired electrons remain and no inner $3d$ orbital is freed.

Step 4 — hybridisation, geometry, magnetism. For $[Ni(CN)_4]^{2-}$ , the freed $3d$ orbital combines with one $4s$ and two $4p$ orbitals for $dsp^2$ hybridisation, giving a square planar geometry; all electrons are paired ( $n = 0$ ), so it is diamagnetic. For $[NiCl_4]^{2-}$ , with no inner $d$ orbital available, nickel uses one $4s$ and three $4p$ orbitals for $sp^3$ hybridisation, giving a tetrahedral geometry; the two unpaired electrons ( $n = 2$ ) make it paramagnetic, with

\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\ \text{BM}

Final answer: $[Ni(CN)_4]^{2-}$ is $dsp^2$ , square planar, diamagnetic; $[NiCl_4]^{2-}$ is $sp^3$ , tetrahedral, paramagnetic with $\mu \approx 2.83$ BM. The same $Ni^{2+}$ ion, same coordination number, diverges purely on ligand field strength.

Competitive stretch

For an octahedral complex, the same logic gives two possible hybridisations rather than one. $[Co(NH_3)_6]^{3+}$ : $Co^{3+}$ is $3d^6$ ; $NH_3$ is a moderately strong ligand and here forces pairing, giving $\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ (\text{empty})\ (\text{empty})$ — two $3d$ orbitals free for $d^2sp^3$ (inner-orbital) hybridisation, octahedral geometry, diamagnetic ( $n = 0$ ). Contrast $[CoF_6]^{3-}$ : $F^-$ is weak-field and does not force pairing, so the free-ion $3d^6$ arrangement $\uparrow\downarrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow$ (four unpaired) is retained, forcing $sp^3d^2$ (outer-orbital) hybridisation with $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM.

Common traps

Assuming every octahedral complex is $d^2sp^3$ — there are two possible octahedral hybridisations, and the choice between $d^2sp^3$ (inner-orbital) and $sp^3d^2$ (outer-orbital) depends entirely on the ligand's field strength.
Skipping straight to hybridisation without first nailing down the oxidation state and $d^n$ configuration — an error at that first step propagates through geometry and magnetism both.
Assuming all four-coordinate complexes are tetrahedral — $d^8$ ions like $Ni^{2+}$ , $Pd^{2+}$ , $Pt^{2+}$ with strong-field ligands commonly go square planar ( $dsp^2$ ) instead.
Equating "paramagnetic" loosely with "high spin" without actually counting unpaired electrons for the specific $d^n$ configuration in front of you — the label follows from the count, not the other way round.

Ask the tutor: pick any $d^n$ metal ion with a strong-field and a weak-field ligand and ask the AI tutor to check your hybridisation, geometry and magnetic-moment prediction step by step.

Why does the oxidation state have to be worked out before the hybridisation

Because the hybridisation prediction depends on how many $d$ electrons the metal ion actually has, and that count only comes from knowing the correct oxidation state. Get the oxidation state wrong and you write the wrong $d^n$ configuration, which can flip whether pairing is even possible, and the whole geometry-and-magnetism prediction collapses. It is the first domino, so it is worth double-checking with the ligand charges before moving on.

How do I know if a ligand is strong-field or weak-field without memorising a huge list

For the ISC syllabus a short list covers nearly every case you will meet: strong-field ligands that force pairing include $CN^-$ , $CO$ and $NO_2^-$ ; weak-field ligands that do not force pairing include $F^-$ , $Cl^-$ , and, generally, $H_2O$ and $NH_3$ (treated as moderate but usually not pairing unless stated otherwise in the question). Learn these examples rather than a rule, since field strength depends on the specific ligand's bonding, which VBT itself does not derive — that ranking comes from the related Crystal Field Theory topic.

Is the number of unpaired electrons always the same as in the free ion

No, and that is the whole point of ligand field strength. In the free ion, electrons spread out across all five $d$ orbitals following Hund's rule, maximising unpaired spins. A strong-field ligand can force some of those electrons to pair up once the complex forms, reducing the unpaired count compared to the free ion — exactly what happens in $[Ni(CN)_4]^{2-}$ , where the free-ion two unpaired electrons become zero after pairing. A weak-field ligand leaves the free-ion count unchanged, as in $[NiCl_4]^{2-}$ .

Valence bond theory

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads