Valence Bond Theory & Magnetic Properties

Valence Bond Theory of Complexes

How does the metal actually bond to its ligands? Valence Bond Theory (VBT) gives one answer: the metal ion makes a set of empty hybrid orbitals of definite geometry, and each ligand donates a lone pair into one of them, forming a coordinate bond.

The type of hybridisation determines the geometry:

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

Inner vs outer orbital: in $d^2sp^3$ the metal uses inner (n-1)d orbitals (inner-orbital complex); in $sp^3d^2$ it uses outer nd orbitals (outer-orbital complex). Which one forms depends on whether the ligand is strong-field or weak-field.

Inner vs Outer Orbital (Low Spin vs High Spin)

When a strong-field ligand approaches, it can cause pairing of the metal ion's d electrons, freeing inner d orbitals for $d^2sp^3$ hybridisation:

• Inner-orbital complex → fewer unpaired electrons → low spin → often diamagnetic or weakly paramagnetic.

When a weak-field ligand approaches, the d electrons usually remain unpaired, so the metal uses outer d orbitals for $sp^3d^2$:

• Outer-orbital complex → more unpaired electrons → high spin → strongly paramagnetic.

Worked example — $[\text{Co(NH}_3)_6]^{3+}$: Co$^{3+}$ is $d^6$. In the presence of NH$_3$, electrons pair in the lower-energy orbitals, leaving two inner d orbitals free for $d^2sp^3$ hybridisation. So the complex is octahedral, has 0 unpaired electrons, and is diamagnetic.

$[\text{CoF}_6]^{3-}$: Co$^{3+}$ ($d^6$) with weak F$^-$ does not pair sufficiently, so it uses $sp^3d^2$ (outer orbital), has 4 unpaired electrons, and is paramagnetic.

Predicting Magnetic Moment with VBT

The spin-only magnetic moment ($\mu = \sqrt{n(n+2)}$ BM) lets us test VBT predictions against experiment. Work through these standard cases:

Complex	Metal ion (d)	Ligand	Hybridisation	Geometry	Unpaired $e^-$	$\mu$ (BM)
$[\text{Ni(CO)}_4]$	Ni$^0$ ($d^{10}$)	strong	$sp^3$	tetrahedral	0	0 (diamag.)
$[\text{NiCl}_4]^{2-}$	Ni$^{2+}$ ($d^8$)	weak	$sp^3$	tetrahedral	2	2.83
$[\text{Ni(CN)}_4]^{2-}$	Ni$^{2+}$ ($d^8$)	strong	$dsp^2$	square planar	0	0 (diamag.)
$[\text{Fe(CN)}_6]^{3-}$	Fe$^{3+}$ ($d^5$)	strong	$d^2sp^3$	octahedral	1	1.73
$[\text{Fe(H}_2\text{O)}_6]^{3+}$	Fe$^{3+}$ ($d^5$)	weak	$sp^3d^2$	octahedral	5	5.92

[JEE Tip] The pair $[\text{NiCl}_4]^{2-}$ (paramagnetic, tetrahedral) vs $[\text{Ni(CN)}_4]^{2-}$ (diamagnetic, square planar) — same Ni$^{2+}$ ($d^8$) but different ligand strength — is the standard VBT comparison. Cl$^-$ is weak (tetrahedral, 2 unpaired); CN$^-$ is strong (square planar, 0 unpaired).

Example 1: Hybridisation of [Co(NH3)6]3+

Predict the hybridisation, geometry and magnetic nature of $[\text{Co(NH}_3)_6]^{3+}$.

Solution: Co$^{3+}$ is $d^6$. NH$_3$ causes pairing of d electrons in the lower-energy orbitals, so two inner d orbitals are available for $d^2sp^3$ hybridisation. Therefore the complex is octahedral, has 0 unpaired electrons, and is diamagnetic (inner-orbital complex).

Example 2: [CoF6]3- vs [Co(NH3)6]3+

Why is $[\text{CoF}_6]^{3-}$ paramagnetic while $[\text{Co(NH}_3)_6]^{3+}$ is diamagnetic, though both are Co$^{3+}$?

Solution: F$^-$ is a weak-field ligand, so pairing is not favoured; the complex uses $sp^3d^2$ (outer orbital), giving 4 unpaired electrons and paramagnetism. NH$_3$ causes pairing to a greater extent, so the complex uses $d^2sp^3$ (inner orbital), has 0 unpaired electrons, and is diamagnetic.

Example 3: [NiCl4]2- vs [Ni(CN)4]2-

Both contain Ni$^{2+}$ ($d^8$). Explain why $[\text{NiCl}_4]^{2-}$ is paramagnetic (tetrahedral) but $[\text{Ni(CN)}_4]^{2-}$ is diamagnetic (square planar).

Solution: Cl$^-$ is a weak-field ligand, so the complex remains $sp^3$ (tetrahedral) with 2 unpaired electrons and is paramagnetic. CN$^-$ is a strong-field ligand, so the d electrons pair and the complex becomes $dsp^2$ (square planar) with 0 unpaired electrons, making it diamagnetic.

Example 4: Magnetic moment of [NiCl4]2-

Calculate the spin-only magnetic moment of $[\text{NiCl}_4]^{2-}$.

Solution: Ni$^{2+}$ ($d^8$) with weak Cl$^-$ gives 2 unpaired electrons. So $$\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.83\ \text{BM}.$$

Example 5: [Fe(CN)6]3- magnetic moment

Predict the hybridisation and spin-only moment of $[\text{Fe(CN)}_6]^{3-}$.

Solution: Fe$^{3+}$ is $d^5$. CN$^-$ is strong field, so pairing occurs and the complex uses $d^2sp^3$ (inner, octahedral). This leaves 1 unpaired electron. Therefore, $$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73\ \text{BM}.$$

Example 6: [Fe(H2O)6]3+ magnetic moment

Predict the spin-only moment of $[\text{Fe(H}_2\text{O)}_6]^{3+}$.

Solution: Fe$^{3+}$ ($d^5$) with weak H$_2$O gives no pairing, so the complex is $sp^3d^2$ (outer), with 5 unpaired electrons. Thus $$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\ \text{BM}$$ which indicates a strongly paramagnetic complex.

Example 7: [Ni(CO)4] geometry

Predict the hybridisation, geometry and magnetism of $[\text{Ni(CO)}_4]$.

Solution: In $[\text{Ni(CO)}_4]$, nickel is in the 0 oxidation state and is treated as $d^{10}$. CO is a strong-field ligand, and the complex uses $sp^3$ hybridisation, giving a tetrahedral geometry with 0 unpaired electrons. Hence it is diamagnetic.

Example 8: Inner vs outer orbital

Distinguish inner-orbital and outer-orbital octahedral complexes.

Solution: Inner-orbital ($d^2sp^3$): uses inner (n-1)d orbitals; formed more readily with strong-field ligands; low spin; fewer unpaired electrons. Outer-orbital ($sp^3d^2$): uses outer nd orbitals; formed more readily with weak-field ligands; high spin; more unpaired electrons.

Example 9: Square planar from VBT

On the basis of VBT, explain why $[\text{Ni(CN)}_4]^{2-}$ is square planar.

Solution: Ni$^{2+}$ is $d^8$. The strong-field CN$^-$ ligand causes pairing of the 8 d electrons, so one 3d orbital becomes available for hybridisation with 4s and 4p orbitals. This gives $dsp^2$ hybridisation and a square planar geometry with 0 unpaired electrons (diamagnetic).

Example 10: Predict unpaired electrons in [Pt(CN)4]2-

Predict the number of unpaired electrons in square planar $[\text{Pt(CN)}_4]^{2-}$.

Solution: Pt$^{2+}$ is $d^8$. Strong CN$^-$ gives a $dsp^2$ square planar complex with all d electrons paired, so there are 0 unpaired electrons and the complex is diamagnetic.