VBT: Hybridisation, Geometry & Magnetic Properties

CHEBonding & Colour

Valence Bond Theory predicts a complex's shape and magnetism from the metal ion's $d$ -electron count and whether the ligand forces inner ( $(n-1)d$ ) or outer ( $nd$ ) orbital hybridisation. The standard task is: find $d^n$ , decide strong vs weak field, assign hybridisation, then read off geometry and unpaired electrons.

Spin-only magnetic moment

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

n

= number of unpaired electrons

Hybridisation → geometry map

sp\!:\text{linear}\quad sp^3\!:\text{tetrahedral}\quad dsp^2\!:\text{square planar}\quad d^2sp^3\,/\,sp^3d^2\!:\text{octahedral}

d^2sp^3

uses inner

(n-1)d

(low spin);

sp^3d^2

uses outer

nd

(high spin)

Octahedral: strong-field/low-spin uses inner $(n-1)d$ orbitals $\rightarrow d^2sp^3$ (inner-orbital complex); weak-field/high-spin uses outer $nd$ orbitals $\rightarrow sp^3d^2$ (outer-orbital complex).
Coordination number 4 splits two ways: $dsp^2$ gives square planar (e.g. $[Ni(CN)_4]^{2-}$ ), $sp^3$ gives tetrahedral (e.g. $[NiCl_4]^{2-}$ ); CN 2 is $sp$ linear (e.g. $[Ag(NH_3)_2]^+$ ).
$[Ni(CN)_4]^{2-}$ : $Ni^{2+}$ is $d^8$ ; strong $CN^-$ pairs electrons, freeing one $3d$ orbital for $dsp^2 \rightarrow$ square planar, $0$ unpaired, diamagnetic. $[NiCl_4]^{2-}$ : weak $Cl^-$ , $sp^3$ , tetrahedral, $2$ unpaired, paramagnetic.
$[Fe(CN)_6]^{4-}$ : $Fe^{2+}$ $d^6$ + strong $CN^- \rightarrow d^2sp^3$ , inner-orbital, $0$ unpaired, diamagnetic. $[FeF_6]^{3-}$ : $Fe^{3+}$ $d^5$ + weak $F^- \rightarrow sp^3d^2$ , outer-orbital, $5$ unpaired, strongly paramagnetic.
$[Co(NH_3)_6]^{3+}$ : $Co^{3+}$ $d^6$ + (strong) $NH_3 \rightarrow d^2sp^3$ , diamagnetic; $[CoF_6]^{3-}$ : $d^6$ + weak $F^- \rightarrow sp^3d^2$ , $4$ unpaired, paramagnetic.
Magnetic moment is spin-only $\mu = \sqrt{n(n+2)}$ BM, so $n$ unpaired electrons directly give $\mu$ ; $\mu = 0$ means diamagnetic.
VBT limitation: it does not explain why a given ligand is strong or weak field, predicts no colour, and gives only an approximate, not quantitative, account of magnetic data.
Step method: (1) find the oxidation number and $d^n$ of the metal ion; (2) decide strong vs weak field from the ligand; (3) for a strong field, pair $d$ -electrons to vacate inner $d$ -orbitals; (4) assign hybridisation; (5) read geometry and count unpaired electrons.
Inner-orbital (low-spin) octahedral complexes are diamagnetic or have fewer unpaired electrons and are usually more stable; outer-orbital (high-spin) complexes use $nd$ orbitals and retain the maximum number of unpaired electrons.
$[Co(NH_3)_6]^{3+}$ ( $d^6$ , low spin, all paired) is diamagnetic and especially inert, whereas $[CoF_6]^{3-}$ ( $d^6$ , high spin) keeps 4 unpaired electrons — same metal/oxidation state, different ligand field, opposite magnetism.
Square-planar $d^8$ examples include $[Ni(CN)_4]^{2-}$ , $[Pt(NH_3)_2Cl_2]$ and $[PtCl_4]^{2-}$ ( $dsp^2$ , diamagnetic); $[PtCl_4]^{2-}$ is square planar because the heavier $4d$ and $5d$ metals (Pt is a $5d$ metal) strongly favour low spin even with weaker ligands.
VBT also rationalises the existence of inner vs outer complexes but cannot decide between $dsp^2$ and $sp^3$ for CN 4 without independent magnetic data — a key reason CFT is needed.

Where the marks go

Forgetting to pair electrons BEFORE assigning $d^2sp^3$ — the two inner $d$ -orbitals must be empty, which requires pairing for $d^4$ – $d^7$ strong-field ions.
Treating $Cl^-$ and $F^-$ as strong-field: they are weak field, so $[NiCl_4]^{2-}$ is tetrahedral ( $sp^3$ ) and $[CoF_6]^{3-}$ / $[FeF_6]^{3-}$ are high-spin outer-orbital.
Assuming all CN-4 complexes are tetrahedral — strong-field $d^8$ ions ( $Ni^{2+}$ , $Pt^{2+}$ , $Pd^{2+}$ ) give $dsp^2$ square planar instead.
Computing $\mu$ from the number of $d$ -electrons rather than unpaired electrons, or quoting $\mu \neq 0$ for a diamagnetic low-spin complex.
Saying VBT explains colour or quantitative magnetic moments — it does neither; that is a CFT job.

How the board asks it

Predict the productthe step method: find $d^n$ , decide field strength, assign hybridisation, read geometry and unpaired electrons
On the basis of valence bond theory, predict the hybridisation, geometry and magnetic behaviour of $[Fe(CN)_6]^{4-}$ , given that $CN^-$ is a strong-field ligand.
Give reasonsinner- vs outer-orbital complexes and strong/weak field ligands
Account for the fact that $[Co(NH_3)_6]^{3+}$ is diamagnetic whereas $[CoF_6]^{3-}$ is paramagnetic, although both contain $Co^{3+}$ ( $d^6$ ).
Distinguish $dsp^2$ square planar vs $sp^3$ tetrahedral for coordination number 4
Explain why $[Ni(CN)_4]^{2-}$ is square planar and diamagnetic, while $[NiCl_4]^{2-}$ is tetrahedral and paramagnetic, though both contain $Ni^{2+}$ .
Numericalthe spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ BM
Calculate the spin-only magnetic moment (in BM) of $[FeF_6]^{3-}$ , in which $Fe^{3+}$ has $5$ unpaired electrons.
Define / statethe limitations of valence bond theory
State any two limitations of valence bond theory in explaining the bonding in coordination compounds.
Structure / namingthe hybridisation-to-geometry map and orbital box diagrams
Draw the orbital diagram showing the hybridisation of $Cr^{3+}$ in $[Cr(NH_3)_6]^{3+}$ ( $d^3$ ) and state its geometry and number of unpaired electrons.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.