Orbital hybridisation

Result

Orbital hybridisation

T-010Home CU-102Threads bonding · quantum

Statement

One \(s\) orbital and \((n-1)\) \(p\) orbitals on a central atom can be mixed into \(n\) equivalent, directional hybrid orbitals. Choosing the mixing so all \(n\) hybrids are equally weighted and mutually orthogonal forces the angle \(\theta\) between any two of them to satisfy \(\cos\theta=-\tfrac{1}{n-1}\) — for \(n=2,3,4\) this reproduces the linear, trigonal planar, and tetrahedral angles exactly, giving a quantum-mechanical, orbital-level account of the geometries VSEPR already predicts.

Why it matters

VSEPR predicts molecular shape from electron-pair repulsion alone, without ever specifying which atomic orbitals the bonding electrons actually occupy. Hybridisation fills that gap: it shows that ordinary atomic \(s\) and \(p\) orbitals — which individually point along fixed Cartesian axes or have no directionality at all — can be linearly combined into new orbitals that point exactly where VSEPR says the bonds and lone pairs should go, and that the resulting inter-orbital angle emerges from orthonormality alone, not from an independent fit to the observed angle. It is also what identifies, for each bonded atom, how many \(\sigma\) bonds versus unhybridised \(p\) orbitals available for \(\pi\) bonding are present — the origin of every double and triple bond's extra bond(s).

Hypotheses

The \(n\) hybrid orbitals formed from 1 \(s\)-orbital and \((n-1)\) \(p\)-orbitals are equally weighted (each carries the same fraction \(1/n\) of \(s\)-character) and mutually orthonormal.Equal weighting follows from requiring the \(n\) hybrids to be geometrically and energetically equivalent (a physical requirement for, e.g., methane's four experimentally indistinguishable C–H bonds); since the total \(s\)-character summed over all \(n\) hybrids must equal the one full \(s\) orbital contributed, equal division gives exactly \(1/n\) each. Orthonormality is required because atomic orbitals belonging to the same atom must combine into a new orthonormal basis for the mixing to be a valid change of basis (a unitary transformation) rather than a loss of information. For \(SN=5\) and \(SN=6\), the traditional \(sp^3d\) and \(sp^3d^2\) labels are used, but genuine valence \(d\)-orbital mixing into main-group bonding is now understood to be small; the labels are a useful geometric bookkeeping convention inherited from early valence-bond theory, not a claim of strong \(d\)-orbital participation.This qualification matters because it is the modern, computationally-supported view (see Discussion) — treating \(sp^3d\)/\(sp^3d^2\) as literally accurate orbital compositions, rather than as convenient labels for already-known VSEPR geometries, overstates what quantum-chemical calculations actually find for period 3+ hypervalent molecules.

Proof

\psi_i = c_s\,\phi_s + c_p\left(a_i\phi_{p_x}+b_i\phi_{p_y}+d_i\phi_{p_z}\right), \quad i=1,\dots,n

Each hybrid orbital is a linear combination of the one \(s\) orbital and a unit-vector direction \((a_i,b_i,d_i)\) built from the three \(p\) orbitals, with normalisation \(c_s^2+c_p^2=1\) for each hybrid. A

\sum_{i=1}^{n} c_s^2 = 1 \ \Longrightarrow\ c_s^2=\frac{1}{n}, \quad c_p^2=1-\frac{1}{n}=\frac{n-1}{n}

Equal weighting (Hypotheses) means each hybrid's \(s\)-fraction is the same, \(c_s^2\); since exactly one full \(s\) orbital's worth of character is distributed across all \(n\) hybrids, the total must sum to 1, fixing \(c_s^2=1/n\) for every hybrid, and \(c_p^2\) follows from normalisation. A

\langle\psi_i|\psi_j\rangle = c_s^2 + c_p^2\,\hat{u}_i\cdot\hat{u}_j = 0 \quad (i\ne j)

Orthogonality between two hybrids requires their overlap to vanish; since \(\phi_s\) is orthogonal to every \(\phi_p\) and the three \(\phi_p\) are mutually orthogonal, the overlap reduces to the \(s\)-part (\(c_s^2\), since both hybrids share the same \(s\)-fraction) plus the \(p\)-part, which is \(c_p^2\) times the dot product of the two unit direction vectors \(\hat u_i\cdot\hat u_j=\cos\theta_{ij}\). A

\cos\theta = -\frac{c_s^2}{c_p^2} = -\frac{1/n}{(n-1)/n} = -\frac{1}{n-1}

Solve Step 3 for \(\cos\theta_{ij}\), substituting the values from Step 2. This holds exactly whenever all \(n\) hybrids are pairwise equivalent by symmetry (a single, common value of \(\theta\) for every pair) — true for \(n=2\) (collinear), \(n=3\) (coplanar, threefold symmetric), and \(n=4\) (tetrahedral, fourfold symmetric), the three cases where \(n\) directions can all be mutually equivalent in three dimensions. A

Result

\cos\theta = -\frac{1}{n-1}: \quad n=2\ (sp):\ 180^\circ, \quad n=3\ (sp^2):\ 120^\circ, \quad n=4\ (sp^3):\ 109.47^\circ

Hybridisation	SN	Geometry	Unhybridised \(p\) orbitals left
sp	2	Linear	2 (up to 2 \(\pi\) bonds)
sp²	3	Trigonal planar	1 (up to 1 \(\pi\) bond)
sp³	4	Tetrahedral	0
sp³d†	5	Trigonal bipyramidal	—
sp³d²†	6	Octahedral	—

† Geometric labels retained by convention; see Hypotheses and Discussion regarding limited true \(d\)-orbital participation.

Reading. The bond angles VSEPR obtains from a purely geometric repulsion argument are reproduced exactly, for \(n=2,3,4\), from nothing but orthonormality of a linear combination of atomic orbitals — two independently-motivated arguments (electron repulsion, orbital orthogonality) converging on the same angles.

Scope. The \(\cos\theta=-1/(n-1)\) derivation is exact only where all \(n\) hybrid directions are pairwise equivalent (\(n\le4\)); for \(SN=5,6\) the geometry (unequal axial/equatorial or 90°/180° angles) is taken directly from VSEPR rather than re-derived from a single-angle orbital formula.

Corollaries & converses

An unhybridised \(p\) orbital left over after hybridisation (Result table, right column) is what forms a \(\pi\) bond with a similarly unhybridised \(p\) orbital on an adjacent atom; a double bond is always one \(\sigma\) bond (from a hybrid orbital) plus one \(\pi\) bond (from a leftover, unhybridised \(p\) orbital), and a triple bond is one \(\sigma\) plus two \(\pi\) bonds.
Hybridisation type can be read directly off a correct Lewis structure without any calculation: count \(\sigma\) bonds plus lone pairs on an atom (its \(SN\), from vsepr-geometry) and match to the Result table — \(SN\) and hybridisation type are in exact one-to-one correspondence.
Converse: knowing an atom's hybridisation immediately gives its \(SN\) and hence its VSEPR electron-domain geometry, without needing to separately count \(\sigma\) bonds and lone pairs — the two results are two directions of the same underlying correspondence.

Fails without

Drop equal weighting (Hypotheses), allow unequal \(s\)-character across the \(n\) hybrids: Step 2's \(c_s^2=1/n\) no longer holds uniformly, and Step 4's single angle \(\theta\) for every pair no longer follows — different pairs of hybrids would need different angles, contradicting the experimentally observed equivalence of, for example, methane's four identical C–H bonds (identical bond length, identical bond energy, identical \(109.5^\circ\) angles all six ways).
Attempt to extend the \(\cos\theta=-1/(n-1)\) formula unmodified to \(n=5,6\): it would predict a single common angle of \(\arccos(-\tfrac14)\approx104.5^\circ\) for \(n=5\) or \(\arccos(-\tfrac15)\approx101.5^\circ\) for \(n=6\) — neither matches the actual, well-established trigonal bipyramidal (\(90^\circ/120^\circ/180^\circ\)) or octahedral (\(90^\circ/180^\circ\)) angle sets, because for \(n=5,6\) no arrangement of \(n\) directions in three dimensions can be pairwise equivalent (only \(n\le4\) permits that); this is exactly why the Result restricts the derived formula's scope to \(n\le4\).

Common errors

Assuming hybridisation independently derives molecular shape from scratch, rather than reproducing a shape already established by VSEPR — the correct logical order is Lewis structure → VSEPR geometry → matching hybridisation (see Discussion).
Forgetting that a multiple bond leaves fewer hybrid orbitals available for \(\sigma\) bonds to other atoms; an \(sp^2\) carbon in a double bond has 3 hybrid orbitals total, only 2 of which may be free for bonds to other substituents once one is used for the \(\sigma\) bond of the double bond.
Extending the exact \(\cos\theta=-1/(n-1)\) formula to \(n=5\) or \(n=6\) as if it gave a valid single bond angle (directly addressed in Fails without).
Treating \(sp^3d\)/\(sp^3d^2\) as evidence that period 3+ atoms use their valence \(d\) orbitals substantially in ordinary bonding, rather than as an inherited geometric-labelling convention (Hypotheses, Discussion).

Discussion

Hybridisation theory was introduced by Linus Pauling in 1931, as part of the broader valence-bond approach to relating quantum mechanics to the empirically-known shapes and bond strengths of simple molecules — work that contributed to his 1954 Nobel Prize in Chemistry. The original motivation was exactly the derivation given here: ordinary, non-hybridised atomic \(s\) and \(p\) orbitals do not point toward tetrahedral corners, yet carbon reliably forms four equivalent tetrahedral bonds, and mixing (not physically separating) the orbitals resolves the puzzle.

The status of \(d\)-orbital involvement in hypervalent main-group molecules (period 3 and beyond, e.g. \(\text{PCl}_5\), \(\text{SF}_6\)) has shifted substantially since Pauling's original formulation. Detailed modern computational studies (natural bond orbital and related analyses, from the 1980s onward) generally find that valence \(d\) orbitals on period 3+ atoms contribute only a small fraction of electron density to the bonding — expanded-octet geometries are now more often explained by the larger atomic radius of period 3+ atoms comfortably accommodating more electron domains around them (an ordinary VSEPR argument, no \(d\)-orbital mixing required) than by genuine \(sp^3d\)-type hybrid orbitals. The \(sp^3d\)/\(sp^3d^2\) labels persist in most introductory treatments (including the Result table above) purely as a compact, geometry-matching notation, not as a claim about the underlying orbital physics.

Common misconception: that hybrid orbitals are a physically separate, independently observable species of orbital, distinct from ordinary \(s\) and \(p\) orbitals. They are not — hybridisation is a change of basis, a particular linear combination chosen because it is convenient for describing directional bonding; the total electron density and total energy of the atom are unchanged by the choice of basis, and an equally valid (if less convenient) description of the same physical bonding could in principle be written using unmixed \(s\) and \(p\) orbitals directly.

Worked examples

\text{Ethylene } \text{C}_2\text{H}_4:\ \text{each C has } SN=3\ (\text{2 C–H }\sigma\text{, 1 C–C }\sigma) \Rightarrow sp^2,\ 120^\circ,\ \text{1 unhybridised }p\text{ orbital each}

The two leftover, unhybridised \(p\) orbitals (one per carbon, both perpendicular to the molecular plane) overlap sideways to form one \(\pi\) bond, giving the C=C double bond its second (\(\pi\)) component on top of the \(\sigma\) bond already counted in \(SN\); this is also why ethylene is planar and rigid about the C=C axis — rotating one CH\(_2\) group relative to the other would break the sideways \(p\)-orbital overlap. A

\text{Acetylene } \text{C}_2\text{H}_2:\ \text{each C has } SN=2\ (\text{1 C–H }\sigma\text{, 1 C–C }\sigma) \Rightarrow sp,\ 180^\circ,\ \text{2 unhybridised }p\text{ orbitals each}

Two independent pairs of leftover \(p\) orbitals (mutually perpendicular) form two \(\pi\) bonds, giving the C≡C triple bond its two \(\pi\) components on top of the \(\sigma\) bond; the linear \(180^\circ\) geometry of \(sp\) hybridisation is exactly why acetylene, unlike ethylene, is a straight-line molecule. A

\text{C}_2\text{H}_4:\ sp^2\text{ carbons, one C=C }\pi\text{ bond}; \qquad \text{C}_2\text{H}_2:\ sp\text{ carbons, two C≡C }\pi\text{ bonds}

Reading. Bond order beyond 1 (double, triple) is directly traceable to leftover, unhybridised \(p\) orbitals, whose count is fixed the moment \(SN\), and hence hybridisation type, is known.

Scope. The same leftover-\(p\)-orbital counting applies to any multiply-bonded \(sp\) or \(sp^2\) atom, including heteroatom double bonds such as C=O (relevant to the calorimetry and organic-chemistry units elsewhere in this curriculum).

Problems

Determine the hybridisation of nitrogen in \(\text{NH}_3\) (established as \(SN=4\) in vsepr-geometry) and state how many unhybridised \(p\) orbitals, if any, remain.
Solution
\(SN=4\Rightarrow sp^3\) hybridisation, matching the Result table. All four atomic orbitals (1 \(s\) + 3 \(p\)) are used in the four \(sp^3\) hybrids (three N–H \(\sigma\) bonds plus one lone pair occupying the fourth hybrid), leaving zero unhybridised \(p\) orbitals — consistent with ammonia having no \(\pi\) bonding.
Determine the hybridisation of carbon in \(\text{CO}_2\) (established as \(SN=2\), linear, in vsepr-geometry) and state the total number of \(\sigma\) and \(\pi\) bonds around the central carbon.
Solution
\(SN=2\Rightarrow sp\) hybridisation, two unhybridised \(p\) orbitals remaining (Result table). Each C=O double bond is one \(\sigma\) bond (from an \(sp\) hybrid) plus one \(\pi\) bond (from one of the two leftover \(p\) orbitals, one used per C=O); with two C=O double bonds total, carbon has 2 \(\sigma\) bonds and 2 \(\pi\) bonds, exactly using up both \(sp\) hybrids and both leftover \(p\) orbitals.
Using \(\cos\theta=-\tfrac{1}{n-1}\), verify algebraically (not just by quoting the known value) that \(n=3\) gives exactly \(120^\circ\).
Solution
For \(n=3\): \(\cos\theta=-\tfrac{1}{3-1}=-\tfrac12\). Since \(\cos(120^\circ)=-\tfrac12\) exactly (a standard trigonometric value, from the unit-circle coordinates of \(120^\circ\)), \(\theta=120^\circ\) exactly, with no rounding or approximation involved anywhere in the derivation — unlike the \(n=4\) case (\(109.47^\circ\)), which is an irrational angle with no exact "nice" degree value.
A student claims that since sulfur in \(\text{SF}_6\) is "\(sp^3d^2\) hybridised," sulfur must be using two of its 3\(d\) valence orbitals substantially in bonding. Using the Discussion's account of modern computational findings, explain what is misleading about this claim, and state the alternative explanation for how sulfur accommodates six bonding domains.
Solution
The \(sp^3d^2\) label is retained as a convenient geometric bookkeeping notation matching the observed octahedral \(SN=6\) geometry, but modern computational (natural-bond-orbital-type) studies find that sulfur's 3\(d\) orbitals contribute only a small fraction of the actual bonding electron density — the label overstates genuine \(d\)-orbital participation. The alternative, currently favoured explanation is simply that sulfur, being a period 3 atom with a larger atomic radius and more valence electrons available than period 2 atoms, can geometrically accommodate six electron domains around it without needing to invoke substantial \(d\)-orbital mixing at all; the same VSEPR electron-counting logic used for \(\text{XeF}_4\) and \(\text{PCl}_5\) elsewhere in this unit already predicts \(\text{SF}_6\)'s six-domain, octahedral shape without any hybridisation argument.