Valence-bond theory

Overlap forms the bond. Two atomic orbitals, each holding one electron, overlap between two nuclei; the shared electron pair is the bond. Two conditions govern how strong that overlap — and hence the bond — can be: the orbitals must have the same symmetry along the bond axis, and they must be of comparable energy. Head-on overlap (lobes pointing directly at each other along the internuclear axis) gives a $\sigma$ bond; sideways overlap of parallel $p$ orbitals gives a $\pi$ bond. Head-on overlap puts more orbital density directly between the nuclei, so $\sigma$ overlap is generally greater than $\pi$ overlap between the same pair of orbitals, and $\sigma$ bonds are usually stronger.

Hybridization reshapes the atom's own orbitals. Real molecular geometries — the perfect $109.5^\circ$ angles of methane, the $120^\circ$ trigonal plane of $\text{BF}_3$ — are not what carbon's or boron's bare atomic orbitals would produce on their own. VBT resolves this by mixing an atom's $s$ and $p$ (and, for expanded octets, $d$) orbitals into a new set of equivalent hybrid orbitals, each with a fixed geometry:

Each hybrid then overlaps with an orbital on the neighbouring atom to form one $\sigma$ bond, so the number of hybrid orbitals fixes the number of $\sigma$-bonded (or lone-pair-holding) directions around the central atom.

Resonance is invoked when no single Lewis/VB structure captures the true bonding — the real molecule is a single quantum-mechanical superposition of the contributing structures (they do not physically interconvert), and this superposition sits at lower energy than any one contributor: resonance stabilization.

Where VBT fails. (1) It predicts $\text{O}_2$, drawn as $\text{O}=\text{O}$ with every electron paired, should be diamagnetic — but $\text{O}_2$ is experimentally paramagnetic, carrying unpaired electrons VBT's paired-bond picture cannot produce. (2) It has no mechanism for colour, which comes from electronic transitions between orbitals of different energy — a concept needing delocalized orbital energy levels, which MOT and crystal field theory supply but VBT's localized bond-pairs do not.

Goal. VBT's predictive power rests on a chain of reasoning that starts from an atom's ground-state electron configuration, moves through orbital promotion and mixing, and lands on a molecular geometry you can check against experiment — following that chain for an expanded-octet case such as phosphorus in $\text{PCl}_5$ shows exactly why hybridization is a bookkeeping device that reproduces observed shapes rather than a description of orbitals that exist in the free atom.

Step 1. Start from the ground-state configuration and count how many $\sigma$ bonds (or lone pairs) the central atom must accommodate. Phosphorus's ground state is $3s^2\,3p^3$, giving three unpaired $3p$ electrons — enough for only three bonds by naive counting, but $\text{PCl}_5$ needs five.

Step 2. Promote an electron into an empty low-lying orbital of comparable energy to unpair enough electrons for the required bonding count. One $3s$ electron is promoted into an empty $3d$ orbital: $3s^2 3p^3 \to 3s^1 3p^3 3d^1$, which now has five unpaired electrons, one per bond needed.

Step 3. Mix the now-singly-occupied orbitals into a new set of equivalent hybrid orbitals whose number equals the number of unpaired electrons, and whose spatial arrangement minimizes electron-pair repulsion — this is the same reasoning as VSEPR, expressed in orbital language. One $3s$, three $3p$, and one $3d$ orbital combine into five $sp^3d$ hybrids, which point toward the corners of a trigonal bipyramid (three equatorial at $120^\circ$, two axial at $90^\circ$ to the equatorial plane) because that arrangement keeps five regions of electron density as far apart as possible.

Step 4. Overlap each hybrid with an orbital on the attached atom to form one $\sigma$ bond, and compare the resulting shape against the experimentally observed geometry. Each $sp^3d$ hybrid on phosphorus overlaps head-on with a chlorine $3p$ orbital; the result is exactly the trigonal bipyramid observed for $\text{PCl}_5$, confirming the scheme — the promotion energy paid in Step 2 is more than repaid by the extra bond formed, which is why the expanded octet is favourable rather than forbidden.

Problem. Use VBT to assign the hybridization and predicted shape of the central atom in $\text{SF}_6$, and then explain, using the same overlap logic, why the VBT picture of $\text{O}_2$ as $\text{O}=\text{O}$ with all electrons paired fails to match experiment.

Step 1 (SF₆ configuration and promotion). Sulfur's ground state is $3s^2\,3p^4$, only two unpaired electrons. Six $\sigma$ bonds are needed, so two electrons are promoted into empty $3d$ orbitals: $3s^2 3p^4 \xrightarrow{\text{promote}} 3s^1 3p^3 3d^2$, giving six unpaired electrons.

Step 2 (hybridize and check geometry). One $3s$, three $3p$, and two $3d$ orbitals mix into six equivalent $sp^3d^2$ hybrids, pointing toward the corners of a regular octahedron. Each overlaps with a fluorine $2p$ orbital to give six equivalent $\sigma$ bonds with F–S–F angles of $90^\circ$ and $180^\circ$ — matching the experimentally observed octahedral $\text{SF}_6$ geometry, so the $sp^3d^2$ assignment is confirmed. (Modern quantum-chemical calculations show actual $d$-orbital participation is smaller than this simple picture suggests, but $sp^3d^2$ remains the standard qualitative VBT tool at this level, and it gets the geometry right.)

Step 3 (O₂ by the same overlap logic). Each oxygen atom is $2s^2\,2p_x^2\,2p_y^1\,2p_z^1$ in the ground state, with two unpaired $2p$ electrons available for bonding. VBT pairs one $2p_y$ electron from each atom head-on into a $\sigma$ bond and one $2p_z$ electron from each atom sideways into a $\pi$ bond, giving the double bond $\text{O}=\text{O}$ with every bonding electron paired — this VB structure has zero unpaired electrons, so VBT's prediction is that $\text{O}_2$ is diamagnetic.

Step 4 (compare with experiment). Liquid $\text{O}_2$ is pulled strongly into a magnetic field: it is experimentally paramagnetic, meaning it does carry unpaired electrons. VBT's localized bond-pair model has no slot for an unpaired electron in a stable double bond, so this is a genuine, well-known failure — not a detail VBT gets slightly wrong, but a case it cannot describe at all without help. The fix requires molecular orbital theory: filling delocalized MOs places two electrons in the degenerate $\pi^*_{2p_x}, \pi^*_{2p_y}$ antibonding orbitals, and Hund's rule keeps them unpaired, correctly predicting paramagnetism.

Final answer: SF₆ is $sp^3d^2$-hybridized with octahedral geometry (VBT correct); O₂'s VBT structure predicts zero unpaired electrons (diamagnetic), which contradicts its experimentally observed paramagnetism — VBT fails here and MOT is needed.

• Treating hybrid orbitals as physically real orbitals that exist in the free atom before bonding — they are a bookkeeping construct built to match the observed geometry, justified because the promotion energy is repaid by the extra bonds formed.
• Assuming VBT and MOT are rival, mutually exclusive theories rather than two approximate models of the same reality, each strong where the other is weak.
• Reaching for expanded-octet ($sp^3d$, $sp^3d^2$) hybridization without checking the result against the experimentally observed shape first — always cross-check with VSEPR.
• Describing resonance as rapid oscillation between contributing structures instead of a single, static quantum-mechanical superposition of them.

Ask the tutor: want to walk through why VBT's expanded-octet picture is now considered an approximation, or compare VBT and MOT side by side for a specific molecule like NO or O2?