Valence Bond Theory (VBT) — hybridization (sp, sp², sp³, sp³d, sp³d²): Exercises

Formula The lone-pair formula — single bonds (the everyday case, stated once)

$LP = \frac{V - n _{σ} - q}{2} (only valid when every bond is a single bond)$ where

$V$ = valence electrons of the central atom (neutral atom),
$n_{σ}$ = number of σ-bond neighbours (same count used in SN — one per attached atom, regardless of single/double/triple),
$q$ = the overall charge of the species with its own sign ( $q = + 1$ for a cation, $q = - 1$ for an anion, $q = 0$ for a neutral molecule).

Unified sign rule (memorise this once — no more ad-hoc "+extra"/"–charge"): a cation ( $q = + 1$ ) removes an electron, so subtracting $q$ makes LP smaller; an anion ( $q = - 1$ ) adds an electron, and subtracting a negative $q$ makes LP larger.

Note: for a molecule with only single bonds, $n_{σ}$ equals the number of shared electrons the central atom contributes, so this reduces to the familiar $LP = (V - B) /2$ with $B = n_{σ}$ and $q = 0$ .

Formula The GENERAL lone-pair formula — works for multiple bonds too

The simple formula above breaks for double/triple bonds, because there $n_{σ}$ (one per neighbour) no longer equals the number of electrons the central atom actually spends on bonding. The honest, always-valid version counts the bonding electrons the central atom contributes: $LP = \frac{V - B _{e} - q}{2}$ where $B_{e}$ = total electrons the central atom contributes to all its bonds (1 per single bond, 2 per double, 3 per triple), and $V, q$ are as before.

For all-single-bond species $B_{e} = n_{σ}$ , so this reduces to the boxed everyday formula — nothing to relearn.
For $CO_{2}$ (two C=O double bonds) $B_{e} = 2 + 2 = 4$ , giving $LP = (4 - 4 - 0) /2 = 0$ cleanly — no " $= 1 ?$ then discard" confusion. Use this version whenever a double or triple bond is present.
Reminder: whichever formula gives LP, the SN still uses $n_{σ}$ (one per neighbour), not $B_{e}$ . Multiple bonds add only one direction.

The full lookup table (all seven steric numbers, each with a canonical molecule) — and, so you see each arrangement rather than memorise words, the matching balloon geometries are drawn in the figure below it:

SN	Hybridization	Electron geometry	Ideal angle	Example
2	$s p$	Linear	$180°$	$CO_{2}$
3	$s p^{2}$	Trigonal planar	$120°$	$BF_{3}$
4	$s p^{3}$	Tetrahedral	$109.5°$	$CH_{4}$
5	$s p^{3} d$	Trigonal bipyramidal	$120°/90°$	$PCl_{5}$
6	$s p^{3} d^{2}$	Octahedral	$90°$	$SF_{6}$
7	$s p^{3} d^{3}$	Pentagonal bipyramidal	$72°/90°$	$IF_{7}$

Figure — Valence Bond Theory (VBT) — hybridization (sp, sp², sp³, sp³d, sp³d²)

Prerequisite links if a step feels shaky: VSEPR Theory, Sigma and Pi Bonds, Atomic Orbitals (s, p, d shapes), Formal Charge and Lewis Structures, Bond Angle and Bent's Rule.

Recall Prerequisite recall: why π bonds share only one axis (needed in L3)

A σ bond points straight along the line joining two atoms. A π bond is made from two p orbitals lying sideways, overlapping above and below that same line. So a double bond (1 σ + 1 π) and a triple bond (1 σ + 2 π) all still point along one single direction — the extra π's add no new direction. That is exactly why a multiple bond adds only 1 to $n_{σ}$ . (Full detail: Sigma and Pi Bonds.)

Level 1 — Recognition

Here you only need to read off SN and match the table. No traps in the chemistry yet — just careful counting.

Recall Solution Q1

WHAT: Be is the central atom. Its valence electrons $V = 2$ ; the species is neutral so $q = 0$ .
Be forms 2 single bonds to Cl, so $n_{σ} = 2$ (all single, so $B_{e} = n_{σ} = 2$ ). $LP = \frac{V - n _{σ} - q}{2} = \frac{2 - 2 - 0}{2} = 0$
$SN = n_{σ} + LP = 2 + 0 = 2$ .
Table (and the SN 2 panel in the figure above): $s p, linear, 180°$ .

Recall Solution Q2

C central, $V = 4$ , $q = 0$ , bonded to 4 Cl so $n_{σ} = 4$ (all single). $LP = \frac{4 - 4 - 0}{2} = 0, SN = 4 + 0 = 4$
Table (SN 4 tetrahedron panel): $s p^{3}, tetrahedral, 109.5°$ . Because there are no lone pairs, the molecular shape equals the electron geometry.

Level 2 — Application

Now lone pairs enter. You must apply the one LP formula and remember that lone pairs occupy hybrid orbitals and count toward SN.

Recall Solution Q3

N central, $V = 5$ , $q = 0$ . Bonded to 3 H, so $n_{σ} = 3$ (all single). $LP = \frac{5 - 3 - 0}{2} = 1$
$SN = 3 + 1 = 4 \Rightarrow s p^{3}$ .
Electron geometry (counting the lone pair) = tetrahedral; molecular shape (atoms only) = trigonal pyramidal.
WHY the real angle is $107°$ , not $109.5°$ : the single lone pair sits fatter and closer to N and pushes the three N–H bonds together — see the "fatter balloon" sketch in the Bent's-rule figure at L5, and Bond Angle and Bent's Rule.

Recall Solution Q4

O central, $V = 6$ . The species is a cation, so $q = + 1$ . O forms 3 σ bonds ( $n_{σ} = 3$ , all single).
Apply the everyday formula — the cation charge simply subtracts: $LP = \frac{V - n _{σ} - q}{2} = \frac{6 - 3 - ( + 1 )}{2} = \frac{2}{2} = 1$
$SN = 3 + 1 = 4 \Rightarrow s p^{3}, trigonal pyramidal$ , just like $NH_{3}$ .

Recall Solution Q5

Xe central, $V = 8$ , $q = 0$ . Bonded to 2 F, $n_{σ} = 2$ (all single). $LP = \frac{8 - 2 - 0}{2} = 3$
$SN = 2 + 3 = 5 \Rightarrow s p^{3} d$ .
The 3 lone pairs sit in the equatorial plane (more room), the 2 F go axial, giving a $linear (180°)$ molecule. Electron geometry is trigonal bipyramidal; molecular shape is linear.

Level 3 — Analysis

Multiple bonds and geometry reasoning. The key discipline: a double or triple bond counts as ONE toward $n_{σ}$ (only the σ part sets direction; the π part rides along on unhybridized p orbitals — see the prerequisite recall at the top). Whenever a multiple bond appears, use the general LP formula with $B_{e}$ .

Recall Solution Q6

C central, $V = 4$ , $q = 0$ . Each C=O is a double bond = 1 σ + 1 π. C has 2 neighbours, so $n_{σ} = 2$ (direction count).
Because there are multiple bonds, the everyday formula does not apply — use the general one with $B_{e}$ = electrons C contributes to bonding. Each double bond takes 2 electrons from C, so $B_{e} = 2 + 2 = 4$ : $LP = \frac{V - B _{e} - q}{2} = \frac{4 - 4 - 0}{2} = 0$ This is exactly why we introduced $B_{e}$ : with the single-bond formula you would have (wrongly) written $(4 - 2 - 0) /2 = 1$ , which is not the real lone-pair count. The general formula gives the correct $LP = 0$ directly.
$SN = n_{σ} + LP = 2 + 0 = 2 \Rightarrow s p, linear, 180°$ . (SN uses $n_{σ} = 2$ , not $B_{e}$ .)
WHY the two π bonds are perpendicular (the requested WHY): carbon is $s p$ , so it used its $2 s$ + one p orbital ( $p_{x}$ , along the O–C–O axis) to make the two hybrids. That leaves two untouched p orbitals, $p_{y}$ and $p_{z}$ . By the very shape of p orbitals, $p_{y}$ and $p_{z}$ point along mutually perpendicular axes ( $90°$ apart). One (say $p_{y}$ ) π-bonds to the left O, the other ( $p_{z}$ ) π-bonds to the right O. So the two π bonds are forced $90°$ apart — it is a direct consequence of which p orbitals were left over.
Census: 2 σ bonds (one per O) and 2 π bonds ( $p_{y}$ and $p_{z}$ ). See Sigma and Pi Bonds and the figure.

Recall Solution Q7

Each C is bonded to 1 H (single) and 1 C (triple). Direction count: $n_{σ} = 2$ (one H, one C).
General formula, $B_{e}$ for one C = $1$ (to H) $+ 3$ (to the triple bond) $= 4$ : $LP = (4 - 4 - 0) /2 = 0$ .
$SN = n_{σ} + LP = 2 + 0 = 2 \Rightarrow s p$ on each carbon, linear, $180°$ .
Total census: σ bonds $= 2$ (C–H) $+ 1$ (C–C) $= 3$ ; π bonds $= 2$ (the two leftover p orbitals on each carbon, $p_{y}$ and $p_{z}$ , overlapping sideways — again $90°$ apart, same reason as $CO_{2}$ ).

Recall Solution Q8

S central, $V = 6$ , $q = 0$ , bonded to 4 F, $n_{σ} = 4$ (all single, so $B_{e} = 4$ ). $LP = \frac{6 - 4 - 0}{2} = 1, SN = 4 + 1 = 5 \Rightarrow s p^{3} d$
Electron geometry = trigonal bipyramidal (SN 5 panel of the top figure). The single lone pair takes an equatorial spot (least repulsion), giving the $see-saw$ shape. Angles are compressed slightly below $90°/120°$ by the lone pair.

Level 4 — Synthesis

Here you combine LP counting, the charge sign rule, and lone-pair placement — usually noble-gas or interhalogen species.

Recall Solution Q9

Xe central, $V = 8$ , $q = 0$ , bonded to 4 F, $n_{σ} = 4$ (all single). $LP = \frac{8 - 4 - 0}{2} = 2, SN = 4 + 2 = 6 \Rightarrow s p^{3} d^{2}$
Electron geometry = octahedral (SN 6 panel of top figure). The 2 lone pairs sit trans (opposite, along one axis) to minimise their mutual repulsion, leaving the 4 F in one plane.
Molecular shape = $square planar, 90°$ . See the figure below for the trans lone-pair placement.

Recall Solution Q10

Cl central, $V = 7$ . The species is an anion, so $q = - 1$ . Bonded to 2 F, $n_{σ} = 2$ (all single).
Apply the everyday formula — subtracting a negative charge adds an electron: $LP = \frac{V - n _{σ} - q}{2} = \frac{7 - 2 - ( - 1 )}{2} = \frac{6}{2} = 3$
$SN = 2 + 3 = 5 \Rightarrow s p^{3} d$ (trigonal bipyramidal electron geometry).
WHY the 3 lone pairs go equatorial (the key reason, stated explicitly): in a trigonal bipyramid the three equatorial positions are $120°$ apart, whereas any axial position sits only $90°$ from its equatorial neighbours. Lone pairs, being fat and repulsive, "want" the widest separations. Placing all three lone pairs equatorially gives them mutual $120°$ separations (largest possible) and keeps them $90°$ from just the two axial atoms; placing any lone pair axially would force costly $90°$ lone-pair–lone-pair contacts. So minimising lone-pair–lone-pair repulsion forces the equatorial choice, pushing the 2 F atoms axial.
Result: 2 axial F on a straight line ⇒ $linear, 180°$ . Same family as $XeF_{2}$ .

Recall Solution Q11

Br central, $V = 7$ , $q = 0$ , bonded to 5 F, $n_{σ} = 5$ (all single). $LP = \frac{7 - 5 - 0}{2} = 1, SN = 5 + 1 = 6 \Rightarrow s p^{3} d^{2}$
Octahedral electron geometry; the single lone pair takes one vertex, pushing the 4 planar F's slightly down ⇒ $square pyramidal$ .

Level 5 — Mastery

Full molecules with several different central atoms, or subtle geometry consequences. You must analyse each central atom separately.

Recall Solution Q12

End carbons: each is bonded to 2 H (single) + 1 C (double) = 3 neighbours, so $n_{σ} = 3$ . General $B_{e} = 1 + 1 + 2 = 4$ , LP $= (4 - 4 - 0) /2 = 0$ . $SN = 3 \Rightarrow s p^{2}$ .
Central carbon: bonded to 2 C, both double bonds, so $n_{σ} = 2$ . General $B_{e} = 2 + 2 = 4$ , LP $= 0$ . $SN = 2 \Rightarrow s p$ .
WHY perpendicular planes: the central $s p$ carbon has two unhybridized p orbitals, $p_{y}$ and $p_{z}$ , at $90°$ (same fact used in Q6). One π-bonds to the left carbon, the other to the right. Each end $CH_{2}$ plane must contain the p orbital it π-bonds to — and those two p orbitals are $90°$ apart. So the two ends are twisted $90°$ relative to each other. The figure below shows the twist.

Recall Solution Q13

Central I: $V = 7$ , anion so $q = - 1$ , bonded to 2 I ( $n_{σ} = 2$ , all single). $LP = \frac{7 - 2 - ( - 1 )}{2} = \frac{6}{2} = 3, SN = 2 + 3 = 5 \Rightarrow s p^{3} d$
3 equatorial lone pairs (same $120°$ -maximising argument as Q10), 2 axial I atoms ⇒ $linear$ . Same family as $XeF_{2}$ and $ClF_{2}^{-}$ — recognising the pattern is the mastery skill.

Recall Solution Q14

All three central atoms are $s p^{3}$ ( $SN = 4$ ): $CH_{4}$ (0 LP), $NH_{3}$ (1 LP), $H_{2} O$ (2 LP).
Trend: each extra lone pair squeezes the bonding pairs harder (lone pairs are fatter and sit closer to the nucleus), so the angle drops: $CH_{4} : 109.5° > NH_{3} : 107° > H_{2} O : 104.5°$
WHY (Bent's rule, quantified): lone pairs hog s-character, forcing the bonding hybrids to carry more p-character; higher p-character means a smaller angle. The figure below plots angle against lone-pair count to show the steady drop. See Bond Angle and Bent's Rule.

Recall One-line self-test recap

Steric number ::: SN = number of σ-bond neighbours (n_σ) + lone pairs on the central atom. Everyday lone-pair formula (single bonds only) ::: LP = (V − n_σ − q)/2, with q the signed overall charge. General lone-pair formula (any bonds) ::: LP = (V − B_e − q)/2, where B_e = electrons the central atom contributes to bonding (1/2/3 per single/double/triple). Does a triple bond add 3 to SN? ::: No — only 1 to n_σ (only the σ part sets direction; the two π bonds ride the same axis). It contributes 3 to B_e though. Why are the two π bonds in CO₂ perpendicular? ::: The sp carbon leaves two untouched p orbitals p_y and p_z, which point along axes 90° apart. Why do 3 lone pairs go equatorial in an sp³d? ::: Equatorial positions are 120° apart (widest), minimising lone-pair–lone-pair repulsion; axial would force costly 90° contacts. Why is XeF₄ square planar not tetrahedral? ::: SN = 6 (4 bonds + 2 lone pairs) → sp³d² octahedral; the 2 lone pairs go trans, leaving 4 F in a plane. SN 7 case? ::: sp³d³, pentagonal bipyramidal, e.g. IF₇.