How do you determine hybridisation quickly?

Count total valence electrons around the central atom, divide by 8, and add lone pairs to the quotient. The total maps to hybridisation: 2 gives sp, 3 sp2, 4 sp3, 5 sp3d, 6 sp3d2, 7 sp3d3.

What is the difference between a sigma bond and a pi bond?

A sigma bond is formed by head-on overlap along the internuclear axis with cylindrical symmetry. A pi bond is formed by sideways overlap of parallel p orbitals with density above and below the axis. Sigma bonds are stronger and form first; pi bonds can only form after a sigma bond exists.

In a trigonal bipyramidal molecule with mixed atoms, which goes axial?

The more electronegative atom occupies the axial position because it pulls electron density away, reducing bp-bp repulsion at the strong 90-degree angle. This is why F goes axial in PF2Br3 and Br goes equatorial.

Valence Bond Theory And Hybridisation

Q: Why are hybrid orbitals more effective in bonding than pure atomic orbitals?

Hybrid orbitals have a larger positive lobe and smaller negative lobe, giving them more directional character. This concentrates electron density along the bond axis for stronger overlap and gives stable geometries.

Q: Why is PCl5 trigonal bipyramidal with two different bond lengths?

Axial bonds experience three 90-degree repulsions while equatorial bonds experience only two, so axial bonds are longer (240 pm) and weaker than equatorial bonds (202 pm), explaining why PCl5 is chemically reactive.

Q: Why can PF5 exist but NF5 cannot?

Phosphorus is third period with vacant 3d orbitals, allowing sp3d hybridisation and octet expansion. Nitrogen is second period with no d orbitals in its valence shell, so it cannot expand beyond an octet.

Q: Does hybridisation change when NH3 forms NH4 plus?

No. Nitrogen remains sp3 hybridised in both. The shape changes from trigonal pyramidal to perfectly tetrahedral, but hybridisation is unchanged.

Q: What is the inert pair effect?

The reluctance of outermost ns2 electrons of heavier p-block elements to participate in bonding, making a lower oxidation state (group number minus 2) more stable than the group oxidation state. Seen prominently in Tl, Pb, and Bi.

ChemistryChemical Bonding And Molecular StructureFor NEET aspirants

Valence Bond Theory (VBT), proposed by Heitler and London (1927) and extended by Pauling and Slater, explains covalent bond formation as the overlap of half-filled atomic orbitals of two atoms, with pairing of their electron spins. VBT introduces the concept of hybridisation, the mixing of atomic orbitals of similar energy on the same atom to give new hybrid orbitals of equal energy and shape that point in specific directions. This one idea explains why methane is tetrahedral ( $s p^{3}$ ), why ethene is planar ( $s p^{2}$ ), why acetylene is linear ( $s p$ ), and why $PC l_{5}$ is trigonal bipyramidal ( $s p^{3} d$ ) and $S F_{6}$ is octahedral ( $s p^{3} d^{2}$ ). This concept covers VBT, orbital overlap, $σ$ and $π$ bonds, all six main hybridisation schemes, the numerical rule for determining hybridisation, worked examples for over twenty molecules, back bonding, maximum covalency, and inert pair effect.

Key Ideas - Quick Reference

Number of hybrid orbitals rule: (total valence electrons) $\div 8$ = ( $σ$ -bonds) remainder (non-bonded electrons); lone pairs = non-bonded $\div 2$ ; hybrid orbitals = $σ$ -bonds + lone pairs.
$s p$ : 2 hybrid orbitals, 180°, linear, e.g. $BeC l_{2}$ , $C_{2} H_{2}$ .
$s p^{2}$ : 3 hybrid orbitals, 120°, trigonal planar, e.g. $B F_{3}$ , $C_{2} H_{4}$ .
$s p^{3}$ : 4 hybrid orbitals, 109.5°, tetrahedral, e.g. $C H_{4}$ , $N H_{3}$ , $H_{2} O$ .
$s p^{3} d$ : 5 hybrid orbitals, trigonal bipyramidal, e.g. $PC l_{5}$ , $S F_{4}$ .
$s p^{3} d^{2}$ : 6 hybrid orbitals, octahedral, e.g. $S F_{6}$ , $Xe F_{4}$ .
$s p^{3} d^{3}$ : 7 hybrid orbitals, pentagonal bipyramidal, e.g. $I F_{7}$ .
s-character order (for bond angle and electronegativity): $s p$ (50%) $>$ $s p^{2}$ (33%) $>$ $s p^{3}$ (25%).

1. Valence Bond Theory - The Basic Idea

Lewis theory explains where the electrons go in a molecule but says nothing about why a covalent bond forms or how much energy is released. Valence Bond Theory fills that gap using quantum mechanics.

Consider two hydrogen atoms A and B far apart. Each has one electron in its 1s orbital. As the atoms approach each other, four new interactions come into play:

Attractive: nucleus A - electron B, and nucleus B - electron A.
Repulsive: electron A - electron B, and nucleus A - nucleus B.

Experiment (and calculation) shows that at close approach, the new attractive forces are stronger than the new repulsive forces. The system loses energy as the atoms come closer, reaching a minimum-energy state at an internuclear distance of 74 pm - the bond length of $H_{2}$ . Any closer approach makes nuclear repulsion dominant and pushes the energy up sharply.

Figure 1: Potential energy curve for the formation of an

H_{2}

molecule. As two H atoms approach, attractive forces dominate and the energy falls; at very short distances, nuclear repulsion causes a steep rise. The minimum at 74 pm is the equilibrium bond length, and the well depth (

\sim 436 kJ mo l^{- 1}

) equals the bond dissociation energy.

Orbital overlap concept

A covalent bond forms when a half-filled atomic orbital of one atom overlaps with a half-filled atomic orbital of another atom, and the two electrons pair up with opposite spins. The greater the overlap, the stronger the bond.

2. Sigma ( $σ$ ) and Pi ( $π$ ) Bonds

Depending on how the atomic orbitals overlap, two types of covalent bonds form.

Sigma ( $σ$ ) bond

Formed by head-on (axial) overlap of atomic orbitals along the internuclear axis. The electron density is maximum between the two nuclei and is cylindrically symmetric about the axis. Three types of $σ$ -overlap are possible:

s-s overlap: Two s orbitals overlap along the axis. Example: $H_{2}$ molecule.
s-p overlap: An s orbital of one atom overlaps head-on with a p orbital of another. Example: H-F bond.
p-p overlap: Two p orbitals overlap along their lobes. Example: F-F bond in $F_{2}$ .

Pi ( $π$ ) bond

Formed by sideways (lateral) overlap of parallel p orbitals perpendicular to the internuclear axis. The electron density lies above and below the axis in two lobes, with a nodal plane containing the axis. $π$ -bonds occur only in multiple bonds - always alongside an existing $σ$ -bond.

Strength and formation order

The first bond between two atoms is always a $σ$ -bond.
A double bond = 1 $σ$ + 1 $π$ .
A triple bond = 1 $σ$ + 2 $π$ .
$σ$ -bonds are stronger than $π$ -bonds because axial overlap is more extensive than lateral overlap.
A $π$ -bond can only form after a $σ$ -bond is already in place.

3. The Concept of Hybridisation

VBT with pure atomic orbitals cannot explain observed molecular geometries. For example, carbon's ground state is $[H e] 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1}$ with only two half-filled orbitals - so pure VBT would predict divalent carbon (like $C H_{2}$ ), not the tetravalent methane we observe. Even after promoting one 2s electron to 2p to give four unpaired electrons, the four bonds would be non-equivalent (three from p orbitals at 90° and one from the s orbital in any direction), yet methane has four equivalent C-H bonds at 109.5°.

Pauling resolved this by proposing hybridisation.

Hybridisation is the mixing (intermixing) of atomic orbitals of nearly equal energy belonging to the same atom to produce a new set of orbitals, called hybrid orbitals, having equivalent energies and shapes and definite directional orientations in space.

Salient features of hybridisation

The number of hybrid orbitals formed equals the number of atomic orbitals mixed.
Hybrid orbitals are always equivalent in energy and shape.
Hybrid orbitals form stronger bonds than pure atomic orbitals (more directional character, better overlap).
Hybrid orbitals point in specific directions in space to minimise repulsion, giving the molecule its geometry.

Conditions for hybridisation

Only orbitals of the valence shell hybridise.
Orbitals to be hybridised should have similar (nearly equal) energies.
Promotion of electrons is not always essential - filled orbitals can also participate.
Only half-filled orbitals are needed for bond formation, but hybridisation itself can involve any orbitals of similar energy.

4. Types of Hybridisation

sp hybridisation - linear geometry

One s and one p orbital mix to form two equivalent $s p$ hybrid orbitals, oriented at 180° to each other. Each hybrid has 50% s-character and 50% p-character.

Example: $BeC l_{2}$ . Ground state Be: $1 s^{2} 2 s^{2}$ . In the excited state, one 2s electron is promoted to 2p, giving $1 s^{2} 2 s^{1} 2 p^{1}$ . The 2s and 2p orbitals hybridise into two $s p$ hybrids, each overlapping with a 2p orbital of chlorine to form linear Cl-Be-Cl.

Remaining unhybridised p orbitals can form $π$ -bonds. This is why $s p$ carbon in ethyne ( $HC \equiv CH$ ) forms one $σ$ -bond and two $π$ -bonds with the neighbouring carbon.

Figure 2: (a) A 2s orbital (spherical) and a 2p_z orbital (dumbbell along z) mix to form two equivalent sp hybrid orbitals oriented 180° apart along z, giving a linear geometry (as in

Be

). Each sp hybrid has a large positive-phase lobe and a small negative-phase back-lobe. (b) In

BeC l_{2}

, each Be sp hybrid overlaps end-on with a Cl 3p_z orbital; the dark oval marks the

σ

-bond electron density where the orbitals interlock. Two collinear

σ

-bonds give the linear

Cl - Be - Cl

molecule.

$s p^{2}$ hybridisation - trigonal planar geometry

One s and two p orbitals mix to form three equivalent $s p^{2}$ hybrid orbitals in a plane at 120° to each other. Each hybrid has 33% s-character and 67% p-character.

Example: $B F_{3}$ . Excited-state B: $1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1}$ . Hybridisation of $2 s$ , $2 p_{x}$ , $2 p_{y}$ gives three $s p^{2}$ hybrids in the xy plane at 120°, each overlapping with a 2p orbital of fluorine to form the trigonal planar $B F_{3}$ molecule.

The remaining pure p orbital (perpendicular to the plane) can form a $π$ -bond. This explains why $s p^{2}$ carbon in ethene forms one $σ$ and one $π$ with its neighbour, and why benzene has a delocalised $π$ system above and below the ring.

$s p^{3}$ hybridisation - tetrahedral geometry

One s and three p orbitals mix to form four equivalent $s p^{3}$ hybrid orbitals directed towards the four corners of a regular tetrahedron at 109.5° to each other. Each hybrid has 25% s-character and 75% p-character.

Example: CH₄. Excited-state C: $1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ . All four valence orbitals mix into four $s p^{3}$ hybrids, each overlapping with the 1s of a hydrogen atom.

$N H_{3}$ and $H_{2} O$ also involve $s p^{3}$ hybridisation. In $N H_{3}$ , three hybrids form N-H bonds and one holds a lone pair, giving trigonal pyramidal shape with the H-N-H angle reduced to 107° by lone-pair repulsion. In $H_{2} O$ , two hybrids form O-H bonds and two hold lone pairs, giving bent shape with H-O-H reduced to 104.5°.

Figure 3: The three basic hybridisation geometries:

s p

linear (180°),

s p^{2}

trigonal planar (120°), and

s p^{3}

tetrahedral (109.5°).

$s p^{3} d$ hybridisation - trigonal bipyramidal geometry

One s, three p, and one d orbital mix to form five $s p^{3} d$ hybrid orbitals in a trigonal bipyramidal arrangement: three equatorial hybrids in a plane at 120° and two axial hybrids at 90° to the equatorial plane.

Example: PCl₅. Excited-state P has five unpaired electrons in $3 s^{1} 3 p^{3} 3 d^{1}$ . The five orbitals hybridise into $s p^{3} d$ hybrids that overlap with 3p orbitals of five chlorine atoms.

The three equatorial P-Cl bonds are shorter (202 pm) than the two axial P-Cl bonds (240 pm), because axial bonds experience three 90° repulsions with the equatorial bonds, while equatorial bonds experience only two 90° repulsions (with the axial bonds) and two 120° repulsions. This makes axial bonds weaker and $PC l_{5}$ chemically reactive at those positions.

$s p^{3} d^{2}$ hybridisation - octahedral geometry

One s, three p, and two d orbitals mix to form six $s p^{3} d^{2}$ hybrid orbitals directed to the six corners of a regular octahedron at 90° to each other.

Example: SF₆. Excited-state S has six unpaired electrons in $3 s^{1} 3 p^{3} 3 d^{2}$ . Six equivalent bonds form to six fluorine atoms, all at 90° - no distortion, no reactive positions.

$s p^{3} d^{3}$ hybridisation - pentagonal bipyramidal geometry

One s, three p, and three d orbitals give seven hybrid orbitals in a pentagonal bipyramidal shape - five equatorial hybrids at 72° apart in a plane, and two axial hybrids perpendicular to that plane.

Example: IF₇. The only common example, showing seven I-F bonds.

Hybridisation	Orbitals mixed	Geometry	Bond angle	Example
$s p$	$s + p$	Linear	180°	$BeC l_{2}$ , $C O_{2}$
$s p^{2}$	$s + 2 p$	Trigonal planar	120°	$B F_{3}$ , $C_{2} H_{4}$
$s p^{3}$	$s + 3 p$	Tetrahedral	109.5°	$C H_{4}$ , $N H_{4}^{+}$
$s p^{3} d$	$s + 3 p + d$	Trigonal bipyramidal	90°, 120°	$PC l_{5}$ , $S F_{4}$
$s p^{3} d^{2}$	$s + 3 p + 2 d$	Octahedral	90°	$S F_{6}$ , $Xe F_{4}$
$s p^{3} d^{3}$	$s + 3 p + 3 d$	Pentagonal bipyramidal	72°, 90°	$I F_{7}$

5. The Numerical Rule for Determining Hybridisation

For any main-group molecule or ion, use this quick recipe:

Identify the central atom and the surrounding (peripheral) atoms.
Count total valence electrons. Add electrons for negative charge, subtract for positive charge.
Divide by 8. The quotient gives the number of $σ$ -bonds; the remainder gives the non-bonded electrons.
Lone pairs = non-bonded electrons ÷ 2.
Total hybrid orbitals = $σ$ -bonds + lone pairs. This gives the hybridisation.

Quick lookup: 2 orbitals →

s p

; 3 →

s p^{2}

; 4 →

s p^{3}

; 5 →

s p^{3} d

; 6 →

s p^{3} d^{2}

; 7 →

s p^{3} d^{3}

6. Worked Examples of Hybridisation

Solved Example 1

Predict the hybridisation and shape of

NC l_{3}

Solution:

Valence electrons $= 5 + (3 \times 7) = 26$ . $26 \div 8 = 3$ remainder $2$ . So $3$ $σ$ -bonds + $1$ lone pair, giving $4$ hybrid orbitals $\Rightarrow$ $s p^{3}$ . Shape: trigonal pyramidal.

Solved Example 2

Predict hybridisation of B in

BB r_{3}

Solution:

Valence electrons $= 3 + (3 \times 7) = 24$ . $24 \div 8 = 3$ remainder $0$ . So $3$ $σ$ -bonds + $0$ lone pairs $\Rightarrow$ $s p^{2}$ . Shape: trigonal planar.

Solved Example 3

Predict hybridisation and shape of

SiC l_{4}

and

C I_{4}

Solution:

Both: valence electrons $= 4 + (4 \times 7) = 32$ ; $32 \div 8 = 4$ remainder $0$ ; $4$ $σ$ -bonds + $0$ lone pairs $\Rightarrow$ $s p^{3}$ . Tetrahedral.

Solved Example 4

Predict hybridisation of S in

S F_{6}

Solution:

Valence electrons $= 6 + (6 \times 7) = 48$ ; $48 \div 8 = 6$ remainder $0$ ; $6$ $σ$ -bonds + $0$ lone pairs $\Rightarrow$ $s p^{3} d^{2}$ . Octahedral.

Solved Example 5

Predict hybridisation and shape of

Cl F_{3}

Solution:

Valence electrons $= 7 + (3 \times 7) = 28$ ; $28 \div 8 = 3$ remainder $4$ ; $3$ $σ$ -bonds + $2$ lone pairs $\Rightarrow$ $s p^{3} d$ . In TBP electron geometry, both lone pairs occupy equatorial positions to minimise 90° repulsions, giving a T-shaped molecular geometry.

Solved Example 6

Predict hybridisation and shape of

Xe F_{4}

and

Xe F_{2}

Solution:

$Xe F_{4}$ : valence $= 8 + 28 = 36$ ; $36 \div 8 = 4$ remainder $4$ ; $4$ $σ$ + $2$ lp $\Rightarrow$ $s p^{3} d^{2}$ . Both lone pairs go trans in the octahedron. Shape: square planar.

$Xe F_{2}$ : valence $= 8 + 14 = 22$ ; $22 \div 8 = 2$ remainder $6$ ; $2$ $σ$ + $3$ lp $\Rightarrow$ $s p^{3} d$ . All three lone pairs go equatorial in the TBP. Shape: linear.

Solved Example 7

Predict hybridisation of C in

C O_{2}

C O_{3}^{2 -}

, and

N O_{2}^{-}

Solution:

$C O_{2}$ : valence $= 4 + 12 = 16$ ; $16 \div 8 = 2$ remainder $0$ ; $2$ $σ$ + $0$ lp $\Rightarrow$ $s p$ . Linear. (Two $π$ -bonds also present.)

$C O_{3}^{2 -}$ : valence $= 4 + 18 + 2 = 24$ ; $24 \div 8 = 3$ remainder $0$ ; $3$ $σ$ + $0$ lp $\Rightarrow$ $s p^{2}$ . Trigonal planar. One $π$ -bond delocalised.

$N O_{2}^{-}$ : valence $= 5 + 12 + 1 = 18$ ; $18 \div 8 = 2$ remainder $2$ ; $2$ $σ$ + $1$ lp $\Rightarrow$ $s p^{2}$ . Bent (angular).

Solved Example 8

Predict hybridisation of central atoms in

Xe O_{3}

XeO F_{4}

Xe O_{2} F_{2}

Solution:

$Xe O_{3}$ : valence $= 8 + 18 = 26$ ; $26 \div 8 = 3$ remainder $2$ ; $3$ $σ$ + $1$ lp $\Rightarrow$ $s p^{3}$ . Trigonal pyramidal.

$XeO F_{4}$ : valence $= 8 + 6 + 28 = 42$ ; $42 \div 8 = 5$ remainder $2$ ; $5$ $σ$ + $1$ lp $\Rightarrow$ $s p^{3} d^{2}$ . Square pyramidal.

$Xe O_{2} F_{2}$ : valence $= 8 + 12 + 14 = 34$ ; $34 \div 8 = 4$ remainder $2$ ; $4$ $σ$ + $1$ lp $\Rightarrow$ $s p^{3} d$ . See-saw (distorted TBP).

Solved Example 9

When

N H_{3}

is treated with HCl, does the hybridisation of nitrogen change?

Solution:

In $N H_{3}$ , N is $s p^{3}$ hybridised (3 bond pairs + 1 lone pair). In $N H_{4}^{+}$ formed by protonation, N still has 4 pairs (now all bond pairs), so it remains $s p^{3}$ hybridised. The molecular geometry changes from trigonal pyramidal to perfectly tetrahedral (all H-N-H angles = 109.5°), but the hybridisation is unchanged.

7. Hybridisation in Organic Molecules

Ethane ( $C_{2} H_{6}$ ) - all $s p^{3}$

Both carbons are $s p^{3}$ hybridised. One $s p^{3}$ hybrid of each C overlaps to form the C-C $σ$ -bond; the other three $s p^{3}$ hybrids overlap with 1s of H atoms. C-C bond length 154 pm; C-H bond length 109 pm; free rotation about C-C axis.

Ethene ( $C_{2} H_{4}$ ) - $s p^{2}$

Each C is $s p^{2}$ hybridised. One $s p^{2}$ hybrid of each C overlaps to form the C-C $σ$ -bond; the other two $s p^{2}$ hybrids on each C bond to H. The unhybridised $p_{z}$ orbitals (perpendicular to the molecular plane) overlap sideways to form a $π$ -bond. The C=C double bond has length 134 pm; H-C-H angle 116°; molecule is planar; no rotation possible about C=C.

Ethyne ( $C_{2} H_{2}$ ) - $s p$

Each C is $s p$ hybridised. One $s p$ hybrid of each C overlaps to form the C-C $σ$ -bond; the other $s p$ hybrid on each C bonds to H. Two unhybridised p orbitals on each C ( $p_{y}$ and $p_{z}$ ) overlap sideways to form two $π$ -bonds. The C≡C triple bond has length 120 pm; H-C-C angle 180°; molecule is linear.

8. Back Bonding

Back bonding occurs when a filled orbital of one atom overlaps with a vacant orbital of another to form a $π$ -bond in the opposite direction to the primary $σ$ -bond. Common when a small, highly electronegative donor atom (with a lone pair) is bonded to an electron-deficient acceptor atom.

Classic example: $B F_{3}$ . Boron is $s p^{2}$ hybridised with an empty $p_{z}$ orbital perpendicular to the plane. Each fluorine has three lone pairs. One filled fluorine 2p lone-pair orbital overlaps with the empty boron 2p, giving a partial $p_{π} - p_{π}$ back bond (F $\to$ B).

Effects of back bonding:

Shortens the B-F bond below the expected single-bond length.
Reduces Lewis acidity of $B F_{3}$ (boron's electron deficiency is partly satisfied). This is why $B F_{3}$ is a weaker Lewis acid than $BC l_{3}$ and $BB r_{3}$ (heavier halogens have larger, more diffuse p orbitals that overlap poorly with boron's small 2p, weakening back bonding).
Explains why $(Si H_{3})_{3} N$ (trisilylamine) is planar with N as $s p^{2}$ : Si has empty 3d orbitals that accept lone-pair density from N. Trimethylamine $(C H_{3})_{3} N$ is pyramidal ( $s p^{3}$ ) because carbon has no such empty orbitals.

9. Maximum Covalency and Inert Pair Effect

Maximum covalency

An element's maximum covalency equals its group number. Second-period elements (Li to F) have no d orbitals in the valence shell, so their maximum covalency is 4 (C forms $C H_{4}$ ; N is limited to 4, forming $N H_{4}^{+}$ but never $N F_{5}$ ). Third-period and heavier elements have accessible d orbitals and can expand their octet, so sulphur can form $S F_{6}$ (covalency 6) and iodine can form $I F_{7}$ (covalency 7).

Inert pair effect

Heavier p-block elements show two stable oxidation states: one equal to the group number, and one two units lower. For Pb ( $6 s^{2} 6 p^{2}$ ), both +II and +IV are common, but +II is more stable because the $6 s^{2}$ electrons are reluctant to participate in bonding. This is the inert pair effect.

Reason: bond energy released after forming Pb-X bonds using the $6 s^{2}$ electrons is less than the energy needed to unpair them (larger atoms have longer, weaker bonds). Similar effect in Tl (+I preferred over +III), Bi (+III over +V), Sn (+II grows more common as we go from Sn to Pb).

Solved Example 10

Why does

Pb I_{4}

not exist while

PbC l_{4}

does?

Solution:

Due to the inert pair effect, Pb(+IV) is a strong oxidising state. Iodide ( $I^{-}$ ) is a good reducing agent - it reduces $P b^{4 +}$ to $P b^{2 +}$ while itself being oxidised to $I_{2}$ . So $Pb I_{4}$ would spontaneously decompose to $Pb I_{2} + I_{2}$ . $C l^{-}$ is not a strong enough reducing agent to reduce $P b^{4 +}$ , so $PbC l_{4}$ remains stable at low temperatures.

Common Mistakes to Avoid

Watch out

Confusing hybridisation with molecular shape. $N H_{3}$ is $s p^{3}$ but its shape is trigonal pyramidal (not tetrahedral). Shape depends on visible atoms only; hybridisation counts electron pairs.
Counting multiple bonds as multiple pairs for hybridisation. A double or triple bond counts as one $σ$ -bond for hybridisation. The $π$ -bonds use unhybridised p orbitals.
Forgetting to include lone pairs. Always calculate lone pairs from valence electrons; hybridisation = $σ$ -bonds + lone pairs.
Placing lone pairs in axial positions for $s p^{3} d$ . Lone pairs always go equatorial in TBP. This is why $Cl F_{3}$ is T-shaped, not trigonal pyramidal.
Assuming second-period elements can expand octets. N, O, F have no valence d orbitals - they never form $N F_{5}$ , $O F_{6}$ , or $I F_{7}$ -like structures.
Ignoring back bonding. $B F_{3}$ Lewis acidity, planarity of trisilylamine, and B-F bond shortening all trace back to $p_{π} - p_{π}$ or $p_{π} - d_{π}$ back bonding.
Confusing bond order with number of hybrid orbitals. Bond order counts electron pairs shared between two atoms; hybrid orbitals count how the central atom's orbitals mix.

Frequently Asked Questions

Q1. What is hybridisation and why is it needed?

Hybridisation is the mixing of atomic orbitals of similar energy on the same atom to give new equivalent hybrid orbitals with definite geometries. It is needed because pure atomic orbital overlap cannot explain why methane has four equivalent C-H bonds at 109.5°, or why ammonia is pyramidal with three equal N-H bonds. Hybridisation gives the observed shapes and equivalent bond lengths.

Q2. Why are hybrid orbitals more effective in bonding than pure atomic orbitals?

Hybrid orbitals have a larger positive lobe and a smaller negative lobe, giving them more directional character. This concentrates electron density along the bond axis, allowing stronger, more effective overlap with the orbital of the bonding partner. They also point in specific directions in space, giving stable geometries.

Q3. How do you determine the hybridisation of a central atom quickly?

Count total valence electrons around the central atom (including any charge adjustment), divide by 8, and add the lone pairs to the quotient. The total gives the number of hybrid orbitals, which maps to hybridisation: 2 → $s p$ , 3 → $s p^{2}$ , 4 → $s p^{3}$ , 5 → $s p^{3} d$ , 6 → $s p^{3} d^{2}$ , 7 → $s p^{3} d^{3}$ .

Q4. Why is $PC l_{5}$ trigonal bipyramidal with two different bond lengths?

$PC l_{5}$ is $s p^{3} d$ hybridised. Its three equatorial P-Cl bonds lie in a plane at 120°, and its two axial P-Cl bonds are perpendicular to that plane. The axial bonds experience three 90° repulsions with equatorial bonds, while equatorial bonds only experience two 90° repulsions. This makes axial bonds longer (240 pm) and weaker than equatorial bonds (202 pm), and explains why $PC l_{5}$ is chemically reactive.

Q5. Why can $P F_{5}$ exist but $N F_{5}$ cannot?

Phosphorus is a third-period element with vacant 3d orbitals in its valence shell, allowing $s p^{3} d$ hybridisation and expansion of the octet to 10 electrons. Nitrogen is a second-period element with no d orbitals in its valence shell, so it cannot expand beyond an octet. Its maximum covalency is 4.

Q6. What is the difference between a $σ$ -bond and a $π$ -bond?

A $σ$ -bond is formed by head-on (axial) overlap of orbitals along the internuclear axis, with maximum electron density between the nuclei and cylindrical symmetry about the axis. A $π$ -bond is formed by sideways overlap of parallel p orbitals with electron density above and below the axis and a nodal plane through it. $σ$ -bonds are stronger and can form independently; $π$ -bonds are weaker and can only form after a $σ$ -bond is in place.

Q7. Why is $B F_{3}$ a weaker Lewis acid than $BC l_{3}$ , despite F being more electronegative?

Back bonding. Fluorine's small 2p orbitals overlap effectively with boron's empty 2p orbital, forming a partial $p_{π} - p_{π}$ back bond that donates electron density from F to B. This partially satisfies B's electron deficiency, reducing its Lewis acidity. In $BC l_{3}$ and $BB r_{3}$ , the halogen p orbitals are 3p or 4p, larger and less able to overlap with B's small 2p, so back bonding is weaker and B remains more electron-deficient.

Q8. When $N H_{3}$ combines with H⁺ to form $N H_{4}^{+}$ , does the hybridisation change?

No. Nitrogen remains $s p^{3}$ hybridised in both $N H_{3}$ and $N H_{4}^{+}$ . In ammonia, three hybrids form N-H bonds and one holds a lone pair. When H⁺ bonds to that lone pair, all four hybrids now form N-H bonds. Hybridisation is the same, but the shape changes from trigonal pyramidal to perfectly tetrahedral.

Q9. In a trigonal bipyramidal molecule with mixed halogens, which atom goes axial?

The more electronegative atom occupies the axial position because axial bonds have three 90° repulsions. Placing the more electronegative atom axially pulls electron density away from the central atom, reducing bp-bp repulsion at the 90° angle. This is why in $P F_{2} B r_{3}$ , the two F atoms occupy axial positions and the three Br atoms occupy equatorial.

Q10. What is the inert pair effect and which elements show it?

The inert pair effect is the reluctance of the outermost $n s^{2}$ electrons of heavier p-block elements to participate in bonding, so a lower oxidation state (group number minus 2) becomes more stable than the group oxidation state. Prominent in Tl, Pb, and Bi. So Tl(+I) is more stable than Tl(+III), Pb(+II) more than Pb(+IV), Bi(+III) more than Bi(+V). This explains why $Pb I_{4}$ does not exist and why Bi(V) compounds are strong oxidisers.

Previous year questions on Valence Bond Theory And Hybridisation

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1. Valence Bond Theory - The Basic Idea

Orbital overlap concept

2. Sigma (σ) and Pi (π) Bonds

Sigma (σ) bond

Pi (π) bond

Strength and formation order

3. The Concept of Hybridisation

Salient features of hybridisation

Conditions for hybridisation

4. Types of Hybridisation

sp hybridisation - linear geometry

sp2 hybridisation - trigonal planar geometry

sp3 hybridisation - tetrahedral geometry

sp3d hybridisation - trigonal bipyramidal geometry

sp3d2 hybridisation - octahedral geometry

sp3d3 hybridisation - pentagonal bipyramidal geometry

5. The Numerical Rule for Determining Hybridisation

6. Worked Examples of Hybridisation

7. Hybridisation in Organic Molecules

Ethane (C2​H6​) - all sp3

Ethene (C2​H4​) - sp2

Ethyne (C2​H2​) - sp

8. Back Bonding

9. Maximum Covalency and Inert Pair Effect

Maximum covalency

Inert pair effect

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is hybridisation and why is it needed?

Q2. Why are hybrid orbitals more effective in bonding than pure atomic orbitals?

Q3. How do you determine the hybridisation of a central atom quickly?

Q4. Why is PCl5​ trigonal bipyramidal with two different bond lengths?

Q5. Why can PF5​ exist but NF5​ cannot?

Q6. What is the difference between a σ-bond and a π-bond?

Q7. Why is BF3​ a weaker Lewis acid than BCl3​, despite F being more electronegative?

Q8. When NH3​ combines with H⁺ to form NH4+​, does the hybridisation change?

Q9. In a trigonal bipyramidal molecule with mixed halogens, which atom goes axial?

Q10. What is the inert pair effect and which elements show it?

Previous year questions on Valence Bond Theory And Hybridisation

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2. Sigma ( $σ$ ) and Pi ( $π$ ) Bonds

Sigma ( $σ$ ) bond

Pi ( $π$ ) bond

$s p^{2}$ hybridisation - trigonal planar geometry

$s p^{3}$ hybridisation - tetrahedral geometry

$s p^{3} d$ hybridisation - trigonal bipyramidal geometry

$s p^{3} d^{2}$ hybridisation - octahedral geometry

$s p^{3} d^{3}$ hybridisation - pentagonal bipyramidal geometry

Ethane ( $C_{2} H_{6}$ ) - all $s p^{3}$

Ethene ( $C_{2} H_{4}$ ) - $s p^{2}$

Ethyne ( $C_{2} H_{2}$ ) - $s p$

Q4. Why is $PC l_{5}$ trigonal bipyramidal with two different bond lengths?

Q5. Why can $P F_{5}$ exist but $N F_{5}$ cannot?

Q6. What is the difference between a $σ$ -bond and a $π$ -bond?

Q7. Why is $B F_{3}$ a weaker Lewis acid than $BC l_{3}$ , despite F being more electronegative?

Q8. When $N H_{3}$ combines with H⁺ to form $N H_{4}^{+}$ , does the hybridisation change?