Hybridization — sp, sp², sp³, sp³d, sp³d²

Why Hybridization?

Here's a puzzle VBT alone can't solve. Carbon's ground state is $1s^2\,2s^2\,2p^2$ — only two unpaired electrons. So carbon should form only 2 bonds. Yet methane is $CH_4$, with four identical C–H bonds at $109.5^\circ$. How?

The answer is hybridization — the idea that atomic orbitals mix to form new, equivalent orbitals before bonding.

Definition: Hybridization is the mixing of atomic orbitals of similar energy belonging to the same atom to produce an equal number of new, equivalent hybrid orbitals with identical energy and shape, oriented to minimise repulsion.

For carbon: one $2s$ and three $2p$ orbitals mix into four equivalent $sp^3$ hybrid orbitals, each with one unpaired electron — perfect for four C–H bonds.

Key features of hybridization

1. The number of hybrid orbitals = number of atomic orbitals mixed.
2. Hybrid orbitals are equivalent in energy and shape.
3. They orient in space to give minimum repulsion (hence definite geometry).
4. Only orbitals of comparable energy hybridise.

[JEE Tip] Hybridization is a model applied to explain observed shapes — it does not 'cause' the shape, it rationalises it.

The Five Common Hybridizations

Each hybridization corresponds to a characteristic geometry and bond angle.

Hybridization	Orbitals mixed	Geometry	Angle	Example
$sp$	1s + 1p	Linear	$180^\circ$	$BeCl_2$, $C_2H_2$, $CO_2$
$sp^2$	1s + 2p	Trigonal planar	$120^\circ$	$BCl_3$, $C_2H_4$
$sp^3$	1s + 3p	Tetrahedral	$109.5^\circ$	$CH_4$, $NH_3$, $H_2O$
$sp^3d$	1s + 3p + 1d	Trigonal bipyramidal	$120^\circ, 90^\circ$	$PCl_5$
$sp^3d^2$	1s + 3p + 2d	Octahedral	$90^\circ$	$SF_6$
$sp^3d^3$	1s + 3p + 3d	Pentagonal bipyramidal	$72^\circ, 90^\circ$	$IF_7$

s-character and bond angle

More s-character → larger bond angle and shorter, stronger bonds:

• $sp$ (50% s) → $180^\circ$
• $sp^2$ (33% s) → $120^\circ$
• $sp^3$ (25% s) → $109.5^\circ$

Key Point: Hybridization fixes the electron geometry; lone pairs then modify the actual molecular shape (as in $NH_3$, $H_2O$, which are still $sp^3$).

[NEET Important] Memorise this table — identifying hybridization is one of the single most-asked things in the whole chapter.

The Shortcut Formula for Hybridization

Drawing orbital diagrams every time is slow. Use this reliable steric-number formula:

$$\text{Steric number} (H) = \frac{1}{2}\,[\,V + M - C + A\,]$$

where $V$ = valence electrons of the central atom, $M$ = number of monovalent atoms (like H, halogens) attached, $C$ = charge of cation (subtract), $A$ = charge of anion (add).

Then map the number to hybridization:

H value	Hybridization	Shape (no lone pairs)
2	$sp$	linear
3	$sp^2$	trigonal planar
4	$sp^3$	tetrahedral
5	$sp^3d$	trigonal bipyramidal
6	$sp^3d^2$	octahedral
7	$sp^3d^3$	pentagonal bipyramidal

Worked use

• $SF_6$: $H = \tfrac{1}{2}(6 + 6) = 6 \Rightarrow sp^3d^2$.
• $NH_3$: $H = \tfrac{1}{2}(5 + 3) = 4 \Rightarrow sp^3$.
• $NH_4^+$: $H = \tfrac{1}{2}(5 + 4 - 1) = 4 \Rightarrow sp^3$.
• $CO_3^{2-}$: $H = \tfrac{1}{2}(4 + 0 + 2) = 3 \Rightarrow sp^2$ (treat double-bonded O as not adding to M).

Key Point: For oxygen/double-bonded atoms, only count monovalent atoms in M; the steric number from this formula already includes lone pairs.

[JEE Tip] This formula gives the steric number (bonds + lone pairs) directly — the fastest route to hybridization in an exam.

Hybridization in Multiple-Bonded & Lone-Pair Molecules

sp carbon — ethyne ($C_2H_2$)

Each carbon is $sp$ hybridised: two $sp$ hybrids (one to H, one to the other C as a σ bond) and two unhybridised p-orbitals that form two π bonds. Linear, $180^\circ$, C≡C triple bond.

sp² carbon — ethene ($C_2H_4$)

Each carbon is $sp^2$: three $sp^2$ hybrids (two C–H σ + one C–C σ) and one unhybridised p-orbital forming one π bond. Trigonal planar, $120^\circ$.

sp³ with lone pairs — NH₃ and H₂O

Both are $sp^3$ hybridised (steric number 4). The lone pairs occupy hybrid orbitals, compressing the bond angles ($107^\circ$ and $104.5^\circ$) below the ideal $109.5^\circ$.

Expanded-octet examples

• $PCl_5$: $sp^3d$, trigonal bipyramidal.
• $SF_6$: $sp^3d^2$, octahedral.
• $IF_7$: $sp^3d^3$, pentagonal bipyramidal.

Key Point: Only σ bonds and lone pairs are made from hybrid orbitals; π bonds always come from unhybridised p-orbitals.

[NEET Important] When counting hybridization, ignore π bonds — they use leftover unhybridised p-orbitals, not hybrid orbitals.

Solved Examples

Example 1: Hybridization of carbon in methane

Determine the hybridization of carbon in $CH_4$.

Solution:

1. Steric number: $H = \tfrac{1}{2}(4 + 4) = 4$.
2. Hybridization: $sp^3$.
3. Geometry: tetrahedral, $109.5^\circ$, four equivalent C–H bonds.

Takeaway: Four σ bonds, no lone pairs → $sp^3$, tetrahedral.

Example 2: Hybridization in BeCl₂

Find the hybridization of beryllium in $BeCl_2$ (gas phase).

Solution:

1. Steric number: $H = \tfrac{1}{2}(2 + 2) = 2$.
2. Hybridization: $sp$.
3. Geometry: linear, $180^\circ$.

Takeaway: Two σ bonds, no lone pairs → $sp$, linear.

Example 3: Hybridization in BCl₃

Find the hybridization of boron in $BCl_3$.

Solution:

1. Steric number: $H = \tfrac{1}{2}(3 + 3) = 3$.
2. Hybridization: $sp^2$.
3. Geometry: trigonal planar, $120^\circ$.

Takeaway: Three σ bonds, no lone pairs → $sp^2$, trigonal planar.

Example 4: Hybridization in PCl₅

Determine the hybridization of phosphorus in $PCl_5$.

Solution:

1. Steric number: $H = \tfrac{1}{2}(5 + 5) = 5$.
2. Hybridization: $sp^3d$.
3. Geometry: trigonal bipyramidal ($120^\circ$ and $90^\circ$ angles).

Takeaway: Five σ bonds → $sp^3d$, trigonal bipyramidal.

Example 5: Hybridization in SF₆

Determine the hybridization of sulphur in $SF_6$.

Solution:

1. Steric number: $H = \tfrac{1}{2}(6 + 6) = 6$.
2. Hybridization: $sp^3d^2$.
3. Geometry: octahedral, all $90^\circ$.

Takeaway: Six σ bonds → $sp^3d^2$, octahedral.

Example 6: Hybridization of carbon in ethyne

What is the hybridization of each carbon in $C_2H_2$?

Solution:

1. Each carbon forms 2 σ bonds (one to H, one to C) and no lone pairs → steric number 2.
2. Hybridization: $sp$.
3. Remaining two p-orbitals on each carbon form the two π bonds of the triple bond.
4. Geometry: linear, $180^\circ$.

Takeaway: sp carbon → linear; the triple bond's two π bonds come from unhybridised p-orbitals.

Example 7: Hybridization in NH₃ despite lone pair

What is the hybridization of nitrogen in $NH_3$?

Solution:

1. Steric number: $H = \tfrac{1}{2}(5 + 3) = 4$ (3 bonds + 1 lone pair).
2. Hybridization: $sp^3$.
3. Shape: the lone pair occupies one $sp^3$ orbital → pyramidal, $107^\circ$.

Takeaway: Lone pairs count toward hybridization; $NH_3$ is $sp^3$ even though it is not tetrahedral in shape.

Example 8: Hybridization in the ammonium ion

Find the hybridization of nitrogen in $NH_4^+$.

Solution:

1. Steric number: $H = \tfrac{1}{2}(5 + 4 - 1) = 4$.
2. Hybridization: $sp^3$.
3. Shape: four N–H bonds, no lone pair → regular tetrahedral, $109.5^\circ$.

Takeaway: Subtract 1 for the positive charge; $NH_4^+$ is $sp^3$ and perfectly tetrahedral.

Example 9: Hybridization in the carbonate ion

Determine the hybridization of carbon in $CO_3^{2-}$.

Solution:

1. Steric number: $H = \tfrac{1}{2}(4 + 0 + 2) = 3$ (3 σ bonds to O, no lone pair on C).
2. Hybridization: $sp^2$.
3. Geometry: trigonal planar, $120^\circ$; the π bond is delocalised (resonance).

Takeaway: Carbonate carbon is $sp^2$; ignore the π bonds when finding hybridization.

Example 10: Hybridization in XeF₄

Find the hybridization of xenon in $XeF_4$.

Solution:

1. Xe has 8 valence electrons: 4 form Xe–F bonds, leaving 4 → 2 lone pairs.
2. Steric number: 4 bonds + 2 lone pairs = 6.
3. Hybridization: $sp^3d^2$.
4. Shape: square planar (two lone pairs trans).

Takeaway: Lone pairs count: $XeF_4$ is $sp^3d^2$ with a square planar shape.

Example 11: s-character and bond angle ordering

Arrange $sp$, $sp^2$, $sp^3$ carbons by bond angle and explain.

Solution:

1. s-character: $sp$ (50%) > $sp^2$ (33%) > $sp^3$ (25%).
2. More s-character → larger bond angle (orbitals spread further apart).
3. Order: $sp\ (180^\circ) > sp^2\ (120^\circ) > sp^3\ (109.5^\circ)$.

Takeaway: Bond angle rises with s-character of the hybrid orbital.

Example 12: Hybridization in SF₄

Determine the hybridization of sulphur in $SF_4$ and relate it to the shape.

Solution:

1. S has 6 valence electrons: 4 form S–F bonds, leaving 2 → 1 lone pair.
2. Steric number: 4 bonds + 1 lone pair = 5.
3. Hybridization: $sp^3d$.
4. Shape: trigonal bipyramidal electron geometry with one equatorial lone pair → see-saw.

Takeaway: $SF_4$ is $sp^3d$; the lone pair turns the trigonal bipyramid into a see-saw shape.