Case: In order to explain the characteristic geometrical shape of polyatomic molecules like CH4, NH3, H2O etc. Pauling introduced the concept of hybridisation. According to him the atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals. Unlike pur

Question

Case: In order to explain the characteristic geometrical shape of polyatomic molecules like CH4, NH3, H2O etc. Pauling introduced the concept of hybridisation. According to him the atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation which can be defined as the process of intermixing of atomic orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape. For example one 2s and three 2p orbitals of carbon hybridise, there is the formation of four new sp3 orbitals. Questions: (i) Which hybrid orbital are used by carbon atoms (1, 2 and 3) in the following molecule? CH3–CH2–COOH (atoms numbered 1, 2, 3 left to right over CH3, CH2, COOH respectively) [1] (ii) Is there any change in hybridisation of B and N as a result of the following reaction? BF3 + NH3 → F3B.NH3 [1] (iii) In PCl5 molecule, why are the axial bonds longer than equatorial bonds? [2]

**OR**

Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Accepted Answer

>

> [!TLDR]
> The two $CH_3$/$CH_2$ carbons in $CH_3CH_2COOH$ are $sp^3$ while the –COOH carbon is $sp^2$; boron's hybridisation changes on forming the $BF_3\cdot NH_3$ adduct but nitrogen's doesn't; and axial bonds in $PCl_5$ are longer because they suffer more 90° bond-pair repulsions than equatorial bonds.
>
> **(i) Hybridisation in $CH_3	ext{-}CH_2	ext{-}COOH$:**
> - **C1** ($CH_3$): four single (sigma) bonds only → **$sp^3$**.
> - **C2** ($CH_2$): four single (sigma) bonds only → **$sp^3$**.
> - **C3** (–COOH carbon): one C=O double bond + two single bonds (to C2 and to –OH), trigonal planar → **$sp^2$**.
>
> **(ii) $BF_3 + NH_3 ightarrow F_3B	ext{-}NH_3$:**
> - **Boron:** in $BF_3$, B is $sp^2$ hybridised (trigonal planar, 3 bond pairs, empty p-orbital). On accepting the lone pair from $NH_3$ to form a 4th (coordinate) bond, B becomes 4-coordinate/tetrahedral, so its hybridisation **changes to $sp^3$**.
> - **Nitrogen:** in $NH_3$, N is already $sp^3$ hybridised (pyramidal, 3 bond pairs + 1 lone pair). After donating its lone pair as the coordinate bond to B, N still has 4 electron pairs around it in a tetrahedral arrangement, so its hybridisation **stays $sp^3$** — it does not change.
> …

Hybridization	Orbitals Mixed	Number of Hybrids	Geometry	Bond Angle	Example
$sp$	one $s$ + one $p$	2	Linear	$180^\circ$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	3	Trigonal planar	$120^\circ$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	4	Tetrahedral	$109.5^\circ$	$\text{CH}_4$

Concept understanding — Orbital Hybridization Theory

Orbital Hybridization Theory – From Intuition to Precision

The Problem That Started It All

The Core Intuition

The Precise Statement

The Three Common Hybridizations

How It Works: The Methane Example

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