Hybridization
Mixing of Orbitals & Molecular Geometry | Chemical Bonding
1. What is Hybridization?
Introduced by Linus Pauling to explain the equivalent nature of bonds in molecules like Methane ($CH_4$).
- Number of Hybrid Orbitals = Number of Atomic Orbitals mixed.
- Hybrid orbitals are identical in shape and energy.
- They form stronger $\sigma$ bonds than pure atomic orbitals.
2. Calculating Hybridization (The Magic Formula)
To find the hybridization state of the central atom, calculate the value $H$ (Steric Number):
Where:
- V = Number of Valence electrons of the central atom.
- M = Number of Monovalent atoms linked ($H, F, Cl, Br, I$). Ignore Divalent ($O, S$).
- C = Charge on Cation (subtract).
- A = Charge on Anion (add).
Note: $H$ also equals (Number of $\sigma$ bonds + Number of Lone Pairs).
3. Types of Hybridization
| Value of H | Hybridization | Geometry | Bond Angle | Example |
|---|---|---|---|---|
| 2 | $sp$ | Linear | $180^\circ$ | $BeCl_2, CO_2$ |
| 3 | $sp^2$ | Trigonal Planar | $120^\circ$ | $BF_3, SO_3$ |
| 4 | $sp^3$ | Tetrahedral | $109.5^\circ$ | $CH_4, NH_3$ |
| 5 | $sp^3d$ | Trigonal Bipyramidal | $90^\circ, 120^\circ$ | $PCl_5$ |
| 6 | $sp^3d^2$ | Octahedral | $90^\circ$ | $SF_6$ |
| 7 | $sp^3d^3$ | Pentagonal Bipyramidal | $72^\circ, 90^\circ$ | $IF_7$ |
4. Percentage s-Character
The % contribution of the s-orbital affects the properties of the hybrid orbital.
- $sp$: 50% s-character. (Shortest, strongest, most electronegative).
- $sp^2$: 33.3% s-character.
- $sp^3$: 25% s-character. (Longest, weakest).
Rule: More s-character $\rightarrow$ More Electronegativity $\rightarrow$ Larger Bond Angle.
5. Coordinate Compounds (Inner d-orbitals)
In coordination compounds, $(n-1)d$ orbitals can mix.
- $dsp^2$: Square Planar Geometry (e.g., $[Ni(CN)_4]^{2-}$).
- $d^2sp^3$: Inner Orbital Octahedral Complex.
Practice Quiz
Test your calculation skills on Hybridization.
Perfect teaching.
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