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Chemical Bonding and Molecular Structure - VSEPR Theory

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The basic postulate of VSEPR theory is that the geometry of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom.

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Electron pairs in the valence shell repel one another because their electron clouds are negatively charged.

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The order of repulsive interactions is: lp−lp>lp−bp>bp−bplp - lp > lp - bp > bp - bp (where lplp is lone pair and bpbp is bond pair).

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Lone pairs occupy more space than bond pairs because they are attracted to only one nucleus, whereas bond pairs are shared between two nuclei.

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If the central atom is surrounded by only bond pairs of similar atoms, the molecule has a regular geometry. If lone pairs are present, the geometry becomes distorted.

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Molecules with 2, 3, 4, 5, and 6 electron pairs around the central atom exhibit Linear, Trigonal Planar, Tetrahedral, Trigonal Bipyramidal, and Octahedral geometries respectively.

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For AB5AB_5 molecules (PCl5PCl_5), axial bonds are longer than equatorial bonds due to greater repulsion from equatorial bond pairs.

📐Formulae

X=12[V+M−C+A]X = \frac{1}{2} [V + M - C + A]

Where: V=Valence electrons of central atom\text{Where: } V = \text{Valence electrons of central atom}

M=Number of monovalent atoms (H, F, Cl, Br, I)M = \text{Number of monovalent atoms (H, F, Cl, Br, I)}

C=Charge on cation, A=Charge on anionC = \text{Charge on cation, } A = \text{Charge on anion}

Number of Lone Pairs (lp)=X−Number of Bond Pairs (bp)\text{Number of Lone Pairs (lp)} = X - \text{Number of Bond Pairs (bp)}

💡Examples

Problem 1:

Predict the geometry and shape of NH3NH_3 (Ammonia) using VSEPR theory.

Solution:

  1. Central atom is NN (V=5V=5). 2. Number of monovalent atoms H=3H=3. 3. X=12(5+3)=4X = \frac{1}{2}(5 + 3) = 4. 4. Total electron pairs = 4 (Tetrahedral geometry). 5. Bond pairs = 3, Lone pairs = 4−3=14 - 3 = 1.

Explanation:

Due to 44 electron pairs, the geometry is tetrahedral. However, the presence of one lone pair causes lp−bplp - bp repulsion, reducing the H−N−HH-N-H bond angle from 109.5∘109.5^\circ to 107∘107^\circ. The shape is Pyramidal.

Problem 2:

Explain the shape of SF4SF_4 using VSEPR theory.

Solution:

  1. Central atom SS (V=6V=6). 2. Monovalent atoms F=4F=4. 3. X=12(6+4)=5X = \frac{1}{2}(6 + 4) = 5. 4. Total pairs = 5 (Trigonal Bipyramidal geometry). 5. Bond pairs = 4, Lone pairs = 5−4=15 - 4 = 1.

Explanation:

With 5 electron pairs, the geometry is Trigonal Bipyramidal. The lone pair occupies an equatorial position to minimize repulsion. The resulting shape is See-saw.

Problem 3:

Determine the shape of H2OH_2O.

Solution:

  1. Central atom OO (V=6V=6). 2. Monovalent atoms H=2H=2. 3. X=12(6+2)=4X = \frac{1}{2}(6 + 2) = 4. 4. Bond pairs = 2, Lone pairs = 4−2=24 - 2 = 2.

Explanation:

The geometry is tetrahedral. Due to two lone pairs, lp−lplp - lp repulsion is high, squeezing the H−O−HH-O-H bond angle to 104.5∘104.5^\circ. The shape is Bent or V-shaped.

VSEPR Theory - Revision Notes & Key Formulas | CBSE Class 11 Chemistry