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Chemical Bonding and Molecular Structure - VSEPR Theory

Grade 11ICSEChemistry

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

🔑Concepts

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The VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the geometry of individual molecules from the number of electron pairs surrounding their central atoms.

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The shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom.

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Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.

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The order of repulsion between electron pairs 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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A lone pair occupies more space than a bond pair because it is attracted to only one nucleus, whereas a bond pair is shared between two nuclei.

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The presence of lone pairs causes a deviation from ideal bond angles. For example, in CH4CH_4 the angle is 109.5∘109.5^\circ, but in NH3NH_3 (one lone pair), it reduces to 107∘107^\circ.

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Multiple bonds (double or triple) are treated as a single electron pair for the purpose of predicting molecular geometry.

📐Formulae

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

Number of Lone Pairs (lp)=H−Number of surrounding atoms (B)\text{Number of Lone Pairs (lp)} = H - \text{Number of surrounding atoms (B)}

Bond Angle in Tetrahedral Geometry≈109.5∘\text{Bond Angle in Tetrahedral Geometry} \approx 109.5^\circ

Bond Angle in Trigonal Planar Geometry=120∘\text{Bond Angle in Trigonal Planar Geometry} = 120^\circ

💡Examples

Problem 1:

Predict the shape and bond angle of the Ammonia molecule (NH3NH_3) using VSEPR theory.

Solution:

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

Explanation:

Since there are 4 electron pairs, the geometry is tetrahedral. However, due to the presence of 1 lone pair, the shape becomes Trigonal Pyramidal. The lp−bplp-bp repulsion reduces the bond angle from 109.5∘109.5^\circ to 107∘107^\circ.

Problem 2:

Explain the geometry of SF6SF_6 (Sulfur Hexafluoride).

Solution:

V=6V=6 (for Sulfur), M=6M=6 (Fluorine atoms). H=12[6+6]=6H = \frac{1}{2}[6 + 6] = 6. Lone pairs = 6−6=06 - 6 = 0.

Explanation:

With 6 bond pairs and 0 lone pairs, the molecule adopts an Octahedral geometry. All F−S−FF-S-F bond angles are 90∘90^\circ.

Problem 3:

Determine the shape of the H2OH_2O molecule.

Solution:

V=6V=6 (Oxygen), M=2M=2 (Hydrogen). H=12[6+2]=4H = \frac{1}{2}[6 + 2] = 4. Lone pairs = 4−2=24 - 2 = 2.

Explanation:

With 2 bond pairs and 2 lone pairs, the electron pair geometry is tetrahedral. Due to strong lp−lplp-lp repulsion followed by lp−bplp-bp repulsion, the shape is 'Bent' or 'V-shaped' with a bond angle of 104.5∘104.5^\circ.

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