Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent a

NCERT Exemplar · Q2

Q. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?

(i) CO2

(ii) HI

(iii) H2O

(iv) SO2

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Concept understanding — VSEPR Theory

VSEPR Theory: Why Molecules Have the Shapes They Do

Imagine you're in a crowded room. Everyone wants their personal space. If you're standing with a few friends, you'll naturally spread out so no one is too close to anyone else. That's exactly what happens inside a molecule.

The Core Intuition

Electron pairs are negatively charged. They repel each other. In a molecule, the electron pairs around a central atom will arrange themselves as far apart as possible — just like those people in the room. This simple idea is the entire foundation of VSEPR (pronounced "ves-per") Theory.

Note

VSEPR stands for Valence Shell Electron Pair Repulsion. The name tells you exactly what it's about: the repulsion between electron pairs in the valence shell.

The Precise Statement

VSEPR Theory states that the geometry around a central atom is determined by minimizing the repulsion between all electron pairs (both bonding and lone pairs) in its valence shell.

Two key points to hold onto:

All electron pairs repel — whether they are shared (bonding pairs) or unshared (lone pairs).
Lone pairs repel more strongly than bonding pairs. A lone pair is "fatter" — it's only attracted to one nucleus, so it spreads out more and pushes harder on its neighbours.

How to Predict Shape in 3 Steps

Step 1: Count the total electron pairs around the central atom.

Add the number of atoms bonded to the central atom plus the number of lone pairs on it. This gives you the steric number.

Step 2: Arrange those pairs as far apart as possible.

This gives you the electron-pair geometry — the shape if you pretend all pairs are identical.

Step 3: Replace lone pairs with "invisible" space.

The actual molecular geometry is the shape formed by the atoms alone, ignoring lone pairs.

The Common Geometries at a Glance

Steric Number	Electron-Pair Geometry	Lone Pairs	Molecular Geometry	Example	Bond Angle
2	Linear	0	Linear	$\text{CO}_2$	180°
3	Trigonal planar	0	Trigonal planar	$\text{BF}_3$	120°
3	Trigonal planar	1	Bent	$\text{SO}_2$	~119°
4	Tetrahedral	0	Tetrahedral	$\text{CH}_4$	109.5°
4	Tetrahedral	1	Trigonal pyramidal	$\text{NH}_3$	~107°
4	Tetrahedral	2	Bent	$\text{H}_2\text{O}$	~104.5°
5	Trigonal bipyramidal	0	Trigonal bipyramidal	$\text{PCl}_5$	90°, 120°
6	Octahedral	0	Octahedral	$\text{SF}_6$	90°

Watch out

A common mistake: thinking that $\text{NH}_3$ is tetrahedral. It has tetrahedral electron-pair geometry, but because one position is a lone pair, the molecular shape is trigonal pyramidal. The bond angle is 107°, not 109.5°.

Why Lone Pairs Squeeze Bond Angles

Take water ( $\text{H}_2\text{O}$ ). The central oxygen has 4 electron pairs: 2 bonding (to H atoms) and 2 lone pairs. The ideal tetrahedral angle is 109.5°. But the two lone pairs push harder on the bonding pairs, compressing the H–O–H angle to about 104.5°.

In ammonia ( $\text{NH}_3$ ), there's only one lone pair, so the compression is less — the H–N–H angle is about 107°.

Important

The order of repulsion strength: Lone pair–lone pair > Lone pair–bonding pair > Bonding pair–bonding pair. This is why angles shrink as lone pairs increase.

A Quick Worked Example: $\text{CH}_4$ vs $\text{NH}_3$ vs $\text{H}_2\text{O}$

Molecule	Central atom	Bonding pairs	Lone pairs	Steric number	Electron geometry	Molecular geometry	Angle
$\text{CH}_4$	C	4	0	4	Tetrahedral	Tetrahedral	109.5°
$\text{NH}_3$	N	3	1	4	Tetrahedral	Trigonal pyramidal	~107°
$\text{H}_2\text{O}$	O	2	2	4	Tetrahedral	Bent	~104.5°

Notice: all three have the same steric number (4) and the same electron-pair geometry (tetrahedral). Only the molecular geometry changes as lone pairs replace bonding pairs.

The Bottom Line

VSEPR is not about memorising shapes — it's about understanding that electron pairs want distance. Count the pairs, push them apart, then see where the atoms end up. That's the whole theory.

VSEPR Theory is a major section of the NCERT Class 11 Chemistry chapter on Chemical Bonding and Molecular Structure, matching searches like "VSEPR theory: molecular shapes class 11 chemistry" or "chemical bonding important questions". Predicting molecular geometry from electron-pair repulsion is one of the most consistently tested topics in CBSE boards, and it remains an important topic for JEE Main and NEET chemistry papers.

The dipole moment of a molecule depends on the polarity of its individual bonds and its overall molecular geometry. Polar bonds arise from differences in electronegativity between bonded atoms. If a molecule has polar bonds, its net dipole moment is determined by the vector sum of these bond dipoles.

CO $_2$ : Carbon dioxide has a linear geometry ( $AX_2$ type according to VSEPR theory). The C-O bonds are polar, but the two bond dipoles are equal in magnitude and opposite in direction, causing them to cancel out. Thus, CO $_2$ has a net dipole moment of zero.
HI: Hydrogen iodide is a diatomic molecule. The H-I bond is polar due to the electronegativity difference between hydrogen and iodine. Since it's a diatomic molecule, there is no cancellation, resulting in a net dipole moment.
H $_2$ O: Water has a bent geometry ( $AX_2E_2$ type). The O-H bonds are highly polar due to the large electronegativity difference between oxygen and hydrogen. Because of the bent shape, the bond dipoles do not cancel out but add up vectorially, leading to a significant net dipole moment.
SO $_2$ : Sulfur dioxide has a bent geometry ( $AX_2E_1$ type). The S-O bonds are polar due to the electronegativity difference between sulfur and oxygen. Similar to water, the bent shape prevents the bond dipoles from cancelling, resulting in a net dipole moment.

Comparing the molecules with non-zero dipole moments (HI, H $_2$ O, SO $_2$ ):

The electronegativity difference for O-H bonds in H $_2$ O ( $\Delta EN \approx 1.24$ ) is significantly greater than for H-I bonds in HI ( $\Delta EN \approx 0.46$ ) or S-O bonds in SO $_2$ ( $\Delta EN \approx 0.86$ ). This indicates that O-H bonds are more polar.
Both H $_2$ O and SO $_2$ are bent molecules. However, the higher bond polarity in H $_2$ O, combined with its specific bond angle ( $\approx 104.5^\circ$ ), results in a larger net dipole moment compared to SO $_2$ (bond angle $\approx 119^\circ$ ) and HI. Water is known for its exceptionally high dipole moment among small molecules.

✓Final answer

\boxed{\text{H}_2\text{O}}

The dipole moment of a molecule depends on the polarity of its bonds and its molecular geometry. Water (H $_2$ O) has highly polar O-H bonds and a bent molecular geometry, causing its bond dipoles and lone pair contributions to add up significantly, resulting in the highest dipole moment among the given options. The correct option is (C).

The polarity of a molecule, quantified by its dipole moment, arises from two main factors: the polarity of individual bonds and the overall molecular geometry.

Bond Polarity: A bond is polar if there is a significant difference in electronegativity between the two bonded atoms. The more electronegative atom pulls the shared electron pair closer to itself, creating a partial negative charge ( $\delta^-$ ) on that atom and a partial positive charge ( $\delta^+$ ) on the less electronegative atom. This separation of charge creates a bond dipole, which is a vector quantity pointing from the positive to the negative end.
Molecular Geometry: The individual bond dipoles in a molecule add up vectorially to give the net molecular dipole moment. If the molecule is symmetrical and the bond dipoles cancel each other out, the net dipole moment is zero. If the molecule is asymmetrical, or if the bond dipoles do not cancel due to their orientation, the molecule will have a net non-zero dipole moment. VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict the molecular geometry.

Let's analyze each molecule:

Carbon Dioxide (CO $_2$ )
- Lewis Structure and VSEPR: The central carbon atom is double-bonded to two oxygen atoms (O=C=O). There are two electron domains around the central carbon, both bonding pairs. According to VSEPR theory, this leads to a linear molecular geometry.
- Bond Polarity: Oxygen is significantly more electronegative than carbon ( $\Delta EN_{C-O} = 0.89$ ). Therefore, each C=O bond is polar, with the dipole pointing towards the oxygen atoms.
- Net Dipole Moment: In a linear CO $_2$ molecule, the two C=O bond dipoles are equal in magnitude and point in opposite directions. They cancel each other out completely.

\stackrel{\delta^+}{\text{C}} \stackrel{\delta^-}{=}\text{O} \quad \leftarrow \text{Dipole} \quad \text{O}\stackrel{\delta^-}{=}\stackrel{\delta^+}{\text{C}} \quad \rightarrow \text{Dipole}

    The net dipole moment of CO$_2$ is **zero**.

2. Hydrogen Iodide (HI)

* Lewis Structure and VSEPR: This is a diatomic molecule, so its geometry is inherently linear.

* Bond Polarity: Iodine is slightly more electronegative than hydrogen ( $\Delta EN_{H-I} = 0.46$ ). Thus, the H-I bond is polar, with the dipole pointing towards iodine.

* Net Dipole Moment: Since it's a diatomic molecule with a polar bond, there is no cancellation. The molecule has a non-zero dipole moment. However, the electronegativity difference is relatively small compared to other options.

Water (H $_2$ O)
- Lewis Structure and VSEPR: The central oxygen atom is bonded to two hydrogen atoms and has two lone pairs of electrons. This gives four electron domains (two bonding, two non-bonding) around the oxygen. According to VSEPR theory, the electron geometry is tetrahedral, but the molecular geometry is bent (or V-shaped). The H-O-H bond angle is approximately $104.5^\circ$ .
- Bond Polarity: Oxygen is significantly more electronegative than hydrogen ( $\Delta EN_{O-H} = 1.24$ ). Each O-H bond is highly polar, with the dipole pointing towards the oxygen atom.
- Net Dipole Moment: Due to the bent geometry, the two O-H bond dipoles do not cancel out. Instead, they add up vectorially, resulting in a significant net dipole moment. The lone pairs on oxygen also contribute to the overall dipole moment, further enhancing it.
  Important
  
  Water has a very high dipole moment (approximately $1.85$ Debye) due to the high electronegativity of oxygen, the bent molecular geometry, and the contribution from its lone pairs.
Sulfur Dioxide (SO $_2$ )
- Lewis Structure and VSEPR: The central sulfur atom is double-bonded to two oxygen atoms and has one lone pair of electrons. This gives three electron domains (two bonding, one non-bonding) around the sulfur. According to VSEPR theory, the electron geometry is trigonal planar, but the molecular geometry is bent. The O-S-O bond angle is approximately $119^\circ$ .
- Bond Polarity: Oxygen is more electronegative than sulfur ( $\Delta EN_{S-O} = 0.86$ ). Each S=O bond is polar, with the dipole pointing towards the oxygen atoms.
- Net Dipole Moment: Similar to water, the bent geometry means the two S=O bond dipoles do not cancel. They add up vectorially, and the lone pair on sulfur also contributes, resulting in a non-zero dipole moment. The dipole moment of SO $_2$ is approximately $1.62$ Debye.

Comparison:

CO $_2$ has a zero dipole moment.
HI has a small non-zero dipole moment.
Both H $_2$ O and SO $_2$ have non-zero dipole moments due to their bent geometries and polar bonds.
Comparing H $_2$ $_{2}$ O and SO $_2$ $_{2}$ :
- The electronegativity difference for O-H bonds ( $\Delta EN = 1.24$ ) is greater than for S-O bonds ( $\Delta EN = 0.86$ ), meaning O-H bonds are more polar.
- The bond angle in H $_2$ O ( $104.5^\circ$ ) is smaller than in SO $_2$ ( $119^\circ$ ). For two equal bond dipoles, a smaller angle between them generally leads to a larger resultant vector (net dipole moment).
- The combined effect of stronger bond polarity and a more acute bond angle in H $_2$ O, along with the significant lone pair contribution, makes its net dipole moment higher than that of SO $_2$ .

Therefore, H $_2$ O has the highest dipole moment among the given options.

✓Final answer

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Q.Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?

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