Polar Covalent Bond, Dipole Moment & Percentage Ionic Character

Polar vs Non-polar Covalent Bonds

Not all covalent bonds share electrons equally. When the two bonded atoms are identical (like $H-H$ or $Cl-Cl$), the shared pair sits exactly in the middle — a non-polar covalent bond. But when the atoms differ in electronegativity (like $H-Cl$), the more electronegative atom pulls the shared pair toward itself.

Definition: A polar covalent bond is a covalent bond between two atoms of different electronegativity, in which the shared electron pair is displaced toward the more electronegative atom, creating partial charges $\delta^+$ and $\delta^-$.

In $H-Cl$, chlorine is more electronegative, so it becomes $\delta^-$ and hydrogen $\delta^+$. This separation of charge is what makes the bond polar.

Think of it this way: ionic and non-polar covalent bonds are the two extremes, and polar covalent bonds lie in between — partial, incomplete electron transfer.

[NEET Important] Greater electronegativity difference → more polar bond → larger partial charges.

Dipole Moment

To measure polarity, we use the dipole moment.

Definition: The dipole moment ($\mu$) is the product of the magnitude of the charge ($q$) and the distance of separation ($d$) between the centres of positive and negative charge. $$\mu = q \times d$$

• Unit: the Debye (D), where $1\ \text{D} = 3.33564 \times 10^{-30}\ \text{C m}$.
• It is a vector quantity — it has both magnitude and direction (conventionally pointing from $\delta^+$ to $\delta^-$).

For a diatomic molecule, the bond dipole is the molecular dipole. For polyatomic molecules, the net dipole is the vector sum of all the individual bond dipoles.

Key Point: A molecule can have polar bonds yet be non-polar overall if the bond dipoles cancel by symmetry (e.g. $CO_2$, $BF_3$, $CH_4$).

[JEE Tip] Net dipole moment = vector sum of bond dipoles. Symmetry is the key to deciding whether they cancel.

Vector Addition: Why Some Symmetric Molecules Are Non-polar

Let's see how the vectors add up in important molecules.

Carbon dioxide ($CO_2$) — $\mu = 0$

Linear $O=C=O$. The two C=O bond dipoles are equal and point in opposite directions, so they cancel exactly. $CO_2$ is non-polar despite polar bonds.

Water ($H_2O$) — $\mu = 1.85\ D$

Bent shape ($104.5°$). The two O–H dipoles do not cancel; they add to a net dipole pointing from the H's toward the O. Water is strongly polar.

Ammonia ($NH_3$) — $\mu = 1.47\ D$

Pyramidal. The three N–H dipoles and the lone-pair effect add to a net dipole. Polar.

Boron trifluoride ($BF_3$) — $\mu = 0$

Trigonal planar, symmetric. The three B–F dipoles at $120°$ cancel. Non-polar.

Comparing $NH_3$ and $NF_3$

Both are pyramidal, but $NH_3$ has $\mu = 1.47\ D$ while $NF_3$ has only $0.23\ D$. In $NH_3$ the lone-pair effect and the bond dipoles point the same way (reinforce); in $NF_3$ they point in opposite directions (partly cancel). A classic JEE favourite.

[JEE Tip] $\mu(NH_3) > \mu(NF_3)$ because of the direction of the lone-pair effect relative to the bond dipoles.

Percentage Ionic Character

Real bonds are neither 100% covalent nor 100% ionic — they have partial ionic character. We quantify this by comparing the observed dipole moment with the theoretical dipole moment that the bond would have if it were 100% ionic.

$$\% \text{ ionic character} = \frac{\mu_{observed}}{\mu_{ionic\ (100\%)}} \times 100$$

where $\mu_{ionic} = e \times d$ (one full electronic charge separated by the bond length).

Pauling's relation

Pauling related ionic character to the electronegativity difference $\Delta(EN)$: $$\% \text{ ionic character} = 16\,|\chi_A - \chi_B| + 3.5\,|\chi_A - \chi_B|^2$$

A larger EN difference ⇒ greater ionic character. As a rule of thumb, $\Delta(EN) \approx 1.7$ corresponds to about 50% ionic character.

Key Point: $\% \text{ ionic} = \dfrac{\mu_{obs}}{\mu_{calc\ for\ 100\%\ ionic}} \times 100$. Memorise this; it is the most common numerical in this section.

[NEET Important] Higher % ionic character often correlates with higher melting point and greater water-solubility.

Solved Examples

Example 1: Why is CO$_2$ non-polar but H$_2$O polar?

Both contain polar bonds. Explain the difference in net dipole moment.

Solution:

1. $CO_2$ is linear: the two C=O bond dipoles are equal and opposite → they cancel → $\mu = 0$.
2. $H_2O$ is bent ($104.5°$): the two O–H bond dipoles do not point oppositely; their vector sum is non-zero → $\mu = 1.85\,D$.
3. Conclusion: geometry (symmetry) decides whether polar bonds cancel.

Takeaway: Polar bonds + symmetric shape = non-polar molecule; polar bonds + bent/asymmetric shape = polar molecule.

Example 2: Calculating % ionic character

The observed dipole moment of $HCl$ is $1.03\,D$ and the bond length is $127\,pm$. The dipole moment for 100% ionic $HCl$ is $6.09\,D$. Find the % ionic character.

Solution:

1. Formula: $\% \text{ ionic} = \dfrac{\mu_{obs}}{\mu_{ionic}} \times 100$.
2. Substitute: $\dfrac{1.03}{6.09} \times 100$.
3. Compute: $\approx 16.9\%$.

Takeaway: $HCl$ is about 17% ionic — predominantly covalent but with significant polarity.

Example 3: Theoretical (100% ionic) dipole moment

Calculate the dipole moment of $HCl$ assuming 100% ionic character. Charge $e = 1.6 \times 10^{-19}\,C$, bond length $d = 127\,pm = 1.27 \times 10^{-10}\,m$.

Solution:

1. Formula: $\mu = q \times d = e \times d$.
2. Substitute: $\mu = (1.6 \times 10^{-19})(1.27 \times 10^{-10}) = 2.03 \times 10^{-29}\,C\,m$.
3. Convert to Debye: $\dfrac{2.03 \times 10^{-29}}{3.336 \times 10^{-30}} \approx 6.09\,D$.

Takeaway: Multiply full charge by bond length, then divide by $3.336\times10^{-30}$ to get Debye.

Example 4: Why μ(NH₃) > μ(NF₃)

Both are pyramidal, yet $NH_3$ ($1.47\,D$) has a much larger dipole moment than $NF_3$ ($0.23\,D$). Explain.

Solution:

1. In $NH_3$: N is more electronegative than H, so each N–H dipole points toward N — the same direction as the lone-pair effect. They reinforce → large $\mu$.
2. In $NF_3$: F is more electronegative than N, so each N–F dipole points toward F — opposite to the lone-pair effect. They partially cancel → small $\mu$.
3. Conclusion: direction of the bond dipoles relative to the lone pair explains the difference.

Takeaway: A classic exam trap — same shape, very different dipole moments, all due to dipole direction.

Example 5: Predicting polarity of BF₃

Is $BF_3$ polar or non-polar?

Solution:

1. Shape: trigonal planar, symmetric, bond angle $120°$.
2. Bond dipoles: three equal B–F dipoles at $120°$ to each other.
3. Vector sum: three equal vectors at $120°$ cancel exactly → $\mu = 0$.
4. Conclusion: $BF_3$ is non-polar.

Takeaway: Symmetric $AX_3$ (planar) and $AX_4$ (tetrahedral) molecules with identical bonds are non-polar.

Example 6: Comparing dipole moments of cis and trans dichloroethene

Which has a larger dipole moment: cis-1,2-dichloroethene or trans?

Solution:

1. trans isomer: the two C–Cl dipoles point in opposite directions and cancel → $\mu \approx 0$.
2. cis isomer: the two C–Cl dipoles are on the same side; their vector sum is non-zero → $\mu > 0$.
3. Conclusion: the cis isomer is more polar.

Takeaway: Geometry/symmetry of substituents decides molecular polarity even with identical bonds.

Example 7: Dipole moment order of hydrogen halides

Arrange $HF, HCl, HBr, HI$ in decreasing order of dipole moment.

Solution:

1. Electronegativity difference with H decreases down the group: $HF > HCl > HBr > HI$.
2. Dipole moment depends mainly on this EN difference (charge separation).
3. Order: $HF\ (1.78) > HCl\ (1.07) > HBr\ (0.79) > HI\ (0.38)$ D.

Takeaway: Down the halogen group, decreasing EN difference lowers the dipole moment.

Example 8: Using Pauling's formula

Estimate the % ionic character of a bond where $\Delta(EN) = 1.0$, using $\% = 16\,\Delta + 3.5\,\Delta^2$.

Solution:

1. Substitute $\Delta = 1.0$: $\% = 16(1.0) + 3.5(1.0)^2$.
2. Compute: $16 + 3.5 = 19.5\%$.
3. Conclusion: about 19.5% ionic character.

Takeaway: Pauling's empirical formula links electronegativity difference directly to ionic character.

Example 9: Net dipole of CCl₄

Why is $CCl_4$ non-polar even though each C–Cl bond is polar?

Solution:

1. Shape: tetrahedral, with four identical C–Cl bonds symmetrically arranged.
2. Vector sum: the four equal bond dipoles point to the corners of a tetrahedron and cancel completely.
3. Conclusion: net $\mu = 0$; $CCl_4$ is non-polar.

Takeaway: Symmetric tetrahedral $AX_4$ molecules have zero net dipole.

Example 10: Polarity of CHCl₃ vs CCl₄

Explain why $CHCl_3$ (chloroform) is polar but $CCl_4$ is not.

Solution:

1. $CCl_4$: four identical C–Cl dipoles cancel by tetrahedral symmetry → $\mu = 0$.
2. $CHCl_3$: replacing one Cl with H breaks the symmetry; the C–H dipole differs from the three C–Cl dipoles, so they no longer cancel → net $\mu \neq 0$.
3. Conclusion: $CHCl_3$ is polar.

Takeaway: Breaking the symmetry of a non-polar molecule generally makes it polar.

Example 11: From dipole moment to charge separation

A diatomic molecule has $\mu = 1.5\,D$ and bond length $d = 1.0 \times 10^{-10}\,m$. Find the fraction of an electronic charge on each atom.

Solution:

1. Convert $\mu$ to SI: $1.5\,D = 1.5 \times 3.336 \times 10^{-30} = 5.0 \times 10^{-30}\,C\,m$.
2. Charge $q = \mu / d = (5.0 \times 10^{-30}) / (1.0 \times 10^{-10}) = 5.0 \times 10^{-20}\,C$.
3. Fraction of e: $q/e = (5.0 \times 10^{-20}) / (1.6 \times 10^{-19}) \approx 0.31$.
4. Conclusion: about $0.31\,e$ (i.e. ~31% of a full charge) on each atom.

Takeaway: Rearranging $\mu = q d$ gives the partial charge, a measure of ionic character.

Example 12: Identify the non-polar molecule

Among $H_2O$, $NH_3$, $CO_2$ and $SO_2$, which is non-polar?

Solution:

1. $H_2O$ (bent), $NH_3$ (pyramidal), $SO_2$ (bent): asymmetric, bond dipoles do not cancel → polar.
2. $CO_2$ (linear, symmetric): the two C=O dipoles cancel → $\mu = 0$.
3. Conclusion: $CO_2$ is the non-polar molecule.

Takeaway: Linear $AX_2$ with identical bonds (like $CO_2$) is non-polar; bent $AX_2$ (like $SO_2$, $H_2O$) is polar.