Electronegativity

What Is Electronegativity?

We've studied three "gains and losses" so far — ionization enthalpy tells us how hard it is to lose an electron, electron gain enthalpy tells us how much energy it releases when it accepts one, and atomic radius tells us how big it is. Electronegativity ties these together in a single number that represents an atom's willingness to pull bonding electrons toward itself within a chemical bond.

Definition

The electronegativity ($\chi$) of an element is the relative tendency of its atom to attract the shared pair of electrons toward itself in a covalent bond with another atom.

It is a dimensionless relative quantity, not an energy. Unlike IE or $\Delta_{\text{eg}}H$, electronegativity does not belong to a free, isolated atom — it describes how that atom behaves when bonded.

Key properties

• Dimensionless: Electronegativity is expressed on a relative scale (Pauling, Mulliken, Allred-Rochow). There are no units, just numbers.
• Depends on hybridization: An $sp$-hybridized C is more electronegative than an $sp^{3}$-hybridized C. Higher $s$-character $\Rightarrow$ electrons held closer to the nucleus $\Rightarrow$ higher $\chi$.
• Depends on oxidation state: A higher positive oxidation state draws bonding electrons harder, increasing the effective electronegativity (e.g., $\chi(\text{Fe}^{3+}) > \chi(\text{Fe}^{2+})$).
• Depends on the bonded partner: Strictly speaking, $\chi$ is defined relative to a particular bonding environment.

The single most important fact

Fluorine has the highest electronegativity of any element. On the Pauling scale, $\chi(\text{F}) = 3.98$. Caesium has the lowest, at $\chi(\text{Cs}) = 0.79$. Remembering this single pair is enough for most comparison questions.

[Board + JEE/NEET essential] — Order of top-5 electronegative elements (Pauling scale):

$$\text{F}(3.98) \;>\; \text{O}(3.44) \;>\; \text{Cl}(3.16) \;>\; \text{N}(3.04) \;>\; \text{Br}(2.96)$$

This order appears in many comparison problems in Class 11 Chemistry.

The Three Main Electronegativity Scales

Different chemists have proposed different ways to quantify electronegativity. The three most famous are:

1. Pauling scale (1932) — the NCERT standard

Linus Pauling based his scale on bond-dissociation energies. He noticed that for a heteronuclear bond A–B, the bond energy is consistently larger than the geometric mean of the homonuclear bond energies A–A and B–B:

$$\Delta \;=\; E_{AB} - \sqrt{E_{AA} \cdot E_{BB}}$$

Pauling attributed this extra stability to the "ionic resonance" that arises from differing electronegativities. His quantitative relationship (with bond energies in kcal/mol) is:

$$|\chi_{A} - \chi_{B}| \;=\; 0.208\,\sqrt{\Delta}$$

He then fixed F near 4.0 (later refined to 3.98) to establish absolute values.

This is the scale that NCERT and most Indian board/entrance exams use. Memorise Pauling values.

2. Mulliken scale (1934)

Robert Mulliken argued that electronegativity should equal the average of an atom's "grabbing power" (ionization enthalpy) and "holding power" (electron affinity):

$$\chi_{M} \;=\; \dfrac{\Delta_{i}H + EA}{2}$$

This definition is conceptually elegant because it directly uses measurable atomic quantities. In practice, however, accurate electron affinity values are not always easy to obtain.

3. Allred-Rochow scale (1958)

Allred and Rochow defined electronegativity as the electrostatic force experienced by an electron at the surface of the atom, giving:

$$\chi_{AR} \;=\; \dfrac{0.359\,Z_{\text{eff}}}{r^{2}} \;+\; 0.744$$

where $r$ is the covalent radius in angstroms. This scale tightly links electronegativity to two fundamental atomic parameters and agrees numerically with Pauling values in many cases.

Which scale should you use in exams?

Always Pauling, unless the question specifically asks for Mulliken or Allred-Rochow. The Pauling value of F is 3.98 — this should become instinct.

[JEE Advanced tip] — The formula $\chi_{M} = \tfrac{1}{2}(IE + EA)$ is a classic derivation question. If an atom has $IE = 1681$ kJ/mol and $EA = 328$ kJ/mol, then $\chi_{M} = (1681 + 328)/2 = 1004.5$ kJ/mol in Mulliken's energy-based form.

Periodic Trends in Electronegativity

Across a period (left to right)

Electronegativity increases across a period because:

1. The nuclear charge $Z$ rises,
2. Atomic size decreases (electrons added to same shell),
3. Effective nuclear charge on the bonding electrons increases.

All three effects pull the bonding pair tighter toward the atom. Examples:

• Period 2: Li (0.98) → Be (1.57) → B (2.04) → C (2.55) → N (3.04) → O (3.44) → F (3.98)
• Period 3: Na (0.93) → Mg (1.31) → Al (1.61) → Si (1.90) → P (2.19) → S (2.58) → Cl (3.16)

[Exam trap] — Noble gases (He, Ne, Ar…) do not have universally agreed electronegativity values because they normally do not form bonds. Some tables give values for Xe in specific compounds, but mainstream school-level exam practice is to skip noble gases in EN comparisons.

Down a group

Electronegativity decreases down a group because:

1. The outermost electrons move into larger, more distant shells,
2. Increased shielding by inner electrons reduces effective nuclear pull,
3. Atomic size increases.

Examples:

• Group 17: F (3.98) > Cl (3.16) > Br (2.96) > I (2.66)
• Group 1: Li (0.98) > Na (0.93) > K (0.82) > Rb (0.82) > Cs (0.79)

The "upper-right corner" rule

If you stare at the heat-map, a simple pattern emerges: the most electronegative atoms cluster in the upper-right corner (F, O, Cl, N). The least electronegative ones sit in the lower-left corner (Cs, Fr, Rb, K). The gradient runs diagonally from lower-left (metallic, low $\chi$) to upper-right (non-metallic, high $\chi$).

Hydrogen's oddity

Hydrogen ($\chi = 2.20$) sits in Group 1 on most periodic tables but its electronegativity is surprisingly high — close to carbon's 2.55. This reflects the fact that H is intermediate between alkali metals (very electropositive) and halogens (very electronegative). It is the reason H can either donate an electron (as in HCl giving H$^{+}$) or accept one (as in NaH containing H$^{-}$).

Comparison of related trends

All four major periodic properties move in coordinated fashion:

Property	Across period	Down group
Atomic radius	decreases	increases
Ionization enthalpy	increases (mostly)	decreases
Electron gain enthalpy	more negative	less negative
Electronegativity	increases	decreases

[NEET Important] — Electronegativity rises toward the top-right because all the helpful factors (smaller size, higher $Z_{\text{eff}}$, less shielding) combine in that corner.

Electronegativity and Bond Character

The real power of electronegativity is that it predicts bond character — whether a bond is nonpolar, polar covalent, or largely ionic. The single controlling number is the electronegativity difference:

$$\Delta\chi \;=\; |\chi_{A} - \chi_{B}|$$

The rule of thumb (NCERT)

$\Delta\chi$	Bond character
0	Pure covalent (nonpolar) — e.g., H–H, Cl–Cl, O=O
0 – 0.4	Essentially nonpolar covalent
0.4 – 1.7	Polar covalent
$>$ 1.7	Predominantly ionic

Any ionic/covalent classification question in Class 11 hinges on this rough cut-off.

Pauling's empirical relation for % ionic character

$$\%\text{ ionic character} \;=\; \left[1 - e^{-(\Delta\chi)^{2}/4}\right] \times 100$$

Plugging in $\Delta\chi = 1.7$ gives roughly 51% ionic character — that is the origin of the 1.7 cut-off.

Some useful points from this relation:

• $\Delta\chi = 0.5 \Rightarrow \sim 6\%$ ionic
• $\Delta\chi = 1.0 \Rightarrow \sim 22\%$ ionic
• $\Delta\chi = 1.7 \Rightarrow \sim 51\%$ ionic
• $\Delta\chi = 2.0 \Rightarrow \sim 63\%$ ionic
• $\Delta\chi = 3.0 \Rightarrow \sim 89\%$ ionic

Dipole moment direction

In a polar covalent bond A–B where B is more electronegative, the bonding electron density sits closer to B, giving:

• B acquires a partial negative charge ($\delta^{-}$)
• A acquires a partial positive charge ($\delta^{+}$)
• The dipole moment points from $\delta^{+}$ to $\delta^{-}$

For example, in H–F: H is $\delta^{+}$, F is $\delta^{-}$; the dipole moment points from H toward F.

Three classic examples

• H–H ($\Delta\chi = 0$) — pure covalent, no dipole.
• H–F ($\Delta\chi = 1.78$) — very strongly polar covalent bond.
• Na–Cl ($\Delta\chi = 2.23$) — predominantly ionic.

Diagonal relationship

A neat consequence: elements on the diagonal of the periodic table (e.g., Li–Mg, Be–Al, B–Si) have similar chemistry in many ways. This is called the diagonal relationship.

[JEE Main classic] — The covalent character in an ionic compound follows Fajan's rules: small cation + large anion + high cation charge $\Rightarrow$ more covalent. Electronegativity difference is the starting point, but Fajan's rules fine-tune the picture.

Applications and Connections to Other Concepts

1. Predicting metal vs non-metal behaviour

• Metals have $\chi$ typically below about 2.0. They tend to lose electrons.
• Non-metals usually have higher $\chi$ values. They tend to attract electrons strongly.
• Metalloids (B, Si, Ge, As, Sb, Te) sit in the intermediate region.

This is why metals reside on the left/bottom of the periodic table and non-metals on the upper right.

2. Predicting oxide character

$\chi$ of element	Oxide character
Very low (e.g., Na, K)	Strongly basic ($Na_{2}O + H_{2}O \rightarrow 2\,NaOH$)
Intermediate (e.g., Al, Zn)	Amphoteric
High (e.g., Cl, S, N)	Strongly acidic ($SO_{3} + H_{2}O \rightarrow H_{2}SO_{4}$)

3. Predicting hydride character

• With electronegative atoms (F, O, N), hydrogen ends up as $\delta^{+}$.
• With very electropositive metals, hydrogen can behave as H$^{-}$ (NaH, CaH$_{2}$).

4. Acid-base strength

For the oxoacids H–O–X, greater EN of X weakens the O–H bond and stabilises the conjugate base. This is why HClO$_{4}$ is a strong acid but HClO is weak.

5. Bond length estimation (Schomaker–Stevenson)

$$d_{A-B} \;=\; r_{A} + r_{B} - 0.09\,|\chi_{A} - \chi_{B}|$$

Greater $\Delta\chi$ contracts the bond because the partial ionic character pulls the atoms closer.

6. The diagonal relationship

The diagonal pairs Li–Mg, Be–Al, and B–Si are important because they show similar chemistry due to a balance of size, charge density, and other periodic factors.

7. Connection back to other periodic properties

• Small size + high $Z_{\text{eff}}$: both boost ionization enthalpy, electron gain enthalpy magnitude, and electronegativity.
• High EN often means greater bond polarity.
• High EN usually correlates with stronger attraction for electrons, though compact-shell effects can create exceptions in electron gain enthalpy.

Rule of thumb for exam questions

"The element with the highest electronegativity in the periodic table is fluorine" is one of the most frequently asked single-statement MCQs in NEET and JEE Main.

[JEE Main exam tip] — When a question asks to identify an unknown given its electronegativity, remember these benchmarks:

• $\chi = 4.0$ → F
• $\chi = 3.5$ → O
• $\chi = 3.0$ → N or Cl (use context)
• $\chi = 2.5$ → C
• $\chi = 2.2$ → H
• $\chi = 1.0$ → Li/Na region

Exam Strategy & Cheat Sheet

Seven landmark facts to commit to memory

1. Fluorine is the most electronegative element, with $\chi = 3.98$.
2. Caesium is the least electronegative stable element, $\chi = 0.79$.
3. Hydrogen's electronegativity is 2.20 — close to carbon's 2.55.
4. The top-5 order is F > O > Cl > N > Br.
5. $\Delta\chi < 0.4$ → nonpolar covalent; $0.4 \le \Delta\chi \le 1.7$ → polar covalent; $\Delta\chi > 1.7$ → predominantly ionic.
6. Pauling scale is dimensionless.
7. Electronegativity increases across a period and decreases down a group.

Ten common traps

1. "Electronegativity is the same as electron affinity." Wrong.
2. "Pauling scale values have units of kJ/mol." Wrong.
3. "The highest electronegativity belongs to chlorine because it has the most negative electron gain enthalpy." Wrong.
4. "Electronegativity increases with atomic number." Wrong.
5. "Noble gases are the most electronegative." Wrong.
6. "Metals have higher electronegativity than non-metals." Wrong.
7. "NaCl has $\Delta\chi = 2.23$ and is therefore 100% ionic." Wrong — real bonds are not 100% ionic.
8. "Electronegativity depends only on the element, not on the bonding environment." Wrong.
9. "Order F > Cl > Br > I holds for both electron gain enthalpy and electronegativity." Wrong — for EN yes, for electron gain enthalpy no.
10. "Electronegativity is the same for all isotopes of an element." Correct.

Quick identification table

$\chi$	Likely element
3.98	F
3.44	O
3.16	Cl
3.04	N
2.58	S
2.55	C
2.20	H
2.04	B
1.57	Be
0.98	Li
0.93	Na
0.79	Cs

Five-second reasoning template

To compare EN of two elements, run through:

1. Are they in the same period? If yes, the more right one has higher $\chi$.
2. Are they in the same group? If yes, the higher one has higher $\chi$.
3. Are they diagonal? Then compare actual values.
4. Is one a noble gas? Usually skip it.
5. Is one in a high oxidation state? Its effective electronegativity is often higher.

With these five tests, you can rank almost any pair quickly.

Solved Examples

Example 1: Identify the most electronegative element

Which of the following has the highest electronegativity: O, Cl, F, or N?

Solution:

Pauling electronegativities:

• F: 3.98
• O: 3.44
• Cl: 3.16
• N: 3.04

Fluorine (F) has the highest electronegativity, not only among these four but in the entire periodic table.

Takeaway: F's small size and high effective nuclear charge combine to make it the electronegativity champion.

Example 2: Predict bond character from $\Delta\chi$

Classify the bond in HCl as nonpolar covalent, polar covalent, or ionic. Given $\chi(H) = 2.20$, $\chi(Cl) = 3.16$.

Solution:

$$\Delta\chi = |2.20 - 3.16| = 0.96$$

Since $0.4 < 0.96 < 1.7$, the bond is polar covalent.

The more electronegative Cl carries $\delta^{-}$, H carries $\delta^{+}$.

Takeaway: Many common molecular bonds fall in the polar covalent range.

Example 3: Compute % ionic character

For a bond with $\Delta\chi = 1.5$, compute the percentage ionic character using Pauling's relation.

Solution:

Pauling's relation:

$$\%\text{ ionic} = \left[1 - e^{-(\Delta\chi)^{2}/4}\right] \times 100$$

With $\Delta\chi = 1.5$:

$$\frac{(\Delta\chi)^{2}}{4} = \frac{2.25}{4} = 0.5625$$ $$e^{-0.5625} \approx 0.5698$$ $$\%\text{ ionic} = (1 - 0.5698) \times 100 \approx \mathbf{43.0\%}$$

Takeaway: $\Delta\chi = 1.5$ gives about 43% ionic character; the 50% region lies near $\Delta\chi \approx 1.7$.

Example 4: Order on the Pauling scale

Arrange Li, Be, B, C, N in order of increasing electronegativity.

Solution:

Moving left-to-right in Period 2 increases EN:

• Li (0.98) < Be (1.57) < B (2.04) < C (2.55) < N (3.04)

So the order is:

$$\boxed{\text{Li} < \text{Be} < \text{B} < \text{C} < \text{N}}$$

Takeaway: This is the standard across-period increase.

Example 5: Mulliken electronegativity computation

Given $\Delta_{i}H = 1680$ kJ/mol and electron affinity = 328 kJ/mol for fluorine, compute the Mulliken electronegativity in energy units.

Solution:

The Mulliken definition is:

$$\chi_{M} = \frac{IE + EA}{2}$$

Substituting:

$$\chi_{M} = \frac{1680 + 328}{2} = \frac{2008}{2} = 1004\ \text{kJ/mol}$$

Takeaway: Mulliken electronegativity is directly tied to ionization enthalpy and electron affinity.

Example 6: Dipole direction

In the C–O bond, which atom is $\delta^{+}$ and which is $\delta^{-}$? Use Pauling EN values.

Solution:

$\chi(C) = 2.55$, $\chi(O) = 3.44$.

Since O is more electronegative, it pulls the bonding electrons closer to itself:

• O acquires $\delta^{-}$
• C acquires $\delta^{+}$
• Dipole arrow points from C to O.

$$\delta^{+}\ \text{C} \longrightarrow \text{O}\ \delta^{-}$$

Takeaway: In a polar covalent bond, the dipole points toward the more electronegative atom.

Example 7: Identifying a bond as mostly ionic

Is the bond in CsF mostly ionic, polar covalent, or nonpolar? Use $\chi(Cs) = 0.79$, $\chi(F) = 3.98$.

Solution:

$$\Delta\chi = |0.79 - 3.98| = 3.19$$

Since $\Delta\chi > 1.7$, the bond is predominantly ionic.

Its ionic character is very high by Pauling's relation.

Takeaway: A very large electronegativity difference indicates a largely ionic bond.

Example 8: Comparing oxides' acidic-basic nature

Which oxide is more acidic: $Na_{2}O$ or $Cl_{2}O_{7}$? Use electronegativity arguments.

Solution:

$\chi(Na)$ is very low, so sodium forms strongly basic oxides.

$\chi(Cl)$ is high, so chlorine forms strongly acidic oxides.

Therefore:

• $Na_{2}O$ is strongly basic.
• $Cl_{2}O_{7}$ is strongly acidic.

Takeaway: Acidic character of an oxide increases with the electronegativity of the central element.

Example 9: Hybridization effect on electronegativity

Why is the electronegativity of an $sp$-hybridized carbon greater than that of an $sp^{3}$-hybridized carbon?

Solution:

The $s$ orbital penetrates closer to the nucleus than the $p$ orbital.

• $sp$-hybrid: 50% $s$ character
• $sp^{2}$-hybrid: 33% $s$ character
• $sp^{3}$-hybrid: 25% $s$ character

More $s$ character means the electrons are held closer to the nucleus, so effective electronegativity increases.

Takeaway: $sp$ > $sp^{2}$ > $sp^{3}$ in electronegativity.

Example 10: Diagonal relationship

Why is the chemistry of Li similar to Mg, even though Li is in Group 1 and Mg is in Group 2?

Solution:

Li and Mg form a classic diagonal relationship pair. This similarity does not arise because their electronegativities are nearly equal; in fact, Li (0.98) is much closer numerically to Na (0.93) than to Mg (1.31). Instead, Li and Mg show similar chemistry because of a combination of factors such as relatively small ionic size, comparatively high charge density or polarising power, and the diagonal balance of periodic trends.

As a result, they share several chemical features:

• Both form normal oxides (Li$_2$O, MgO).
• Both form nitrides directly with nitrogen.
• Both show greater covalent character in some of their compounds than their neighbouring group members.

Takeaway: The diagonal relationship is a multi-factor periodic effect, not a simple one-number electronegativity match.

Example 11: Bond length estimation

Using the Schomaker-Stevenson relation, estimate the H–F bond length. Given $r_{H} = 0.32$ Å, $r_{F} = 0.72$ Å, $\chi_{H} = 2.20$, $\chi_{F} = 3.98$.

Solution:

Schomaker–Stevenson equation:

$$d_{A-B} = r_{A} + r_{B} - 0.09\,|\chi_{A} - \chi_{B}|$$

Substituting:

$$d_{H-F} = 0.32 + 0.72 - 0.09 \times |2.20 - 3.98|$$ $$= 1.04 - 0.09 \times 1.78$$ $$= 1.04 - 0.160$$ $$\approx \mathbf{0.88\ Å}$$

Takeaway: Greater electronegativity difference shortens a bond slightly because partial ionic character pulls the atoms closer.

Example 12: Rank compounds by ionic character

Using electronegativity values, rank the following bonds by increasing ionic character: H–Cl, Na–Cl, Li–F, Cs–F, O–H.

Solution:

Compute $\Delta\chi$ for each:

Bond	$\Delta\chi$
H–Cl	0.96
O–H	1.24
Na–Cl	2.23
Li–F	3.00
Cs–F	3.19

Ranking from least to most ionic:

$$H\text{-}Cl < O\text{-}H < Na\text{-}Cl < Li\text{-}F < Cs\text{-}F$$

Takeaway: Larger electronegativity difference generally means greater ionic character.