Covalent Bond — Lewis Structures & Octet Rule

The Covalent Bond: Sharing is Caring

When two atoms are both non-metals, neither wants to give up electrons — so they compromise and share.

Definition: A covalent bond is formed by the mutual sharing of one or more pairs of electrons between two atoms, so that each attains a stable noble-gas configuration.

The shared pair is called a bond pair; electrons not involved in bonding are lone pairs.

• Single bond — one shared pair (e.g., $H-H$, $Cl-Cl$).
• Double bond — two shared pairs (e.g., $O=O$ in $O_2$, $C=O$ in $CO_2$).
• Triple bond — three shared pairs (e.g., $N \equiv N$, $C \equiv C$ in ethyne).

More shared pairs means a stronger, shorter bond: triple > double > single in strength, and triple < double < single in length.

[NEET Important] A covalent bond is directional (it points in a specific direction between the two atoms), which is why covalent molecules have definite shapes — unlike non-directional ionic bonds.

How to Draw a Lewis (Electron-Dot) Structure

Lewis structures show every valence electron as a dot (or a line for a bond pair). Here's the reliable step-by-step recipe used throughout the chapter:

1. Count total valence electrons. Add the valence electrons of all atoms. For an anion, add one electron per negative charge; for a cation, subtract one per positive charge.
2. Pick the central atom. Usually the least electronegative atom (except H, which is always terminal). Carbon is almost always central.
3. Draw a skeleton with single bonds connecting the central atom to the surrounding atoms. Each single bond uses 2 electrons.
4. Complete the octets of the outer atoms first by adding lone pairs.
5. Place any leftover electrons on the central atom.
6. If the central atom lacks an octet, form multiple bonds by converting lone pairs of outer atoms into shared pairs.

Key Point: Hydrogen needs only 2 electrons (duplet); every other main-group atom aims for 8.

[JEE Tip] Always verify your structure by recounting: total electrons used (bond pairs × 2 + lone-pair electrons) must equal the total you started with.

Worked Skeletons: H$_2$, Cl$_2$, O$_2$, N$_2$

Let's apply the recipe to the simple diatomics.

Hydrogen ($H_2$)

Total valence electrons $= 1 + 1 = 2$. One shared pair: $H-H$. Each H now has a duplet. ✓

Chlorine ($Cl_2$)

Total $= 7 + 7 = 14$. One bond pair + 3 lone pairs on each Cl. Each Cl counts 8. ✓

Oxygen ($O_2$)

Total $= 6 + 6 = 12$. A single bond leaves each O with only 7 — so we form a double bond ($O=O$). Each O: 2 lone pairs + 2 shared pairs = 8. ✓

Nitrogen ($N_2$)

Total $= 5 + 5 = 10$. A single or double bond leaves N short of an octet, so we form a triple bond ($:N \equiv N:$). Each N: 1 lone pair + 3 shared pairs = 8. ✓

Key Point: When the central/terminal atom falls short of an octet after single bonds, upgrade to double or triple bonds using the outer atom's lone pairs.

Worked Skeletons: CO$_2$, H$_2$O, NH$_3$, CH$_4$

Carbon dioxide ($CO_2$)

Total $= 4 + 2(6) = 16$. Carbon is central. Single C–O bonds leave carbon with only 4 electrons, so form two double bonds: $O = C = O$. Carbon: 4 shared pairs = 8; each O: 2 lone pairs + 2 shared = 8. ✓ Linear molecule.

Water ($H_2O$)

Total $= 6 + 2(1) = 8$. Oxygen central, two O–H single bonds (2 bond pairs) + 2 lone pairs on O. Bent shape.

Ammonia ($NH_3$)

Total $= 5 + 3(1) = 8$. Nitrogen central, three N–H bonds (3 bond pairs) + 1 lone pair on N. Pyramidal shape.

Methane ($CH_4$)

Total $= 4 + 4(1) = 8$. Carbon central, four C–H single bonds. Carbon octet complete, no lone pairs. Tetrahedral shape.

Key Point: The lone pairs left on the central atom (2 in water, 1 in ammonia, 0 in methane) will decide the molecular shape in VSEPR (Section 8).

[NEET Important] Memorise these four 'reference molecules' — they appear again in VSEPR, hybridization and dipole-moment questions.

Solved Examples

Example 1: Lewis structure of HCl

Draw the Lewis structure of hydrogen chloride and count lone pairs.

Solution:

1. Total valence electrons: $H(1) + Cl(7) = 8$.
2. Bond: one H–Cl single bond (2 electrons shared).
3. Remaining 6 electrons go on Cl as 3 lone pairs.
4. Check: H has a duplet; Cl has $6 + 2 = 8$. ✓

Takeaway: Hydrogen is always terminal and aims for a duplet, not an octet.

Example 2: Number of bonds in $CO_2$

How many sigma and pi bonds are present in carbon dioxide?

Solution:

1. Structure: $O = C = O$ (two C=O double bonds).
2. Each double bond = 1 sigma + 1 pi.
3. Count: 2 sigma bonds + 2 pi bonds in total.

Takeaway: A single bond = 1 σ; a double bond = 1 σ + 1 π; a triple bond = 1 σ + 2 π.

Example 3: Lewis structure of ammonia

Draw the Lewis structure of $NH_3$ and state the number of bond pairs and lone pairs on nitrogen.

Solution:

1. Total valence electrons: $N(5) + 3H(3) = 8$.
2. Bonds: three N–H single bonds → 3 bond pairs (6 electrons).
3. Remaining 2 electrons form 1 lone pair on N.
4. Result: 3 bond pairs + 1 lone pair; nitrogen octet complete.

Takeaway: The single lone pair on N makes ammonia pyramidal and a good electron-pair donor (Lewis base).

Example 4: Triple bond in N$_2$

Why does $N_2$ contain a triple bond and what is its bond order?

Solution:

1. Each N has 5 valence electrons; needs 3 more.
2. Single/double bonds leave nitrogen short of an octet.
3. Three shared pairs (triple bond) complete both octets: $:N \equiv N:$.
4. Bond order = 3 — one of the strongest bonds known, explaining $N_2$'s inertness.

Takeaway: The very strong triple bond (bond order 3) is why atmospheric nitrogen is so unreactive.

Example 5: Counting total valence electrons in an ion

How many valence electrons must the Lewis structure of $NO_3^-$ account for?

Solution:

1. N: 5. Three O: $3 \times 6 = 18$. Subtotal: 23.
2. Charge: add 1 electron for the single negative charge → 24.
3. Pairs to place: $24/2 = 12$ pairs.

Takeaway: Always adjust the total for the ionic charge before drawing the structure.

Example 6: Lewis structure of ethyne (C$_2$H$_2$)

Draw the Lewis structure of acetylene and identify the carbon-carbon bond order.

Solution:

1. Total valence electrons: $2C(8) + 2H(2) = 10$.
2. Skeleton: $H-C-C-H$ uses 3 single bonds (6 electrons), leaving 4 electrons.
3. Complete carbon octets: add the remaining 4 electrons as 2 more C–C bonds → $H-C \equiv C-H$.
4. C–C bond order = 3 (triple bond); each carbon counts 8.

Takeaway: Build the σ skeleton first, then upgrade bonds until every central atom reaches an octet.

Example 7: Why is the C–C bond in ethene a double bond?

In ethene $C_2H_4$, justify the C=C double bond.

Solution:

1. Total valence electrons: $2C(8) + 4H(4) = 12$.
2. Skeleton: each carbon bonds to 2 H (4 C–H bonds, 8 electrons) and to each other.
3. Remaining 4 electrons: placed as a C–C double bond so both carbons reach an octet.
4. Result: $H_2C = CH_2$, C=C bond order 2.

Takeaway: Distribute electrons to give every carbon exactly 4 bonds (its covalence).

Example 8: Coordinate bond in ozone (preview)

Ozone $O_3$ has one O–O single bond and one O=O double bond, with a coordinate bond involved. Briefly explain.

Solution:

1. Central O forms a double bond with one terminal O and a single (coordinate) bond with the other.
2. The coordinate bond: both electrons are donated by the central oxygen to the electron-deficient terminal oxygen.
3. Reality: the two O–O bonds are actually identical due to resonance (Section 7), so the bond order is 1.5 each.

Takeaway: A first hint that a single Lewis structure can be inadequate — resonance fixes it.

Example 9: Identifying sigma and pi bonds in HCN

How many sigma and pi bonds are in hydrogen cyanide $HCN$?

Solution:

1. Structure: $H-C \equiv N$ (one C–H single bond, one C≡N triple bond).
2. C–H: 1 sigma. C≡N: 1 sigma + 2 pi.
3. Totals: 2 sigma bonds and 2 pi bonds.

Takeaway: Count one σ per bonded pair of atoms; the extra bonds in double/triple bonds are π.

Example 10: Lewis structure of CCl$_4$

Draw the Lewis structure of carbon tetrachloride and check the octets.

Solution:

1. Total valence electrons: $C(4) + 4Cl(28) = 32$.
2. Skeleton: four C–Cl single bonds (8 electrons).
3. Remaining 24 electrons: distributed as 3 lone pairs on each of the 4 chlorines ($4 \times 6 = 24$).
4. Check: C has 4 bond pairs = 8; each Cl has $6 + 2 = 8$. ✓

Takeaway: After bonding the central atom, fill outer-atom octets with lone pairs.

Example 11: Bond order and bond length link

Without exact numbers, predict which is shorter: the C–C bond in ethane or the C=C bond in ethene.

Solution:

1. Ethane: C–C single bond (bond order 1).
2. Ethene: C=C double bond (bond order 2).
3. Rule: higher bond order → more shared electron density → shorter bond.
4. Conclusion: the C=C bond in ethene is shorter (and stronger).

Takeaway: Bond length decreases as bond order increases (single > double > triple in length).

Example 12: Lewis structure of the hydronium ion H$_3$O$^+$

Draw the Lewis structure of $H_3O^+$.

Solution:

1. Total valence electrons: $O(6) + 3H(3) = 9$; subtract 1 for the $+$ charge → 8.
2. Bonds: three O–H single bonds (6 electrons).
3. Remaining 2 electrons: 1 lone pair on oxygen.
4. Result: oxygen has 3 bond pairs + 1 lone pair (octet complete); the positive charge sits on O.

Takeaway: For cations, subtract one electron per positive charge before drawing — the mirror of the anion rule.