Ionic and Covalent Bonding

25min Part 1 / Ch1 / Lesson 4

Prerequisites: 1-1-3

Objectives

Explain ionic bonding using electron configurations
Describe covalent bonding and write electron/structural formulas
Understand coordinate (dative) bonding
Explain the relationship between electronegativity and bond polarity

Overview of Chemical Bonds

Three ways atoms achieve stable electron configurations (noble gas configuration)

All chemical bonds aim to achieve a stable electron configuration (noble gas type: 8 electrons in the outer shell). If atoms transfer electrons → ionic bond. If atoms share electrons → covalent bond.

Ionic Bonding

Na (2, 8, 1) → loses 1 electron → $\mathrm{Na^+}$ (2, 8) = Ne configuration

Cl (2, 8, 7) → gains 1 electron → $\mathrm{Cl^-}$ (2, 8, 8) = Ar configuration

$\mathrm{Na^+}$ and $\mathrm{Cl^-}$ attract via electrostatic force (Coulomb force) → ionic bond

Covalent Bonding

Nonmetal atoms share electrons to form bonds.

Bond type	Shared pairs	Notation	Example
Single bond	1 pair	─	$\mathrm{H-H,\ C-H}$
Double bond	2 pairs	═	$\mathrm{O=O,\ C=O}$
Triple bond	3 pairs	≡	$\mathrm{N≡N}$

Electronegativity and Polarity

Electronegativity = the ability of an atom to attract shared electrons. Greater difference → more polar the bond.

Element	F	O	N, Cl	C	H
Electronegativity	4.0	3.4	3.0	2.6	2.2

Nonpolar covalent: same atoms ( $\mathrm{H_2, O_2}$ ) → difference = 0
Polar covalent: different atoms ( $\mathrm{H-Cl}$ ) → electrons shifted toward Cl
Ionic: very large difference ( $\mathrm{NaCl}$ ) → electrons effectively transferred

Coordinate (Dative) Bonding

Coordinate bond: a covalent bond where one atom donates both electrons (a lone pair). Once formed, it’s indistinguishable from a regular covalent bond. Examples: $\mathrm{NH_4^+}$ , $\mathrm{H_3O^+}$ .

Check Your Understanding

Q1 What type of bond does NaCl have?

Q2 The bond between nitrogen atoms in N₂ is:

Q3 Which element has the highest electronegativity?

Exercises

Q1. Write the structural formula of $\mathrm{CO_2}$ . (C: 4 valence e⁻, O: 6 valence e⁻)

Solution

C needs to share 4 electrons; each O shares 2. Two double bonds between C and O:

O=C=O

C has 4 shared pairs (2 double bonds). Each O has 2 shared pairs + 2 lone pairs. All atoms satisfy the octet rule.

Q2. Explain why $\mathrm{MgCl_2}$ is an ionic compound, using electron configurations.

Solution

Mg (2, 8, 2): loses 2 electrons → $\mathrm{Mg^{2+}}$ (2, 8) = Ne configuration
Cl (2, 8, 7): gains 1 electron → $\mathrm{Cl^-}$ (2, 8, 8) = Ar configuration

One Mg gives 2 electrons to 2 Cl atoms. The resulting $\mathrm{Mg^{2+}}$ and 2 $\mathrm{Cl^-}$ ions bond ionically → formula $\mathrm{MgCl_2}$ .