Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and

Exercises · 4.14

Q. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:

(a) K and S

(b) Ca and O

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Concept understanding — Lewis Dot Structures

Why do we need Lewis dot structures?

Atoms are held together in molecules by chemical bonds — but what exactly is a bond? In the early 20th century, Gilbert N. Lewis realised that the key lies in the valence electrons (the outermost electrons). He noticed that atoms of noble gases (like Ne, Ar) are extremely stable and unreactive, and they all have 8 electrons in their outermost shell (except helium, which has 2). This led to the octet rule: atoms tend to gain, lose, or share electrons to achieve a full outer shell of 8 electrons (or 2 for hydrogen).

Lewis dot structures are simply a shorthand picture of this idea. They show:

Which atoms are connected to which
How many valence electrons each atom contributes
How those electrons are arranged as bonding pairs (shared) or lone pairs (unshared)

The precise statement

A Lewis dot structure (or electron dot structure) represents the valence electrons of an atom or molecule using dots placed around the element's symbol. Each dot stands for one valence electron. Shared pairs (bonds) are shown as lines, and unshared pairs as pairs of dots.

For a single atom, you write the element symbol and place dots on its four sides (top, bottom, left, right) — up to 8 dots. The order of filling doesn't matter for the final picture, but conventionally you place one dot on each side first, then pair them up.

For example:

Carbon (group 14, 4 valence electrons): $\cdot \text{C} \cdot$ (four single dots)
Oxygen (group 16, 6 valence electrons): $\cdot \ddot{\text{O}} \cdot$ (two single dots and two pairs)

How to draw a Lewis structure for a molecule

Here's the step-by-step method you'll use in exams:

Count total valence electrons — add up valence electrons from all atoms. For ions, add 1 electron for each negative charge, subtract 1 for each positive charge.
Identify the central atom — usually the least electronegative element (not hydrogen or fluorine). Place it in the centre.
Connect atoms with single bonds — each bond uses 2 electrons. Subtract these from your total.
Complete octets of outer atoms — place remaining electrons as lone pairs on terminal atoms (except hydrogen, which only needs 2).
Place leftover electrons on the central atom — if any remain.
If the central atom has fewer than 8 electrons, form multiple bonds — move lone pairs from outer atoms to create double or triple bonds until the central atom has an octet.

Watch out

A common mistake: forgetting that hydrogen only needs 2 electrons (a duet), not 8. Never put more than 2 electrons around H.

A concrete example: water (H₂O)

Total valence electrons: O has 6, each H has 1 → $6 + 1 + 1 = 8$ electrons.
Central atom: oxygen (least electronegative after H).
Connect: O—H bonds (2 bonds × 2 electrons = 4 electrons used).
Remaining: $8 - 4 = 4$ electrons → place as two lone pairs on oxygen.
Check: O has 2 bonds (4 electrons) + 2 lone pairs (4 electrons) = 8. Each H has 1 bond (2 electrons) = 2. Done.

The structure: $\text{H} - \ddot{\text{O}} - \text{H}$

What the structure tells you

Once drawn, a Lewis structure reveals:

Bond order (single, double, triple)
Lone pairs (which affect molecular shape and reactivity)
Formal charge (a bookkeeping tool to check which structure is most stable) …

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Lewis symbols track valence electrons as dots around element symbols. Metals lose electrons to form cations, nonmetals gain them to form anions, and the transfer continues until both achieve noble-gas configurations. For (a) K and S: two $\text{K}$ atoms each lose 1e⁻ to form $\text{K}^+$ , one $\text{S}$ gains 2e⁻ to form $\text{S}^{2-}$ , giving $\text{K}_2\text{S}$ . For (b) Ca and O: $\text{Ca}$ loses 2e⁻ to form $\text{Ca}^{2+}$ , $\text{O}$ gains 2e⁻ to form $\text{O}^{2-}$ , giving $\text{CaO}$ . For (c) Al and N: two $\text{Al}$ atoms each lose 3e⁻ to form $\text{Al}^{3+}$ , one $\text{N}$ gains 3e⁻ to form $\text{N}^{3-}$ , giving $\text{AlN}$ .

Why Lewis symbols work for ionic bonding

Lewis symbols represent only the valence electrons—the outermost shell electrons that participate in bonding. Each dot stands for one valence electron. When a metal meets a nonmetal, the metal's low ionization energy makes it energetically favorable to lose electrons, while the nonmetal's high electron affinity drives it to gain them. The transfer stops when both species reach a stable noble-gas electron configuration (an octet for most elements, a duet for those near helium).

The number of electrons transferred determines the charges on the resulting ions, and those charges dictate the stoichiometry of the ionic compound through charge balance.

(a) Potassium (K) and Sulfur (S)

1. Write the Lewis symbols for the neutral atoms.

Potassium is in Group 1, so it has 1 valence electron:

\text{K}^\cdot

Sulfur is in Group 16, so it has 6 valence electrons:

\cdot\!\!\overset{..}{\text{S}}\!\!\underset{..}{\cdot}\quad\text{or}\quad \overset{..}{\underset{..}{\text{S}}}\!\!\cdot\!\cdot

(The exact arrangement of dots doesn't matter; what counts is the total.)

2. Determine the electron transfer.

Potassium wants to lose its single valence electron to achieve the argon configuration (empty valence shell, stable $3s^2 3p^6$ core). Sulfur needs 2 more electrons to complete an octet (to match argon's $3s^2 3p^6$ ). Therefore, two potassium atoms each donate 1 electron to one sulfur atom.

3. Show the transfer with arrows.

2\,\text{K}^\cdot \quad + \quad \cdot\!\!\overset{..}{\text{S}}\!\!\underset{..}{\cdot} \quad \xrightarrow{\text{transfer}} \quad 2\,\text{K}^+ \quad + \quad \left[\overset{..}{\underset{..}{\text{S}}}\!\!\overset{..}{\cdot\!\cdot}\right]^{2-}

Each potassium loses its dot (1e⁻), becoming $\text{K}^+$ with no dots (stable noble-gas core). Sulfur gains two dots (2e⁻), completing an octet and becoming $\text{S}^{2-}$ .

4. Write the resulting ionic compound.

The formula is $\text{K}_2\text{S}$ (potassium sulfide), with charge balance: $2(+1) + 1(-2) = 0$ .

(b) Calcium (Ca) and Oxygen (O)

1. Write the Lewis symbols.

Calcium is in Group 2, so it has 2 valence electrons:

\cdot\!\text{Ca}\!\cdot

Oxygen is in Group 16, so it has 6 valence electrons:

\cdot\!\!\overset{..}{\text{O}}\!\!\underset{..}{\cdot}

2. Determine the electron transfer.

Calcium loses both valence electrons to achieve the argon configuration. Oxygen needs 2 electrons to complete an octet (to match neon's $2s^2 2p^6$ ). The numbers match perfectly: one calcium atom donates 2 electrons to one oxygen atom.

3. Show the transfer.

\cdot\!\text{Ca}\!\cdot \quad + \quad \cdot\!\!\overset{..}{\text{O}}\!\!\underset{..}{\cdot} \quad \xrightarrow{\text{transfer}} \quad \text{Ca}^{2+} \quad + \quad \left[\overset{..}{\underset{..}{\text{O}}}\!\!\overset{..}{\cdot\!\cdot}\right]^{2-}

Calcium loses both dots (2e⁻) to become $\text{Ca}^{2+}$ . Oxygen gains two dots (2e⁻) to complete an octet and become $\text{O}^{2-}$ .

4. Write the resulting ionic compound.

The formula is $\text{CaO}$ (calcium oxide), with charge balance: $(+2) + (-2) = 0$ .

(c) Aluminum (Al) and Nitrogen (N)

1. Write the Lewis symbols.

Aluminum is in Group 13, so it has 3 valence electrons:

\cdot\!\overset{\cdot}{\text{Al}}\!\cdot

Nitrogen is in Group 15, so it has 5 valence electrons:

\cdot\!\!\overset{..}{\text{N}}\!\!\underset{..}{\cdot}

2. Determine the electron transfer. …

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Q.Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:

Why Lewis symbols work for ionic bonding

(a) Potassium (K) and Sulfur (S)

(b) Calcium (Ca) and Oxygen (O)

(c) Aluminum (Al) and Nitrogen (N)

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