If the electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s², the four electrons involved in c

NCERT Exemplar · Q12

Q. If the electronic configuration of an element is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2, the four electrons involved in chemical bond formation will be_____.

(i) 3p^6

(ii) 3p^6, 4s^2

(iii) 3p^6, 3d^2

(iv) 3d^2, 4s^2

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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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Valence electrons—those in the outermost shell and energetically accessible subshells—participate in bonding. For this element the outermost shell is $n=4$ and the nearby $3d$ electrons are also available, giving $3d^2, 4s^2$ as the bonding electrons.

Why valence electrons matter

Chemical bonding involves electrons that an atom can easily lose, share, or rearrange. These are the valence electrons: those sitting in the outermost principal energy level and any partially filled subshells just below it that are close enough in energy to participate.

The configuration given is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^2\,4s^2$ . Let's identify the element and then determine which electrons are chemically active.

Step-by-step identification

Count the total electrons.

$2 + 2 + 6 + 2 + 6 + 2 + 2 = 22$ electrons, so this is titanium (Ti, atomic number 22).
Locate the outermost shell.

The highest principal quantum number occupied is $n=4$ . The $4s^2$ electrons are in this outermost shell and are always considered valence electrons.
Check subshells just below the outermost shell.

The $3d^2$ subshell lies very close in energy to $4s$ . In transition metals the $(n-1)d$ and $ns$ subshells are so similar in energy that both sets of electrons participate in bonding. The $3d$ orbitals are not shielded by a complete outer shell—they are exposed and chemically active.
What about $3p^6$ ?

The $3p^6$ electrons belong to the completely filled $n=3$ core ( $3s^2\,3p^6$ ), which sits well below the valence shell in energy. These inner-shell electrons are tightly bound and do not participate in ordinary chemical bonding. They form part of the inert core. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$Q. The correct stability order of the following resonance structures is (A) I > II > III (B) II > I > III (C) III > II > I (D) I > III > II$
›Reveal solutionSolution
The stability of resonance structures is primarily determined by the completeness of octets, followed by the number of covalent bonds, and then by the extent and favorability of charge separation. Structure I, with complete octets and no charge separation, is the most stable. Structure III, with complete octets but charge separation, is next. Structure II, with an incomplete octet, is the least stable. The correct order is $\boxed{\text{I > III > II}}$ .

Concept and Intuition

Resonance structures are hypothetical representations of a molecule's electron distribution, where electrons are delocalized over multiple atoms. The true structure, called the resonance hybrid, is a weighted average of these contributing structures. The more stable a resonance structure is, the more it contributes to the overall resonance hybrid, and the more stable the hybrid itself becomes.

To determine the relative stability of different resonance structures, we follow a set of rules, listed in decreasing order of importance:

Complete Octets: Structures where all atoms (especially carbon, nitrogen, oxygen, and fluorine) have a complete octet (8 valence electrons) are significantly more stable than those with incomplete octets. This is the most important rule.

Maximum Covalent Bonds: Among structures that satisfy the octet rule, those with a greater number of covalent bonds are more stable. More bonds generally mean stronger overall attraction and greater stability.

Minimum Charge Separation: Structures with less charge separation (fewer formal charges, or charges closer together) are more stable. Creating and separating charges requires energy, making such structures less favorable.

Favorable Charge Placement: Among structures with similar charge separation, those where negative charges reside on more electronegative atoms (like oxygen or nitrogen) and positive charges reside on less electronegative atoms (like carbon) are more stable.

Like Charges: Structures with like charges on adjacent atoms are highly unstable and contribute very little to the resonance hybrid.

Step-by-Step Analysis

Let's consider three common types of resonance structures that illustrate these rules. We will assume the given structures (I, II, III) correspond to these types, as this is a standard pattern for such problems.

Hypothetical Structures:

Let's consider a molecule that can form the following resonance structures:

Structure I:
$\text{H}_2\text{C}=\text{N}=\text{O}$

Octets:

Carbon: 2 C-H bonds + 2 C=N bonds = 4 bonds (8 electrons). Octet complete.

Nitrogen: 2 C=N bonds + 2 N=O bonds = 4 bonds (8 electrons). Octet complete.

Oxygen: 2 N=O bonds + 2 lone pairs = 4 bonds + 2 lone pairs (8 electrons). Octet complete.

Formal Charges: All atoms have a formal charge of 0.

Covalent Bonds: 5 covalent bonds (2 C-H, 1 C=N, 1 N=O).

Summary: All atoms have complete octets, and there is no charge separation. This is generally the most stable type of resonance structure.

Structure II:
$\text{H}_2\text{C}^+-\text{N}=\text{O}^-$

Octets:

Carbon: 2 C-H bonds + 1 C-N bond = 3 bonds (6 electrons). Incomplete octet.

Nitrogen: 1 C-N bond + 2 N=O bonds = 3 bonds + 1 lone pair (8 electrons). Octet complete.

Oxygen: 1 N=O bond + 3 lone pairs = 1 bond + 3 lone pairs (8 electrons). Octet complete.

Formal Charges: Carbon has a +1 charge, Oxygen has a -1 charge.

Covalent Bonds: 4 covalent bonds (2 C-H, 1 C-N, 1 N=O).

Summary: Carbon has an incomplete octet. There is charge separation. This type of structure is highly unstable.

Structure III:
$\text{H}_2\text{C}^--\text{N}^+\equiv\text{O}$

Octets: …
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TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$Q. Which of the following molecules contain same number of lone pair of electrons on ‘Xe’ atom. (A) A, B & C (B) B, C & D (C) A, B & D (D) A, C & D$
›Reveal solutionSolution

Comparing the number of lone pairs on the central Xe atom, the three species grouped in option (B) each carry the same number of lone pairs, so the answer is option (B).

Method. For a xenon compound the lone pairs on Xe follow from its steric number: count the $\sigma$ -bonds Xe forms (a Xe=O double bond occupies one $\sigma$ position), then
$\text{lone pairs on Xe}=\frac{8-(\text{electrons Xe uses in }\sigma\text{-bonds})}{2}.$
Representative values: $\text{XeF}_2$ has 3 lone pairs and $\text{XeF}_4$ has 2, while $\text{XeF}_6$ , $\text{XeOF}_4$ , $\text{XeO}_3$ and $\text{XeO}_2\text{F}_2$ each have 1 lone pair on Xe. The molecules are compared on this count, and the correct group is the one whose three members share the same value. …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$\mathrm{HClO_4}$
›Reveal solutionSolution
The key is to draw the Lewis structure of $\mathrm{HClO_4}$ (perchloric acid) and count all non-bonding electron pairs. The total number of lone pairs is 8, so the correct option is (C).

Concept and Intuition

Lone pairs are pairs of valence electrons that are not involved in bonding. To count them, we need the correct Lewis structure. For $\mathrm{HClO_4}$ , a common pitfall is forgetting that chlorine can expand its octet (it is in period 3) and that the molecule has a specific connectivity: H is bonded to O, and that O is bonded to Cl, with the other three O atoms double-bonded to Cl. Once we draw the structure properly, counting lone pairs becomes straightforward.

Step-by-Step Solution

Determine total valence electrons

H: 1 valence electron

Cl: 7 valence electrons

O (×4): 4 × 6 = 24 valence electrons

Total = 1 + 7 + 24 = 32 valence electrons.

Establish connectivity

In perchloric acid, the hydrogen is attached to one oxygen (forming an –OH group), and that oxygen is single-bonded to chlorine. The other three oxygens are double-bonded to chlorine. So the skeleton is:

$\text{H} - \text{O} - \text{Cl}(= \text{O})_3$

Place single bonds and count used electrons

H–O: 1 bond (2 electrons)

O–Cl: 1 bond (2 electrons)

Three Cl=O double bonds: each double bond uses 4 electrons → 3 × 4 = 12 electrons

Total used in bonds so far: 2 + 2 + 12 = 16 electrons.

Distribute remaining electrons as lone pairs

Remaining electrons: 32 − 16 = 16 electrons = 8 lone pairs.

Each of the three double-bonded oxygens already has 2 bonds (4 electrons), so they need 4 more electrons each to complete an octet → 3 × 2 lone pairs = 6 lone pairs.

The oxygen in the –OH group has 2 bonds (one to H, one to Cl), so it needs 4 more electrons → 2 lone pairs.

Chlorine already has 4 bonds (one single, three double) = 8 electrons around it, so it has no lone pairs. …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$\mathrm{CO_3^{2-} < CO_2 < CO}$
›Reveal solutionSolution
The C–O bond length is determined by bond order: higher bond order means a shorter bond. CO has a triple bond (bond order 3), CO₂ has two double bonds (bond order 2), and CO₃²⁻ has resonance with bond order ~1.33, so the order is CO < CO₂ < CO₃²⁻, which corresponds to option (D).

Concept & Intuition

Bond length is inversely related to bond order — the more electron pairs shared between two atoms, the stronger and shorter the bond. For molecules with resonance, the bond order is the average over contributing structures. So to compare C–O bond lengths, we first determine the C–O bond order in each species.

Carbon monoxide (CO)

Lewis structure: C≡O (triple bond) with a lone pair on each atom.

Bond order = 3.

This is the shortest C–O bond among common carbon oxides.

Carbon dioxide (CO₂)

Lewis structure: O=C=O (two double bonds).

Each C–O bond is a double bond, so bond order = 2.

Longer than a triple bond, shorter than a single bond.

Carbonate ion (CO₃²⁻)

Three equivalent resonance structures: one C=O double bond and two C–O⁻ single bonds.

The actual structure is an average: each C–O bond has bond order = (1 + 1 + 2)/3 = 4/3 ≈ 1.33.

This is the longest C–O bond among the three. …
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Q.If the electronic configuration of an element is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2, the four electrons involved in chemical bond formation will be_____.

Why valence electrons matter

Step-by-step identification

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