Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can

NCERT Exemplar · Q66

Q. Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?

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Concept understanding — Orbital Hybridization Theory

Orbital Hybridization Theory – From Intuition to Precision

The Problem That Started It All

Imagine you are looking at a methane molecule, $\text{CH}_4$ . Carbon has four valence electrons: two in the $2s$ orbital and two in the $2p$ orbitals. If carbon used its pure atomic orbitals to bond, you would expect two bonds from the $2s$ (identical, but one direction) and two from the $2p$ (at $90^\circ$ to each other). That would give you three different bond types and bond angles of $90^\circ$ and something else.

But experiment says methane is perfectly tetrahedral: all four bonds are identical in length, strength, and energy, and the bond angle is $109.5^\circ$ , not $90^\circ$ . Something is fundamentally wrong with the "pure orbital" picture.

This is the puzzle that hybridization theory solves.

The Core Intuition

Think of atomic orbitals as shapes that an electron can occupy. The $s$ orbital is a sphere. The $p$ orbitals are dumbbells along the $x$ , $y$ , and $z$ axes. When an atom forms bonds, it wants to mix these shapes together to create new, hybrid shapes that point in directions that maximise bond strength and minimise repulsion.

It is like mixing primary colours to get new colours. You don't have to use red, blue, and yellow separately — you can blend them to get green, orange, or purple. Similarly, an atom can blend its $s$ and $p$ orbitals to get new hybrid orbitals that are better suited for bonding.

The key insight: hybridization is a mathematical mixing of atomic orbitals on the same atom to produce an equal number of new, equivalent hybrid orbitals. The number of hybrid orbitals formed always equals the number of atomic orbitals mixed.

The Precise Statement

s

The theory rests on three pillars:

Conservation of orbitals: Mixing $n$ atomic orbitals gives exactly $n$ hybrid orbitals. No orbitals are created or destroyed.
Energy averaging: The hybrid orbitals have energies that are intermediate between the original $s$ and $p$ energies.
Directionality: Hybrid orbitals point in specific directions to minimise electron pair repulsion, which directly determines molecular shape.

The Three Common Hybridizations

Hybridization	Orbitals Mixed	Number of Hybrids	Geometry	Bond Angle	Example
$sp$	one $s$ + one $p$	2	Linear	$180^\circ$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	3	Trigonal planar	$120^\circ$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	4	Tetrahedral	$109.5^\circ$	$\text{CH}_4$

Note

The superscript in $sp^2$ or $sp^3$ tells you how many $p$ orbitals were mixed. $sp^3$ means one $s$ and three $p$ orbitals were blended. It does not mean there are three $s$ orbitals — there is only one $s$ orbital per shell.

How It Works: The Methane Example

Carbon in its ground state has the configuration $1s^2 2s^2 2p_x^1 2p_y^1$ . Only two unpaired electrons — it should form only two bonds. But we know carbon forms four bonds.

Step 1: Promotion. One electron from the $2s$ orbital is promoted (excited) to the empty $2p_z$ orbital. This costs a small amount of energy, but it is more than compensated by the energy released when four strong bonds form instead of two.

Step 2: Hybridization. The one $2s$ orbital and three $2p$ orbitals mix to form four equivalent $sp^3$ hybrid orbitals. Each hybrid has 25% $s$ character and 75% $p$ character.

Step 3: Bonding. Each $sp^3$ hybrid overlaps with the $1s$ orbital of a hydrogen atom, forming four identical $\sigma$ bonds. The hybrids point to the corners of a tetrahedron, giving the $109.5^\circ$ angle. …

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Valence Bond Theory explains covalent bond formation as the overlap of atomic orbitals containing unpaired electrons, leading to electron pairing and energy stabilization; for dihydrogen, two hydrogen 1s orbitals overlap, and the system's potential energy decreases to a minimum at the bond length, signifying bond formation and energy release.

Valence Bond Theory (VBT) is a fundamental concept in chemistry that explains the formation of covalent bonds based on the overlap of atomic orbitals. The core idea is that a covalent bond forms when two atomic orbitals, each containing an unpaired electron, overlap. This overlap allows the electrons to pair up, leading to a region of increased electron density between the nuclei, which stabilizes the molecule. The greater the overlap, the stronger the bond.

Valence Bond Theory of Covalent Bond Formation (Example: Hydrogen)

Let's use the formation of a dihydrogen molecule ( $\text{H}_2$ ) from two hydrogen atoms to understand VBT and the associated energy changes.

Isolated Hydrogen Atoms:

Each hydrogen atom has one electron in its 1s atomic orbital. The electron configuration is $1s^1$ . The 1s orbital is spherical. When two hydrogen atoms are far apart, they do not interact, and their potential energy is considered zero (a reference point).
Approach of Hydrogen Atoms:

As two hydrogen atoms begin to approach each other, several forces come into play:
- Attractive Forces: The nucleus of one hydrogen atom is attracted to the electron of the other hydrogen atom, and vice-versa.
- Repulsive Forces: The two nuclei repel each other, and the two electrons repel each other.
Initially, as the atoms get closer, the attractive forces are stronger than the repulsive forces.
Orbital Overlap and Bond Formation:

When the two hydrogen atoms come close enough, their 1s atomic orbitals begin to overlap. This overlap allows the two unpaired electrons (one from each hydrogen atom) to pair up and occupy the overlapping region. This pairing of electrons in the overlapping orbitals is the essence of covalent bond formation according to VBT. The bond formed by the direct, head-on overlap of atomic orbitals along the internuclear axis is called a sigma ( $\sigma$ ) bond.

Important

A covalent bond forms when atomic orbitals containing unpaired electrons overlap, and the electrons in the overlapping region become paired, leading to a decrease in the system's potential energy.

Interpretation of Energy Changes in Dihydrogen Formation

The energy changes during the formation of $\text{H}_2$ can be visualized by plotting the potential energy of the system against the internuclear distance between the two hydrogen atoms.

Initial State (Infinite Separation):

When the two hydrogen atoms are infinitely far apart, there is no interaction between them. The potential energy of the system is conventionally taken as zero.
Decreasing Potential Energy (Attraction Dominates):

As the two hydrogen atoms approach each other, the attractive forces (nucleus-electron attractions) start to dominate over the repulsive forces (nucleus-nucleus and electron-electron repulsions). This net attraction leads to a decrease in the potential energy of the system. The system becomes more stable as energy is released.
Minimum Potential Energy (Equilibrium Bond Length): …

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

Showing the 12 most recent of 28 on this concept.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$Q. In graphite the C-C bond length with in the layer is X pm and the distance between two adjacent layers is Y pm. X and Y respectively are (A) 340, 141.5 (B) 141.5, 340 (C) 141.5, 154 (D) 143.5, 340$
›Reveal solutionSolution
Graphite has strong covalent bonds within layers (short C–C bond length ≈ 141.5 pm) and weak van der Waals forces between layers (large interlayer spacing ≈ 340 pm). The correct pair is (141.5, 340), which is option (B).

Graphite’s structure is the key: it consists of flat sheets (graphene layers) where each carbon is bonded to three others in a hexagonal honeycomb. The in‑plane C–C bonds are strong and short (like in benzene or graphene), while the layers stack loosely, held only by weak dispersion forces, so the distance between layers is much larger.

Identify the in‑plane bond length (X)

In a graphene layer, each carbon is sp²‑hybridized, forming σ bonds 120° apart. The C–C bond length in graphite is well‑known to be about 141.5 pm (1.415 Å). This is similar to the bond length in benzene (140 pm) and much shorter than a typical single bond (154 pm). So X = 141.5 pm.

Identify the interlayer distance (Y)

Adjacent layers are held together by weak van der Waals forces, not covalent bonds. The typical spacing between layers in graphite is about 340 pm (3.4 Å). This is much larger than any covalent bond length, reflecting the weak, non‑bonded interaction.

Match to the options

(A) 340, 141.5 → swaps the values (wrong order).

(B) 141.5, 340 → correct. …
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TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$sp^3$
›Reveal solutionSolution
The reaction of XeF₄ with O₂F₂ yields XeF₆ (compound A), which hydrolyzes to XeO₃ (compound B). The central Xe in XeO₃ has $sp^3$ hybridization, so the correct option is (A).

Concept & Intuition

This problem tests your understanding of noble gas chemistry and the link between molecular structure and hybridization. The key is to track the xenon atom through two reactions: first, an oxidation/fluorination step that increases the number of fluorine atoms around xenon; second, a hydrolysis that replaces fluorine with oxygen. The final compound’s geometry (and thus hybridization) is determined by counting electron domains around xenon using VSEPR theory.

Step-by-step reasoning

Identify the first reaction product (compound A). XeF₄ reacts with O₂F₂ at 143 K. O₂F₂ is a strong fluorinating agent (it contains an O–O bond and readily provides fluorine radicals). The reaction is known to produce XeF₆:

$\text{XeF}_4 + \text{O}_2\text{F}_2 \rightarrow \text{XeF}_6 + \text{O}_2$
So compound A is XeF₆.

Determine the hydrolysis product (compound B). Complete hydrolysis of XeF₆ gives HF and a xenon oxide. The reaction is:

$\text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF}$
Thus compound B is XeO₃ (xenon trioxide).

Find the hybridization of the central atom in XeO₃.

Count valence electrons: Xe has 8, each O contributes 0 (as a ligand), and the molecule is neutral.

Draw the Lewis structure: Xe forms three double bonds with three oxygen atoms (Xe=O). This uses 6 electrons. Xenon also has one lone pair (the remaining 2 electrons).

Electron domain count: 3 double bonds + 1 lone pair = 4 electron domains around Xe. …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$_6$
›Reveal solutionSolution
The hydrolysis of XeF $_6$ with one mole of water yields XeOF $_4$ (A) with $sp^3d^2$ hybridization, while hydrolysis with two moles of water yields XeO $_2$ F $_2$ (B) with $sp^3d$ hybridization. The correct option is (B).

The problem asks us to determine the hybridization of the central xenon atom in the products formed from the partial hydrolysis of XeF $_6$ . Xenon hexafluoride (XeF $_6$ ) undergoes hydrolysis in a stepwise manner, where fluorine atoms are progressively replaced by oxygen atoms from water molecules. The extent of hydrolysis depends on the stoichiometry of the reaction, specifically the molar ratio of XeF $_6$ to H $_2$ O.

To determine the hybridization of the central atom (Xenon) in the products, we will use the VSEPR theory, which relies on calculating the steric number. The steric number is the sum of the number of sigma bonds formed by the central atom and the number of lone pairs on the central atom.
$sp$
Let's break down the reactions and determine the hybridization for each product.

Identify Product A and its Hybridization:

The first reaction given is:

$\text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{A} + 2\text{HF}$

This is a partial hydrolysis reaction where one molecule of water reacts with XeF $_6$ . In this process, one oxygen atom from water replaces two fluorine atoms from XeF $_6$ .

Therefore, product A is $\text{XeOF}_4$ .

Now, let's determine the hybridization of Xe in $\text{XeOF}_4$ :

Central atom: Xenon (Xe)

Valence electrons of Xe: 8

Atoms bonded to Xe: 4 Fluorine atoms and 1 Oxygen atom.

Bonds formed: Each F forms a single bond (1 electron from Xe), and O forms a double bond (2 electrons from Xe).

Electrons used in bonding: $(4 \times 1) + (1 \times 2) = 4 + 2 = 6$ electrons.

Remaining valence electrons: $8 - 6 = 2$ electrons.

Number of lone pairs: $2 / 2 = 1$ lone pair.

Number of sigma bonds: 4 (to F atoms) + 1 (to O atom) = 5 sigma bonds.

Steric number: (Number of sigma bonds) + (Number of lone pairs) = $5 + 1 = 6$ .

A steric number of 6 corresponds to $\mathbf{sp^3d^2}$ hybridization.

The geometry of XeOF $_4$ is square pyramidal (due to one lone pair distorting the octahedral arrangement).

Identify Product B and its Hybridization:

The second reaction given is:

$\text{XeF}_6 + 2\text{H}_2\text{O} \rightarrow \text{B} + 4\text{HF}$

This is a further hydrolysis reaction where two molecules of water react with XeF $_6$ . Here, two oxygen atoms from water replace four fluorine atoms from XeF $_6$ .

Therefore, product B is $\text{XeO}_2\text{F}_2$ .

Now, let's determine the hybridization of Xe in $\text{XeO}_2\text{F}_2$ :

Central atom: Xenon (Xe) …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$\mathrm{NH_3(g) + H_2O(l) \rightarrow}$
›Reveal solutionSolution
Only in $\text{SO}_2+\text{H}_2\text{O}$ does the central atom change hybridisation, going from $sp^2$ in $\text{SO}_2$ to $sp^3$ in $\text{H}_2\text{SO}_3$ — option (C).

Check the hybridisation of the central atom on each side of every reaction.

(A) $\text{NH}_3\to\text{NH}_4^+$ (with water): N is $sp^3$ in $\text{NH}_3$ and remains $sp^3$ in $\text{NH}_4^+$ . No change.

(B) $\text{POCl}_3\to\text{H}_3\text{PO}_4$ : P is $sp^3$ in $\text{POCl}_3$ and $sp^3$ in $\text{H}_3\text{PO}_4$ . No change. …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$-\mathrm{CH}_3, -\mathrm{NO}_2$
›Reveal solutionSolution
The key is to identify one substituent that donates electrons via hyperconjugation (only alkyl groups with $\alpha$ -H do this) and another that withdraws electrons via resonance (groups with a $\pi$ -system that pulls electron density). The correct pair is $-\mathrm{CH}_3$ (hyperconjugation donor) and $-\mathrm{NO}_2$ (resonance withdrawer), which matches option (A).

The question tests your ability to distinguish between two common electronic effects: hyperconjugation and resonance. Hyperconjugation is a sigma-bond donation — it requires a $\mathrm{C-H}$ bond adjacent to the ring to overlap with the $\pi$ -system. Only alkyl groups like $-\mathrm{CH}_3$ can do this. Resonance, on the other hand, involves pi-electron delocalization; a group that withdraws by resonance must have an electronegative atom or a $\pi$ -bond that can pull electron density away from the ring. $-\mathrm{NO}_2$ is the classic example.

Let’s check each option carefully.

Option (A): $-\mathrm{CH}_3, -\mathrm{NO}_2$

$-\mathrm{CH}_3$ has three $\mathrm{C-H}$ bonds whose $\sigma$ electrons can delocalize into the ring — this is hyperconjugation, which increases electron density. $-\mathrm{NO}_2$ has a nitrogen with a positive formal charge and two oxygen atoms; its $\pi$ -system pulls electrons from the ring via resonance, decreasing electron density. This fits perfectly.

Option (B): $-\mathrm{OCH}_3, -\mathrm{COCH}_3$

$-\mathrm{OCH}_3$ donates electrons by resonance (lone pairs on oxygen), not by hyperconjugation. $-\mathrm{COCH}_3$ withdraws by resonance (carbonyl group), so the second part is correct, but the first is not.

Option (C): $-\mathrm{OH}, -\mathrm{NO}_2$

$-\mathrm{OH}$ donates by resonance (lone pairs), not hyperconjugation. $-\mathrm{NO}_2$ is correct for the second, but again the first fails.

Option (D): $-\mathrm{CH}_3, -\mathrm{NHCH}_3$ …
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TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$_2$
›Reveal solutionSolution
The dehydration of an organic acid with concentrated H₂SO₄ at 373 K typically produces carbon monoxide (CO), where the carbon is sp‑hybridized and the gas is neutral. The correct option is (B).

Concept & Intuition

When an organic acid (like formic acid, HCOOH) is heated with concentrated sulfuric acid, it undergoes dehydration — the acid loses a water molecule. The sulfuric acid acts as a powerful dehydrating agent. The product gas is not CO₂ (which would be acidic) but CO, which is neutral and has a triple bond between carbon and oxygen. The carbon in CO is sp‑hybridized because it forms two sigma bonds (one to oxygen, one lone pair) and two pi bonds, giving a linear geometry.

Step‑by‑Step Reasoning

Identify the reaction type The problem says “dehydration of an organic acid X with conc. H₂SO₄ at 373 K”. The classic example is formic acid (HCOOH):

$\text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 373\text{ K}} \text{H}_2\text{O} + \text{CO}$
This is a standard laboratory preparation of carbon monoxide.

Determine the gas Y

The gas produced is carbon monoxide (CO). It is not CO₂ because dehydration of a carboxylic acid at this temperature (with conc. H₂SO₄) removes water from the –COOH group, leaving CO. (Oxalic acid would give CO₂ + CO, but a simple mono‑carboxylic acid gives CO.)

Hybridisation of carbon in CO

In CO, the carbon is bonded to oxygen by a triple bond (one σ, two π). The carbon has two regions of electron density: the triple bond counts as one region, and the lone pair on carbon counts as another. Two regions → sp hybridisation.

$\text{Hybridisation: sp}$

Nature of CO (acidic or neutral?) …
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TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$Q. Choose the correct statements about allotropes of carbon I. Graphite has layered structure II. Buckminster fullerene is not aromatic in nature III. The distance between two adjacent layers in graphite is 141.5 pm IV. The hybridization of carbons in graphite and Buckminster fullerene is same (A) I & IV (B) I & II (C) II & III (D) III & IV$
›Reveal solutionSolution
Graphite has a layered structure (I true), buckminsterfullerene is aromatic (II false), the interlayer distance in graphite is 335 pm not 141.5 pm (III false), and both allotropes use sp² hybridization (IV true). So statements I and IV are correct.

Concept & Intuition

Carbon allotropes differ in bonding geometry. Graphite’s layers are held by weak forces, while buckminsterfullerene (C₆₀) is a closed cage with delocalized π‑electrons — that makes it aromatic. The key is to recall the actual interlayer spacing in graphite and the hybridization in each structure.

Step‑by‑step reasoning

Statement I: Graphite has a layered structure

Graphite consists of flat sheets of carbon atoms arranged in hexagonal rings. Within each sheet, strong covalent bonds hold atoms; between sheets, only weak van der Waals forces exist. This layered structure is why graphite is a lubricant and conducts electricity only along the sheets.

→ True

Statement II: Buckminster fullerene is not aromatic in nature

C₆₀ has 60 carbon atoms, each sp² hybridized, forming a soccer‑ball shape. The π‑electrons are delocalized over the entire surface, satisfying Hückel’s rule for spherical aromaticity (2(N+1)² π‑electrons with N = 2 gives 18 π‑electrons, but C₆₀ actually has 60 π‑electrons; however, it is considered aromatic due to three‑dimensional delocalization). Chemically, it undergoes addition reactions typical of aromatic systems.

→ False (it is aromatic)

Statement III: The distance between two adjacent layers in graphite is 141.5 pm

The in‑plane C–C bond length in graphite is about 141.5 pm. The distance between layers (interlayer spacing) is much larger, approximately 335 pm. The given value confuses the bond length with the interlayer distance.

→ False …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$Q. Choose the correct statements about allotropes of carbon I. Graphite has layered structure II. Buckminster fullerene is not aromatic in nature III. The distance between two adjacent layers in graphite is 141.5 pm IV. The hybridization of carbons in graphite and Buckminster fullerene is same (A) I & II (B) I & IV (C) II & III (D) III & IV$
›Reveal solutionSolution
Graphite has a layered structure (true), Buckminster fullerene is aromatic (false), the interlayer distance in graphite is 335 pm (not 141.5 pm), and both graphite and fullerene use sp² hybridization (true). So only statements I and IV are correct.

Concept & Intuition

Allotropes of carbon differ in how carbon atoms bond and arrange themselves. Graphite forms flat sheets of sp²-hybridized carbons in a hexagonal lattice, with weak forces between layers. Buckminster fullerene (C₆₀) is a soccer-ball-shaped molecule where each carbon is also sp²-hybridized, but the curved surface creates a delocalized π-system that is aromatic. The key is to recall the exact interlayer spacing in graphite (not the in-plane bond length) and the hybridization common to both.

Step-by-step reasoning

Statement I: Graphite has a layered structure

Graphite consists of stacked, two-dimensional sheets of carbon atoms arranged in a honeycomb lattice. Within each layer, strong covalent bonds hold atoms together; between layers, only weak van der Waals forces exist. This is a defining property of graphite.

→ True

Statement II: Buckminster fullerene is not aromatic in nature

C₆₀ has 60 π-electrons, which satisfies Hückel’s rule for spherical aromaticity (2(N+1)² for N=2 gives 18, but C₆₀’s π-system is delocalized over the entire sphere and exhibits aromatic character). It undergoes reactions typical of aromatic compounds and is considered aromatic.

→ False (it is aromatic)

Statement III: The distance between two adjacent layers in graphite is 141.5 pm …
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$_3$
›Reveal solutionSolution
The number of lone pairs on the central Xe atom is determined by counting total valence electrons, subtracting those used in bonding and in the terminal atoms' own lone pairs, and dividing the remainder by two. For XeO $_3$ , XeOF $_4$ , and XeF $_6$ , the lone pairs are 1, 1, and 1 respectively — option (D).

The key to this problem is understanding that xenon is a noble gas with 8 valence electrons. In its compounds, it forms bonds by expanding its octet, using d-orbitals. The number of lone pairs on Xe is simply the leftover electrons after accounting for all bonds and the terminal atoms' own octets.

Let's work through each molecule step by step.

XeO $_3$

Xenon has 8 valence electrons; each oxygen contributes 6, and there are 3 oxygens.

Total valence electrons available: $8 + 3 \times 6 = 26$ .

In the Lewis structure, each Xe=O double bond uses 4 electrons, so the three double bonds use $3 \times 4 = 12$ electrons. Each double-bonded oxygen already has 4 electrons from the bond and needs 4 more to complete its octet — that's 2 lone pairs (4 electrons) per oxygen, or $3 \times 4 = 12$ electrons total for the three oxygens. Adding the bonding electrons and the oxygens' own lone-pair electrons: $12 + 12 = 24$ . The remaining $26 - 24 = 2$ electrons go on xenon as 1 lone pair.

XeOF $_4$

Xenon: 8 valence electrons. Oxygen: 6. Each fluorine: 7, and there are 4 fluorines.

Total valence electrons: $8 + 6 + 4 \times 7 = 8 + 6 + 28 = 42$ .

The structure: Xe forms a double bond with O (4 electrons) and single bonds with each F (2 electrons each, total 8). So bonding uses $4 + 8 = 12$ electrons.

Each fluorine gets 3 lone pairs (6 electrons each, total $4 \times 6 = 24$ ). Oxygen gets 2 lone pairs (4 electrons). That accounts for $24 + 4 = 28$ electrons in lone pairs on terminal atoms. …
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TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$\mathrm{SF_4},\ \mathrm{XeF_4}$
›Reveal solutionSolution
The key is to count the lone pairs on the central atom using the VSEPR formula (total valence electrons minus bonding electrons, divided by 2). The pair with the same lone-pair count is $\mathrm{ClF_3}$ and $\mathrm{XeF_4}$ , each having 2 lone pairs on the central atom.

The question asks which two molecules have the same number of lone pairs on their central atom. This is a classic VSEPR (Valence Shell Electron Pair Repulsion) problem. The central idea is simple: count the total valence electrons around the central atom, subtract the electrons used in bonding (each bond uses 2 electrons), and the remainder, divided by 2, gives the number of lone pairs.

Let’s work through each option systematically.

Option (A): $\mathrm{SF_4}$ and $\mathrm{XeF_4}$

For $\mathrm{SF_4}$ : Sulfur has 6 valence electrons. It forms 4 S–F bonds, using $4 \times 2 = 8$ electrons. But wait — sulfur only has 6 valence electrons, so it uses 4 of its own for bonding and gets 4 more from fluorine atoms? Actually, the correct method: total valence electrons = $6 + 4 \times 7 = 34$ . Bonding pairs = 4 bonds × 2 electrons = 8 electrons used in bonds. Remaining electrons = $34 - 8 = 26$ . These are distributed as lone pairs on the central atom and on fluorines. Each fluorine gets 3 lone pairs (6 electrons), so 4 fluorines take $4 \times 6 = 24$ electrons. That leaves $26 - 24 = 2$ electrons on sulfur, which is 1 lone pair.

For $\mathrm{XeF_4}$ : Xenon has 8 valence electrons. Total valence = $8 + 4 \times 7 = 36$ . Bonding uses 8 electrons. Remaining = 28. Each fluorine takes 6 electrons (3 lone pairs), so 4 fluorines take 24. Leftover = $28 - 24 = 4$ electrons on xenon, which is 2 lone pairs.

So $\mathrm{SF_4}$ has 1 lone pair, $\mathrm{XeF_4}$ has 2 — not the same.

Option (B): $\mathrm{ClF_3}$ and $\mathrm{BrF_5}$

$\mathrm{ClF_3}$ : Chlorine has 7 valence electrons. Total valence = $7 + 3 \times 7 = 28$ . Bonding uses 6 electrons. Remaining = 22. Each fluorine takes 6 electrons, so 3 fluorines take 18. Leftover = $22 - 18 = 4$ electrons on chlorine, which is 2 lone pairs.

$\mathrm{BrF_5}$ : Bromine has 7 valence electrons. Total valence = $7 + 5 \times 7 = 42$ . Bonding uses 10 electrons. Remaining = 32. Each fluorine takes 6 electrons, so 5 fluorines take 30. Leftover = $32 - 30 = 2$ electrons on bromine, which is 1 lone pair.

Not the same (2 vs. 1). …
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TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$_2$
›Reveal solutionSolution
Hydrogen peroxide decomposes in the presence of catalysts or impurities; stabilizers slow this decomposition. The correct stabilizer among the options is acetanilide-like urea (H₂NCONH₂), which acts as a negative catalyst. The answer is (B).

Concept & Intuition

Hydrogen peroxide (H₂O₂) is inherently unstable — it slowly decomposes into water and oxygen:
$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$
This decomposition is accelerated by heat, light, and especially by catalysts like metal ions (e.g., Fe³⁺, Cu²⁺) or solid surfaces (e.g., MnO₂). To make H₂O₂ safe for storage and transport, we add stabilizers — substances that inhibit decomposition. A stabilizer works by either:

Chelating (binding) trace metal ions that catalyze decomposition, or

Acting as a negative catalyst (inhibitor) that slows the reaction.

Now, let’s examine each option:

Option (A): CO(CH₃)₂ (acetone)

Acetone is a common organic solvent. It does not chelate metal ions effectively nor does it inhibit H₂O₂ decomposition. In fact, acetone can react with H₂O₂ under certain conditions to form explosive peroxides (e.g., acetone peroxide). So it is not a stabilizer — it’s a hazard.

Option (B): H₂NCONH₂ (urea)

Urea is a well-known stabilizer for hydrogen peroxide. It works by forming weak hydrogen bonds with H₂O₂ molecules and by sequestering trace metal impurities. Urea is often added to commercial H₂O₂ solutions to slow decomposition. This is the correct choice.

Option (C): MnO₂ (manganese dioxide) …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$\mathrm{CCl_4, NF_3, H_2S, HBr, SF_4, SiF_4, XeF_4, BeCl_2, SnCl_2, BrF_5, SO_2}$
›Reveal solutionSolution
Seven of the eleven molecules are polar: $\text{NF}_3,\ \text{H}_2\text{S},\ \text{HBr},\ \text{SF}_4,\ \text{SnCl}_2,\ \text{BrF}_5,\ \text{SO}_2$ .

A molecule has a permanent dipole moment only if its bond dipoles do not cancel by symmetry:

$\text{CCl}_4$ — regular tetrahedral, symmetric → non-polar

$\text{NF}_3$ — pyramidal (lone pair) → polar ✓

$\text{H}_2\text{S}$ — bent → polar ✓

$\text{HBr}$ — polar diatomic → polar ✓

$\text{SF}_4$ — see-saw (lone pair) → polar ✓

$\text{SiF}_4$ — regular tetrahedral, symmetric → non-polar

$\text{XeF}_4$ — square planar, symmetric → non-polar …
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Q.Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?

Valence Bond Theory of Covalent Bond Formation (Example: Hydrogen)

Interpretation of Energy Changes in Dihydrogen Formation

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