Q.Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?

Showing the 12 most recent of 27 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The set of elements which can form electron rich hydrides is (A) B, N, O (B) N, O, F (C) C, Si, N (D) Be, B, C

   ›Reveal solutionSolution

   Electron-rich hydrides are formed by elements from groups 15, 16, and 17 that have lone pairs after bonding with hydrogen. The correct set is N, O, F — option (B).

   The idea of "electron-rich" hydrides comes from how many electrons are left on the central atom after it has formed bonds with hydrogen. In a hydride, each hydrogen contributes one electron to a bond, and the central atom contributes one. Once all bonds are made, any remaining valence electrons on the central atom sit as lone pairs. If the central atom has more electrons than needed for an octet (or duet for hydrogen), the hydride is electron-rich. This happens when the central atom belongs to groups 15, 16, or 17 — because these elements have 5, 6, or 7 valence electrons respectively, and after bonding with enough hydrogens to satisfy their normal valency, they still have lone pairs left over.

   For example, nitrogen in NH₃ has 5 valence electrons; it uses 3 to bond with three hydrogens, leaving one lone pair — that's an electron-rich hydride. Oxygen in H₂O has 6 valence electrons; it uses 2 to bond, leaving two lone pairs. Fluorine in HF has 7 valence electrons; it uses 1 to bond, leaving three lone pairs. In contrast, elements from groups 13 and 14 (like boron, carbon, silicon) form electron-deficient or electron-precise hydrides — they either have too few electrons (boron in BH₃) or exactly enough (carbon in CH₄) with no lone pairs.

   Now let's check each option:

   1. Option (A): B, N, O — Boron (group 13) forms electron-deficient hydrides (e.g., BH₃, B₂H₆), not electron-rich. So this set is wrong.

   2. Option (B): N, O, F — Nitrogen (group 15), oxygen (group 16), and fluorine (group 17) all form hydrides (NH₃, H₂O, HF) with lone pairs on the central atom. These are classic examples of electron-rich hydrides. This set is correct.

   3. Option (C): C, Si, N — Carbon and silicon (group 14) form electron-precise hydrides (CH₄, SiH₄) with no lone pairs — they are not electron-rich. Nitrogen is electron-rich, but the presence of C and Si makes this set incorrect.

   4. Option (D): Be, B, C — Beryllium (group 2) and boron (group 13) form electron-deficient hydrides; carbon forms an electron-precise hydride. None are electron-rich. So this set is wrong.

   Watch out

   A common mistake is to think that any hydride with a lone pair is electron-rich — but that's exactly right. The trap is including elements from groups 13 or 14, which either lack lone pairs or have too few electrons. Always check the group number: groups 15, 16, 17 give electron-rich hydrides.

   ✓Final answer

   The correct option is (B) N, O, F.

2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.SO$_2$ reacts with chlorine in the presence of charcoal and gives a compound X. X on reaction with phosphorus gives SO$_2$ and Y. The shape of Y is (A) Pyramidal (B) Tetrahedral (C) Trigonal bipyramidal (D) Angular

   ›Reveal solutionSolution

   SO₂ reacts with Cl₂ to form SO₂Cl₂ (compound X), which then reacts with phosphorus to give SO₂ and PCl₅ (compound Y). PCl₅ has a trigonal bipyramidal shape.

   Understanding the Reaction Sequence

   Let's trace through this chemistry puzzle step by step, identifying each compound and then determining the molecular geometry.

   First Reaction: SO₂ + Cl₂ → Compound X

   When sulfur dioxide reacts with chlorine gas in the presence of charcoal (which acts as a catalyst), we get sulfuryl chloride:

   $\text{SO}_2 + \text{Cl}_2 \xrightarrow{\text{charcoal}} \text{SO}_2\text{Cl}_2$

   So compound X is SO₂Cl₂ (sulfuryl chloride).

   Second Reaction: X + P → SO₂ + Y

   When sulfuryl chloride reacts with phosphorus, it acts as a chlorinating agent. The phosphorus gets chlorinated while SO₂ is regenerated:

   $\text{SO}_2\text{Cl}_2 + \text{P} \rightarrow \text{SO}_2 + \text{PCl}_5$

   (In practice, multiple molecules of SO₂Cl₂ are needed, but the key product is that compound Y is PCl₅, phosphorus pentachloride.)

   Determining the Shape of PCl₅

   Now we need to find the molecular geometry of PCl₅.

   1. Count valence electrons: Phosphorus is in Group 15, so it has 5 valence electrons. Each of the 5 chlorine atoms contributes 1 electron to bonding.

   2. Apply VSEPR theory: PCl₅ has 5 bonding pairs and 0 lone pairs around the central phosphorus atom.

   3. Determine electron geometry: With 5 electron pairs, the electron geometry that minimizes repulsion is trigonal bipyramidal.

   4. Determine molecular geometry: Since there are no lone pairs, the molecular geometry is the same as the electron geometry.

   Trigonal Bipyramidal Structure of PCl₅

   • 3 chlorine atoms occupy equatorial positions (in a triangular plane)
   • 2 chlorine atoms occupy axial positions (above and below the plane)
   • Bond angles: 120° (equatorial-equatorial), 90° (axial-equatorial), 180° (axial-axial)

   Tip

   PCl₅ is one of the classic examples of expanded octet compounds where the central atom (phosphorus) can accommodate more than 8 electrons by using its d-orbitals (though modern theory attributes this to other factors).

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Match the following List-1 (Molecule) A BrF$_5$ B XeF$_4$ C ClF$_3$ D SF$_4$ List-2 (Geometry) I T-Shape II See-Saw III Square planar IV Square pyramid Options : (A) A - IV, B - III, C - II, D - I (B) A - III, B - IV, C - II, D - I (C) A - IV, B - I, C - III, D - II (D) A - IV, B - III, C - I, D - II

   ›Reveal solutionSolution

   The key idea is to use VSEPR theory to predict molecular geometry from the number of bond pairs and lone pairs on the central atom. The correct matches are: A–IV, B–III, C–I, D–II, so the answer is option (D).

   Concept & Intuition

   VSEPR (Valence Shell Electron Pair Repulsion) theory tells us that electron pairs (bonding and lone) arrange themselves as far apart as possible around a central atom. The geometry is determined by the total number of electron domains, while the shape (the arrangement of atoms only) depends on how many of those domains are lone pairs. Lone pairs “push” more strongly than bonding pairs, distorting ideal geometries. Here, we count domains for each molecule, then deduce the shape.

   Step-by-step reasoning

   1. A: BrF₅

      • Central atom: Br (Group 17). It has 7 valence electrons.
      • Five F atoms each contribute 1 electron for bonding → 5 bond pairs.
      • Remaining electrons: $7 - 5 = 2$ → 1 lone pair.
      • Total domains = 6 (5 bond + 1 lone) → octahedral electron geometry.
      • With one lone pair, the shape is square pyramidal (like a pyramid with a square base).
      • So A matches IV.

   2. B: XeF₄

      • Central atom: Xe (Group 18). It has 8 valence electrons.
      • Four F atoms → 4 bond pairs.
      • Remaining electrons: $8 - 4 = 4$ → 2 lone pairs.
      • Total domains = 6 (4 bond + 2 lone) → octahedral electron geometry.
      • Two lone pairs occupy opposite positions (to minimize repulsion), leaving the four F atoms in a plane → square planar shape.
      • So B matches III.

   3. C: ClF₃

      • Central atom: Cl (Group 17). It has 7 valence electrons.
      • Three F atoms → 3 bond pairs.
      • Remaining electrons: $7 - 3 = 4$ → 2 lone pairs.
      • Total domains = 5 (3 bond + 2 lone) → trigonal bipyramidal electron geometry.
      • Lone pairs go to equatorial positions (120° apart) to minimize repulsion, leaving the three F atoms in a T-shape (two axial, one equatorial).
      • So C matches I.

   4. D: SF₄

      • Central atom: S (Group 16). It has 6 valence electrons.
      • Four F atoms → 4 bond pairs.
      • Remaining electrons: $6 - 4 = 2$ → 1 lone pair.
      • Total domains = 5 (4 bond + 1 lone) → trigonal bipyramidal electron geometry.
      • One lone pair occupies an equatorial position, distorting the shape into a see-saw (or “distorted tetrahedron”).
      • So D matches II.

   Watch out

   A common mistake is to confuse “electron geometry” with “molecular shape.” For example, XeF₄ has octahedral electron geometry but square planar shape. Always count lone pairs separately.

   Tip

   For molecules with 5 or 6 domains, remember: lone pairs always prefer equatorial positions in trigonal bipyramidal (to minimize 90° repulsions), and opposite positions in octahedral.

   Final matches: A–IV, B–III, C–I, D–II. This corresponds to option (D).

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Which one of the following molecules has maximum number of lone pairs of electrons? (A) $\mathrm{XeF_2}$ (B) $\mathrm{SF_4}$ (C) $\mathrm{CF_4}$ (D) $\mathrm{ClF_3}$

   ›Reveal solutionSolution

   To find the molecule with the maximum number of lone pairs, we systematically determine the Lewis structure for each option and count all non-bonding electron pairs. $\mathrm{SF_4}$ emerges with the highest count of 13 lone pairs.

   The Concept: Lone Pairs and Lewis Structures

   At the heart of this problem lies the concept of lone pairs of electrons and the ability to construct Lewis structures.

   • Lone Pairs: These are pairs of valence electrons that are not shared with another atom in a covalent bond. They belong exclusively to one atom and play a crucial role in determining a molecule's shape and reactivity.
   • Lewis Structures: These diagrams represent the valence electrons of atoms within a molecule, showing how they are arranged into bonding pairs (shared electrons) and lone pairs (unshared electrons). They are our primary tool for visualizing and counting these electron pairs.

   Why Lewis Structures Work:

   Every atom strives for stability, often by achieving a full outer electron shell (an octet, or a duet for hydrogen). In covalent compounds, atoms share electrons to reach this goal. Lewis structures help us account for all valence electrons in a molecule and distribute them according to bonding and non-bonding roles, ensuring each atom (where possible) satisfies the octet rule. By following a systematic approach, we can accurately determine the number of lone pairs.

   The general approach to drawing Lewis structures and counting lone pairs involves these steps:

   1. Calculate Total Valence Electrons: Sum the valence electrons of all atoms in the molecule.
   2. Identify Central Atom: Usually the least electronegative atom (excluding hydrogen), or the unique atom.
   3. Form Single Bonds: Connect the central atom to terminal atoms with single bonds. Subtract these bonding electrons from the total.
   4. Complete Octets of Terminal Atoms: Distribute remaining electrons as lone pairs to terminal atoms until their octets are full.
   5. Place Remaining Electrons on Central Atom: Any leftover electrons are placed on the central atom as lone pairs. (For elements in Period 3 and beyond, the central atom can accommodate more than an octet).
   6. Check Octets (and expand if necessary): Ensure all atoms have a stable electron configuration. If the central atom lacks an octet and there are still electrons available on terminal atoms, consider forming multiple bonds (though not needed for this problem).

   Watch out

   A common pitfall is to only count lone pairs on the central atom. The question asks for the "maximum number of lone pairs of electrons" in the molecule, which implies counting all lone pairs, both on the central atom and on the terminal atoms. We will be thorough and count all of them.

   Let's apply this systematic approach to each molecule.

   ---

   Step-by-Step Analysis of Each Molecule

   We'll calculate the total valence electrons, form bonds, and then distribute lone pairs, keeping track of both central and terminal atom lone pairs.

   1. Molecule (A): $\mathrm{XeF_2}$

   1. Identify atoms and valence electrons:
      • Central atom: Xenon (Xe), a noble gas from Group 18, has 8 valence electrons.
      • Terminal atoms: Fluorine (F), a halogen from Group 17, has 7 valence electrons each.
   2. Calculate total valence electrons:
      • Total valence electrons = $1 \times (\text{valence e}^- \text{ of Xe}) + 2 \times (\text{valence e}^- \text{ of F})$
      • Total valence electrons = $1 \times 8 + 2 \times 7 = 8 + 14 = 22$ electrons.
   3. Form single bonds:
      • Connect the central Xe atom to the two F atoms with single bonds.
      • Number of bonding electrons = $2 \text{ bonds} \times 2 \text{ electrons/bond} = 4$ electrons.
   4. Distribute remaining electrons to terminal atoms:
      • Remaining electrons = $22 - 4 = 18$ electrons.
      • Each F atom needs 6 more electrons to complete its octet (it already has 2 from the bond). This means 3 lone pairs per F atom.
      • Electrons used on terminal F atoms = $2 \text{ F atoms} \times 6 \text{ electrons/F} = 12$ electrons.
   5. Place leftover electrons on the central atom:
      • Electrons remaining for the central atom = $18 - 12 = 6$ electrons.
      • These 6 electrons form lone pairs on Xe: $6 \text{ electrons} / 2 \text{ electrons/pair} = 3$ lone pairs on Xe.
   6. Count total lone pairs:
      • Lone pairs on central atom (Xe) = 3
      • Lone pairs on terminal atoms (F) = $2 \text{ F atoms} \times 3 \text{ lone pairs/F} = 6$
      • Total lone pairs in $\mathrm{XeF_2}$ = $3 + 6 = 9$.

   2. Molecule (B): $\mathrm{SF_4}$

   1. Identify atoms and valence electrons:
      • Central atom: Sulfur (S) from Group 16, has 6 valence electrons.
      • Terminal atoms: Fluorine (F) from Group 17, has 7 valence electrons each.
   2. Calculate total valence electrons:
      • Total valence electrons = $1 \times (\text{valence e}^- \text{ of S}) + 4 \times (\text{valence e}^- \text{ of F})$
      • Total valence electrons = $1 \times 6 + 4 \times 7 = 6 + 28 = 34$ electrons.
   3. Form single bonds:
      • Connect the central S atom to the four F atoms with single bonds.
      • Number of bonding electrons = $4 \text{ bonds} \times 2 \text{ electrons/bond} = 8$ electrons.
   4. Distribute remaining electrons to terminal atoms:
      • Remaining electrons = $34 - 8 = 26$ electrons.
      • Each F atom needs 6 more electrons (3 lone pairs) to complete its octet.
      • Electrons used on terminal F atoms = $4 \text{ F atoms} \times 6 \text{ electrons/F} = 24$ electrons.
   5. Place leftover electrons on the central atom:
      • Electrons remaining for the central atom = $26 - 24 = 2$ electrons.
      • These 2 electrons form lone pairs on S: $2 \text{ electrons} / 2 \text{ electrons/pair} = 1$ lone pair on S.
   6. Count total lone pairs:
      • Lone pairs on central atom (S) = 1
      • Lone pairs on terminal atoms (F) = $4 \text{ F atoms} \times 3 \text{ lone pairs/F} = 12$
      • Total lone pairs in $\mathrm{SF_4}$ = $1 + 12 = 13$.

   3. Molecule (C): $\mathrm{CF_4}$

   1. Identify atoms and valence electrons:
      • Central atom: Carbon (C) from Group 14, has 4 valence electrons.
      • Terminal atoms: Fluorine (F) from Group 17, has 7 valence electrons each.
   2. Calculate total valence electrons:
      • Total valence electrons = $1 \times (\text{valence e}^- \text{ of C}) + 4 \times (\text{valence e}^- \text{ of F})$
      • Total valence electrons = $1 \times 4 + 4 \times 7 = 4 + 28 = 32$ electrons.
   3. Form single bonds:
      • Connect the central C atom to the four F atoms with single bonds.
      • Number of bonding electrons = $4 \text{ bonds} \times 2 \text{ electrons/bond} = 8$ electrons.
   4. Distribute remaining electrons to terminal atoms:
      • Remaining electrons = $32 - 8 = 24$ electrons.
      • Each F atom needs 6 more electrons (3 lone pairs) to complete its octet.
      • Electrons used on terminal F atoms = $4 \text{ F atoms} \times 6 \text{ electrons/F} = 24$ electrons.
   5. Place leftover electrons on the central atom:
      • Electrons remaining for the central atom = $24 - 24 = 0$ electrons.
      • There are no lone pairs on the central atom C.
   6. Count total lone pairs:
      • Lone pairs on central atom (C) = 0
      • Lone pairs on terminal atoms (F) = $4 \text{ F atoms} \times 3 \text{ lone pairs/F} = 12$
      • Total lone pairs in $\mathrm{CF_4}$ = $0 + 12 = 12$.

   4. Molecule (D): $\mathrm{ClF_3}$

   1. Identify atoms and valence electrons:
      • Central atom: Chlorine (Cl) from Group 17, has 7 valence electrons.
      • Terminal atoms: Fluorine (F) from Group 17, has 7 valence electrons each.
   2. Calculate total valence electrons:
      • Total valence electrons = $1 \times (\text{valence e}^- \text{ of Cl}) + 3 \times (\text{valence e}^- \text{ of F})$
      • Total valence electrons = $1 \times 7 + 3 \times 7 = 7 + 21 = 28$ electrons.
   3. Form single bonds:
      • Connect the central Cl atom to the three F atoms with single bonds.
      • Number of bonding electrons = $3 \text{ bonds} \times 2 \text{ electrons/bond} = 6$ electrons.
   4. Distribute remaining electrons to terminal atoms:
      • Remaining electrons = $28 - 6 = 22$ electrons.
      • Each F atom needs 6 more electrons (3 lone pairs) to complete its octet.
      • Electrons used on terminal F atoms = $3 \text{ F atoms} \times 6 \text{ electrons/F} = 18$ electrons.
   5. Place leftover electrons on the central atom:
      • Electrons remaining for the central atom = $22 - 18 = 4$ electrons.
      • These 4 electrons form lone pairs on Cl: $4 \text{ electrons} / 2 \text{ electrons/pair} = 2$ lone pairs on Cl.
   6. Count total lone pairs:
      • Lone pairs on central atom (Cl) = 2
      • Lone pairs on terminal atoms (F) = $3 \text{ F atoms} \times 3 \text{ lone pairs/F} = 9$
      • Total lone pairs in $\mathrm{ClF_3}$ = $2 + 9 = 11$.

   ---

   Comparison and Conclusion

   Let's summarize our findings in a clear table:

   Molecule	Lone Pairs on Central Atom	Lone Pairs on Terminal Atoms	Total Lone Pairs in Molecule
   (A) $\mathrm{XeF_2}$	3	$2 \times 3 = 6$	9
   (B) $\mathrm{SF_4}$	1	$4 \times 3 = 12$	13
   (C) $\mathrm{CF_4}$	0	$4 \times 3 = 12$	12
   (D) $\mathrm{ClF_3}$	2	$3 \times 3 = 9$	11

   Comparing the total lone pairs, $\mathrm{SF_4}$ has 13 lone pairs, which is the highest number among the given options.

   ---

   ✓Final answer

   The molecule with the maximum number of lone pairs of electrons is $\mathrm{SF_4}$, with a total of 13 lone pairs. The correct option is (B).

   ANSWER: B

5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Which of the following pairs contain isostructural species? I. $\mathrm{H_3PO_3, H_2SO_4}$ II. $\mathrm{SO_3, BF_3}$ III. $\mathrm{O_3, NO_2}$ The correct answer is (only = మాత్రమే) (A) I, II only (B) I, III only (C) I, II, III (D) II, III only

   ›Reveal solutionSolution

   Isostructural species share the same molecular geometry and hybridization; by applying VSEPR theory, we find that all three pairs ($\mathrm{H_3PO_3}$ and $\mathrm{H_2SO_4}$, $\mathrm{SO_3}$ and $\mathrm{BF_3}$, $\mathrm{O_3}$ and $\mathrm{NO_2}$) are indeed isostructural, making (C) the correct option.

   The Concept: What Does "Isostructural" Mean?

   When we say two chemical species are isostructural, we mean they possess the same molecular geometry (shape) and the same hybridization of their central atom. This doesn't necessarily mean they have the exact same bond angles down to the decimal, but rather the same fundamental arrangement of atoms and electron domains around the central atom. To determine if species are isostructural, we primarily rely on the Valence Shell Electron Pair Repulsion (VSEPR) theory.

   Why VSEPR Theory Works

   VSEPR theory is a powerful tool because it's based on a simple, intuitive idea: electron pairs (both bonding and non-bonding, i.e., lone pairs) around a central atom will arrange themselves as far apart as possible to minimize repulsion. This arrangement dictates the electron domain geometry, which in turn determines the molecular geometry and the hybridization of the central atom.

   How to Apply VSEPR:

   1. Draw the Lewis Structure: This helps identify the central atom and count valence electrons.
   2. Count Electron Domains (Steric Number, SN): An electron domain can be a single bond, a double bond, a triple bond, or a lone pair of electrons. Each multiple bond counts as one domain for geometry purposes.
      • $\text{SN} = (\text{Number of sigma bonds}) + (\text{Number of lone pairs})$
   3. Determine Electron Domain Geometry: This is based solely on the SN.
      • SN = 2: Linear
      • SN = 3: Trigonal Planar
      • SN = 4: Tetrahedral
      • SN = 5: Trigonal Bipyramidal
      • SN = 6: Octahedral
   4. Determine Hybridization: This directly corresponds to the SN.
      • SN = 2: $\mathrm{sp}$
      • SN = 3: $\mathrm{sp^2}$
      • SN = 4: $\mathrm{sp^3}$
      • SN = 5: $\mathrm{sp^3d}$
      • SN = 6: $\mathrm{sp^3d^2}$
   5. Determine Molecular Geometry: This is the shape formed by the atoms only, taking into account that lone pairs occupy space but are not "seen" as part of the molecular shape.

   Let's apply this framework to each pair.

   ---

   Step-by-Step Analysis of Each Pair

   1. Pair I: $\mathrm{H_3PO_3}$ and $\mathrm{H_2SO_4}$

   Let's analyze the central atom in each molecule.

   • $\mathrm{H_3PO_3}$ (Phosphorous acid)

     • Central Atom: Phosphorus (P).
     • Lewis Structure: Phosphorous acid is an oxyacid with a direct P-H bond. The P atom is bonded to one oxygen via a double bond ($\mathrm{P=O}$), two hydroxyl groups ($\mathrm{P-OH}$), and one hydrogen atom ($\mathrm{P-H}$).
       • Total valence electrons: P(5) + O(36) + H(31) = 5 + 18 + 3 = 26.
       • The structure is: $\mathrm{O=P(OH)_2H}$.
     • Electron Domains around P:
       • One P=O bond (counts as 1 domain).
       • Two P-OH bonds (each counts as 1 domain, total 2 domains).
       • One P-H bond (counts as 1 domain).
       • No lone pairs on P.
       • Steric Number (SN) = 1 + 2 + 1 = 4.
     • Electron Domain Geometry: Tetrahedral.
     • Molecular Geometry: Since there are no lone pairs on the central atom, the molecular geometry is also Tetrahedral.
     • Hybridization: $\mathrm{sp^3}$.

   • $\mathrm{H_2SO_4}$ (Sulfuric acid)

     • Central Atom: Sulfur (S).
     • Lewis Structure: Sulfur is bonded to two oxygen atoms via double bonds ($\mathrm{S=O}$) and two hydroxyl groups ($\mathrm{S-OH}$).
       • Total valence electrons: S(6) + O(46) + H(21) = 6 + 24 + 2 = 32.
       • The structure is: $\mathrm{O_2S(OH)_2}$.
     • Electron Domains around S:
       • Two S=O bonds (each counts as 1 domain, total 2 domains).
       • Two S-OH bonds (each counts as 1 domain, total 2 domains).
       • No lone pairs on S.
       • Steric Number (SN) = 2 + 2 = 4.
     • Electron Domain Geometry: Tetrahedral.
     • Molecular Geometry: Since there are no lone pairs on the central atom, the molecular geometry is also Tetrahedral.
     • Hybridization: $\mathrm{sp^3}$.

   • Conclusion for Pair I: Both $\mathrm{H_3PO_3}$ and $\mathrm{H_2SO_4}$ have a tetrahedral molecular geometry and $\mathrm{sp^3}$ hybridization. Therefore, they are isostructural.

   2. Pair II: $\mathrm{SO_3}$ and $\mathrm{BF_3}$

   • $\mathrm{SO_3}$ (Sulfur trioxide)

     • Central Atom: Sulfur (S).
     • Lewis Structure: Sulfur is bonded to three oxygen atoms. Due to resonance, it can be represented as one S=O double bond and two S-O single bonds, or more accurately, three equivalent S-O bonds with partial double bond character. For VSEPR, each bond (single or multiple) to a different atom counts as one domain.
       • Total valence electrons: S(6) + O(3*6) = 6 + 18 = 24.
     • Electron Domains around S:
       • Three S-O bonds (each counts as 1 domain, total 3 domains).
       • No lone pairs on S.
       • Steric Number (SN) = 3.
     • Electron Domain Geometry: Trigonal Planar.
     • Molecular Geometry: Since there are no lone pairs on the central atom, the molecular geometry is also Trigonal Planar.
     • Hybridization: $\mathrm{sp^2}$.

   • $\mathrm{BF_3}$ (Boron trifluoride)

     • Central Atom: Boron (B).
     • Lewis Structure: Boron is bonded to three fluorine atoms via single bonds. Boron is an exception to the octet rule, being electron deficient.
       • Total valence electrons: B(3) + F(3*7) = 3 + 21 = 24.
     • Electron Domains around B:
       • Three B-F single bonds (each counts as 1 domain, total 3 domains).
       • No lone pairs on B.
       • Steric Number (SN) = 3.
     • Electron Domain Geometry: Trigonal Planar.
     • Molecular Geometry: Since there are no lone pairs on the central atom, the molecular geometry is also Trigonal Planar.
     • Hybridization: $\mathrm{sp^2}$.

   • Conclusion for Pair II: Both $\mathrm{SO_3}$ and $\mathrm{BF_3}$ have a trigonal planar molecular geometry and $\mathrm{sp^2}$ hybridization. Therefore, they are isostructural.

   3. Pair III: $\mathrm{O_3}$ and $\mathrm{NO_2}$

   • $\mathrm{O_3}$ (Ozone)

     • Central Atom: Oxygen (O).
     • Lewis Structure: The central oxygen atom is bonded to one oxygen via a double bond and to another oxygen via a single bond. It also has one lone pair of electrons. (Resonance structures exist, making the two O-O bonds equivalent with partial double bond character).
       • Total valence electrons: O(3*6) = 18.
     • Electron Domains around Central O:
       • Two bonding domains (one double bond, one single bond, each counts as 1 domain, total 2 domains).
       • One lone pair.
       • Steric Number (SN) = 2 + 1 = 3.
     • Electron Domain Geometry: Trigonal Planar.
     • Molecular Geometry: Due to the presence of one lone pair, the molecular geometry is Bent.
     • Hybridization: $\mathrm{sp^2}$.

   • $\mathrm{NO_2}$ (Nitrogen dioxide)

     • Central Atom: Nitrogen (N).
     • Lewis Structure: Nitrogen dioxide is an odd-electron species (a radical). The central nitrogen atom is bonded to two oxygen atoms (one double bond, one single bond, with resonance). It has one unpaired electron.
       • Total valence electrons: N(5) + O(2*6) = 17.
     • Electron Domains around N:
       • Two bonding domains (one double bond, one single bond, total 2 domains).
       • One unpaired electron.
       • Steric Number (SN): For odd-electron species, the unpaired electron is often treated as a "half-lone pair" or a domain that influences the geometry, similar to a lone pair but with less repulsive force. Thus, we consider 2 bonding domains + 1 unpaired electron domain, leading to an effective SN of 3.
     • Electron Domain Geometry: Approximately Trigonal Planar.
     • Molecular Geometry: Due to the presence of the unpaired electron, the molecular geometry is Bent.
     • Hybridization: $\mathrm{sp^2}$.
     • Tip

       While the bond angles in $\mathrm{O_3}$ (~117°) and $\mathrm{NO_2}$ (~134°) are different due to the varying repulsive forces of a lone pair versus an unpaired electron (lone pair repulsion > unpaired electron repulsion), both molecules share the same fundamental bent molecular geometry and $\mathrm{sp^2}$ hybridization. In the context of "isostructural," this similarity in geometry type and hybridization is what matters.

   • Conclusion for Pair III: Both $\mathrm{O_3}$ and $\mathrm{NO_2}$ have a bent molecular geometry and $\mathrm{sp^2}$ hybridization. Therefore, they are isostructural.

   ---

   Since all three pairs (I, II, and III) contain isostructural species, the correct option is (C).

   ✓Final answer

   All three pairs contain isostructural species. The correct option is (C).

   ANSWER: C

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Xenon hexafluoride on partial hydrolysis gives ‘X’ and HF. The shape of ‘X’ is (A) Pyramidal (B) Tetrahedral (C) Square pyramidal (D) Linear

   ›Reveal solutionSolution

   Partial hydrolysis of $\ce{XeF6}$ yields $\ce{XeOF4}$, which has a square pyramidal shape due to a lone pair on xenon; the correct option is (C).

   The key here is to recall the hydrolysis behavior of xenon hexafluoride. Xenon hexafluoride ($\ce{XeF6}$) is a noble gas compound with a distorted octahedral geometry (one lone pair on Xe). When it undergoes partial hydrolysis (limited water), it does not fully convert to $\ce{XeO3}$; instead, it forms xenon oxytetrafluoride, $\ce{XeOF4}$, along with HF. The shape of $\ce{XeOF4}$ is determined by VSEPR theory: xenon is the central atom with 8 valence electrons, four fluorine atoms (each single-bonded), one double-bonded oxygen, and one lone pair. This gives a steric number of 6 (5 bonds + 1 lone pair), leading to an octahedral electron-pair geometry. The lone pair occupies one position, making the molecular shape square pyramidal.

   Let’s walk through the reasoning step by step:

   1. Identify the product of partial hydrolysis. $\ce{XeF6}$ reacts with water in a controlled manner:

   $\ce{XeF6 + H2O -> XeOF4 + 2 HF}$

   This is a well-known reaction. The product ‘X’ is $\ce{XeOF4}$.

   2. Determine the central atom’s valence electrons and bonding.

      Xenon (Xe) has 8 valence electrons. In $\ce{XeOF4}$:

      • Four single bonds to F (each uses 1 Xe electron, total 4).
      • One double bond to O (uses 2 Xe electrons).
      • This accounts for 6 electrons in bonds. The remaining 2 electrons form one lone pair on Xe.

   3. Apply VSEPR theory to find the electron-pair geometry.

      The steric number = number of atoms bonded + number of lone pairs = 5 (4 F + 1 O) + 1 lone pair = 6.

      For steric number 6, the electron-pair geometry is octahedral.

   4. Determine the molecular shape (arrangement of atoms only).

      In an octahedral arrangement, lone pairs occupy one of the six positions. The remaining five positions (four F atoms and one O atom) form a square pyramidal shape. The oxygen is typically at the apex (due to double bond repulsion), but the overall shape is square pyramidal regardless of which atom is at the apex.

   5. Eliminate other options.

      • (A) Pyramidal: This is for steric number 4 (e.g., $\ce{NH3}$), not applicable here.
      • (B) Tetrahedral: Steric number 4, no lone pair on central atom (e.g., $\ce{CH4}$).
      • (D) Linear: Steric number 2 (e.g., $\ce{CO2}$). Only square pyramidal fits the VSEPR prediction for $\ce{XeOF4}$.

   Watch out

   A common mistake is to think $\ce{XeOF4}$ is octahedral because it has six electron pairs. But the molecular shape (positions of atoms only) is square pyramidal — the lone pair is not counted as a “corner” of the shape.

   Tip

   Remember: For $\ce{XeF6}$ itself, the shape is distorted octahedral (or “capped octahedral”) due to the lone pair. Hydrolysis replaces one F with O, but the lone pair remains, so the shape stays square pyramidal.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Observe the following list of molecules. Number of polar and non polar molecules are respectively NH$_3$, BF$_3$, NF$_3$, H$_2$S, CO$_2$, CH$_4$, CHCl$_3$, H$_2$O (A) 4, 4 (B) 3, 5 (C) 5, 3 (D) 2, 6

   ›Reveal solutionSolution

   The polarity of a molecule depends on both bond polarity and molecular geometry; after checking each molecule, we find 5 polar and 3 non‑polar, so the correct option is (C).

   Concept & Intuition

   A molecule is polar if it has a net dipole moment. That happens when:

   • The bonds are polar (atoms have different electronegativities).
   • The molecular shape does not cancel those bond dipoles symmetrically.

   So we must check each molecule’s geometry (VSEPR theory) and the symmetry of its polar bonds. A common mistake is to assume that just because a molecule contains polar bonds, it is automatically polar — but symmetric shapes like linear, trigonal planar, or tetrahedral can cancel dipoles perfectly.

   ---

   Step‑by‑step analysis

   1. NH₃ (ammonia)

      • N–H bonds are polar (N more electronegative).
      • Shape: trigonal pyramidal (lone pair on N).
      • The bond dipoles do not cancel → polar.

   2. BF₃ (boron trifluoride)

      • B–F bonds are polar (F very electronegative).
      • Shape: trigonal planar, symmetric.
      • The three bond dipoles cancel exactly → non‑polar.

   3. NF₃ (nitrogen trifluoride)

      • N–F bonds are polar (F more electronegative than N).
      • Shape: trigonal pyramidal (lone pair on N).
      • The dipoles do not cancel → polar.
      Watch out

      NF₃ is often mistaken as non‑polar because it looks like BF₃, but the lone pair breaks symmetry.

   4. H₂S (hydrogen sulfide)

      • S–H bonds are slightly polar (S more electronegative).
      • Shape: bent (two lone pairs on S, similar to H₂O).
      • Bond dipoles add up → polar.

   5. CO₂ (carbon dioxide)

      • C=O bonds are polar.
      • Shape: linear, symmetric (O=C=O).
      • The two bond dipoles point opposite directions and cancel → non‑polar.

   6. CH₄ (methane)

      • C–H bonds are very weakly polar (almost non‑polar).
      • Shape: perfect tetrahedron.
      • Even if bonds were polar, the tetrahedral symmetry cancels all dipoles → non‑polar.

   7. CHCl₃ (chloroform)

      • C–H, C–Cl bonds; Cl is much more electronegative than H.
      • Shape: tetrahedral, but not symmetric (three Cl atoms vs one H).
      • The net dipole points toward the three Cl atoms → polar.

   8. H₂O (water)

      • O–H bonds are very polar.
      • Shape: bent (two lone pairs on O).
      • Bond dipoles reinforce each other → polar.

   ---

   Counting

   Polar: NH₃, NF₃, H₂S, CHCl₃, H₂O → 5

   Non‑polar: BF₃, CO₂, CH₄ → 3

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The molecule ‘X’ has see-saw shape with central atom in $\mathrm{sp^3d}$ hybridization. What is ‘X’? (A) $\mathrm{ClF_3}$ (B) $\mathrm{XeF_4}$ (C) $\mathrm{SF_4}$ (D) $\mathrm{BrF_5}$

   ›Reveal solutionSolution

   A see-saw shape arises from an $\mathrm{sp^3d}$ hybridized central atom with one lone pair, giving a trigonal bipyramidal electron geometry. Among the options, only $\mathrm{SF_4}$ fits this description, so the correct answer is (C).

   Concept & Intuition

   The shape of a molecule is determined by its electron-pair geometry (from VSEPR theory) and the number of lone pairs on the central atom. A "see-saw" shape is a specific molecular geometry that occurs when the central atom has five regions of electron density (trigonal bipyramidal electron geometry) and one of those regions is a lone pair. The lone pair occupies an equatorial position, pushing the other atoms into a distorted, see-saw arrangement. The hybridization for five electron domains is $\mathrm{sp^3d}$. So we need a molecule whose central atom is $\mathrm{sp^3d}$ hybridized, has one lone pair, and four bonded atoms.

   Step-by-step reasoning

   1. Identify the electron-domain count for each option

      • (A) $\mathrm{ClF_3}$: Central Cl has 7 valence electrons. It forms 3 bonds with F atoms, leaving 4 electrons (2 lone pairs). Total electron domains = 3 bonds + 2 lone pairs = 5. Hybridization = $\mathrm{sp^3d}$.
      • (B) $\mathrm{XeF_4}$: Central Xe has 8 valence electrons. It forms 4 bonds with F atoms, leaving 4 electrons (2 lone pairs). Total electron domains = 4 bonds + 2 lone pairs = 6. Hybridization = $\mathrm{sp^3d^2}$.
      • (C) $\mathrm{SF_4}$: Central S has 6 valence electrons. It forms 4 bonds with F atoms, leaving 2 electrons (1 lone pair). Total electron domains = 4 bonds + 1 lone pair = 5. Hybridization = $\mathrm{sp^3d}$.
      • (D) $\mathrm{BrF_5}$: Central Br has 7 valence electrons. It forms 5 bonds with F atoms, leaving 2 electrons (1 lone pair). Total electron domains = 5 bonds + 1 lone pair = 6. Hybridization = $\mathrm{sp^3d^2}$.

   2. Eliminate molecules that do not have $\mathrm{sp^3d}$ hybridization

      • $\mathrm{XeF_4}$ and $\mathrm{BrF_5}$ both have 6 electron domains, so they are $\mathrm{sp^3d^2}$ hybridized. They cannot be see-saw shaped (XeF₄ is square planar; BrF₅ is square pyramidal). So (B) and (D) are out.

   3. Compare the two $\mathrm{sp^3d}$ candidates: $\mathrm{ClF_3}$ and $\mathrm{SF_4}$

      • $\mathrm{ClF_3}$ has 3 bonded atoms and 2 lone pairs. With 5 electron domains and 2 lone pairs, the lone pairs both occupy equatorial positions (to minimize repulsion), giving a T-shaped molecular geometry — not see-saw.
      • $\mathrm{SF_4}$ has 4 bonded atoms and 1 lone pair. With 5 electron domains and 1 lone pair, the lone pair occupies an equatorial position, leaving the other four atoms in a see-saw shape (two axial and two equatorial, but the equatorial lone pair distorts the angles).

   4. Confirm the shape of $\mathrm{SF_4}$

      • In $\mathrm{SF_4}$, the sulfur is $\mathrm{sp^3d}$ hybridized. The lone pair sits in the equatorial plane, causing the axial F–S–F bond angle to be less than 180° (about 173°) and the equatorial F–S–F angle to be less than 120° (about 101°). This is the classic see-saw geometry.

   Watch out

   A common mistake is to confuse "see-saw" with "T-shaped." Both come from $\mathrm{sp^3d}$ hybridization, but see-saw has one lone pair (4 bonds), while T-shaped has two lone pairs (3 bonds). $\mathrm{ClF_3}$ is T-shaped, not see-saw.

   Tip

   A quick memory aid: For $\mathrm{sp^3d}$ molecules, the number of lone pairs tells the shape:

   • 0 lone pairs → trigonal bipyramidal (e.g., $\mathrm{PCl_5}$)
   • 1 lone pair → see-saw (e.g., $\mathrm{SF_4}$)
   • 2 lone pairs → T-shaped (e.g., $\mathrm{ClF_3}$)
   • 3 lone pairs → linear (e.g., $\mathrm{XeF_2}$)

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Identify the pairs in which both molecules do not possess same type of hybridization. I. H$_2$O, SO$_2$ II. PCl$_3$, NH$_3$ III. SO$_2$, SO$_3$ IV. BCl$_3$, NCl$_3$ (A) I, III only (B) II, IV only (C) II, III only (D) I, IV only

   ›Reveal solutionSolution

   The question asks for pairs where the two molecules have different hybridization. We check each pair by counting electron domains (steric number) around the central atom. The pairs that differ are I (H₂O vs SO₂) and IV (BCl₃ vs NCl₃), so the correct option is (D).

   The core idea is simple: hybridization is determined by the number of electron domains (bond pairs + lone pairs) around the central atom. Two molecules have the same type of hybridization only if their central atoms have the same steric number and similar domain geometry. Let’s check each pair one by one.

   1. Pair I: H₂O and SO₂

      • In H₂O, oxygen has 2 bond pairs and 2 lone pairs → steric number = 4 → sp³ hybridization.
      • In SO₂, sulfur has 2 bond pairs (to two oxygens) and 1 lone pair → steric number = 3 → sp² hybridization. These are different. So pair I is a mismatch.

   2. Pair II: PCl₃ and NH₃

      • In PCl₃, phosphorus has 3 bond pairs and 1 lone pair → steric number = 4 → sp³.
      • In NH₃, nitrogen has 3 bond pairs and 1 lone pair → steric number = 4 → sp³. Both are sp³ hybridized. So pair II is not a mismatch.

   3. Pair III: SO₂ and SO₃

      • SO₂: sulfur has 2 bond pairs + 1 lone pair → steric number = 3 → sp².
      • SO₃: sulfur has 3 bond pairs (double bonds count as one domain each) and no lone pairs → steric number = 3 → sp². Both are sp² hybridized. So pair III is not a mismatch.

   4. Pair IV: BCl₃ and NCl₃

      • In BCl₃, boron has 3 bond pairs and no lone pairs → steric number = 3 → sp².
      • In NCl₃, nitrogen has 3 bond pairs and 1 lone pair → steric number = 4 → sp³. These are different. So pair IV is a mismatch.

   Watch out

   A common mistake is to think that SO₃ has sp³ hybridization because sulfur can expand its octet. But remember: hybridization depends only on the number of electron domains, not on the number of atoms. In SO₃, sulfur has three double bonds — each double bond is still one domain — so steric number = 3, giving sp².

   Thus, the pairs where both molecules do not have the same hybridization are I and IV.

   ✓Final answer

   The correct option is (D) I, IV only.

10. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.The bond angles $\widehat{\mathrm{H-O-N}}$ and $\widehat{\mathrm{O-N-O}}$ in the planar structure of nitric acid molecule are respectively (A) $130^\circ, 102^\circ$ (B) $102^\circ, 130^\circ$ (C) $134^\circ, 100^\circ$ (D) $100^\circ, 134^\circ$

    ›Reveal solutionSolution

    The bond angles in planar nitric acid (HNO₃) are determined by VSEPR theory and resonance: the H–O–N angle is about 102° (bent at oxygen with two lone pairs) and the O–N–O angle is about 130° (trigonal planar nitrogen with one lone pair in resonance). The correct choice is (B).

    The key is to recognize that nitric acid (HNO₃) has a planar structure due to resonance, and we must apply VSEPR (Valence Shell Electron Pair Repulsion) theory to both the central nitrogen and the oxygen bonded to hydrogen. Many students mistakenly treat all bonds as single/double without considering lone pairs and resonance, leading to wrong angle estimates.

    1. Draw the Lewis structure with resonance.

       Nitric acid has the connectivity H–O–NO₂. The nitrogen is central, bonded to one –OH group and two =O groups. However, the two terminal oxygens are equivalent by resonance: the N–O bonds are intermediate between single and double (bond order ~1.5), and the nitrogen has no lone pair because it uses all its valence electrons in bonding and resonance. The oxygen in the –OH group has two lone pairs and is bonded to H and N.

    2. Determine the geometry around nitrogen.

       Nitrogen in HNO₃ is sp² hybridized (three sigma bonds, no lone pair, and a π system delocalized over the three oxygens). This gives a trigonal planar arrangement with ideal angles of 120°. But the actual O–N–O angle is larger than 120° because the two terminal oxygens repel each other more strongly than they repel the –OH oxygen (the –OH oxygen has less double-bond character, so its electron density is slightly less repulsive). Experimental data show the O–N–O angle is about 130°.

    3. Determine the geometry around the –OH oxygen.

       The oxygen in the –OH group has two sigma bonds (to H and N) and two lone pairs. According to VSEPR, this is an AX₂E₂ arrangement (bent/angular), with ideal tetrahedral angles (~109.5°). Lone pairs repel more strongly than bonding pairs, so the H–O–N bond angle is compressed to about 102° (typical for water-like bent molecules).

    4. Match to the options.

       We have H–O–N ≈ 102° and O–N–O ≈ 130°. This corresponds exactly to option (B).

    Watch out

    A common mistake is to assume all angles are 120° because the molecule is planar. But the oxygen in the –OH group is not sp² hybridized — it is sp³-like with two lone pairs, giving a much smaller angle.

    Tip

    Remember: In VSEPR, lone pairs “take up more space” than bonding pairs. For the –OH oxygen, two lone pairs push the H and N closer together, reducing the angle below 109.5°.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.Identify the option in which the molecules are arranged in the correct order of their dipole moments (A) $\ce{CO2 < H2O < H2S}$ (B) $\ce{HF < HCl < HBr}$ (C) $\ce{BF3 < NF3 < NH3}$ (D) $\ce{CH4 < CHCl3 < CCl4}$

    ›Reveal solutionSolution

    The key idea is that molecular dipole moments depend on both bond polarity and molecular geometry; by analyzing symmetry and vector cancellation, the correct order is found in option (C).

    Concept and Intuition

    A molecule’s dipole moment is the vector sum of its individual bond dipoles. Symmetry can cause cancellation (e.g., linear CO₂ or planar BF₃), while lone pairs and bent shapes create net dipoles. The question tests whether you can predict relative dipole strengths from structure and electronegativity trends.

    Step-by-step reasoning

    1. Option (A): CO₂ < H₂O < H₂S

       • CO₂ is linear and symmetric: bond dipoles cancel → dipole moment = 0.
       • H₂O is bent (≈104.5°), with two O–H bonds and two lone pairs; net dipole is large (~1.85 D).
       • H₂S is also bent but S is less electronegative than O, and the bond angle is smaller (≈92°), so the net dipole is smaller (~0.97 D).
       • Thus the correct order should be CO₂ < H₂S < H₂O, not CO₂ < H₂O < H₂S.
       • Conclusion: (A) is wrong.

    2. Option (B): HF < HCl < HBr

       • Dipole moment depends on both electronegativity difference and bond length.
       • HF has the largest electronegativity difference but also the shortest bond; its dipole is ~1.91 D.
       • HCl: ~1.08 D; HBr: ~0.82 D.
       • The actual order is HF > HCl > HBr, decreasing down the group.
       • Conclusion: (B) is reversed, so wrong.

    3. Option (C): BF₃ < NF₃ < NH₃

       • BF₃ is trigonal planar and symmetric: bond dipoles cancel → dipole moment = 0.
       • NF₃ is pyramidal (like NH₃) but N–F bonds are polar toward F, and the lone pair on N opposes the net bond dipole, giving a small net dipole (~0.24 D).
       • NH₃ is pyramidal with N–H bonds polar toward N; the lone pair reinforces the bond dipoles, giving a large net dipole (~1.47 D).
       • So 0 < 0.24 < 1.47 → BF₃ < NF₃ < NH₃ is correct.
       • Conclusion: (C) is correct.

    4. Option (D): CH₄ < CHCl₃ < CCl₄

       • CH₄ is tetrahedral and symmetric → dipole = 0.
       • CHCl₃ (chloroform) is tetrahedral but not symmetric; the three C–Cl bond dipoles partially cancel but leave a net dipole (~1.04 D).
       • CCl₄ is tetrahedral and symmetric → dipole = 0.
       • So the order should be CH₄ = CCl₄ < CHCl₃, not CH₄ < CHCl₃ < CCl₄.
       • Conclusion: (D) is wrong.

    Watch out

    A common mistake is to assume that more polar bonds always mean a larger dipole moment, ignoring geometry. For example, CCl₄ has four polar C–Cl bonds but zero net dipole due to symmetry.

    Tip

    For molecules like NH₃ vs NF₃, remember that the lone pair direction matters: in NH₃, bond dipoles and lone pair point the same way; in NF₃, they oppose, drastically reducing the net dipole.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

12. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.In which of the following options, molecules are correctly arranged with respect to their bond angles. (A) $S_6 < O_3 < S_8 < P_4$ (B) $P_4 < O_3 < S_6 < S_8$ (C) $P_4 < S_6 < S_8 < O_3$ (D) $S_6 < P_4 < S_8 < O_3$

    ›Reveal solutionSolution

    The key idea is to recall the bond angles in each molecule: $P_4$ (60°), $S_6$ (90°), $S_8$ (108°), and $O_3$ (≈117°). The correct increasing order is $P_4 < S_6 < S_8 < O_3$, which matches option (C).

    Concept & Intuition

    Bond angles depend on molecular geometry and the presence of lone pairs. For these allotropes and simple molecules, we need to know their shapes:

    • $P_4$ (white phosphorus) is a tetrahedral cage with each P–P–P angle forced to 60°.
    • $S_6$ (hexasulfur) forms a chair-like six-membered ring; each S–S–S angle is 90°.
    • $S_8$ (cyclooctasulfur) is a puckered eight-membered ring with angles near 108°.
    • $O_3$ (ozone) is bent with a bond angle of about 117° due to resonance and lone-pair repulsion.

    Thus the order from smallest to largest bond angle is $P_4$ (60°) < $S_6$ (90°) < $S_8$ (108°) < $O_3$ (117°).

    Step-by-step reasoning

    1. Bond angle in $P_4$

       White phosphorus consists of four P atoms at the vertices of a regular tetrahedron, but the molecule is actually a cage where each P is bonded to three others. The P–P–P bond angle is exactly 60° (the angle in an equilateral triangle face). This is the smallest possible bond angle among these molecules.

    2. Bond angle in $S_6$

       $S_6$ is a cyclic molecule with six sulfur atoms in a chair conformation (like cyclohexane but with S–S bonds). The S–S–S bond angle is 90°, because each sulfur uses nearly pure p-orbitals for bonding, giving right-angle geometry.

    3. Bond angle in $S_8$

       The most stable form of sulfur is $S_8$, a puckered ring of eight atoms. The S–S–S bond angle is approximately 108°, close to the tetrahedral angle, due to some s-orbital mixing in the bonding.

    4. Bond angle in $O_3$

       Ozone is a bent molecule with a central oxygen atom having one lone pair and one double bond (resonance). The O–O–O angle is about 117°, larger than the others because of greater repulsion from the lone pair and the double-bond character.

    5. Arranging in increasing order

       From smallest to largest:

       $P_4$ (60°) < $S_6$ (90°) < $S_8$ (108°) < $O_3$ (117°).

       This corresponds exactly to option (C).

    Watch out

    A common mistake is to think $S_8$ has a 90° angle like $S_6$, but the larger ring relieves strain, allowing angles closer to 109°. Also, $O_3$ is often mistakenly thought to be 120° (like $SO_2$), but it’s slightly less due to lone-pair compression.

    Tip

    Remember the mnemonic: “P4 is 60 (like a triangle), S6 is 90 (like a square), S8 is 108 (like a pentagon’s interior angle), O3 is 117 (bent like water but wider).”

    ✓Final answer

    The correct option is (C).

    ANSWER: C