Chemical Bonding and Molecular Structure questions
Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Given below are two statements:
Statement (I): The correct sequence of bond lengths in the following species is:
$O_2^+ < O_2 < O_2^- < O_2^{2-}$
Statement (II): The correct sequence of number of unpaired electrons in the following species is:
$O_2 > O_2^+ > O_2^- > O_2^{2-}$
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (3)
Key idea: Molecular Orbital Theory - Bond Order and Magnetic Properties
Calculate bond order for each species using: Bond Order = $\frac{\text{bonding e}^- - \text{antibonding e}^-}{2}$
For $O_2^+$ (15e⁻): BO = 2.5; $O_2$ (16e⁻): BO = 2; $O_2^-$ (17e⁻): BO = 1.5; $O_2^{2-}$ (18e⁻): BO = 1
Since bond length ∝ $\frac{1}{\text{Bond Order}}$, Statement I gives correct sequence: $O_2^+ < O_2 < O_2^- < O_2^{2-}$ ✓
Count unpaired electrons from MO configuration: $O_2$ has 2 (π*²), $O_2^+$ has 1 (π*¹), $O_2^-$ has 1 (π*³), $O_2^{2-}$ has 0 (π*⁴)
Statement II is incorrect because $O_2^+$ and $O_2^-$ both have 1 unpaired electron (not $O_2^+ > O_2^-$)
Therefore, Statement I is true but Statement II is false
Which of the following pictorial diagram most correctly represents the $\pi^*$ ($\pi$ - antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the z-direction ($\xrightarrow{z\text{-axis}}$)?
(1) [Figure: Two lobes stacked vertically with shading on top lobe, two dots in middle representing nuclei]
(2) [Figure: Two lobes side by side horizontally, left lobe shaded, right lobe unshaded]
(3) [Figure: Two lobes arranged diagonally, top-left shaded, bottom-right shaded]
(4) [Figure: Four lobes in a row alternating shaded and unshaded]
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: π* (pi antibonding) molecular orbital formation and nodal plane characteristics
Consider the following species:
$BrF_5$, $XeF_5^-$, $BF_4^-$, $ICl_4^-$, $XeF_4$, $SF_4$, $NH_4^+$, $ClF_3$, $XeF_2$, $ICl_2^-$
Number of species having $sp^3d$ hybridized central atom is _______.
Show answer and solution
Answer: 4
Key idea: Hybridization and VSEPR Theory
For $sp^3d$ hybridization, steric number = 5 (bonding pairs + lone pairs)
Calculate steric number for each species using: Steric number = $\frac{1}{2}$(valence electrons + monovalent atoms ± charge)
$SF_4$: S has 6 valence e⁻, 4 bonds, 1 lone pair → Steric number = 5 → $sp^3d$ ✓
$ClF_3$: Cl has 7 valence e⁻, 3 bonds, 2 lone pairs → Steric number = 5 → $sp^3d$ ✓
$XeF_2$: Xe has 8 valence e⁻, 2 bonds, 3 lone pairs → Steric number = 5 → $sp^3d$ ✓
$ICl_2^-$: I has 8 valence e⁻ (with charge), 2 bonds, 3 lone pairs → Steric number = 5 → $sp^3d$ ✓
Given below are two statements:
**Statement I:** Aluminium upon reaction with NaOH forms $[Al(OH)_{6}]^{3-}$ ion.
**Statement II:** The geometry of $ICl_{4}^{-}$, $ClO_{3}^{-}$ and $IBr_{2}^{-}$ is square planar, pyramidal and linear respectively.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (4)
Key idea: VSEPR Theory and Coordination Chemistry
**Statement I Analysis:** Al reacts with NaOH to form $[Al(OH)_4]^-$ (tetrahedral), NOT $[Al(OH)_6]^{3-}$. Statement I is **FALSE**.
**For $ICl_4^-$:** I has 7 valence e⁻, 4 bonds, 1 negative charge → Total = 12 e⁻ = 6 pairs. With 4 bonding pairs and 2 lone pairs, geometry is **square planar** ✓
**For $ClO_3^-$:** Cl has 7 valence e⁻, 3 bonds, 1 negative charge → 8 e⁻ total. With 3 bonding pairs and 1 lone pair (steric number = 4), geometry is **pyramidal** ✓
**For $IBr_2^-$:** I has 7 valence e⁻, 2 bonds, 1 negative charge → 10 e⁻ = 5 pairs. With 2 bonding pairs and 3 lone pairs, geometry is **linear** ✓
**Statement II Analysis:** All three geometries are correctly stated. Statement II is **TRUE**.
**Conclusion:** Statement I is false, Statement II is true → Answer is option **(4)**
The pairs among
$A = [SO_3^{2-}, CO_3^{2-}]$, $B = [O_2^{2-}, I_2]$, $C = [CN^-, CO]$, $D = [NH_3, H_3O^+]$ and
$E = [MnO_4^-, CrO_4^{2-}]$ that do not have similar Lewis dot structure are
(1) A, B and E
(2) A and E
(3) B, C and D
(4) C and D
Show answer and solution
Answer: Option (2)
Key idea: Lewis Structures and Isoelectronic Species
Analyze **Pair A**: $SO_3^{2-}$ has trigonal pyramidal geometry (lone pair on S), while $CO_3^{2-}$ has trigonal planar geometry (no lone pair on C) - **DIFFERENT**
Analyze **Pair B**: $O_2^{2-}$ and $I_2$ both have single bonds with 3 lone pairs on each atom, 14 valence electrons - **SIMILAR**
Analyze **Pair C**: $CN^-$ and $CO$ are isoelectronic (10 electrons), both have triple bonds - **SIMILAR**
Analyze **Pair D**: $NH_3$ and $H_3O^+$ are isoelectronic (8 electrons), both trigonal pyramidal with 1 lone pair - **SIMILAR**
Analyze **Pair E**: $MnO_4^-$ (Mn in +7) and $CrO_4^{2-}$ (Cr in +6) have different charges and oxidation states despite both being tetrahedral - **DIFFERENT**
Pairs with dissimilar Lewis structures are **A and E**
Given below are two statements:
**Statement I:** The number of compounds among $SO_2$, $SO_3$, $SF_4$, $SF_6$ and $H_2S$ in which sulphur does not obey the Octet rule is 3.
**Statement II:** Among $[H_2O, ClF_3, SF_4]$, $[NH_3, BrF_5, SF_4]$, $[BrF_5, ClF_3, XeF_4]$ and $[XeF_4, ClF_3, H_2O]$, the number of sets in which all the molecules have one lone pair of electrons on the central atom is 1.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both **Statement I** and **Statement II** are true
(2) Both **Statement I** and **Statement II** are false
(3) **Statement I** is true but **Statement II** is false
(4) **Statement I** is false but **Statement II** is true
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Octet Rule, VSEPR Theory, and Lone Pair Determination
Given below are two statements:
**Statement I:** $F_2 O < H_2 O < Cl_2 O$ is the correct trend in terms of bond angle.
**Statement II:** $SiF_4$, $SnF_4$ and $PbF_4$ are ionic in nature.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (3)
Key idea: Bond angles in bent molecules and ionic vs covalent character in fluorides
**Statement I Analysis:** Compare bond angles in $F_2O$, $H_2O$, and $Cl_2O$ (all bent with 2 lone pairs on O)
In $F_2O$: F is more electronegative than O, pulls bonding electrons away, increases lone pair repulsion → bond angle $\approx 103°$
In $H_2O$: Intermediate electronegativity → bond angle $\approx 104.5°$
In $Cl_2O$: Cl is less electronegative than O, bonding electrons closer to O, reduces lone pair repulsion; Cl is larger → bond angle $\approx 110.9°$
Therefore $F_2O < H_2O < Cl_2O$ is correct. **Statement I is TRUE**
**Statement II Analysis:** $SiF_4$, $SnF_4$, and $PbF_4$ all have high oxidation state (+4) and form covalent bonds due to high charge density and polarization. **Statement II is FALSE**
Answer: Statement I is true but Statement II is false → Option (3)
Bromine trifluoride autoionizes to form $\text{BrF}_2^+$ and $\text{BrF}_4^-$. The shapes of the cation and anion are respectively __________, and __________.
(1) bent, square planar
(2) ...
(3) ...
(4) ...
Show answer and solution
Answer: Option (1)
Key idea: VSEPR Theory and Molecular Geometry
For $\text{BrF}_2^+$: Count valence electrons: Br has 7, subtract 1 for positive charge = 6 electrons total
$\text{BrF}_2^+$: 2 bond pairs (to F atoms) + 2 lone pairs. Steric number = 4, giving $sp^3$ hybridization
With 2 bond pairs and 2 lone pairs, $\text{BrF}_2^+$ has **bent** shape
For $\text{BrF}_4^-$: Count valence electrons: Br has 7, add 1 for negative charge = 8 electrons total
$\text{BrF}_4^-$: 4 bond pairs (to F atoms) + 2 lone pairs. Steric number = 6, giving $sp^3d^2$ hybridization
With 4 bond pairs and 2 lone pairs (opposite positions), $\text{BrF}_4^-$ has **square planar** shape
FAQs
Frequently asked questions
How many questions come from Chemical Bonding and Molecular Structure in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 10 questions on it, about 1.1 per shift.
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On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
