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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 181-210 of 342 questions

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181 Medium
181 of 342
Understanding +4 | -1
(A):A water molecule can form at the most four hydrogen bonds

(R): ice pices on pressing with each other join to give a single piece due to formation of hydrogen bonds in them
View answer and explanation
Correct option: A
Both (A)(A) and (R)(R) are true and (R)(R) is the correct explnation of (A)(A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because water molecules can form up to four hydrogen bonds, and these bonds are responsible for ice pieces joining together under pressure.
182 Medium
182 of 342
Analyzing +4 | -1
Assertion : The O−O{O}-{O} bond length in H2O2{H}_2 {O}_2 is shorter than that of O2 F2{O}_2 {~F}_2.

Reason: H2O2{H}_2 {O}_2 is an ionic compound.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D is correct because both the assertion and the reason are incorrect. The O-O bond length in H2O2 is longer than in O2F2, and H2O2 is a covalent compound.
183 Medium
183 of 342
Application +4 | -1
The bond order and magnetic behaviour of O2−{O}_2^{-}ion are, respectively
View answer and explanation
Correct option: A
1.5 and paramagnetic
Option A (1.5 and paramagnetic) is correct because the O2- ion has a bond order of 1.5 and unpaired electrons, making it paramagnetic.
184 Medium
184 of 342
Analyzing +4 | -1
Dipole moment is highest for:
View answer and explanation
Correct option: C
CHF3\mathrm{CH}{\mathrm{F}_3}
Option C (CHF3) is correct because the electronegativity difference between C and F is greater, and the dipole moments of the C-F bonds add up constructively.
185 Medium
185 of 342
Remembering +4 | -1
A π\pi-bond is formed by side way overlapping of
View answer and explanation
Correct option: C
p-p orbitals
Option C (p-p orbitals) is correct because a π\pi-bond is formed by the sideways or lateral overlapping of p-p orbitals.
186 Medium
186 of 342
Analyzing +4 | -1
Molecule having largest number of 90∘90^{\circ} bond angles is
View answer and explanation
Correct option: A
SF6{SF}_6
Option A (SF6{SF}_6) has the largest number of 90° bond angles because it has an octahedral geometry, which features twelve 90° bond angles.
187 Medium
187 of 342
Understanding +4 | -1
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with almost similar energy, position of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately. This concept is known as
View answer and explanation
Correct option: A
Resonance
Option A (Resonance) is correct because resonance describes the phenomenon where multiple Lewis structures are needed to accurately represent a molecule.
188 Medium
188 of 342
Remembering +4 | -1
The bond energies of F2,Cl2,Br2\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2 and I2\mathrm{I}_2 are 155 , 244,193 and 151 kJ/mol151 \mathrm{~kJ} / \mathrm{mol}. The weakest bond will be in
View answer and explanation
Correct option: B
I2\mathrm{I}_2
Option B (I2\mathrm{I}_2) is correct because iodine has the weakest bond due to its large size and long bond length, leading to lower bond energy.
189 Medium
189 of 342
Understanding +4 | -1
Which among the following is not an exception
to Lewis octet rule ?
View answer and explanation
Correct option: C
MgF2{MgF}_2
The correct answer is MgF2{MgF}_2 because it is an ionic compound and both ions achieve a stable octet configuration.
190 Medium
190 of 342
Understanding +4 | -1
During change of O2 to  O2−{\mathrm{O}_2}\,\mathrm{to}\,\,\mathrm{O}_2^ - ion, the electron adds on which one of the following orbitals?
View answer and explanation
Correct option: A
π∗\pi * orbital
Option A (π∗\pi * orbital) is correct because the electron adds to the antibonding π∗\pi * orbital during the change of O2{\mathrm{O}_2} to O2−{\mathrm{O}_2}^ - ion.
191 Medium
191 of 342
Understanding +4 | -1
The correct decreasing order of the boiling points of H2O,  HF{H_2}O,\,\,HF and NH3N{H_3} is
View answer and explanation
Correct option: A
H2O>HF>NH3{H_2}O > HF > N{H_3}
Option A is correct because water has the highest boiling point due to extensive hydrogen bonding, followed by HF, and then NH3.
192 Medium
192 of 342
Analyzing +4 | -1
Which of the following diatomic molecules would be stabilized by the removal of an electron?
View answer and explanation
Correct option: D
O2{\mathrm{O}_2}
Option D (O2{\mathrm{O}_2}) is correct because removing an electron from O2, which has electrons in antibonding orbitals, increases the bond order and stabilizes the molecule.
193 Medium
193 of 342
Analyzing +4 | -1
Read the following statements and choose the correct option. Here T stands for True and F stands for False statement.

(i) The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.

(ii) The smaller the charge on the cation, the greater the covalent character of the ionic bond.

(iii) For cations of the same size and charge, the one, with electronic configuration (n−1)dnns0(n-1) d^n n s^0, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2np6n s^2 n p^6, typical of alkali and alkaline earth metal cations.
View answer and explanation
Correct option: C
TFT
Option C (TFT) is correct because the first statement is true, the second is false (covalent character increases with charge), and the third is true (transition metals are more polarizing).
194 Medium
194 of 342
Analyzing +4 | -1
Statement-1: The number of bonding electrons in N2N_2 and N2−N_2^{-}is same.

Statement-2: The bond order and bond length of N2+N_2{ }^{+} and N2−N_2^{-}are same.
View answer and explanation
Correct option: C
Statement-1 is true, statement-2 is false.
Statement-1 is true, statement-2 is false because while the bond order is the same for N2+N_2^+ and N2−N_2^-, their bond lengths are not necessarily the same due to differing electron configurations.
195 Medium
195 of 342
Understanding +4 | -1
Assertion : Ionic compounds tend to be non-volalite.

Reason : Inter ionic forces in these compounds are strong.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because ionic compounds are non-volatile due to strong interionic forces.
196 Medium
196 of 342
Understanding +4 | -1
Consider the following statements.
I. Bond enthalpy increases as bond order decreases
II. Bond length decreases as bond order increases
III. There is a effect of electronegativity on bond energy
Select the correct statement (s).
View answer and explanation
Correct option: B
II & III
Option B (II & III) is correct because bond length decreases with increasing bond order and electronegativity difference affects bond energy.
197 Medium
197 of 342
Analyzing +4 | -1
The correct order of increasing bond angles in the following species is:
View answer and explanation
Correct option: D
ClO2−<Cl2O<ClO2\mathrm{ClO}_2^ - < C{\mathrm{l}_2}\mathrm{O} < Cl{\mathrm{O}_2}
Option D (ClO2−<Cl2O<ClO2\mathrm{ClO}_2^ - < C{\mathrm{l}_2}\mathrm{O} < Cl{\mathrm{O}_2}) is correct because the bond angle depends on the number of lone pairs and the electronegativity of the surrounding atoms.
198 Medium
198 of 342
Understanding +4 | -1
Bond energy of Cl2,Br2{Cl}_2, {Br}_2 and I2{I}_2 follow the order
View answer and explanation
Correct option: A
Cl2>Br2>I2{Cl}_2>{Br}_2>{I}_2
Option A (Cl2>Br2>I2{Cl}_2>{Br}_2>{I}_2) is correct because bond energy decreases down the halogen group due to increasing atomic size and bond length.
199 Medium
199 of 342
Analyzing +4 | -1
Statement-1: C2,F2,B2,N2{C}_2, {F}_2, {B}_2, {N}_2 are diamagnetic species
Statement-2: Species with unpaired electrons only in anti bonding orbitals are paramagnetic
View answer and explanation
Correct option: C
STATEMENT–1 is False, STATEMENT–2 is False
The correct answer is C (STATEMENT–1 is False, STATEMENT–2 is False) because both statements about the magnetic properties of the given species and the origin of paramagnetism are incorrect.
200 Medium
200 of 342
Understanding +4 | -1
In He2{He}_2, the electrons in bonding and antibonding orbitals are
View answer and explanation
Correct option: A
2,2
Option A (2,2) is correct because in He2{He}_2, there are 4 electrons, with 2 filling the bonding molecular orbital and 2 filling the antibonding molecular orbital.
201 Medium
201 of 342
Analyzing +4 | -1
The pair of species having identical shape of both species:
View answer and explanation
Correct option: D
XeF2,CO2\mathrm{Xe}{\mathrm{F}_2},\mathrm{C}{\mathrm{O}_2}
Option D is correct. Both XeF2 and CO2 are linear molecules.
202 Medium
202 of 342
Analyzing +4 | -1
Which of the following is correct order of σ\sigma bond strength
I) 2S – 2S II) 2S – 2P III) 2P – 2P IV) 3S – 3S
View answer and explanation
Correct option: B
III > II > I > IV
Option B (III > II > I > IV) is correct because the sigma bond strength depends on the energy levels of the orbitals involved, with 2p-2p being the strongest and 3s-3s being the weakest.
203 Medium
203 of 342
Analyzing +4 | -1
Which of the following molecules does not possess a permanent electric dipole moment?
View answer and explanation
Correct option: D
CS2\mathrm{C}{\mathrm{S}_2}
Option D (CS2\mathrm{C}{\mathrm{S}_2}) is correct because it is a linear molecule with symmetrical bond dipoles that cancel each other out, resulting in a zero dipole moment.
204 Medium
204 of 342
Analyzing +4 | -1
Which of the following statements are correct?
(A) Both LiCl  and  MgCl2\mathrm{LiCl}\,\,\mathrm{and}\,\,\mathrm{MgC}{\mathrm{l}_\mathrm{2}} are soluble in ethanol.
(B) The oxides Li2O   and  MgO\mathrm{L}{\mathrm{i}_\mathrm{2}}\mathrm{O}\,\,\,\mathrm{and}\,\,\mathrm{MgO} combine with excess of oxygen to give superoxide.
(C) LiF\mathrm{LiF} is less soluble in water than other alkali metal fluorides.
(D) Li2O\mathrm{L}{\mathrm{i}_2}\mathrm{O} is more soluble in water than other alkali metal oxides.
Choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: A
(A) and (C) only
Option A ((A) and (C) only) is correct because LiCl and MgCl₂ are soluble in ethanol due to their covalent nature, and LiF is less soluble in water due to its high lattice energy.
205 Medium
205 of 342
Analyzing +4 | -1
Among the following species, identify the isostructural pairs

NF3,NO3−,BF3,H3O+,HN3\mathrm{NF}_3, \mathrm{NO}_3^{-}, \mathrm{BF}_3, \mathrm{H}_3 \mathrm{O}^{+}, \mathrm{HN}_3
View answer and explanation
Correct option: C
[NF3,H3O+]\left[\mathrm{NF}_3, \mathrm{H}_3 \mathrm{O}^{+}\right]and [NO3−,BF3]\left[\mathrm{NO}_3^{-}, \mathrm{BF}_3\right]
The correct answer is C because NF3 and H3O+ are both pyramidal (sp3 hybridization), while NO3- and BF3 are trigonal planar (sp2 hybridization).
206 Medium
206 of 342
Understanding +4 | -1
Bond angles of NH3,PH3,AsH3\mathrm{N}{\mathrm{H}_3},\mathrm{P}{\mathrm{H}_3},\mathrm{As}{\mathrm{H}_3} and SbH3\mathrm{Sb}{\mathrm{H}_3} is in the order
View answer and explanation
Correct option: D
NH3>PH3>AsH3>SbH3\mathrm{N}{\mathrm{H}_3} > P{\mathrm{H}_3} > As{\mathrm{H}_3} > Sb{\mathrm{H}_3}
Option D (NH3>PH3>AsH3>SbH3\mathrm{N}{\mathrm{H}_3} > P{\mathrm{H}_3} > As{\mathrm{H}_3} > Sb{\mathrm{H}_3}) is correct because as the size of the central atom increases down the group, the bond angle decreases due to decreasing electronegativity and increasing bond length.
207 Medium
207 of 342
Understanding +4 | -1
Which of the following has the lowest bond angle?
View answer and explanation
Correct option: C
H2S{H}_2 {S}
Option C (H2S{H}_2 {S}) is correct because sulfur's larger size compared to oxygen leads to less repulsion between the bonding pairs, resulting in a smaller bond angle.
208 Medium
208 of 342
Understanding +4 | -1
The hybrid state of 1 and 3 carbon atoms in CH2=C=CH2C{H_2} = C = C{H_2}
View answer and explanation
Correct option: B
sp2s{p^2}
Option B (sp2s{p^2}) is correct because the terminal carbon atoms in allene (CH2=C=CH2C{H_2} = C = C{H_2}) are sp2s{p^2} hybridized.
209 Medium
209 of 342
Remembering +4 | -1
In a crystal cations and anions are held together by
View answer and explanation
Correct option: B
Electrostatic forces
Option B (Electrostatic forces) is correct because cations and anions in a crystal are held together by electrostatic forces of attraction.
210 Medium
210 of 342
Remembering +4 | -1
Among the following the weakest one is
View answer and explanation
Correct option: C
Van der Waal's force
Option C (Van der Waal's force) is correct because Van der Waal's forces are the weakest intermolecular forces compared to metallic, ionic, and covalent bonds.
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