Back to CHEMICAL BONDING & MOLECULAR STRUCTURE

Your Progress

0 / 494 Answered

Correct

Incorrect

Accuracy

JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →

Showing 241-270 of 494 questions

Preparing practice and saved progress…

241 Easy

241 of 494

Analyzing +4 | -1

Which of the following statements are not correct?
A. The electron gain enthalpy of

F

is more negative than that of

Cl.

B. Ionization enthalpy decreases in a group of periodic table.
C. The electronegativity of an atom depends upon the atoms bonded to it.
D.

Al_2O_3

and

NO

are examples of amphoteric oxides.
Choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: D

A, C and D only

Option D (A, C and D only) is correct because statements A, C, and D are incorrect regarding electron gain enthalpy, electronegativity, and amphoteric oxides.

242 Medium

242 of 494

Understanding +4 | -1

Statement I:

\mathrm{NaCl}

forms ionic crystal

Statement II: In

\mathrm{NaCl}

atoms have electrostatic attractive forces

View answer and explanation

Correct option: C

If statement I is correct but statement II is false

Option D (If statement I is false but statement II is correct) is correct because NaCl forms ionic crystal and atoms in NaCl have electrostatic attractive forces.

243 Medium

243 of 494

Analyzing +4 | -1

Which of the following pair contains 2 lone pair of electrons on the central atom?

View answer and explanation

Correct option: A

{I}_3^{+}, {H}_2 {O}

Option A (

{I}_3^{+}, {H}_2 {O}

) is correct because both

{I}_3^{+}

and

{H}_2 {O}

have 2 lone pairs on the central atom.

244 Medium

244 of 494

Analyzing +4 | -1

Consider the following statements
Statement I : The

HF_2^ -

ion exists in the solid state and also in the liquid state but not in aqueous solution
Statement II : The magnitude of hydrogen bonds in between HF-HF molecules is weaker than in between HF and

{H_2}O

molecules

View answer and explanation

Correct option: C

Both the statements are correct

Option C (Both the statements are correct) is correct because both statements regarding the existence of

HF_2^-

and the strength of hydrogen bonds are true.

245 Medium

245 of 494

Understanding +4 | -1

The correct order of bond angles is

View answer and explanation

Correct option: C

{H_2}S < N{H_3} < Si{H_4} < B{F_3}

Option D is correct. The bond angles increase in the order H2S < SiH4 < NH3 < BF3 due to differences in electronegativity and lone pair repulsion.

246 Medium

246 of 494

Understanding +4 | -1

The correct sequence of hybridisation of methane, ethene and acetylene is

View answer and explanation

Correct option: C

s{p^3},s{p^2},sp

The correct answer is C (

s{p^3}, s{p^2}, s

). The order of electronegativity of methane, ethene and ethyne is

sp^3, sp^2, sp

247 Medium

247 of 494

Analyzing +4 | -1

The correct order against indicated property is

View answer and explanation

Correct option: A

\mathrm{N}_2{ }^{+}>\mathrm{H}_2{ }^{+}

(bond order)

The correct answer is A because the bond order of N2+ is higher than that of H2+.

248 Easy

248 of 494

Application +4 | -1

The hybridization of phosphorus in

\mathrm{POC}{\mathrm{l}_3}

is same as in:

View answer and explanation

Correct option: A

\mathrm{P in PC}{\mathrm{l}_3}

Option A (P in

PCl_3

) is correct because both

POCl_3

and

PCl_3

have a central phosphorus atom with

sp^3

hybridization.

249 Medium

249 of 494

Remembering +4 | -1

Which of the following has the highest melting point

View answer and explanation

Correct option: B

Diamond

Option B (Diamond) is correct because diamond has a giant covalent network structure, requiring a large amount of energy to break, resulting in a very high melting point.

250 Medium

250 of 494

Understanding +4 | -1

The decreasing order of bond angle is

View answer and explanation

Correct option: C

\mathrm{NO}_2^ + > N{\mathrm{O}_2} > \mathrm{NO}_2^ -

Option C (

\mathrm{NO}_2^ + > N{\mathrm{O}_2} > \mathrm{NO}_2^ -

) is correct because the bond angle decreases as the number of lone pairs increases.

251 Easy

251 of 494

Application +4 | -1

Two hybrid orbitals have a bond angle of

120^\circ

. The percentage ofs character in the hybrid orbital is nearly

View answer and explanation

Correct option: B

33\%

Option B (33%) is correct because a bond angle of 120° corresponds to sp² hybridization, which has 33% s character.

252 Medium

252 of 494

Understanding +4 | -1

In case of bonding molecular orbital, the electron density is located

View answer and explanation

Correct option: B

Between the nuclei of the bonded atoms

Option B is correct. In a bonding molecular orbital, the electron density is concentrated between the nuclei of the bonded atoms, leading to increased stability.

253 Easy

253 of 494

Understanding +4 | -1

The bond in the formation of fluorine molecule is

View answer and explanation

Correct option: C

Due to p - p overlapping

The correct answer is C (Due to p - p overlapping) because fluorine forms a single bond through the overlap of its p-orbitals.

254 Medium

254 of 494

Remembering +4 | -1

Hybridization involves:

View answer and explanation

Correct option: B

Mixing of atomic orbitals

Option B (Mixing of atomic orbitals) is correct because hybridization is the process of mixing atomic orbitals to form new hybrid orbitals.

255 Easy

255 of 494

Application +4 | -1

Which is not paramagnetic

View answer and explanation

Correct option: C

{O}_2^{2-}

Option C (

{O}_2^{2-}

) is correct because it has all its electrons paired and is therefore diamagnetic, while the others have unpaired electrons.

Explore Other Chapters in CHEMISTRY

View All →

P-BLOCK ELEMENTS 736 MCQs Equilibrium 696 MCQs STRUCTURE OF ATOM 634 MCQs THERMODYNAMICS 536 MCQs SOLUTIONS 480 MCQs CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 455 MCQs

256 Medium

256 of 494

Application +4 | -1

Chlorine atom, in its third excited state, reacts with fluorine to form a compound X. The formula and shape of X are

View answer and explanation

Correct option: D

\mathrm{Cl}{\mathrm{F}_7}

pentagonal bipyramidal

Option D (

\mathrm{Cl}{\mathrm{F}_7}

pentagonal bipyramidal) is correct because in the third excited state, chlorine can form 7 bonds with fluorine, resulting in a pentagonal bipyramidal shape.

257 Medium

257 of 494

Understanding +4 | -1

The shape of gaseous

\mathrm{SnC}{\mathrm{l}_2}

View answer and explanation

Correct option: C

Angular

Option C (Angular) is correct because SnCl2 has sp2 hybridization with one lone pair, resulting in a bent or angular shape.

258 Medium

258 of 494

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)

and

(R)

are true and

(R)

is the correct explnation of

(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.

259 Medium

259 of 494

Analyzing +4 | -1

Assertion : The

{O}-{O}

bond length in

{H}_2 {O}_2

is shorter than that of

{O}_2 {~F}_2

.

Reason:

{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.

260 Easy

260 of 494

Understanding +4 | -1

The geometry of the molecule with

s p^3 d^2

hybridised central atom is

View answer and explanation

Correct option: C

Octahedral

Option C (Octahedral) is correct because

sp^3d^2

hybridization results in six hybrid orbitals directed towards the corners of an octahedron.

261 Medium

261 of 494

Application +4 | -1

The bond order and magnetic behaviour of

{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.

262 Medium

262 of 494

Analyzing +4 | -1

Dipole moment is highest for:

View answer and explanation

Correct option: C

\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.

263 Medium

263 of 494

Remembering +4 | -1

\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

\pi

-bond is formed by the sideways or lateral overlapping of p-p orbitals.

264 Easy

264 of 494

Understanding +4 | -1

Strongest hydrogen bonding is present in

View answer and explanation

Correct option: A

HF

Option A (HF) is correct because fluorine is the most electronegative element, leading to the strongest hydrogen bonding.

265 Medium

265 of 494

Analyzing +4 | -1

Molecule having largest number of

90^{\circ}

bond angles is

View answer and explanation

Correct option: A

{SF}_6

Option A (

{SF}_6

) has the largest number of 90° bond angles because it has an octahedral geometry, which features twelve 90° bond angles.

266 Easy

266 of 494

Understanding +4 | -1

Formation of

\pi

-bond:

View answer and explanation

Correct option: B

Decreases bond length

Option B is correct because the formation of a

\pi

-bond results in increased electron density between the atoms, leading to a shorter bond length.

267 Medium

267 of 494

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.

268 Medium

268 of 494

Remembering +4 | -1

The bond energies of

\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2

and

\mathrm{I}_2

are 155 , 244,193 and

151 \mathrm{~kJ} / \mathrm{mol}

. The weakest bond will be in

View answer and explanation

Correct option: B

\mathrm{I}_2

Option B (

\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.

269 Medium

269 of 494

Understanding +4 | -1

Which among the following is not an exception
to Lewis octet rule ?

View answer and explanation

Correct option: C

{MgF}_2

The correct answer is

{MgF}_2

because it is an ionic compound and both ions achieve a stable octet configuration.

270 Medium

270 of 494

Understanding +4 | -1

During change of

{\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

\pi *

orbital

Option A (

\pi *

orbital) is correct because the electron adds to the antibonding

\pi *

orbital during the change of

{\mathrm{O}_2}

{\mathrm{O}_2}^ -

ion.

Revision notes and key takeaways

Review CHEMICAL BONDING & MOLECULAR STRUCTURE in your textbook or class notes, then use this set to check your understanding.

Showing 241-270 of 494 questions

Your Progress

Chemical bonding & molecular structure practice · Mixed difficulty

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice