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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Mixed difficulty

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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.

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Showing 181-210 of 494 questions

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181 Hard

181 of 494

Application +4 | -1

The ion that has

s{p^3}{d^2}

hybridisation for the central atom, is

View answer and explanation

Correct option: C

{[IC{l_4}]^ - }

Option C (

{[IC{l_4}]^ - }

) is correct because its central iodine atom has 7 valence electrons, 4 bonded chlorine atoms, and a -1 charge, leading to

s{p^3}{d^2}

hybridization.

182 Easy

182 of 494

Understanding +4 | -1

Select the compound of highest covalent character.

View answer and explanation

Correct option: D

NaI

Option D (NaI) is correct because, according to Fajan's rule, larger anions and smaller cations favor covalent character.

183 Easy

183 of 494

Remembering +4 | -1

According to Fajan’s rule polarization is more when:

View answer and explanation

Correct option: A

Small cation and large anion

Option A (Small cation and large anion) is correct because according to Fajan's rule, polarization is favored by a small, highly charged cation and a large, highly charged anion.

184 Medium

184 of 494

Analyzing +4 | -1

Maximum bond angle at nitrogen is present in which of the following

View answer and explanation

Correct option: C

N{O_2}^ +

Option C (

N{O_2}^ +

) is correct because it has a linear structure with a bond angle of 180 degrees.

185 Easy

185 of 494

Understanding +4 | -1

The angle between two covalent bonds is minimum in

View answer and explanation

Correct option: A

{H_2}O

The correct answer is A (

{H_2}O

) because the bond angle in water is the smallest due to the presence of two lone pairs on the oxygen atom, which cause greater repulsion and reduce the bond angle to 104.5°.

186 Medium

186 of 494

Analyzing +4 | -1

Polar molecule among the following is

View answer and explanation

Correct option: A

Xe{O_3}

Option A (

Xe{O_3}

) is correct because it has a non-zero dipole moment due to its pyramidal shape.

187 Easy

187 of 494

Analyzing +4 | -1

The hybrid orbital having only 20%

d

– character

View answer and explanation

Correct option: C

S{F_4}

Option B (

S{F_6}

) is incorrect, Option A (

CC{l_4}

) is incorrect, Option D (

C{l_2}O

) is incorrect and Option C (

S{F_4}

) is correct because

S{F_4}

has

s{p^3}d

hybridization, which means it has 20% d-character.

188 Medium

188 of 494

Understanding +4 | -1

Which one has highest melting point?

View answer and explanation

Correct option: C

{NaF}

Option C (NaF) is correct because the melting point of ionic compounds increases as the size of the ions decreases, and fluoride is the smallest halide.

189 Medium

189 of 494

Analyzing +4 | -1

For two ionic solids

CaO

and

KI

, identify the wrong statement among the following?

View answer and explanation

Correct option: D

KI

has high M.P

Option D (KI has high M.P) is the wrong statement because KI has a lower melting point compared to CaO due to weaker electrostatic forces.

190 Medium

190 of 494

Analyzing +4 | -1

Which of the following has lowest boiling point?

View answer and explanation

Correct option: C

\mathrm{CuC}{\mathrm{l}_2}

Option D (

\mathrm{CuC}{\mathrm{l}_2}

) is correct because, according to Fajan’s rule, smaller cation size and higher charge increase covalent character, lowering the boiling point.

191 Medium

191 of 494

Understanding +4 | -1

The compound possessing most strongly ionic nature is:

View answer and explanation

Correct option: D

\mathrm{CsCl}

Option D (CsCl) is correct because Cesium Chloride has the largest difference in electronegativity between the cation and anion, leading to the most ionic character.

192 Medium

192 of 494

Understanding +4 | -1

Which of the following has largest bond angle?

View answer and explanation

Correct option: A

C{H_4}\,

Option A (

C{H_4}\,

) is correct because

CH_4

has a tetrahedral geometry with bond angles of 109.5°, while

H_2O

and

NH_3

have bent and pyramidal geometries, respectively, due to the presence of lone pairs, which reduce the bond angles.

193 Medium

193 of 494

Understanding +4 | -1

In which compound, the hydrogen bonding is the strongest in its liquid phase

View answer and explanation

Correct option: A

{HF}

Option A (

{HF}

) is correct because fluorine is the most electronegative element and has a small size, leading to the strongest hydrogen bonding.

194 Medium

194 of 494

Remembering +4 | -1

Electron deficient species are known as:

View answer and explanation

Correct option: A

Lewis acids

Option A (Lewis acids) is correct because electron-deficient species can accept electron pairs and are therefore classified as Lewis acids.

195 Medium

195 of 494

Analyzing +4 | -1

Which of the following is the correct order of dipole moment?

View answer and explanation

Correct option: C

{PCl}_3 {F}_2 < {PF}_3 {Cl}_2

Option D is correct. The dipole moment of PF3Cl2 is non-zero, while PCl3F2 has a dipole moment of zero due to its symmetrical structure.

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196 Medium

196 of 494

Application +4 | -1

Which of the following species has four lone pairs in its outer shell?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{l}^ - }

Option C (Cl-) is correct because it has 7 valence electrons plus 1 from the negative charge, resulting in 8 electrons or 4 lone pairs.

197 Medium

197 of 494

Understanding +4 | -1

Which one of the following arrangements shows the bonds

\mathrm{H}-\mathrm{H}, \mathrm{C}-\mathrm{C}

and

\mathrm{Si}-\mathrm{Si}

in order of increasing bond energy?

View answer and explanation

Correct option: A

\mathrm{Si}-\mathrm{Si}<\mathrm{C}-\mathrm{C}<\mathrm{H}-\mathrm{H}

The correct answer is A. Bond energy increases with decreasing atomic size, making Si-Si < C-C < H-H the correct order.

198 Medium

198 of 494

Analyzing +4 | -1

Which of the following statement is incorrect?

View answer and explanation

Correct option: A

Low solubility of LiF in water is due to its small hydration enthalpy.

Option A is correct. The low solubility of LiF in water is primarily due to its high lattice enthalpy, not small hydration enthalpy. While Li⁺ has high hydration enthalpy, the lattice enthalpy dominates.

199 Medium

199 of 494

Understanding +4 | -1

The correct increasing order off polarising power is:

View answer and explanation

Correct option: D

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

Option D (

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

) is correct because polarising power increases with increasing charge and decreasing size.

200 Easy

200 of 494

Understanding +4 | -1

The geometry of

\mathrm{IF}_7

View answer and explanation

Correct option: D

Pentagonal bipyramidal

Option D (Pentagonal bipyramidal) is correct because IF7 has 7 bond pairs and 0 lone pairs, resulting in sp3d3 hybridization and a pentagonal bipyramidal geometry.

201 Easy

201 of 494

Analyzing +4 | -1

In which of the following bond angle is maximum?

View answer and explanation

Correct option: B

\mathrm{NH}_4^ +

Option B (

\mathrm{NH}_4^+

) is correct because it has a tetrahedral geometry with bond angles of 109.5 degrees, which is the largest among the given options.

202 Medium

202 of 494

Analyzing +4 | -1

Find the pair that has the same bond order with diamagnetic and paramagnetic properties respectively

View answer and explanation

Correct option: C

L{i_2}\,\,\mathrm{and}\,\,{B_2}

The correct answer is C because Li2 has a bond order of 1 and is diamagnetic, while B2 has a bond order of 1 and is paramagnetic.

203 Medium

203 of 494

Understanding +4 | -1

The species analogous to paramagnetic behaviour like

{O}_2

View answer and explanation

Correct option: D

Gaseous sulphur.

Option D (Gaseous sulphur) is correct because gaseous sulfur exists as

S_2

, which has unpaired electrons similar to

O_2

, leading to paramagnetic behavior.

204 Medium

204 of 494

Analyzing +4 | -1

Arrange the following in order of increasing dipole moment (I) Toluene (II)

m

- dichlorobenzene (III)

o

- dichlorobenzene (IV)

p

- dichlorobenzene :

View answer and explanation

Correct option: B

IV < I < II < III

The correct answer is B (IV < I < II < III) because dipole moment increases as follows: para < toluene < meta < ortho due to vector addition of bond dipoles.

205 Hard

205 of 494

Understanding +4 | -1

Choose the correct sequence of

{T}

and

{F}

for following statements. Here

{T}

stands for true statement and

{F}

stands for false statement.

(i) Formal charge in the Lewis structure helps in keeping track of the valence electrons in the molecule.

(ii) Formal charge indicates the actual charge separation within the molecule.

(iii) Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures.

View answer and explanation

Correct option: B

{T} F {T}

Option B (

{T} F {T}

) is correct because formal charge helps track valence electrons and select the lowest energy structure, but it doesn't represent actual charge separation.

206 Easy

206 of 494

Remembering +4 | -1

Dipole moment is usually expressed in Debye Units (D). The conversion factor is

View answer and explanation

Correct option: D

1 \mathrm{~D}=3.33564 \times 10^{-30} \mathrm{~C} \mathrm{m}

Option D (

1 \mathrm{~D}=3.33564 \times 10^{-30} \mathrm{~C} \mathrm{m}

) is correct because it represents the standard conversion factor between Debye units and Coulomb-meters.

207 Medium

207 of 494

Remembering +4 | -1

Polarization of an anion is maximum by high charged cation

View answer and explanation

Correct option: A

Polarization of an anion is maximum by high charged cation

Option A (Polarization of an anion is maximum by high charged cation) is correct because, according to Fajans' rules, a higher charge on the cation increases the polarization of the anion.

208 Easy

208 of 494

Analyzing +4 | -1

The correct order of strength of H-bond

View answer and explanation

Correct option: B

HF > {H_2}{O_2} > {H_2}O > {H_2}s

Option B is correct because the strength of the H-bond depends on the electronegativity of the atom bonded to hydrogen. Higher electronegativity leads to stronger H-bonds.

209 Medium

209 of 494

Analyzing +4 | -1

Consider the following statements.

I. In

F_2

molecule

F - F\,\,\,\sigma - {\text{bond}}

is formed by axial overlapping of two p-orbitals.

II. In

O_2

molecule one

\sigma - {\text{bond}}

is formed by axial overlap of p-orbitals and one

\pi - {\text{bond}}

is formed by lateral overlap of p-orbitals.

III. In

N_2

molecule one

\sigma

and two

\pi - {\text{bonds,}}

i.e., all the three bonds are formed by lateral overlap of p-orbitals.

Which of the above statements are incorrect?

View answer and explanation

Correct option: C

Only III

Option C (Only III) is correct because in

N_2

molecule, one sigma bond is formed by axial overlap and two pi bonds are formed by lateral overlap of p-orbitals.

210 Medium

210 of 494

Analyzing +4 | -1

The bond order and magnetic property of acetylide ion are same as that of

View answer and explanation

Correct option: C

\mathrm{NO}^{+}

The correct answer is NO⁺ because both acetylide ion (C_2^2-) and NO⁺ have a bond order of 3 and are diamagnetic.

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

Showing 181-210 of 494 questions

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