The $C{\rm{ }}-{\rm{ }}H$ bond distance is the longest in

${{\rm{C}}_2}{{\rm{H}}_2}{\rm{B}}{{\rm{r}}_2}$

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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Mixed difficulty

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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 121-150 of 494 questions

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121 Medium

121 of 494

Understanding +4 | -1

Among the following, the molecule that will have the highest dipole moment is

View answer and explanation

Correct option: D

{HF}

Option D (HF) is correct because dipole moment depends on the electronegativity difference between atoms, and HF has the largest electronegativity difference.

122 Medium

122 of 494

Understanding +4 | -1

Bonded electron pairs present in octahedral

\mathrm{S}{\mathrm{F}_6}

molecule:

View answer and explanation

Correct option: C

6

Option C (6) is correct because in

\mathrm{S}{\mathrm{F}_6}

, sulfur forms six single bonds with six fluorine atoms.

123 Medium

123 of 494

Understanding +4 | -1

Which of the species follows octet rule

View answer and explanation

Correct option: B

\mathrm{N}^{3-}

Option B (

\mathrm{N}^{3-}

) is correct because it has a complete octet (8 valence electrons) in its outermost shell.

124 Easy

124 of 494

Understanding +4 | -1

The high density of water compared to ice is due to

View answer and explanation

Correct option: A

Hydrogen bonding interactions

The correct answer is A. Hydrogen bonding in water leads to a more compact structure than in ice, resulting in higher density.

125 Easy

125 of 494

Understanding +4 | -1

Boron form covalent compound due to

View answer and explanation

Correct option: D

Both (1) and (2)

Option D (Both (1) and (2)) is correct because boron's small size and high ionization energy contribute to its tendency to form covalent compounds.

126 Easy

126 of 494

Understanding +4 | -1

Of the following compounds, the one having a linear structure is

View answer and explanation

Correct option: C

{C}_2 {H}_2

Option C is correct because acetylene (C2H2) has a linear structure due to sp hybridization of the carbon atoms.

127 Medium

127 of 494

Analyzing +4 | -1

Which of the following has minimum bond angle?

View answer and explanation

Correct option: B

O{F_2}

Option B (

OF_2

) has the minimum bond angle because the electronegativity of the side atom (F) is greater than that of the central atom (O), leading to a smaller bond angle.

128 Medium

128 of 494

Analyzing +4 | -1

Which of the following have same shape:

(i)

{ICl}_2^{-}

(ii)

{SO}_2

(iii)

{SnCl}_2

(iv)

{XeF}_2

View answer and explanation

Correct option: C

(i) and (iv)

Option C is correct because both

{ICl}_2^{-}

and

{XeF}_2

have a linear shape.

129 Medium

129 of 494

Remembering +4 | -1

The half of the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals is known as:

View answer and explanation

Correct option: A

Bond order

Option A (Bond order) is correct because bond order is defined as half the difference between the number of electrons in bonding and antibonding molecular orbitals.

130 Medium

130 of 494

Remembering +4 | -1

One would expect the elemental form of

\mathrm{Cs}

at room temperature to be:

View answer and explanation

Correct option: B

A metallic solid

Option B (A metallic solid) is correct because Cesium (Cs) is a metal and exists as a solid at room temperature due to metallic bonding.

131 Medium

131 of 494

Analyzing +4 | -1

Assuming

2 s-2 p

mixing is NOT operative, the paramagnetic species among the following is

View answer and explanation

Correct option: B

{B}_2

Option B (

{B}_2

) is correct. Without 2s-2p mixing, the electronic configuration of

{B}_2

results in two unpaired electrons in the

\pi_{2p}

orbitals, making it paramagnetic.

132 Medium

132 of 494

Analyzing +4 | -1

Octet rule is not valid for the molecule:

View answer and explanation

Correct option: B

{\mathrm{H}_2}\mathrm{O}

Option B (

{\mathrm{H}_2}\mathrm{O}

) is correct because the H-atom in

{\mathrm{H}_2}\mathrm{O}

contains only two electrons, violating the octet rule.

133 Medium

133 of 494

Understanding +4 | -1

Which of the following corresponds unstable molecule? Here

{N}_{{b}}

is number of bonding electrons and

{N}_{{a}}

is number of antibonding electrons.

View answer and explanation

Correct option: D

Both (2) and (3)

Option D (Both (2) and (3)) is correct because an unstable molecule corresponds to a negative or zero bond order, which occurs when

{N}_{{b}} < {N}_{{a}}

{N}_{{a}}={N}_{{b}}

134 Medium

134 of 494

Understanding +4 | -1

Amongst

\mathrm{LiCl}, \mathrm{RbCl}, \mathrm{BeCl}_2

and

\mathrm{MgCl}_2

the compounds with the greatest and the least ionic character, respectively are:

View answer and explanation

Correct option: B

\mathrm{RbCl}

and

\mathrm{BeCl}_2

Option B (

\mathrm{RbCl}

and

\mathrm{BeCl}_2

) is correct because RbCl has the greatest ionic character due to the large size and low charge of Rb+, while BeCl2 has the least due to the small size and high polarizing power of Be2+.

135 Medium

135 of 494

Understanding +4 | -1

Of the following hydrides which has the lowest boiling point

View answer and explanation

Correct option: B

{PH}_3

Option B (

{PH}_3

) is correct because

{PH}_3

does not form hydrogen bonds, and its molecular weight is lower than

{SbH}_3

and

{AsH}_3

, resulting in weaker van der Waals forces and a lower boiling point.

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136 Medium

136 of 494

Analyzing +4 | -1

Which one of the following conversion involve change in both by hybridisation and shape?

View answer and explanation

Correct option: C

{BF}_3 \longrightarrow {BF}_4^{-}

Option C (

{BF}_3 \longrightarrow {BF}_4^{-}

) is correct because the hybridization changes from

sp^2

sp^3

, and the shape changes from trigonal planar to tetrahedral.

137 Medium

137 of 494

Understanding +4 | -1

Which hybridization results non-polar orbitals?

View answer and explanation

Correct option: C

s{p^3}

Option C (

s{p^3}

) is correct because

s{p^3}

hybridization leads to a tetrahedral geometry, which, if all substituents are the same, results in a non-polar molecule due to symmetrical distribution of electron density.

138 Medium

138 of 494

Analyzing +4 | -1

Which one of the following is not paramagetic?

View answer and explanation

Correct option: C

{CO}

Option C is correct because CO has 14 electrons, all of which are paired according to molecular orbital theory, making it diamagnetic, not paramagnetic. The other options (NO, N2+, O2-) have unpaired electrons and are thus paramagnetic.

139 Medium

139 of 494

Analyzing +4 | -1

Statement I:

\mathrm{ClO}_3^{-}

and

\mathrm{SO}_3{ }^{2-}

has three bond pair so according to VSEPR theory it has tetrahedral geometry and

\mathrm{sp}^3

hybridization.

Statement II:

\mathrm{ClO}_3{ }^{-}

and

\mathrm{SO}_3{ }^{2-}

has tetrahedral shape.

View answer and explanation

Correct option: D

Statement I and Statement II both are incorrect.

Option D is correct because both statements regarding the geometry and shape of ClO3- and SO32- are incorrect based on VSEPR theory.

140 Medium

140 of 494

Understanding +4 | -1

Identify the non-polar molecule in the set of compounds given

HCl

HF

{\mathrm{H}_2}

HBr

View answer and explanation

Correct option: A

{\mathrm{H}_2}

Option A (

{\mathrm{H}_2}

) is correct because it is a diatomic molecule composed of two identical atoms, resulting in an equal sharing of electrons and no net dipole moment.

141 Medium

141 of 494

Understanding +4 | -1

The polarising ability of which one of the following is highest?

View answer and explanation

Correct option: A

Small highly positive ion

Option A (Small highly positive ion) is correct because polarizing ability increases with increasing charge and decreasing size of the cation.

142 Medium

142 of 494

Understanding +4 | -1

Glycerol is more viscous than ethanol due to

View answer and explanation

Correct option: C

Many hydrogen bonds per molecule

Option C (Many hydrogen bonds per molecule) is correct because glycerol has three hydroxyl groups, allowing it to form more hydrogen bonds than ethanol, which has only one, leading to higher viscosity.

143 Easy

143 of 494

Understanding +4 | -1

The extent of hydrogen bonding is maximum in:

View answer and explanation

Correct option: C

\mathrm{H}_2 \mathrm{O}

Option C (

\mathrm{H}_2 \mathrm{O}

) is correct because it has the highest extent of hydrogen bonding due to both having two hydrogen atoms and two lone pairs available for bonding.

144 Easy

144 of 494

Remembering +4 | -1

Bond angle in

{H}_2 {O}

is:

View answer and explanation

Correct option: C

104.5^0

Option C (

104.5^0

) is correct because the bond angle in water is reduced from the tetrahedral angle due to lone pair-lone pair repulsion.

145 Hard

145 of 494

Understanding +4 | -1

The

C\mathrm{ }-\mathrm{ }H

bond distance is the longest in

View answer and explanation

Correct option: A

{\mathrm{C}_2}{\mathrm{H}_6}

The correct answer is A (

{\mathrm{C}_2}{\mathrm{H}_6}

) because single bonds (C-H in ethane) are longer than double or triple bonds due to lower s-character.

146 Medium

146 of 494

Understanding +4 | -1

Most covalent halide of aluminium is:

View answer and explanation

Correct option: B

\mathrm{Al}{\mathrm{I}_3}

Option B (

\mathrm{Al}{\mathrm{I}_3}

) is correct because larger anions are easily deformed, making it more covalent according to Fajan's rule.

147 Easy

147 of 494

Analyzing +4 | -1

Which of the following statement is incorrect

View answer and explanation

Correct option: D

{F_2}

molecule, the energy of

\sigma 2{P_z}

is more than

\pi 2{P_x}\,and\,\pi 2{P_y}

The correct answer is D (In

{F_2}

molecule, the energy of

\sigma 2{P_z}

is more than

\pi 2{P_x}\,and\,\pi 2{P_y}

) because for

{O_2}

and

{F_2}

, the energy of

\pi 2{P_x}

and

\pi 2{P_y}

is greater than

\sigma 2{P_z}

148 Medium

148 of 494

Understanding +4 | -1

Which of the following is least ionic?

View answer and explanation

Correct option: C

\mathrm{Ca}{\mathrm{I}_2}

Option C (

\mathrm{CaI_2}

) is correct because as the size of the anion increases, the covalent character increases, and thus ionic character decreases according to Fajan's rules.

149 Medium

149 of 494

Analyzing +4 | -1

In which of the following pairs molecules have bond order three and are isoelectronics

View answer and explanation

Correct option: A

{CN}^{-}, {CO}

Option A is correct because both

{CN}^-

and

{CO}

have a bond order of 3 and are isoelectronic, each having 14 electrons.

150 Medium

150 of 494

Analyzing +4 | -1

Assertion : Nitrogen is unreactive at room temperature but becomes reactive at elevated temperature (on heating or in the presence of catalysts).

Reason : In nitrogen molecule, there is extensive delocalization of electrons.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False

Option C (If Assertion is True but the Reason is False) is correct because while nitrogen is unreactive at room temperature, the reason is due to the high bond dissociation energy of the N≡N triple bond, not electron delocalization.

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