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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Easy

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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 31-60 of 124 questions

31 Easy

31 of 124

Understanding +4 | -1

The electronic configuration of the outermost shell of the most electronegative elements is

View answer and explanation

Correct option: A

2 s^2 2 p^5

Option A is correct because the most electronegative element, fluorine, has an electronic configuration of

2s^22p^5

in its outermost shell.

32 Easy

32 of 124

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.

33 Easy

33 of 124

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.

34 Easy

34 of 124

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.

35 Easy

35 of 124

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.

36 Easy

36 of 124

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.

37 Easy

37 of 124

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}

38 Easy

38 of 124

Analyzing +4 | -1

Which among the following has only two

\pi

-bonds and no

\sigma

-bond?

View answer and explanation

Correct option: A

\mathrm{C}_2

Option D is correct because the cyclic hydrocarbon with equal C-C bond lengths between 1.34 and 1.54 Å is benzene, which has a C-C-C bond angle of 120° due to sp2 hybridization.

39 Easy

39 of 124

Understanding +4 | -1

The correct order of repulsion is

View answer and explanation

Correct option: B

L.P - L.P > L.P - B.P > B.P - B.P

Option B (

L.P - L.P > L.P - B.P > B.P - B.P

) is correct because lone pair-lone pair repulsions are the strongest, followed by lone pair-bond pair, and then bond pair-bond pair repulsions.

40 Easy

40 of 124

Analyzing +4 | -1

Which of the following orders are correct?

I. B.E. :

{C}-{C}<{C}={C}<{C} \equiv {C}

II. B.E. :

{H}-{F}>{H}-{Cl}>{H}-{Br}>\mathrm{H}-

I

III. B.E.

: s p-s p>s p^2-s p^2>s p^3-s p^3

IV.

B . E .: {C} \equiv {C}>{C} \equiv {N}>{N} \equiv {N}

View answer and explanation

Correct option: D

I, II and III only

Option D (I, II and III only) is correct because the bond energy order for C≡C, C≡N, and N≡N is C≡C < C≡N < N≡N.

41 Easy

41 of 124

Understanding +4 | -1

Which one has a pyramidal structure

View answer and explanation

Correct option: B

{NH}_3

Option B is correct because ammonia (

{NH}_3

) has a pyramidal structure due to the presence of a lone pair of electrons on the nitrogen atom.

42 Easy

42 of 124

Understanding +4 | -1

The unequal sharing of bonded pair of electrons between two atoms in a molecule gives rise to:

View answer and explanation

Correct option: B

Polar covalent bond

The correct answer is B (Polar covalent bond) because unequal sharing of electrons leads to a dipole moment and a polar covalent bond.

43 Easy

43 of 124

Remembering +4 | -1

Which is present in peroxides?

View answer and explanation

Correct option: C

\mathrm{O}_2^{2 - }

Option C (

\mathrm{O}_2^{2 - }

) is correct because the peroxide ion is represented as O2^2-.

44 Easy

44 of 124

Application +4 | -1

Using MO theory predict which of the following species has the shortest bond length

View answer and explanation

Correct option: A

{O_2}^ +

Option A (

{O_2}^ +

) is correct because it has the highest bond order, leading to the shortest bond length.

45 Easy

45 of 124

Remembering +4 | -1

The

d

-orbital used in

{sp}^3 d

hybridisation forming square pyramidal geometry is:

View answer and explanation

Correct option: B

d_{x^2-y^2}

Option B (

d_{x^2-y^2}

) is correct because in

{sp}^3 d

hybridization forming square pyramidal geometry, the

d_{x^2-y^2}

orbital is involved.

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46 Easy

46 of 124

Application +4 | -1

The order of increasing lattice energy of the following salts is

View answer and explanation

Correct option: B

NaBr < NaCl < BaO < CaO

Option D (

NaBr < NaCl < CaO < BaO

) is incorrect. Lattice energy depends on charge and size; higher charge and smaller size lead to greater lattice energy. The correct order considers both factors.

47 Easy

47 of 124

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.

48 Easy

48 of 124

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.

49 Easy

49 of 124

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°.

50 Easy

50 of 124

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.

51 Easy

51 of 124

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.

52 Easy

52 of 124

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.

53 Easy

53 of 124

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.

54 Easy

54 of 124

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.

55 Easy

55 of 124

Analyzing +4 | -1

Which of the following is the most polar

View answer and explanation

Correct option: C

{CH}_3 {OH}

Option C (

{CH}_3 {OH}

) is the most polar because it has the largest difference in electronegativity between its atoms and a bent molecular geometry, resulting in a significant dipole moment.

56 Easy

56 of 124

Understanding +4 | -1

The shape of

\mathrm{NO}_3^ -

is planar. It is formed by the overlapping of oxygen orbitals with orbitals of nitrogen.

View answer and explanation

Correct option: B

s{p^2}

-hybridized

Option B (

s{p^2}

-hybridized) is correct because in

\mathrm{NO}_3^ -

, the nitrogen atom is

s{p^2}

hybridized, leading to a trigonal planar geometry.

57 Easy

57 of 124

Remembering +4 | -1

\pi

-bond is formed by sideways overlapping of:

View answer and explanation

Correct option: B

p-p orbitals

Option B (p-p orbitals) is correct because a

\pi

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

58 Easy

58 of 124

Remembering +4 | -1

Which type of compounds show high melting and boiling points

View answer and explanation

Correct option: A

Electrovalent compounds

Option A (Electrovalent compounds) is correct because ionic compounds have strong electrostatic forces, leading to high melting and boiling points.

59 Easy

59 of 124

Analyzing +4 | -1

Which of the following molecule stabilizes by removal of electron?

View answer and explanation

Correct option: B

{O_2}

Option B is correct because removing an electron from O₂ increases its bond order from 2 to 2.5, thus increasing stability.

60 Easy

60 of 124

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.

Revision notes and key takeaways

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

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