According to molecular orbital theory for ${\rm{O}}_2^ + $:

Bond order is more than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is paramagnetic

Bond order is less than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is paramagnetic

Bond order is less than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is diamagnetic

Bond order is more than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is diamagnetic

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Chemical bonding & molecular structure practice · Mixed difficulty

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361 Hard

361 of 579

Application +4 | -1

The isoelectronic pair is

View answer and explanation

Correct option: D

\mathrm{ClO}_2^ - ,\mathrm{ClF}_2^ +

Option D (

\mathrm{ClO}_2^ - ,\mathrm{ClF}_2^ +

) is correct because both species have the same number of electrons (20).

362 Medium

362 of 579

Understanding +4 | -1

The molecule which has zero moment is

View answer and explanation

Correct option: C

\mathrm{B}{\mathrm{F}_3}

Option C (

\mathrm{B}{\mathrm{F}_3}

) is correct because its trigonal planar geometry causes the bond dipoles to cancel out, resulting in a zero net dipole moment.

363 Easy

363 of 579

Understanding +4 | -1

Among HX, the maximum dipole moment is of:

View answer and explanation

Correct option: A

Option A is correct because HF has the highest electronegativity difference between H and F, leading to the largest dipole moment.

364 Hard

364 of 579

Understanding +4 | -1

Which species does not exist?

View answer and explanation

Correct option: C

{\left( {\mathrm{CC}{\mathrm{l}_6}} \right)^{2 - }}

Option C (

{\left( {\mathrm{CC}{\mathrm{l}_6}} \right)^{2 - }}

) is correct because carbon cannot expand its octet due to the absence of available d-orbitals.

365 Easy

365 of 579

Remembering +4 | -1

The hybridisation state of carbon in fullerene is

View answer and explanation

Correct option: B

s p^2

Option B (

s p^2

) is correct because each carbon atom in fullerene is bonded to three other carbon atoms, similar to graphite, resulting in

sp^2

hybridization.

366 Easy

366 of 579

Remembering +4 | -1

According to Pauling, the atomic orbitals combine to form new set of equivalent atomic orbitals known as

View answer and explanation

Correct option: B

Hybrid orbitals

Option B (Hybrid orbitals) is correct because, according to Pauling's theory, atomic orbitals combine to form a new set of equivalent orbitals called hybrid orbitals.

367 Medium

367 of 579

Understanding +4 | -1

Which among the following contains only two pi bonds ?

View answer and explanation

Correct option: D

{C_2}

Option D (

{C_2}

) is correct because the diatomic carbon molecule (C₂) has two pi bonds and no sigma bonds.

368 Hard

368 of 579

Application +4 | -1

The bond order of

\mathrm{C}_2^ +

is:

View answer and explanation

Correct option: C

3/2

Option C (

3/2

) is correct because

\mathrm{C}_2^ +

has 11 electrons, resulting in a bond order of 1.5.

369 Hard

369 of 579

Analyzing +4 | -1

The correct order of increasing intermolecular hydrogen bond strength is

View answer and explanation

Correct option: C

C{H_4} < HCN < N{H_3}

Option C (

C{H_4} < HCN < N{H_3}

) is correct because

CH_4

cannot form hydrogen bonds,

HCN

can form relatively weak hydrogen bonds, and

NH_3

forms stronger hydrogen bonds.

370 Hard

370 of 579

Understanding +4 | -1

In which of the following pairs attractive force work?

\mathop {{N_A} - {e_A}}\limits_{\text{I}} ,\mathop {{N_A} - {e_B}}\limits_{{\text{II}}} ,\mathop {{N_B} - {e_A}}\limits_{{\text{III}}} ,\mathop {{N_B} - {e_B}}\limits_{{\text{IV}}} ,\mathop {{N_A} - {N_B}}\limits_{\text{V}} ,\mathop {{e_A} - {e_B}}\limits_{{\text{VI}}}

Here,

N_A

and

N_B

represent two nuclei and

e_A

and

e_B

refer to electrons surrounding

N_A

and

N_B

nuclei.

View answer and explanation

Correct option: C

I, II, III and IV

Option C (I, II, III and IV) is correct because attractive forces exist between nuclei and electrons, specifically between

N_A

and

e_A

N_A

and

e_B

N_B

and

e_A

, and

N_B

and

e_B

371 Easy

371 of 579

Understanding +4 | -1

Specify the geometry and hybridisation of N and B atoms in a

1:1

complex of

B{F_3}\,\& \,N{H_3}

View answer and explanation

Correct option: A

N : tetrahedral ,

s{p^3}

B : tetrahedral ,

s{p^3}

Option A (N : tetrahedral ,

sp^3

B : tetrahedral ,

sp^3

) is correct because after forming the complex, both Nitrogen and Boron atoms become tetrahedrally coordinated and

sp^3

hybridized.

372 Easy

372 of 579

Remembering +4 | -1

Shape of methane molecule is

View answer and explanation

Correct option: A

Tetrahedral

Option A (Tetrahedral) is correct because methane (

CH_4

) has a tetrahedral shape due to

sp^3

hybridization of the carbon atom.

373 Medium

373 of 579

Analyzing +4 | -1

In which molecule the central atom does not use

s{p^3}

-hybrid orbitals in its bonding?

View answer and explanation

Correct option: B

\mathrm{BeF}_3^ -

Option B is correct because

\mathrm{BeF}_3^ -

has

sp^2

hybridization, while the others have

sp^3

hybridization.

374 Easy

374 of 579

Analyzing +4 | -1

Which of the following has least bond energy?

View answer and explanation

Correct option: A

Xe{F_2}

Option A (

Xe{F_2}

) is correct because noble gas compounds generally have lower bond energies compared to covalent bonds in other molecules.

375 Medium

375 of 579

Application +4 | -1

The pair of species having identical shape for molecules of both species is

View answer and explanation

Correct option: A

Xe{F_2},IF_2^ -

Option A (

XeF_2, IF_2^ -

) is correct because both species have a linear shape due to the same number of bond pairs and lone pairs around the central atom.

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376 Easy

376 of 579

Understanding +4 | -1

When metals react with non-metals, the metal atoms tend to

View answer and explanation

Correct option: B

Lose electrons

Option B (Lose electrons) is correct because metals tend to lose electrons to achieve a stable electron configuration, forming positive ions in the process.

377 Medium

377 of 579

Analyzing +4 | -1

Consider the following statement
I. When

\mathrm{O}_2

is converted into

\mathrm{O}_2^{2+}

bond order decreases.
II.

\mathrm{O}_2

molecule is paramagnetic because it contains two unpaired electrons in

\mathop \pi \limits^ * 2{p_x}

and

\mathop \pi \limits^ * 2{p_y}

molecular orbitals.
III. The bond length in

\mathrm{NO}

is greater than in

{\text{N}}\mathop {\text{O}}\limits^ +

.
Choose the correct statements.

View answer and explanation

Correct option: C

II and III

Option C (II and III) is correct. Statement II is true because O2 has two unpaired electrons, and statement III is true because NO+ has a higher bond order and thus a shorter bond length than NO.

378 Hard

378 of 579

Application +4 | -1

If the electronegativity difference between two atoms A and B is 2.0, then the percentage of covalent character in the molecule is

View answer and explanation

Correct option: A

0.54

Option A (0.54) is correct because it represents the approximate percentage of covalent character when the electronegativity difference is 2.0, based on the Pauling scale and related empirical formulas.

379 Easy

379 of 579

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 is correct because for a bond angle of 120°, the hybridization is sp2, which has 33% s character.

380 Easy

380 of 579

Understanding +4 | -1

N{H_4}CN

contains

View answer and explanation

Correct option: D

All

Option D (All) is correct because

N{H_4}CN

contains an ionic bond between

N{H_4}^+

and

CN^-

, covalent bonds within the

CN^-

ion, and a dative bond within the

N{H_4}^+

ion.

381 Easy

381 of 579

Understanding +4 | -1

Compounds formed by

s{p^3}{d^2}

-hybridization will have configuration:

View answer and explanation

Correct option: B

Octahedral

Option B (Octahedral) is correct because

s{p^3}{d^2}

hybridization results in six hybrid orbitals, which are arranged in an octahedral geometry.

382 Medium

382 of 579

Understanding +4 | -1

Hydrogen bonding is maximum in

View answer and explanation

Correct option: A

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}

Option A (C2H5OH) is correct because ethanol has a hydroxyl group (-OH) that allows for strong hydrogen bonding.

383 Hard

383 of 579

Analyzing +4 | -1

The pair of molecules forming strongest hydrogen bonds are

View answer and explanation

Correct option: C

\mathrm{H} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}{\mkern 1mu} {\mkern 1mu} and{\mkern 1mu} {\mkern 1mu} \mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}

Option C is correct because both formic acid (H-C(=O)-OH) and acetic acid (CH3-C(=O)-OH) have both a hydrogen bond donor (-OH group) and a hydrogen bond acceptor (carbonyl oxygen), allowing for strong intermolecular hydrogen bonding.

384 Easy

384 of 579

Analyzing +4 | -1

Which one of the following arrangements of molecules is correct on the basis of their dipole moments

View answer and explanation

Correct option: D

N{H_3} > N{F_3} > B{F_3}

Option D (

N{H_3} > N{F_3} > B{F_3}

) is correct because the dipole moment of

NH_3

is higher than

NF_3

due to the direction of bond dipoles and lone pair dipole, and

BF_3

has zero dipole moment due to its symmetrical trigonal planar structure.

385 Easy

385 of 579

Application +4 | -1

According to molecular orbital theory for

\mathrm{O}_2^ +

View answer and explanation

Correct option: B

Bond order is more than

{\mathrm{O}_2}

and

\mathrm{O}_2^ +

is paramagnetic

Option B (Bond order is more than

{\mathrm{O}_2}

and

\mathrm{O}_2^ +

is paramagnetic) is correct because removing an electron from

O_2

to form

O_2^+

increases the bond order and leaves an unpaired electron, making it paramagnetic.

386 Medium

386 of 579

Application +4 | -1

The shape of

\mathrm{ClO}_3^ -

according to VSEPR model is:

View answer and explanation

Correct option: B

Pyramidal

Option B (Pyramidal) is correct because in

ClO_3^-

, Chlorine has 3 bond pairs and 1 lone pair, resulting in a pyramidal shape.

387 Medium

387 of 579

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 a highly charged cation has a greater ability to distort the electron cloud of the anion, leading to maximum polarization.

388 Medium

388 of 579

Remembering +4 | -1

Dipole-dipole interaction energy between stationary polar molecules is proportional to

View answer and explanation

Correct option: C

\frac{1}{{{r^3}}}

Option C is correct because the dipole-dipole interaction energy between stationary polar molecules is inversely proportional to the cube of the distance (r) between them, hence 1/r³.

389 Medium

389 of 579

Remembering +4 | -1

The structure of

P F_5

molecule is

View answer and explanation

Correct option: B

Trigonal bipyramidal

Option B (Trigonal bipyramidal) is correct because

PF_5

has five bonding pairs and no lone pairs, leading to a trigonal bipyramidal geometry.

390 Easy

390 of 579

Analyzing +4 | -1

The hybrid orbital having only 20%

d

– character

View answer and explanation

Correct option: C

S{F_4}

Option C (

S{F_4}

) is correct because it has

sp^3d

hybridization, which means one d orbital is involved in the hybridization, resulting in approximately 20% d-character.

Revision notes and key takeaways

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