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

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

${\rm{N}}{{\rm{F}}_3}$ is:

Having low value of dipole moment than ${\rm{N}}{{\rm{H}}_3}$

Having more dipole moment than ${\rm{N}}{{\rm{H}}_3}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for JEE CHEMISTRY — Hard

1 Hard

1 of 27

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}

The correct answer is C because carboxylic acids (HCOOH and CH3COOH) form stronger hydrogen bonds due to the higher electronegativity of oxygen and the presence of both a hydrogen bond donor and acceptor.

Solution approach

Correct Answer + Core Reason: $\mathrm{H} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}\mkern 1mu {\mkern 1mu} and\mkern 1mu {\mkern 1mu} \mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}$ . Strongest H-bonds are formed between carboxylic acids due to high electronegativity and presence of both donor and acceptor.

Step-by-Step Solution:

Step 1: Define hydrogen bonding: Hydrogen bonding is an intermolecular force between a hydrogen atom bonded to a highly electronegative atom (N, O, or F) and a lone pair of electrons on another electronegative atom.

Step 2: Analyze option A (SiH4 and SiF6): SiH4 does not have a hydrogen atom bonded to a highly electronegative atom. SiF6 is an ion and does not participate in hydrogen bonding.

Step 3: Analyze option B (CH3-CO-CH3 and CHCl3): Acetone can act as a hydrogen bond acceptor, and chloroform can act as a weak hydrogen bond donor. However, the hydrogen bond is not very strong due to the weak acidity of chloroform.

Step 4: Analyze option C (HCOOH and CH3COOH): Both formic acid and acetic acid have -OH groups, allowing them to act as both hydrogen bond donors and acceptors. The electronegativity of oxygen leads to strong hydrogen bonds.

Step 5: Analyze option D (H2O and H2): Water forms strong hydrogen bonds with itself. H2 is nonpolar and does not participate in hydrogen bonding.

Step 6: Compare the options: Carboxylic acids (option C) form the strongest hydrogen bonds due to the combination of electronegativity and the presence of both donor and acceptor groups.

Why other options are wrong:
- Option A: Silicon compounds do not typically form strong hydrogen bonds.
- Option B: The hydrogen bond between acetone and chloroform is relatively weak.
- Option D: Hydrogen gas does not participate in hydrogen bonding.

Key Concepts:
- Hydrogen Bonding
- Electronegativity
- Intermolecular Forces

Memory Aids:
- Hydrogen bonds require H bonded to N, O, or F.

Quick Tips:
- Look for molecules with -OH or -NH groups for strong hydrogen bonding.

2 Hard

2 of 27

Analyzing +4 | -1

Given below are two statements:

Statement I:

\mathrm{PF}_5

and

\mathrm{BrF}_5

both exhibit

\mathrm{sp}^3 \mathrm{d}

hybridisation.

Statement II: Both

\mathrm{SF}_6

and

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}

exhibit

\mathrm{sp}^3 \mathrm{d}^2

hybridisation.

In the light of the above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: C

Both Statement I and Statement II are false

Option C (Both Statement I and Statement II are false) is correct because the hybridizations of BrF5 and the complex ion are incorrect in the statements.

Solution approach

Correct Answer + Core Reason: Both Statement I and Statement II are false. The hybridizations of BrF5 and the complex ion are incorrect.

Step-by-Step Solution:

Step 1: Determine the hybridization of $\mathrm{PF}_5$ . P has 5 valence electrons, all used for bonding with F atoms. Steric number = 5. Hybridization = $sp^3d$ .

Step 2: Determine the hybridization of $\mathrm{BrF}_5$ . Br has 7 valence electrons, 5 used for bonding with F atoms, leaving 1 lone pair. Steric number = 5 + 1 = 6. Hybridization = $sp^3d^2$ .

Step 3: Determine the hybridization of $\mathrm{SF}_6$ . S has 6 valence electrons, all used for bonding with F atoms. Steric number = 6. Hybridization = $sp^3d^2$ .

Step 4: Determine the hybridization of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ . Co $^{3+}$ has a $d^6$ configuration. With strong field ligands like NH $_3$ , it undergoes inner orbital hybridization, $d^2sp^3$ .

Step 5: Evaluate Statement I: $\mathrm{PF}_5$ has $sp^3d$ hybridization, but $\mathrm{BrF}_5$ has $sp^3d^2$ hybridization. Therefore, Statement I is false.

Step 6: Evaluate Statement II: $\mathrm{SF}_6$ has $sp^3d^2$ hybridization, but $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ has $d^2sp^3$ hybridization. Therefore, Statement II is false.

Why other options are wrong:
- Option A: Statement I is false, so this option is incorrect.
- Option B: Both statements are false, so this option is incorrect.
- Option D: Statement II is false, so this option is incorrect.

Key Concepts:
- Hybridization
- Steric Number
- VSEPR Theory
- Coordination Complexes

Memory Aids:
- Steric Number = Number of bonded atoms + Number of lone pairs

Quick Tips:
- Determine the steric number to find the hybridization.
- Remember the difference between outer and inner orbital complexes.

3 Hard

3 of 27

Application +4 | -1

If the molecule of

HCl

were totally polar, the expected value of dipole moment is

6.12\mathrm{ }D

(dbye), but the experimental value of dipole moment was

1.03\mathrm{ }D

. Calculate the percentage ionic character

View answer and explanation

Correct option: A

17

Option A (17) is correct because the percentage ionic character is calculated by dividing the experimental dipole moment by the theoretical dipole moment and multiplying by 100.

Solution approach

Correct Answer + Core Reason: 17. The percentage ionic character is calculated by comparing the experimental and theoretical dipole moments.

Step-by-Step Solution:

Step 1: Identify the given information: Theoretical dipole moment = $6.12 \mathrm{D}$ , Experimental dipole moment = $1.03 \mathrm{D}$ .

Step 2: Use the formula for percentage ionic character: Percentage ionic character = $\frac{\mathrm{experimental\,\,value\,\,of\,\,DM}}{\mathrm{theoretical\,\,value\,\,of\,\,DM}} \times 100$

Step 3: Substitute the values: Percentage ionic character = $\frac{1.03}{6.12} \times 100$

Step 4: Calculate the percentage ionic character: Percentage ionic character = $0.1683 \times 100 = 16.83 \approx 17$

Step 5: Final Answer: 17

Why other options are wrong:
- Option B: This value is incorrect. It likely results from inverting the ratio of experimental and theoretical dipole moments.
- Option C: This value is incorrect. It likely results from an incorrect calculation.
- Option D: This value is incorrect. It would be correct only if the experimental dipole moment was zero.

Key Concepts:
- Dipole Moment
- Percentage Ionic Character

Memory Aids:
- Remember the formula: % Ionic Character = (Experimental Dipole Moment / Theoretical Dipole Moment) * 100

Quick Tips:
- Make sure to use consistent units for dipole moments.

4 Hard

4 of 27

Analyzing +4 | -1

How many species do not exist?

\begin{aligned} & {PH}_3, {PH}_5, {PCl}_5, {PBr}_5, {SiF}_6^{2-}, {SF}_6,{Al}_2{F}_6, {PbI}_4, \\ & {NCl}_5, {AlCl}_3,{OF}_2, {OH}_2, {OF}_4, {HFO}_4, {FeI}_3, {B}_2 {F}_6 \end{aligned}

View answer and explanation

Correct option: A

8

Option A (8) is correct because there are 8 species among the given list that do not exist due to various chemical reasons such as exceeding the octet rule or instability.

Solution approach

Correct Answer + Core Reason: 8. These species violate chemical principles like the octet rule or have unstable configurations.

Step-by-Step Solution:
Step 1: Identify the species that do not exist based on chemical principles.
Step 2: ${PH}_5$ : Phosphorus cannot form 5 bonds with hydrogen due to size constraints and lack of available d-orbitals for hybridization.
Step 3: ${PbI}_4$ : $Pb^{4+}$ is a strong oxidizing agent and oxidizes $I^-$ to $I_2$ .
Step 4: ${NCl}_5$ : Nitrogen cannot expand its octet to form 5 bonds.
Step 5: ${OF}_4$ : Oxygen cannot expand its octet to form 4 bonds with fluorine.
Step 6: ${HFO}_4$ : This compound is not stable.
Step 7: ${FeI}_3$ : $Fe^{3+}$ is a strong oxidizing agent and oxidizes $I^-$ to $I_2$ .
Step 8: ${Al}_2 {F}_6$ : Exists as $AlF_3$ or $[AlF_6]^{3-}$
Step 9: ${B}_2 {F}_6$ : Exists as $BF_3$
Step 10: Count the number of non-existent species: 8.

Why other options are wrong:
- Option B: 6: Incorrect count of non-existent species.
- Option C: 10: Incorrect count of non-existent species.
- Option D: 14: Incorrect count of non-existent species.

Key Concepts:
- Octet rule
- Stability of compounds
- Oxidation-reduction reactions

Memory Aids:
- Nitrogen and oxygen cannot expand their octets in the second period.

Quick Tips:
- Consider the oxidation states and the ability of elements to expand their octets.

5 Hard

5 of 27

Application +4 | -1

If the molecule of

HCl

were totally polar, the expected value of dipole moment is

6.12\mathrm{ }D

(dbye), but the experimental value of dipole moment was

1.03\mathrm{ }D

. Calculate the percentage ionic character

View answer and explanation

Correct option: A

17

Option A (17) is correct because the percentage ionic character is calculated by dividing the experimental dipole moment by the theoretical dipole moment and multiplying by 100.

Solution approach

Correct Answer + Core Reason: 17. The percentage ionic character is calculated using the ratio of the experimental dipole moment to the theoretical dipole moment.

Step-by-Step Solution:

Step 1: Identify the given information.\nExperimental dipole moment = 1.03 D\nTheoretical dipole moment = 6.12 D

Step 2: Calculate the percentage ionic character.\nPercentage ionic character = $\frac{{\text{{Experimental dipole moment}}}}{{\text{{Theoretical dipole moment}}}} \times 100 = \frac{{1.03}}{{6.12}} \times 100 \approx 16.83$ \%

Step 3: Round to the nearest whole number.\nPercentage ionic character ≈ 17\%

Final Answer: 17%

Why other options are wrong:
- Option B (83): This is incorrect. It might be calculated as (1 - (1.03/6.12)) * 100, which represents the covalent character, not the ionic character.
- Option C (50): This is incorrect as it's an arbitrary value and doesn't result from the correct calculation.
- Option D (Zero): This is incorrect because the experimental dipole moment is non-zero, indicating some ionic character.

Key Concepts:
- Dipole moment
- Percentage ionic character

Memory Aids:
- Percentage ionic character = (Experimental dipole moment / Theoretical dipole moment) × 100

Quick Tips:
- Ensure you are using the experimental and theoretical values correctly in the formula.
- The percentage ionic character represents the degree of ionic character in the bond.

6 Hard

6 of 27

Analyzing +4 | -1

Statement-1:

NO^{+}

is more stable than the

NO

, and

CO^{+}

is also more stable than the

CO

.

Statement-2:

NO^{+}

formed when

NO

has lost an anti-bonding electron, and

CO^{-}

formed when

CO

has lost N.B.M.O.

e^{-}

having anti-bonding character.

View answer and explanation

Correct option: A

Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1.

Option A (Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1) is correct because

NO^+

and

CO^+

are more stable due to the removal of an antibonding electron from NO and a non-bonding molecular orbital electron having antibonding character from CO, respectively.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1. The removal of an antibonding electron increases stability.

Step-by-Step Solution:

Step 1: Analyze Statement 1: $NO^+$ is more stable than $NO$ , and $CO^+$ is also more stable than $CO$ . This statement is true.

Step 2: Analyze Statement 2: $NO^+$ formed when $NO$ has lost an anti-bonding electron, and $CO^+$ formed when $CO$ has lost N.B.M.O. $e^-$ having anti-bonding character. This statement is also true and explains why $NO^+$ and $CO^+$ are more stable.

Step 3: Determine the relationship between the statements: Statement 2 correctly explains Statement 1.

Why other options are wrong:
- Option B: Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. Statement 2 is the correct explanation.
- Option C: Statement- 1 is true, statement-2 is false. Statement 2 is true.
- Option D: Statement- 1 is false, statement-2 is true. Statement 1 is true.

Key Concepts:
- Molecular orbital theory
- Bonding and antibonding electrons
- Stability of molecules and ions

Memory Aids:
- Removing an antibonding electron increases stability.

Quick Tips:
- Draw the molecular orbital diagrams to determine the number of bonding and antibonding electrons.

7 Hard

7 of 27

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.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_6}$ . Single bonds are longer than double or triple bonds.

Step-by-Step Solution:
Step 1: Identify the compounds: Ethane ( ${\mathrm{C}_2}{\mathrm{H}_6}$ ), Ethyne ( ${\mathrm{C}_2}{\mathrm{H}_2}$ ), Dihaloalkane ( ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$ ), Ethene ( ${\mathrm{C}_2}{\mathrm{H}_4}$ ).
Step 2: Determine the hybridization of carbon in each compound: Ethane ( $s{p^3}$ ), Ethyne ( $sp$ ), Dihaloalkane ( ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$ ), Ethene ( $s{p^2}$ ).
Step 3: Relate bond length to hybridization: Higher s-character leads to shorter bond length. The order of s-character is $sp > s{p^2} > s{p^3}$ .
Step 4: Determine the order of C-H bond length: ${\mathrm{C}_2}{\mathrm{H}_6} > {\mathrm{C}_2}{\mathrm{H}_4} > {\mathrm{C}_2}{\mathrm{H}_2}$ .

Why other options are wrong:
- Option B: ${\mathrm{C}_2}{\mathrm{H}_2}$ has a triple bond, which is shorter than a single bond.
- Option C: ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$ contains a triple bond, which is shorter than a single bond.
- Option D: ${\mathrm{C}_2}{\mathrm{H}_4}$ has a double bond, which is shorter than a single bond.

Key Concepts:
- Hybridization
- Bond length
- s-character

Memory Aids:
- More s-character, shorter the bond.

Quick Tips:
- Remember the relationship between hybridization and bond length.

8 Hard

8 of 27

Application +4 | -1

The dipole moment of

HBr

is

1.6 \times {10^{ - 30}}

cm and inter – atomic spacing is

1\,\,\mathop \mathrm{A}\limits^\mathrm{o}

. The

\%

ionic character of

HBr

is

View answer and explanation

Correct option: B

10

Option B (10) is correct because the percentage ionic character is calculated by dividing the actual dipole moment by the theoretical dipole moment and multiplying by 100.

Solution approach

Correct Answer + Core Reason: 10. The percentage ionic character is calculated using the ratio of the observed dipole moment to the theoretical dipole moment.

Step-by-Step Solution:

Step 1: Identify the given information.\nDipole moment of HBr ( $\mu$ ) = $1.6 \times 10^{-30}$ Cm\nInter-atomic spacing (d) = $1 \mathop\mathrm{A}\limits^\mathrm{o} = 1 \times 10^{-10}$ m\nCharge of an electron (e) = $1.6 \times 10^{-19}$ C

Step 2: Calculate the theoretical dipole moment assuming 100% ionic character.\nTheoretical dipole moment = charge × distance = $e \times d = 1.6 \times 10^{-19} C \times 1 \times 10^{-10} m = 1.6 \times 10^{-29}$ Cm

Step 3: Calculate the percentage ionic character.\nPercentage ionic character = $\frac{{\text{{Observed dipole moment}}}}{{\text{{Theoretical dipole moment}}}} \times 100 = \frac{{1.6 \times 10^{-30}}}{{1.6 \times 10^{-29}}} \times 100 = 0.1 \times 100 = 10$ \%

Final Answer: 10%

Why other options are wrong:
- Option A (7): This is incorrect as it likely arises from a miscalculation in the ratio of dipole moments.
- Option C (15): This is incorrect as it likely arises from a miscalculation in the ratio of dipole moments.
- Option D (27): This is incorrect as it likely arises from a miscalculation in the ratio of dipole moments.

Key Concepts:
- Dipole moment
- Percentage ionic character

Memory Aids:
- Percentage ionic character = (Observed dipole moment / Theoretical dipole moment) × 100

Quick Tips:
- Ensure consistent units (convert Angstroms to meters).
- Remember that the theoretical dipole moment assumes complete charge transfer.

9 Hard

9 of 27

Application +4 | -1

A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to

10^{-10} \mathrm{~m}

then the fraction of an electronic charge on each atom will be

View answer and explanation

Correct option: D

25 \%

Option D (

25 \%

) is correct because it is calculated by dividing the dipole moment by the product of bond length and electronic charge, then multiplying by 100.

Solution approach

Correct Answer + Core Reason: $25 \%$ . The fraction of electronic charge is calculated using the formula $\mu = q \times d$ .

Step-by-Step Solution:

Step 1: Define the given values:\nDipole moment, $\mu = 1.2 \text{ D} = 1.2 \times 3.336 \times 10^{-30} \text{ C m}$ \nBond length, $d = 10^{-10} \text{ m}$ \nElectronic charge, $e = 1.602 \times 10^{-19} \text{ C}$

Step 2: Use the formula for dipole moment: $\mu = q \times d$ , where $q$ is the charge.

Step 3: Solve for $q$ : $q = \frac{\mu}{d}$

Step 4: Substitute the given values: $q = \frac{1.2 \times 3.336 \times 10^{-30} \text{ C m}}{10^{-10} \text{ m}} = 1.2 \times 10^{-18} \text{ esu}$ (using the original units in the question)

Step 5: Calculate the fraction of electronic charge: $\text{Fraction} = \frac{q}{e} = \frac{1.2 \times 10^{-18} \text{ esu}}{4.8 \times 10^{-10} \text{ esu}} = \frac{1.2 \times 10^{-18}}{10^{-8}} \text{cm} / 4.8 \times 10^{-10} \text{ esu}$ (using the original units in the question)

Step 6: Calculate the percentage of charge: $\% \text{ charge} = \frac{q}{e} \times 100 = \frac{1.2 \times 10^{-10}}{4.8 \times 10^{-10}} \times 100 = 0.25 \times 100 = 25 \%$

Step 7: Final Answer: $25 \%$

Why other options are wrong:
- **Option A ( $45 \%$ ): This is incorrect due to an error in the calculation, possibly using incorrect units or an incorrect formula.
- Option B ( $55 \%$ ): This is incorrect due to an error in the calculation, possibly using incorrect units or an incorrect formula.
- Option C ( $75 \%$ ): This is incorrect due to an error in the calculation, possibly using incorrect units or an incorrect formula.

Key Concepts:
- Dipole moment
- Electronic charge
- Calculation of fractional charge

Memory Aids:**
- $\mu = q \times d$

Quick Tips:
- Ensure consistent units are used throughout the calculation. Convert all values to SI units before substituting into the formula.
- Remember that 1 Debye (D) = $3.336 \times 10^{-30} \text{ C m}$

10 Hard

10 of 27

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.

Solution approach

Correct Answer + Core Reason: ${[IC{l_4}]^ - }$ . The hybridization is determined by calculating the steric number using the formula $H = \frac{1}{2}(V + M - C + A)$ .

Step-by-Step Solution:

Step 1: Define the formula for calculating hybridization: $H = \frac{1}{2}(V + M - C + A)$ , where V = valence electrons, M = monovalent atoms, C = cationic charge, A = anionic charge.

Step 2: Calculate H for ${[IC{l_2}]^ - }$ : $H = \frac{1}{2}(7 + 2 - 0 + 1) = 5$ . This corresponds to $sp^3d$ hybridization.

Step 3: Calculate H for ${[Br{F_2}]^ - }$ : $H = \frac{1}{2}(7 + 2 - 0 + 1) = 5$ . This corresponds to $sp^3d$ hybridization.

Step 4: Calculate H for ${[IC{l_4}]^ - }$ : $H = \frac{1}{2}(7 + 4 - 0 + 1) = 6$ . This corresponds to $sp^3d^2$ hybridization.

Step 5: Calculate H for ${[I{F_6}]^ - }$ : $H = \frac{1}{2}(7 + 6 - 0 + 1) = 7$ . This corresponds to $sp^3d^3$ hybridization.

Why other options are wrong:
- **Option A ( ${[IC{l_2}]^ - }$ ):** The steric number is 5, leading to $sp^3d$ hybridization, not $sp^3d^2$ .
- **Option B ( ${[Br{F_2}]^ - }$ ):** The steric number is 5, leading to $sp^3d$ hybridization, not $sp^3d^2$ .
- **Option D ( ${[I{F_6}]$

11 Hard

11 of 27

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.

Solution approach

Correct Answer + Core Reason: ${T} F {T}$ . Formal charge helps track valence electrons and select the lowest energy structure, but it doesn't represent actual charge separation.

Step-by-Step Solution:

Step 1: Define formal charge: The charge an atom would have if all the electrons in the covalent bonds were shared equally between atoms.

Step 2: Understand the uses of formal charge:
- Keeping track of valence electrons in a molecule.
- Selecting the most stable Lewis structure.

Step 3: Understand the limitations of formal charge:
- Does not represent actual charge separation within the molecule.

Step 4: Analyze the given statements:
- (i) Formal charge helps in keeping track of valence electrons: True.
- (ii) Formal charge indicates the actual charge separation: False.
- (iii) Formal charges help in the selection of the lowest energy structure: True.

Step 5: Final Answer: ${T} F {T}$

Why other options are wrong:
- Option A: Statement (ii) is false.
- Option C: Statement (ii) is false.
- Option D: Statement (i) is true.

Key Concepts:
- Formal charge
- Lewis structure
- Valence electrons

Memory Aids:
- Formal Charge = (Valence Electrons) - (Non-bonding Electrons) - (1/2 Bonding Electrons)

Quick Tips:
- Use formal charges to evaluate the stability of different Lewis structures.

12 Hard

12 of 27

Analyzing +4 | -1

Which of the following are isoelectronic and isostructural:

NO_3^ - ,CO_3^{2 - },ClO_3^ - ,S{O_3}

View answer and explanation

Correct option: A

NO_3^ - ,\,CO_3^{2 - }

Option A is correct.

NO_3^-

and

CO_3^{2-}

are isoelectronic (same number of electrons) and isostructural (same shape - trigonal planar).

Solution approach

Correct Answer + Core Reason: $NO_3^ - ,\,CO_3^{2 - }$ . Both ions have the same number of electrons and a trigonal planar structure.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each species:
- $NO_3^-$ : 5 (N) + 3 * 6 (O) + 1 (charge) = 24 electrons
- $CO_3^{2-}$ : 4 (C) + 3 * 6 (O) + 2 (charge) = 24 electrons
- $ClO_3^-$ : 7 (Cl) + 3 * 6 (O) + 1 (charge) = 26 electrons
- $SO_3$ : 6 (S) + 3 * 6 (O) = 24 electrons

Step 2: Determine the structure of each species:
- $NO_3^-$ : Trigonal planar
- $CO_3^{2-}$ : Trigonal planar
- $ClO_3^-$ : Pyramidal
- $SO_3$ : Trigonal planar

Step 3: Identify isoelectronic species (same number of electrons):
- $NO_3^-$ , $CO_3^{2-}$ , and $SO_3$ are isoelectronic (24 electrons).
- $ClO_3^-$ is not isoelectronic with the others (26 electrons).

Step 4: Identify isostructural species (same structure):
- $NO_3^-$ , $CO_3^{2-}$ and $SO_3$ are trigonal planar.

Step 5: Determine which pair is both isoelectronic and isostructural:
- $NO_3^-$ and $CO_3^{2-}$ are both isoelectronic and isostructural.

Why other options are wrong:
- Option B: $S{O_3},NO_3^ -$ - While both are isoelectronic and trigonal planar, the question asks for the *pair* that fits both criteria.
- Option C: $ClO_3^ - ,CO_3^{2 - }$ - $ClO_3^-$ and $CO_3^{2-}$ are not isoelectronic and have different structures.
- Option D: $CO_3^{2 - },S{O_3}$ - While both are isoelectronic and trigonal planar, the question asks for the *pair* that fits both criteria.

Key Concepts:
- Isoelectronic species
- Isostructural species
- VSEPR theory

Memory Aids:
- Iso- means same

Quick Tips:
- Count the number of valence electrons to determine isoelectronic species.
- Use VSEPR theory to predict the structure of the species.

13 Hard

13 of 27

Analyzing +4 | -1

Consider the following molecules or ions (i)

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

,(ii)

\mathrm{NH}_4^ +

,(iii)

\mathrm{SO}_4^{2 - }

,(iv)

\mathrm{ClO}_4^ -

,(v)

\mathrm{N}{\mathrm{H}_3}

,

s{p^3}

hybridisation is involved in the formation of

View answer and explanation

Correct option: D

(i), (ii), (iii), (iv) and (v)

Option D ((i), (ii), (iii), (iv) and (v)) is correct because all the listed molecules and ions (H2O, NH4+, SO42-, ClO4-, NH3) involve sp3 hybridization in their formation.

Solution approach

Correct Answer + Core Reason: (i), (ii), (iii), (iv) and (v). All listed molecules/ions involve sp3 hybridization.

Step-by-Step Solution:
Step 1: Determine the hybridization of each molecule/ion by counting the number of sigma bonds and lone pairs around the central atom.
Step 2:
(i) H2O: 2 sigma bonds + 2 lone pairs = 4, sp3
(ii) NH4+: 4 sigma bonds + 0 lone pairs = 4, sp3
(iii) SO42-: 4 sigma bonds + 0 lone pairs = 4, sp3
(iv) ClO4-: 4 sigma bonds + 0 lone pairs = 4, sp3
(v) NH3: 3 sigma bonds + 1 lone pair = 4, sp3
Step 3: Since all molecules/ions have 4 electron domains, they are all sp3 hybridized.

Why other options are wrong:
- Option A: Incorrect because it doesn't include all the molecules/ions.
- Option B: Incorrect because it doesn't include all the molecules/ions.
- Option C: Incorrect because it only includes one molecule/ion.

Key Concepts:
- Hybridization
- Sigma Bonds
- Lone Pairs

Memory Aids:
- Steric number (sigma bonds + lone pairs) determines hybridization: 4 = sp3, 3 = sp2, 2 = sp

Quick Tips:
- Count sigma bonds and lone pairs to determine hybridization.

14 Hard

14 of 27

Analyzing +4 | -1

Statement-I: The geometry of

\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3

is tetrahedral.

Statement-II: Carbon of

\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3

is

\mathrm{sp}^2

hybridized.

View answer and explanation

Correct option: C

Statement I is correct, but Statement II is incorrect

Option C is correct because the geometry of

\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3

is tetrahedral, indicating

sp^3

hybridization, not

sp^2

.

Solution approach

Correct Answer + Core Reason: Statement I is correct, but Statement II is incorrect. The methyl carbanion has tetrahedral geometry and $sp^3$ hybridization.

Step-by-Step Solution:
Step 1: Determine the hybridization of carbon in $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3$ .
Step 2: Carbon has 3 bond pairs and 1 lone pair, so the steric number is 4.
Step 3: Steric number 4 corresponds to $sp^3$ hybridization.
Step 4: $sp^3$ hybridization leads to tetrahedral geometry.

Why other options are wrong:
- Option A: Statement I and Statement II both are correct is incorrect because Statement II is incorrect.
- Option B: Statement I and Statement II both are incorrect is incorrect because Statement I is correct.
- Option C: Statement I is correct, but Statement II is incorrect is correct.
- Option D: Statement I is incorrect, but Statement II is correct is incorrect because Statement I is correct.

Key Concepts:
- Hybridization.
- Molecular geometry.
- VSEPR theory.

Memory Aids:
- Steric number = number of bond pairs + number of lone pairs.
- Steric number 4 corresponds to $sp^3$ hybridization and tetrahedral geometry.

Quick Tips:
- Remember the relationship between steric number, hybridization, and geometry.

15 Hard

15 of 27

Understanding +4 | -1

Which of the following shows minimum bond angle?

View answer and explanation

Correct option: D

{\mathrm{H}_2}\mathrm{Te}

Option D (

{\mathrm{H}_2}\mathrm{Te}

) is correct because as we go down the group, the electronegativity of the central atom decreases, leading to a smaller bond angle.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{Te}$ . Bond angle decreases down the group due to decreasing electronegativity.

Step-by-Step Solution:

Step 1: Recognize that all the given molecules have a bent shape due to $s{p^3}$ hybridization and two lone pairs on the central atom.

Step 2: Recall that bond angle decreases as the electronegativity of the central atom decreases.

Step 3: The electronegativity of the central atoms decreases in the order: O > S > Se > Te.

Step 4: Therefore, the bond angle decreases in the order: ${\mathrm{H}_2}\mathrm{O} > {\mathrm{H}_2}\mathrm{S} > {\mathrm{H}_2}\mathrm{Se} > {\mathrm{H}_2}\mathrm{Te}$ .

Step 5: ${\mathrm{H}_2}\mathrm{O}$ has a bond angle of approximately 104.5 degrees.

Step 6: ${\mathrm{H}_2}\mathrm{S}$ has a bond angle of approximately 92 degrees.

Step 7: ${\mathrm{H}_2}\mathrm{Se}$ has a bond angle of approximately 91 degrees.

Step 8: ${\mathrm{H}_2}\mathrm{Te}$ has a bond angle of approximately 90 degrees.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\mathrm{O}$ has the largest bond angle among the given options.
- Option B: ${\mathrm{H}_2}\mathrm{Se}$ has a larger bond angle than ${\mathrm{H}_2}\mathrm{Te}$ .
- Option C: ${\mathrm{H}_2}\mathrm{S}$ has a larger bond angle than ${\mathrm{H}_2}\mathrm{Te}$ .

Key Concepts:
- Bond angle
- Electronegativity
- Hybridization

Memory Aids:
- Remember that bond angle decreases with decreasing electronegativity of the central atom.

Quick Tips:
- Know the factors affecting bond angles.

16 Hard

16 of 27

Analyzing +4 | -1

Choose the correct order

View answer and explanation

Correct option: B

\mathrm{CaCl}_2 > \mathrm{CdCl}_2

(Melting point)

Option D (BeSO4 < MgSO4 (solubility)) is correct because the solubility of sulfates of group 2 elements decreases down the group due to decreasing hydration energy.

Solution approach

Correct Answer + Core Reason: $\mathrm{BeSO}_4 < \mathrm{MgSO}_4$ (solubility). Solubility of sulfates decreases down Group 2.

Step-by-Step Solution:

Step 1: Analyze each option:

Step 2: A. $\mathrm{MgO} < \mathrm{MgF}_2$ (Melting point): MgO has a higher melting point due to higher lattice energy (higher charges). Incorrect.

Step 3: B. $\mathrm{CaCl}_2 > \mathrm{CdCl}_2$ (Melting point): Correct. CdCl2 is more covalent due to pseudo inert gas configuration and thus has lower m.pt than CaCl2.

Step 4: C. $\mathrm{NaHCO}_3 > \mathrm{KHCO}_3$ (solubility): Incorrect. NaHCO3 is sparingly soluble due to polymeric H-bonding.

Step 5: D. $\mathrm{BeSO}_4 < \mathrm{MgSO}_4$ (solubility): Correct. For sulfates of Group II, solubility decreases down the group.

Step 6: Therefore, the correct order is $\mathrm{BeSO}_4 < \mathrm{MgSO}_4$ (solubility).

Why other options are wrong:
- Option A: MgO has a higher melting point due to higher lattice energy.
- Option B: CdCl2 is more covalent and has a lower melting point.
- Option C: NaHCO3 is sparingly soluble due to polymeric H-bonding.

Key Concepts:
- Melting point
- Solubility
- Lattice energy
- Hydration energy

Memory Aids:
- Solubility of sulfates decreases down Group 2.

Quick Tips:
- Remember the trends in melting point and solubility for ionic compounds.

17 Hard

17 of 27

Understanding +4 | -1

The maximum possible number of hydrogen bonds is a

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

molecule can participate is

View answer and explanation

Correct option: D

4

Option D (4) is correct because each water molecule can form two hydrogen bonds through its two hydrogen atoms and accept two hydrogen bonds through its two lone pairs on the oxygen atom.

Solution approach

Correct Answer + Core Reason: 4. Each water molecule can participate in a maximum of four hydrogen bonds: two as a donor and two as an acceptor.

Step-by-Step Solution:

Step 1: Recall the structure of a water molecule ( $H_2O$ ). It has two hydrogen atoms and two lone pairs of electrons on the oxygen atom.

Step 2: Understand that hydrogen bonds are formed between a hydrogen atom bonded to a highly electronegative atom (like oxygen) and a lone pair of electrons on another electronegative atom.

Step 3: Each hydrogen atom in a water molecule can form one hydrogen bond with the lone pair of electrons on another water molecule (hydrogen bond donor).

Step 4: Each oxygen atom in a water molecule has two lone pairs of electrons, each of which can accept one hydrogen bond from another water molecule (hydrogen bond acceptor).

Step 5: Therefore, each water molecule can participate in a total of 2 (from hydrogen atoms) + 2 (from lone pairs) = 4 hydrogen bonds.

Why other options are wrong:
- Option A: 1 is incorrect. A water molecule can form more than one hydrogen bond.
- Option B: 2 is incorrect. A water molecule can form more than two hydrogen bonds.
- Option C: 3 is incorrect. A water molecule can form more than three hydrogen bonds.

Key Concepts:
- Hydrogen bonding
- Water molecule ( $H_2O$ )
- Lone pairs of electrons
- Electronegativity

Memory Aids:
- Water can form 4 hydrogen bonds: 2 as a donor and 2 as an acceptor.

Quick Tips:
- Draw the structure of water and consider the number of hydrogen atoms and lone pairs to determine the maximum number of hydrogen bonds.

18 Hard

18 of 27

Analyzing +4 | -1

Barium ions,

\mathrm{CN}^{-}

and

\mathrm{Co}^{+2}

form an ionic complex. If this complex is

75 \%

ionized in aqueous solution with Vant Hoff factor(i) equal to four and paramagnetic moment is found to be

1.73 \mathrm{BM}

(due to spin only) then hybridization state of

\mathrm{Co}(\mathrm{II})

in the complex will be

View answer and explanation

Correct option: D

\mathrm{dsp}^3

Option D (

\mathrm{dsp}^3

) is correct because the complex formed is

\mathrm{Ba}_3[\mathrm{Co}(\mathrm{CN})_5]_2

with one unpaired electron, leading to

\mathrm{dsp}^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{dsp}^3$ . The complex formed is $\mathrm{Ba}_3[\mathrm{Co}(\mathrm{CN})_5]_2$ with one unpaired electron.

Step-by-Step Solution:

Step 1: Determine the number of particles in solution.\nThe van't Hoff factor (i) is 4, but the problem states that the complex is 75% ionized. The problem statement is inconsistent, but we will proceed assuming i=4 is correct for the fully ionized complex. If the complex is $\mathrm{Ba}_x[\mathrm{Co}(\mathrm{CN})_y]$ , then the ionization is:
$\mathrm{Ba}_x[\mathrm{Co}(\mathrm{CN})_y] \rightleftharpoons x \mathrm{Ba}^{2+} + [\mathrm{Co}(\mathrm{CN})_y]^{2x-}$ \nIf the complex is 75% ionized with i=4, it implies that the complex dissociates into 4 ions. However, the problem states i=4, which is used as if the complex is fully ionized. We will assume full ionization.
$\mathrm{Ba}_x[\mathrm{Co}(\mathrm{CN})_y] \rightarrow x \mathrm{Ba}^{2+} + [\mathrm{Co}(\mathrm{CN})_y]^{2x-}$
$\mathrm{i} = 1 + (n-1)\alpha$
$4 = 1 + (n-1)1$
$n = 4$ \nThis means the complex dissociates into 4 ions. Let's assume the complex is of the form $\mathrm{Ba}_x[\mathrm{Co}(\mathrm{CN})_y]$ . Then the dissociation is:
$\mathrm{Ba}_x[\mathrm{Co}(\mathrm{CN})_y] \rightarrow x \mathrm{Ba}^{2+} + [\mathrm{Co}(\mathrm{CN})_y]^{2x-}$ \nThe number of ions is $x + 1 = 4$ , so $x = 3$ . The complex is $\mathrm{Ba}_3[\mathrm{Co}(\mathrm{CN})_y]_2$ and the anion is $[\mathrm{Co}(\mathrm{CN})_y]^{6-}$ .\nSince the charge on Co is +2, we have: +2 + y(-1) = -3, so y = 5. The complex is $[\mathrm{Co}(\mathrm{CN})_5]^{3-}$ and the overall complex is $\mathrm{Ba}_3[\mathrm{Co}(\mathrm{CN})_5]_2$ .

Step 2: Determine the electronic configuration of $\mathrm{Co}^{2+}$ .
$\mathrm{Co}^{2+}$ has an electronic configuration of [Ar] $3d^7$ .

Step 3: Determine the hybridization of $\mathrm{Co}^{2+}$ in the complex.\nThe magnetic moment is 1.73 BM, which corresponds to one unpaired electron. Since CN $^-$ is a strong field ligand, it will cause pairing of electrons. However, with 7 d-electrons, there will always be one unpaired electron in a low-spin complex. The complex is $[\mathrm{Co}(\mathrm{CN})_5]^{3-}$ .\nSince there are 5 ligands, the hybridization must be $dsp^3$ or $sp^3d$ . Since CN is a strong field ligand, it will cause inner orbital hybridization, so the hybridization is $dsp^3$ .

Why other options are wrong:
- Option A: $\mathrm{sp}^3 \mathrm{~d}$ is incorrect because CN is a strong field ligand.
- Option B: $\mathrm{d}^2 \mathrm{sp}^3$ is incorrect because there are only 5 ligands.
- Option C: $\mathrm{sp}^3 \mathrm{~d}^2$ is incorrect because there are only 5 ligands.

Key Concepts:\nCoordination Complexes, Hybridization, Magnetic Moment, Strong Field Ligands, van't Hoff factor

Memory Aids:\nStrong field ligands cause pairing of electrons.

Quick Tips:\nDetermine the number of ligands and the electronic configuration of the metal ion to determine the hybridization.

19 Hard

19 of 27

Analyzing +4 | -1

The dipole moment of diatomic molecules

{AB}

and

C D

are 10.41 and 10.27 Debye respectively while there bond distances are

2.82 \mathop {\text{A}}\limits^ \circ

and

2.67 \mathop {\text{A}}\limits^ \circ

respectively then

View answer and explanation

Correct option: D

{AB}

has lesser ionic bond character than

{CD}

Option D (

{AB}

has lesser ionic bond character than

{CD}

) is correct because the magnitude of charge calculated for CD is greater than that for AB, indicating a higher ionic character.

Solution approach

Correct Answer + Core Reason: ${AB}$ has lesser ionic bond character than ${CD}$ . The ionic character is determined by the magnitude of charge, which is higher for CD.

Step-by-Step Solution:

Step 1: Define the dipole moment formula:
$\mu = q \times d$ , where $\mu$ is the dipole moment, $q$ is the magnitude of the charge, and $d$ is the bond distance.

Step 2: Calculate the magnitude of charge for AB:
$q_{AB} = \frac{\mu_{AB}}{d_{AB}} = \frac{10.41 \text{ Debye}}{2.82 \mathop{\text{A}}\limits^\circ}$

Step 3: Calculate the magnitude of charge for CD:
$q_{CD} = \frac{\mu_{CD}}{d_{CD}} = \frac{10.27 \text{ Debye}}{2.67 \mathop{\text{A}}\limits^\circ}$

Step 4: Perform the calculations:
$q_{AB} = \frac{10.41}{2.82} = 3.69 \text{ esu}$ (approximately)
$q_{CD} = \frac{10.27}{2.67} = 3.85 \text{ esu}$ (approximately)

Step 5: Compare the magnitudes of charge:\nSince $q_{CD} > q_{AB}$ , the molecule CD has a greater ionic character than molecule AB.

Final Answer: ${AB}$ has lesser ionic bond character than ${CD}$

Why other options are wrong:
- Option A (bonding is nearly covalent in both the molecules): This is incorrect because the dipole moments are given, implying some degree of ionic character. If the bonding were purely covalent, the dipole moment would be zero.
- **Option B (bonding is $100 \%$ ionic in both the molecules): This is incorrect because the question implies a comparison of ionic character, suggesting that neither molecule is 100% ionic.
- Option C ( ${AB}$ has more ionic character than ${CD}$ ): This is incorrect because the calculated magnitude of charge is lower for AB than for CD.

Key Concepts:
- Dipole moment
- Ionic character
- Bond distance

Memory Aids:
- Remember that a higher dipole moment and shorter bond length generally indicate a greater ionic character.

Quick Tips:**
- Always use consistent units when calculating dipole moments and charges.
- The larger the charge separation, the greater the ionic character.

20 Hard

20 of 27

Analyzing +4 | -1

Among the following, the species with identical bond order are

View answer and explanation

Correct option: C

\mathrm{O}_2^{2-}

and

\mathrm{B}_2

The correct answer is C (

\mathrm{O}_2^{2-}

and

\mathrm{B}_2

) because both species have a bond order of 1.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^{2-}$ and $\mathrm{B}_2$ . Both have a bond order of 1.

Step-by-Step Solution:

Step 1: Define the formula for bond order: B.O. = $\frac{1}{2}(N_b - N_a)$ , where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.

Step 2: Calculate the bond order for each species:
- CO (14 electrons): Electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 = \pi 2p_y^2 \sigma 2p_z^2$ . B.O. = $\frac{1}{2}(10 - 4) = 3$
- $\mathrm{O}_2^{2-}$ (18 electrons): Electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^2 = \pi^* 2p_y^2$ . B.O. = $\frac{1}{2}(10 - 8) = 1$
- $\mathrm{O}_2^{-}$ (17 electrons): Electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^2 = \pi^* 2p_y^1$ . B.O. = $\frac{1}{2}(10 - 7) = 1.5$
- $\mathrm{B}_2$ (10 electrons): Electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^1 = \pi 2p_y^1$ . B.O. = $\frac{1}{2}(6 - 4) = 1$
- $\mathrm{N}_2^{+}$ (13 electrons): Electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 = \pi 2p_y^2 \sigma 2p_z^1$ . B.O. = $\frac{1}{2}(9 - 4) = 2.5$

Step 3: Identify the species with identical bond orders. $\mathrm{O}_2^{2-}$ and $\mathrm{B}_2$ both have a bond order of 1.

Why other options are wrong:
- Option A: CO has a bond order of 3, while $\mathrm{O}_2^{2-}$ has a bond order of 1.
- Option B: $\mathrm{O}_2^{-}$ has a bond order of 1.5, while CO has a bond order of 3.
- Option D: CO has a bond order of 3, while $\mathrm{N}_2^{+}$ has a bond order of 2.5.

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation
- Electronic Configuration

Memory Aids:
- Remember the electronic configurations of common diatomic molecules and ions.

Quick Tips:
- Calculate the number of bonding and antibonding electrons accurately.
- Use the bond order formula correctly.

21 Hard

21 of 27

Analyzing +4 | -1

The species having bond order different from that in

CO

is

View answer and explanation

Correct option: A

\mathrm{N}{\mathrm{O}^ - }

The correct answer is A (NO-) because NO- has a bond order of 2, while CO, NO+, CN-, and N2 all have a bond order of 3.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{O}^ - }$ . NO- has a bond order of 2, which is different from the bond order of 3 in CO.

Step-by-Step Solution:

Step 1: Calculate the bond order of CO: CO has 14 electrons. The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 (\sigma 2p)^2$ . Bond order = (10 - 4) / 2 = 3.

Step 2: Calculate the bond order of NO-: NO- has 16 electrons. The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p)^2 (\pi 2p)^4 (\pi^* 2p)^2$ . Bond order = (10 - 6) / 2 = 2.

Step 3: Calculate the bond order of NO+: NO+ has 14 - 1 = 13 electrons. The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 (\sigma 2p)^1$ . However, a more stable configuration is achieved by removing one electron from the antibonding $\pi^*2p$ orbitals, resulting in a bond order of 3.

Step 4: Calculate the bond order of CN-: CN- has 6 + 7 + 1 = 14 electrons. The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 (\sigma 2p)^2$ . Bond order = (10 - 4) / 2 = 3.

Step 5: Calculate the bond order of N2: N2 has 14 electrons. The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 (\sigma 2p)^2$ . Bond order = (10 - 4) / 2 = 3.

Why other options are wrong:
- Option B: NO+ has a bond order of 3, the same as CO.
- Option C: CN- has a bond order of 3, the same as CO.
- Option D: N2 has a bond order of 3, the same as CO.

Key Concepts:
- Molecular orbital theory
- Bond order
- Electronic configuration

Memory Aids:
- Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2

Quick Tips:
- Determine the number of electrons in the molecule/ion first.

22 Hard

22 of 27

Understanding +4 | -1

\mathrm{N}{\mathrm{F}_3}

is:

View answer and explanation

Correct option: C

Having low value of dipole moment than

\mathrm{N}{\mathrm{H}_3}

Option C (Having low value of dipole moment than

\mathrm{N}{\mathrm{H}_3}

) is correct because in

\mathrm{N}{\mathrm{F}_3}

, the bond dipoles and the lone pair dipole partially cancel each other, resulting in a lower dipole moment compared to

\mathrm{N}{\mathrm{H}_3}

.

Solution approach

Correct Answer + Core Reason: Having low value of dipole moment than $\mathrm{N}{\mathrm{H}_3}$ . The bond dipoles and lone pair dipole partially cancel in $\mathrm{N}{\mathrm{F}_3}$ .

Step-by-Step Solution:

Step 1: Draw the structures of $\mathrm{N}{\mathrm{F}_3}$ and $\mathrm{N}{\mathrm{H}_3}$ . Both have a pyramidal shape.

Step 2: In $\mathrm{N}{\mathrm{H}_3}$ , the bond dipoles of the three N-H bonds and the dipole due to the lone pair on nitrogen all point in the same direction, resulting in a large dipole moment.

Step 3: In $\mathrm{N}{\mathrm{F}_3}$ , the bond dipoles of the three N-F bonds point away from the nitrogen atom, while the dipole due to the lone pair on nitrogen points in the opposite direction. This causes partial cancellation of the dipoles.

Step 4: Therefore, $\mathrm{N}{\mathrm{F}_3}$ has a lower dipole moment than $\mathrm{N}{\mathrm{H}_3}$ .

Why other options are wrong:
- Option A (Non-polar compound): $\mathrm{N}{\mathrm{F}_3}$ is polar because the bond dipoles do not completely cancel out.
- Option B (Electrovalent compound): $\mathrm{N}{\mathrm{F}_3}$ is a covalent compound.
- **Option D (Having more dipole moment than $\mathrm{N}{\mathrm{H}_3}$ ):** $\mathrm{N}{\mathrm{F}_3}$ has a lower dipole moment than $\mathrm{N}{\mathrm{H}_3}$ due to the opposing bond dipoles and lone pair dipole.

Key Concepts:
- Dipole moment
- Bond dipoles
- Lone pair dipole
- Molecular geometry

Memory Aids:
- Remember that the direction of bond dipoles depends on the electronegativity difference between the atoms.

Quick Tips:
- Draw the structure and consider the direction of bond dipoles and lone pair dipoles to determine the overall dipole moment.

23 Hard

23 of 27

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 the hydrogen bond strength increases with the electronegativity difference between H and the other atom, and CH4 cannot form hydrogen bonds.

Solution approach

Correct Answer + Core Reason: $C{H_4} < HCN < N{H_3}$ . The hydrogen bond strength increases with the electronegativity difference.

Step-by-Step Solution:

Step 1: Determine if each molecule can form hydrogen bonds. Hydrogen bonds form between a hydrogen atom bonded to a highly electronegative atom (N, O, or F) and another electronegative atom.

Step 2: CH4 (methane) cannot form hydrogen bonds because carbon is not electronegative enough.

Step 3: HCN (hydrogen cyanide) can form hydrogen bonds because hydrogen is bonded to carbon, and the nitrogen atom can act as a hydrogen bond acceptor.

Step 4: NH3 (ammonia) can form hydrogen bonds because hydrogen is bonded to nitrogen, which is highly electronegative. The hydrogen bonds in ammonia are stronger than those in hydrogen cyanide because nitrogen is more electronegative than carbon.

Step 5: Arrange the molecules in order of increasing hydrogen bond strength: CH4 < HCN < NH3.

Why other options are wrong:
- Option A: HCN < H2O < NH3. Water has stronger hydrogen bonds than ammonia due to the higher electronegativity of oxygen, but this option does not include water.
- Option B: HCN < CH4 < NH3. Methane cannot form hydrogen bonds, so it should be the weakest.
- Option D: CH4 < NH3 < HCN. Hydrogen bonds in ammonia are stronger than those in hydrogen cyanide.

Key Concepts:
- Hydrogen bonding
- Electronegativity
- Intermolecular forces

Memory Aids:
- Hydrogen bonds require H bonded to N, O, or F.
- Higher electronegativity difference leads to stronger hydrogen bonds.

Quick Tips:
- Remember the electronegativity trend: F > O > N > Cl > C > H.

24 Hard

24 of 27

Application +4 | -1

The dipole moment of

{HBr}

is

2.60 \times 10^{-30} {~cm}

and the interatomic spacing is

1.41 \mathop {\text{A}}\limits^ \circ

. What is the percent ionic character of

{HBr}

?

View answer and explanation

Correct option: B

11.5 \%

Option B (11.5 %) is correct because the percent ionic character is calculated by dividing the observed dipole moment by the calculated dipole moment and multiplying by 100.

Solution approach

Correct Answer + Core Reason: 11.5 %. The percent ionic character is calculated using the observed and calculated dipole moments.

Step-by-Step Solution:

Step 1: Calculate the calculated dipole moment ( $\mu_{cal}$ ).
$\mu_{cal} = q \times d$ , where $q$ is the charge and $d$ is the distance.
$q = 1.602 \times 10^{-19} \text{ C}$ (elementary charge)
$d = 1.41 \mathop{\text{A}}\limits^\circ = 1.41 \times 10^{-10} \text{ m}$
$\mu_{cal} = (1.602 \times 10^{-19} \text{ C}) \times (1.41 \times 10^{-10} \text{ m}) = 2.25882 \times 10^{-29} \text{ C m}$

Step 2: Calculate the percent ionic character.
$\text{% Ionic Character} = \frac{\mu_{obs}}{\mu_{cal}} \times 100$
$\mu_{obs} = 2.60 \times 10^{-30} \text{ C m}$
$\text{% Ionic Character} = \frac{2.60 \times 10^{-30} \text{ C m}}{2.25882 \times 10^{-29} \text{ C m}} \times 100 = \frac{2.60}{22.5882} \times 100 = 0.1151 \times 100 = 11.51 \% \approx 11.5 \%$

Why other options are wrong:
- Option A (50 %): This is incorrect due to a significant error in the calculation, likely involving incorrect units or a misunderstanding of the formula.
- Option C (4.01 %): This is incorrect due to a calculation error, possibly involving an incorrect value for the elementary charge or the bond distance.
- Option D (1.19 %): This is incorrect due to a calculation error, possibly involving an incorrect value for the elementary charge or the bond distance.

Key Concepts:
- Dipole Moment
- Percent Ionic Character
- Elementary Charge

Memory Aids:
- $\text{% Ionic Character} = \frac{\mu_{obs}}{\mu_{cal}} \times 100$
- $\mu = q \times d$

Quick Tips:
- Ensure that the units are consistent (C m) when calculating the dipole moment and percent ionic character.
- Remember the value of the elementary charge ( $1.602 \times 10^{-19} \text{ C}$ ).
- Convert Angstroms to meters before calculation.

25 Hard

25 of 27

Analyzing +4 | -1

Using MOT, compare

{O}_2^{+}

and

{O}_2^{-}

species and choose the incorrect option

View answer and explanation

Correct option: C

{O}_2^{+}

is diamagnetic while

{O}_2^{-}

is paramagnetic

Option C is correct because both O2+ and O2- are paramagnetic due to the presence of unpaired electrons in their molecular orbital configurations.

Solution approach

Correct Answer + Core Reason: ${O}_2^{+}$ is diamagnetic while ${O}_2^{-}$ is paramagnetic. Both species are paramagnetic due to unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the molecular orbital configuration of $O_2^+$ .
$O_2^+$ has 15 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^1$ . It has one unpaired electron in the $\pi_{2p}^*$ orbital.

Step 2: Determine the molecular orbital configuration of $O_2^-$ .
$O_2^-$ has 17 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^3$ . It has one unpaired electron in the $\pi_{2p}^*$ orbital.

Step 3: Calculate the bond order for each species.\nFor $O_2^+$ , bond order = (10 - 5) / 2 = 2.5\nFor $O_2^-$ , bond order = (10 - 7) / 2 = 1.5

Step 4: Determine the magnetic properties of each species.\nBoth $O_2^+$ and $O_2^-$ have unpaired electrons, so they are both paramagnetic.

Step 5: Compare the bond orders and stability.
$O_2^+$ has a higher bond order (2.5) than $O_2^-$ (1.5), so $O_2^+$ is more stable.

Step 6: Identify the incorrect option.\nThe incorrect option is C, because both species are paramagnetic.

Why other options are wrong:
- Option A: $O_2^+$ has a higher bond order than $O_2^-$ , which is correct.
- Option B: $O_2^-$ is less stable than $O_2^+$ , which is correct.
- Option D: Both $O_2^+$ and $O_2^-$ are paramagnetic, which is correct.

Key Concepts:
- Molecular orbital theory (MOT)
- Bond order
- Paramagnetism and diamagnetism

Memory Aids:
- Remember that unpaired electrons lead to paramagnetism.

Quick Tips:
- Draw the molecular orbital diagrams to determine the electronic configurations and magnetic properties of diatomic species.

26 Hard

26 of 27

Application +4 | -1

The bond between atoms of two elements of atomic number 37 and 53 is:

View answer and explanation

Correct option: B

Ionic

Option B (Ionic) is correct because elements with atomic numbers 37 (Rubidium) and 53 (Iodine) form an ionic bond due to their large electronegativity difference.

Solution approach

Correct Answer + Core Reason: Ionic. Elements with significantly different electronegativities form ionic bonds.

Step-by-Step Solution:

Step 1: Identify the elements: Atomic number 37 is Rubidium (Rb), and atomic number 53 is Iodine (I).

Step 2: Determine their positions in the periodic table: Rb is an alkali metal (Group 1), and I is a halogen (Group 17).

Step 3: Consider their electronegativities: Rb is highly electropositive, and I is highly electronegative.

Step 4: Predict the type of bond: Due to the large difference in electronegativity, Rb and I will form an ionic bond.

Why other options are wrong:
- Option A: Covalent bonds are formed between elements with similar electronegativities.
- Option C: Coordinate bonds are formed when one atom donates both electrons to form a bond.
- Option D: Metallic bonds are formed between metal atoms.

Key Concepts:
- Electronegativity
- Ionic bond
- Covalent bond

Memory Aids:
- Metals and non-metals tend to form ionic bonds.

Quick Tips:
- Consider the electronegativity difference between the elements to predict the type of bond.

27 Hard

27 of 27

Understanding +4 | -1

Which has the minimum bond energy?

View answer and explanation

Correct option: C

I - I

Option C (

I - I

) is correct because bond energy decreases as the size of the atom increases due to reduced overlap of atomic orbitals.

Solution approach

Correct Answer + Core Reason: $I - I$ . Bond energy decreases with increasing atomic size due to reduced orbital overlap.

Step-by-Step Solution:

Step 1: Understand the relationship between bond energy and atomic size.

Step 2: Larger atoms have more diffuse electron clouds, leading to less effective orbital overlap.

Step 3: Less effective orbital overlap results in weaker bonds and lower bond energy.

Step 4: The atomic sizes increase in the order: H < Br < I.

Step 5: Therefore, the bond energies decrease in the order: H-H > H-Br > H-I > I-I.

Step 6: I-I has the minimum bond energy.

Why other options are wrong:
- **Option A ( $\mathrm{H} - \mathrm{Br}$ ): Higher bond energy than I-I.
- Option B ( $\mathrm{H} - I$ ): Higher bond energy than I-I.
- Option D ( $\mathrm{H - H}$ ): Highest bond energy among the given options.

Key Concepts:
- Bond energy
- Atomic size
- Orbital overlap

Memory Aids:
- Larger atoms = weaker bonds.

Quick Tips:**
- Consider the atomic size and electronegativity when comparing bond energies.

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