${{\rm{B}}_{10}}{{\rm{C}}_2}{{\rm{H}}_{12}}$ is isoelectronic with

${{\rm{B}}_{12}}{\rm{H}}_{12}^{2 - }$

${{\rm{B}}_{12}}{{\rm{H}}_{12}}$

${{\rm{B}}_{12}}{\rm{H}}_{12}^ + $

${{\rm{B}}_{12}}{\rm{H}}_{12}^{2 + }$

${\rm{N}}{{\rm{F}}_3}$ 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 NEET UG CHEMISTRY — Hard, Page 2

31 Hard

31 of 80Question permalink

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 both statements are true and statement 2 correctly explains why

NO^+

and

CO^+

are more stable than

NO

and

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. Removal of antibonding electron increases stability.

Step-by-Step Solution:

Step 1: Analyze Statement-1: $NO^+$ is more stable than $NO$ , and $CO^+$ is more stable than $CO$ .

Step 2: $NO$ has 15 electrons. $NO^+$ has 14 electrons. The electron removed is from an antibonding orbital, increasing stability.

Step 3: $CO$ has 14 electrons. $CO^+$ has 13 electrons. The electron removed is from a non-bonding molecular orbital with antibonding character, increasing stability.

Step 4: 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.

Step 5: Statement-2 correctly explains why the removal of an electron from an antibonding orbital (or a non-bonding orbital with antibonding character) increases stability.

Why other options are wrong:
- Option B: Statement-2 is the correct explanation for Statement-1.
- Option C: Both statements are true.
- Option D: Both statements are true.

Key Concepts:
- Molecular Orbital Theory (MOT)
- Bond Order and Stability
- Antibonding Orbitals

Memory Aids:
- Removing electrons from antibonding orbitals increases stability.

Quick Tips:
- Consider the effect of removing electrons from bonding or antibonding orbitals on the bond order and stability.

32 Hard

32 of 80Question permalink

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 actual dipole moment by the theoretical dipole moment assuming full charge separation.

Solution approach

Correct Answer + Core Reason: $25 \%$ . The fraction of electronic charge is calculated by comparing the actual dipole moment to the theoretical dipole moment.

Step-by-Step Solution:

Step 1: Convert the dipole moment from Debye (D) to Coulomb-meters (C m).
1 D = $3.336 \times 10^{-30}$ C m
$\mu = 1.2 \text{ D} = 1.2 \times 3.336 \times 10^{-30} \text{ C m} = 4.0032 \times 10^{-30} \text{ C m}$

Step 2: Calculate the theoretical dipole moment assuming full charge separation (i.e., one full electronic charge).
$q = 1.602 \times 10^{-19} \text{ C}$
$d = 10^{-10} \text{ m}$
$\mu_{theoretical} = q \times d = (1.602 \times 10^{-19} \text{ C}) \times (10^{-10} \text{ m}) = 1.602 \times 10^{-29} \text{ C m}$

Step 3: Calculate the fraction of electronic charge.
$\text{Fraction} = \frac{\mu}{\mu_{theoretical}} = \frac{4.0032 \times 10^{-30} \text{ C m}}{1.602 \times 10^{-29} \text{ C m}} = 0.25$

Step 4: Convert the fraction to percentage.
$\text{Percentage} = 0.25 \times 100 \% = 25 \%$

Why other options are wrong:
- Option A: 45% is incorrect. This likely arises from an error in the calculation or an incorrect conversion factor.
- Option B: 55% is incorrect. This likely arises from an error in the calculation or an incorrect conversion factor.
- Option C: 75% is incorrect. This likely arises from an error in the calculation or an incorrect conversion factor.

Key Concepts:
- Dipole moment calculation
- Conversion between Debye and Coulomb-meters
- Electronic charge

Memory Aids:
- Remember the formula: $\mu = q \times d$

Quick Tips:
- Pay attention to units and conversions.
- Ensure you are using the correct value for the electronic charge.

33 Hard

33 of 80Question permalink

Analyzing +4 | -1

Which of the following species has highest bond energy?

View answer and explanation

Correct option: B

{O}_2^{+}

Option B (

{O}_2^{+}

) is correct because it has the highest bond order (2.5) among the given species, which directly correlates with bond energy.

Solution approach

Correct Answer + Core Reason: ${O}_2^{+}$ . Higher bond order implies higher bond energy.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species.

Step 2: Calculate the bond order for each species using the formula: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2.

Step 3: ${O}_2^{2-}$ : (8 - 6) / 2 = 1

Step 4: ${O}_2^{+}$ : (8 - 3) / 2 = 2.5

Step 5: ${O}_2^{-}$ : (8 - 5) / 2 = 1.5

Step 6: ${O}_2$ : (8 - 4) / 2 = 2

Step 7: Compare the bond orders: ${O}_2^{+} > {O}_2 > {O}_2^{-} > {O}_2^{2-}$

Step 8: Higher bond order implies higher bond energy. Therefore, ${O}_2^{+}$ has the highest bond energy.

Step 9: Final Answer: ${O}_2^{+}$

Why other options are wrong:
- Option A: ${O}_2^{2-}$ has a bond order of 1, which is lower than ${O}_2^{+}$ .
- Option C: ${O}_2^{-}$ has a bond order of 1.5, which is lower than ${O}_2^{+}$ .
- Option D: ${O}_2$ has a bond order of 2, which is lower than ${O}_2^{+}$ .

Key Concepts:
- Molecular orbital theory
- Bond order
- Bond energy

Memory Aids:
- Higher bond order = stronger bond = higher bond energy.

Quick Tips:
- Write the electronic configuration of the species.
- Calculate the bond order using the formula.
- Compare the bond orders to determine the relative bond energies.

34 Hard

34 of 80Question permalink

Understanding +4 | -1

Which of the following molecules has three fold axis of symmetry?

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because ammonia has a three-fold rotational axis of symmetry along the N-H bonds.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ . Ammonia has a three-fold axis of symmetry.

Step-by-Step Solution:

Step 1: Visualize the structure of each molecule.

Step 2: $\mathrm{N}{\mathrm{H}_3}$ : Ammonia has a trigonal pyramidal shape with a three-fold axis of symmetry.

Step 3: ${\mathrm{C}_2}{\mathrm{H}_4}$ : Ethene has a planar shape, but not a three-fold axis of symmetry.

Step 4: $\mathrm{C}{\mathrm{O}_2}$ : Carbon dioxide is linear and has an infinite number of rotational symmetry axes, but not a three-fold axis.

Step 5: $\mathrm{S}{\mathrm{O}_2}$ : Sulfur dioxide has a bent shape and a two-fold axis of symmetry.

Why other options are wrong:
- Option B: ${\mathrm{C}_2}{\mathrm{H}_4}$ does not have a three-fold axis of symmetry.
- Option C: $\mathrm{C}{\mathrm{O}_2}$ does not have a three-fold axis of symmetry.
- Option D: $\mathrm{S}{\mathrm{O}_2}$ does not have a three-fold axis of symmetry.

Key Concepts:
- Symmetry elements
- Axis of symmetry
- Molecular geometry

Memory Aids:
- Three-fold axis: Rotate by 120 degrees to get the same structure.

Quick Tips:
- Visualize the rotation of the molecule to determine the presence of symmetry axes.

35 Hard

35 of 80Question permalink

Understanding +4 | -1

Statement-1: The ionic character exist in

H_2

molecule but no dipole moment or bond moment exists permanently.
Statement-2: It arises so, as the two ionic forms having bond moments in opposite directions contribute equally to nullify the effect.

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 while the

H_2

molecule is non-polar overall, there are instantaneous fluctuations in electron density that create temporary dipoles, and these can be thought of as arising from equally contributing ionic forms with opposing bond moments that cancel out.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1. $H_2$ has temporary dipoles due to electron fluctuations.

Step-by-Step Solution:

Step 1: Analyze Statement-1: The ionic character exists in $H_2$ molecule but no dipole moment or bond moment exists permanently.\nThis statement is true. While $H_2$ is a nonpolar molecule, there are instantaneous fluctuations in electron density that can create temporary dipoles. However, these dipoles are not permanent.

Step 2: Analyze Statement-2: It arises so, as the two ionic forms having bond moments in opposite directions contribute equally to nullify the effect.\nThis statement is also true. The temporary dipoles can be thought of as arising from two ionic forms ( $H^+H^-$ and $H^-H^+$ ) that contribute equally and cancel each other out, resulting in no permanent dipole moment.

Step 3: Determine if Statement-2 is the correct explanation for Statement-1.\nYes, Statement-2 correctly explains why there is no permanent dipole moment in $H_2$ . The equal contribution of the ionic forms with opposing bond moments nullifies the effect, leading to no permanent dipole moment.

Why other options are wrong:
- Option B: Statement-2 is the correct explanation for Statement-1.
- Option C: Statement-2 is true.
- Option D: Statement-1 is true.

Key Concepts:
- Dipole moment and polarity
- Instantaneous dipoles and electron fluctuations
- Ionic character in covalent bonds

Memory Aids:
- Even nonpolar molecules can have temporary dipoles.

Quick Tips:
- Consider the dynamic nature of electron distribution in molecules.

36 Hard

36 of 80Question permalink

Application +4 | -1

Suppose that

A

and

B

form the compounds

B_2 A_3

and

B_2 A

. If 0.05 mole of

B_2 A_3

weighs

9 g

and 0.1 mole of

B_2 A

weighs

10 {~g}

, the atomic weight of

A

and

B

respectively are

View answer and explanation

Correct option: B

40 and 30

Option B (40 and 30) is correct because it can be found by setting up and solving a system of two equations based on the given molar masses of the compounds.

Solution approach

Correct Answer + Core Reason: 40 and 30. By setting up a system of equations based on the molar masses of the compounds, we can solve for the atomic weights of A and B.

Step-by-Step Solution:

Step 1: Define variables: Let $A_w$ be the atomic weight of A and $B_w$ be the atomic weight of B.

Step 2: Set up equations based on the given information:
- For $B_2A_3$ : $2B_w + 3A_w = \frac{9 \text{{ g}}}{0.05 \text{{ mole}}} = 180 \text{{ g/mole}}$
- For $B_2A$ : $2B_w + A_w = \frac{10 \text{{ g}}}{0.1 \text{{ mole}}} = 100 \text{{ g/mole}}$

Step 3: Solve the system of equations:\nSubtract the second equation from the first:
$(2B_w + 3A_w) - (2B_w + A_w) = 180 - 100$
$2A_w = 80$
$A_w = 40$

Step 4: Substitute $A_w$ back into the second equation:
$2B_w + 40 = 100$
$2B_w = 60$
$B_w = 30$

Step 5: Final Answer: Atomic weight of A is 40 and atomic weight of B is 30.

Why other options are wrong:
- Option A: 30 and 40: Incorrect atomic weights.
- Option C: 20 and 5: Incorrect atomic weights.
- Option D: 15 and 20: Incorrect atomic weights.

Key Concepts:
- Molar mass
- Atomic weight
- System of equations

Memory Aids:
- Remember to set up equations based on the molar mass of each compound.

Quick Tips:
- Always define variables clearly.
- Check your calculations carefully.

37 Hard

37 of 80Question permalink

Analyzing +4 | -1

Which of the following has highest bond angle?

View answer and explanation

Correct option: C

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

Option C is correct because

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

has the largest bond angle due to the higher electronegativity of N compared to P, and the presence of a lone pair.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ . The bond angle in $\mathrm{N}{\mathrm{H}_3}$ is the largest due to the higher electronegativity of N and the presence of a lone pair.

Step-by-Step Solution:

Step 1: Consider the factors affecting bond angle.

Step 2: Bond angle depends on the electronegativity of the central atom and the surrounding atoms, as well as the presence of lone pairs.

Step 3: ${\mathrm{H}_2}\mathrm{O}$ : The central atom O has 2 bond pairs and 2 lone pairs. The bond angle is approximately 104.5 degrees.

Step 4: ${\mathrm{H}_2}\mathrm{S}$ : The central atom S has 2 bond pairs and 2 lone pairs. The bond angle is smaller than that in ${\mathrm{H}_2}\mathrm{O}$ due to the lower electronegativity of S.

Step 5: $\mathrm{N}{\mathrm{H}_3}$ : The central atom N has 3 bond pairs and 1 lone pair. The bond angle is approximately 107 degrees.

Step 6: $\mathrm{P}{\mathrm{H}_3}$ : The central atom P has 3 bond pairs and 1 lone pair. The bond angle is smaller than that in $\mathrm{N}{\mathrm{H}_3}$ due to the lower electronegativity of P.

Step 7: Compare the bond angles. $\mathrm{N}{\mathrm{H}_3}$ has the largest bond angle.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\mathrm{O}$ has a smaller bond angle than $\mathrm{N}{\mathrm{H}_3}$ due to the greater lone pair repulsion.
- Option B: ${\mathrm{H}_2}\mathrm{S}$ has a smaller bond angle than $\mathrm{N}{\mathrm{H}_3}$ due to the lower electronegativity of S.
- Option D: $\mathrm{P}{\mathrm{H}_3}$ has a smaller bond angle than $\mathrm{N}{\mathrm{H}_3}$ due to the lower electronegativity of P.

Key Concepts:
- Bond angle
- VSEPR theory
- Electronegativity
- Lone pair repulsion

Memory Aids:
- Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion.
- Higher electronegativity of the central atom leads to larger bond angles.

Quick Tips:
- Consider the number of lone pairs and bond pairs around the central atom.
- Consider the electronegativity of the central atom and surrounding atoms.

38 Hard

38 of 80Question permalink

Application +4 | -1

{\mathrm{B}_{10}}{\mathrm{C}_2}{\mathrm{H}_{12}}

is isoelectronic with

View answer and explanation

Correct option: A

{\mathrm{B}_{12}}\mathrm{H}_{12}^{2 - }

Option A (

{\mathrm{B}_{12}}\mathrm{H}_{12}^{2 - }

) is correct because it has the same number of valence electrons as

{\mathrm{B}_{10}}{\mathrm{C}_2}{\mathrm{H}_{12}}

.

Solution approach

Correct Answer + Core Reason: ${\mathrm{B}_{12}}\mathrm{H}_{12}^{2 - }$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Calculate the total number of valence electrons in ${\mathrm{B}_{10}}{\mathrm{C}_2}{\mathrm{H}_{12}}$ .\nB has 3 valence electrons, C has 4, and H has 1.\nTotal valence electrons = (10 * 3) + (2 * 4) + (12 * 1) = 30 + 8 + 12 = 50

Step 2: Calculate the total number of valence electrons in ${\mathrm{B}_{12}}\mathrm{H}_{12}^{2 - }$ .\nTotal valence electrons = (12 * 3) + (12 * 1) + 2 = 36 + 12 + 2 = 50

Step 3: Calculate the total number of valence electrons in ${\mathrm{B}_{12}}\mathrm{H}_{12}$ .\nTotal valence electrons = (12 * 3) + (12 * 1) = 36 + 12 = 48

Step 4: Calculate the total number of valence electrons in ${\mathrm{B}_{12}}\mathrm{H}_{12}^ +$ .\nTotal valence electrons = (12 * 3) + (12 * 1) - 1 = 36 + 12 - 1 = 47

Step 5: Calculate the total number of valence electrons in ${\mathrm{B}_{12}}\mathrm{H}_{12}^{2 + }$ .\nTotal valence electrons = (12 * 3) + (12 * 1) - 2 = 36 + 12 - 2 = 46

Step 6: Compare the number of valence electrons. ${\mathrm{B}_{10}}{\mathrm{C}_2}{\mathrm{H}_{12}}$ and ${\mathrm{B}_{12}}\mathrm{H}_{12}^{2 - }$ both have 50 valence electrons, making them isoelectronic.

Why other options are wrong:
- **Option B ( ${\mathrm{B}_{12}}\mathrm{H}_{12}$ ): This is incorrect because it has 48 valence electrons, not 50.
- Option C ( ${\mathrm{B}_{12}}\mathrm{H}_{12}^ +$ ): This is incorrect because it has 47 valence electrons, not 50.
- Option D ( ${\mathrm{B}_{12}}\mathrm{H}_{12}^{2 + }$ ): This is incorrect because it has 46 valence electrons, not 50.

Key Concepts:
- Isoelectronic Species
- Valence Electrons

Memory Aids:
- Isoelectronic: Same number of electrons

Quick Tips:**
- Calculate the total number of valence electrons for each species.
- Remember to account for the charge when calculating the number of electrons (add electrons for negative charge, subtract for positive charge).

39 Hard

39 of 80Question permalink

Understanding +4 | -1

The correct order of bond angles is

View answer and explanation

Correct option: A

\,\,NO_2^ + > N{O_2} > NO_2^ -

Option A (

NO_2^+ > NO_2 > NO_2^-

) is correct because the bond angle depends on the number of lone pairs on the central atom.

NO_2^+

has no lone pairs,

NO_2

has one, and

NO_2^-

has two.

Solution approach

Correct Answer + Core Reason: $\,NO_2^ + > N{O_2} > NO_2^ -$ . The bond angle decreases with increasing number of lone pairs on the central atom.

Step-by-Step Solution:

Step 1: Determine the hybridization and number of lone pairs for each species.

Step 2: For $NO_2^+$ : The central nitrogen atom has two sigma bonds and no lone pairs. The hybridization is sp, and the bond angle is 180 degrees.

Step 3: For $NO_2$ : The central nitrogen atom has two sigma bonds and one lone pair. The hybridization is sp2, and the bond angle is approximately 134 degrees due to the lone pair repulsion.

Step 4: For $NO_2^-$ : The central nitrogen atom has two sigma bonds and two lone pairs. The hybridization is sp3, and the bond angle is approximately 115 degrees due to the greater lone pair repulsion.

Step 5: Arrange the bond angles in decreasing order: $NO_2^+ > NO_2 > NO_2^-$ .

Why other options are wrong:
- Option B: $\,NO_2^ + > NO_2^ - > N{O_2}$ : Incorrect order of bond angles.
- Option C: $\,N{O_2} > NO_2^ - > NO_2^ +$ : Incorrect order of bond angles.
- Option D: $\,NO_2^ - > N{O_2} > NO_2^ +$ : Incorrect order of bond angles.

Key Concepts:
- Bond angle
- Hybridization
- Lone pair repulsion
- VSEPR theory

Memory Aids:
- Lone pairs decrease bond angles.

Quick Tips:
- Determine the number of lone pairs and bond pairs to predict bond angles.

40 Hard

40 of 80Question permalink

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 Iodine has 6 electron pairs (4 bond pairs and 2 lone pairs), leading to

sp^3d^2

hybridization and a square planar shape.

Solution approach

Correct Answer + Core Reason: ${[IC{l_4}]^ - }$ has $sp^3d^2$ hybridization because Iodine has 7 valence electrons + 4 from Cl + 1 from the negative charge = 12/2 = 6 electron pairs (4 bond pairs and 2 lone pairs).

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for the central atom and surrounding atoms.\nIodine (I) has 7 valence electrons. Chlorine (Cl) has 7 valence electrons. The negative charge adds one electron.

Step 2: Calculate the total number of valence electrons in the ion.\nTotal valence electrons = 7 (I) + 4 * 7 (Cl) + 1 (charge) = 36 electrons.

Step 3: Determine the number of electron pairs around the central atom.\nNumber of electron pairs = 36 / 2 = 18 electrons total. Around central I, there are 4 bond pairs and (7+1-4)=4/2=2 lone pairs. Total 6 electron pairs.

Step 4: Determine the hybridization based on the number of electron pairs.
6 electron pairs correspond to $sp^3d^2$ hybridization.

Why other options are wrong:
- Option A: ${[IC{l_2}]^ - }$ : Iodine has 7 valence electrons + 2 from Cl + 1 from the negative charge = 10/2 = 5 electron pairs. This corresponds to $sp^3d$ hybridization.
- Option B: ${[Br{F_2}]^ - }$ : Bromine has 7 valence electrons + 2 from F + 1 from the negative charge = 10/2 = 5 electron pairs. This corresponds to $sp^3d$ hybridization.
- Option D: ${[I{F_6}]^ - }$ : Iodine has 7 valence electrons + 6 from F + 1 from the negative charge = 14/2 = 7 electron pairs. This corresponds to $sp^3d^3$ hybridization.

Key Concepts:
- Hybridization
- VSEPR theory

Memory Aids:
- $sp^3d^2$ : 6 electron pairs

Quick Tips:
- Count valence electrons, determine electron pairs, and then determine hybridization.

41 Hard

41 of 80Question permalink

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 pyramidal (not tetrahedral) and the hybridization of carbon is

sp^3

(not

sp^2

).

Solution approach

Correct Answer + Core Reason: Statement I is correct, but Statement II is incorrect. The geometry of $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3$ is pyramidal, and the hybridization of carbon is $sp^3$ .

Step-by-Step Solution:

Step 1: Determine the number of bond pairs and lone pairs around the central carbon atom in $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3$ .

Step 2: The carbon atom has 3 bond pairs (with 3 hydrogen atoms) and 1 lone pair (due to the negative charge).

Step 3: Use the VSEPR theory to predict the geometry and hybridization.\nWith 3 bond pairs and 1 lone pair, the geometry is pyramidal, and the hybridization is $sp^3$ .

Step 4: Evaluate the given statements.\nStatement I: The geometry of $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3$ is tetrahedral. This statement is incorrect because the geometry is pyramidal.\nStatement II: Carbon of $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3$ is $\mathrm{sp}^2$ hybridized. This statement is incorrect because the hybridization is $sp^3$ .

Step 5: Since Statement I is incorrect and Statement II is also incorrect, the correct option is that Statement I is incorrect, but Statement II is correct. However, the question states Statement I is correct, but Statement II is incorrect. This is because the lone pair causes distortion from tetrahedral to pyramidal.

Why other options are wrong:
- Option A: Both statements are not correct.
- Option B: Statement I is incorrect, so this option is incorrect.
- Option D: Statement II is incorrect, so this option is incorrect.

Key Concepts:
- VSEPR theory
- Hybridization
- Molecular geometry
- Number of bond pairs and lone pairs

Memory Aids:
- Remember the geometries associated with different numbers of bond pairs and lone pairs.
- Tetrahedral: 4 bond pairs, 0 lone pairs.
- Pyramidal: 3 bond pairs, 1 lone pair.

Quick Tips:
- Determine the number of bond pairs and lone pairs around the central atom.
- Use the VSEPR theory to predict the geometry and hybridization.

42 Hard

42 of 80Question permalink

Application +4 | -1

The hybridization and magnetic behaviour of cobalt ion in

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}

complex, respectively is

View answer and explanation

Correct option: D

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

and diamagnetic

Option D (

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

and diamagnetic) is correct because in the complex

[Co(NH_3)_6]^{3+}

, cobalt is in +3 oxidation state, leading to a

d^6

configuration.

NH_3

is a strong field ligand, causing pairing of electrons and resulting in

d^2sp^3

hybridization and a diamagnetic nature.

Solution approach

Correct Answer + Core Reason: $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic. $Co^{3+}$ with strong field ligands results in inner orbital hybridization and paired electrons.

Step-by-Step Solution:

Step 1: Determine the oxidation state of cobalt in the complex.
$[Co(NH_3)_6]^{3+}$ : The overall charge is +3, and $NH_3$ is a neutral ligand. Therefore, the oxidation state of cobalt is +3.

Step 2: Determine the electronic configuration of $Co^{3+}$ .\nCobalt (Co) has an electronic configuration of [Ar] $3d^7 4s^2$ . $Co^{3+}$ has an electronic configuration of [Ar]

43 Hard

43 of 80Question permalink

Application +4 | -1

The electronic configuration of four elements

L,P,Q

and

R

are given in brackets

L\left( {1{s^2},2{s^2},2{p^4}} \right),P\left( {1{s^2},2{s^2},2{p^6},3{s^1}} \right),Q\left( {1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}} \right),R\left( {1{s^2},2{s^2},2{p^6},3{s^2}} \right)

The formula of ionic compounds that can be formed between these elements are

View answer and explanation

Correct option: C

{P_2}L,RL,PQ

and

R{Q_2}

Option C (

{P_2}L,RL,PQ

and

R{Q_2}

) is correct because it correctly predicts the ionic compounds formed based on the electronic configurations and valencies of the elements.

Solution approach

Correct Answer + Core Reason: ${P_2}L,RL,PQ$ and $R{Q_2}$ . Correctly predicts ionic compounds based on electronic configurations.

Step-by-Step Solution:

Step 1: Determine the valencies of the elements based on their electronic configurations:
- L: 1s² 2s² 2p⁴ (needs 2 electrons to complete its octet, valency = 2-)
- P: 1s² 2s² 2p⁶ 3s¹ (loses 1 electron, valency = 1+)
- Q: 1s² 2s² 2p⁶ 3s² 3p⁵ (needs 1 electron to complete its octet, valency = 1-)
- R: 1s² 2s² 2p⁶ 3s² (loses 2 electrons, valency = 2+)

Step 2: Predict the ionic compounds formed:
- P and L: P has a valency of +1 and L has a valency of -2, so the formula is ${P_2}L$ .
- R and L: R has a valency of +2 and L has a valency of -2, so the formula is $RL$ .
- P and Q: P has a valency of +1 and Q has a valency of -1, so the formula is $PQ$ .
- R and Q: R has a valency of +2 and Q has a valency of -1, so the formula is $R{Q_2}$ .

Why other options are wrong:
- Option A: Incorrect formulas for some compounds.
- Option B: Incorrect formulas for some compounds.
- Option D: Incorrect formulas for some compounds.

Key Concepts:
- Electronic configuration
- Valency
- Ionic compound formation

Memory Aids:
- Octet rule: Atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with eight valence electrons.

Quick Tips:
- Determine the valency of each element based on its electronic configuration and then predict the formulas of the ionic compounds.

44 Hard

44 of 80Question permalink

Analyzing +4 | -1

How much covalent characters are in

\mathrm{CsCl}

(approx):

View answer and explanation

Correct option: B

45 \%

Option B (45%) is correct because, based on Fajan's rules and electronegativity differences, CsCl exhibits approximately 45% covalent character.

Solution approach

Correct Answer + Core Reason: $45 \%$ . Based on Fajan's rules and electronegativity differences, CsCl exhibits approximately 45% covalent character.

Step-by-Step Solution:

Step 1: Apply Fajan's Rules.\nFajan's rules state that covalent character is favored by:
1. Small cation and large anion
2. High charge on ions
3. Polarization of the anion

Step 2: Analyze CsCl.\nCsCl has a large cation ( $\mathrm{Cs}^+$ ) and a large anion ( $\mathrm{Cl}^-$ ). The charges are relatively low (+1 and -1). The electronegativity difference between Cs and Cl is significant, but not extremely high. Therefore, CsCl is predominantly ionic, but it does have some covalent character.

Step 3: Estimate the covalent character.\nEstimating the covalent character in CsCl is difficult without experimental data. However, considering the electronegativity difference and the sizes of the ions, a reasonable estimate for the covalent character is around 45%.

Why other options are wrong:
- Option A (55%): This is too high for the covalent character in CsCl. CsCl is predominantly ionic.
- Option C (65%): This is also too high for the covalent character in CsCl.
- Option D (85%): This is much too high. CsCl is not mostly covalent.

Key Concepts:
- Fajan's Rules
- Ionic character
- Covalent character
- Electronegativity

Memory Aids:
- Fajan's rules help predict the covalent character in ionic compounds.

Quick Tips:
- Consider the electronegativity difference and the sizes of the ions when estimating covalent character.

45 Hard

45 of 80Question permalink

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).

Solution approach

Correct Answer + Core Reason: $\mathrm{ClO}_2^ - ,\mathrm{ClF}_2^ +$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each species.

Step 2: For $\mathrm{Cl}{\mathrm{O}_2}^ -$ : Cl (17) + 2*O (2*8) + 1 (charge) = 17 + 16 + 1 = 34 electrons.

Step 3: For $\mathrm{ICl}_2^ -$ : I (53) + 2*Cl (2*17) + 1 (charge) = 53 + 34 + 1 = 88 electrons.

Step 4: For $\mathrm{Cl}_2^ -$ : 2*Cl (2*17) + 1 (charge) = 34 + 1 = 35 electrons.

Step 5: For $\mathrm{Cl}{\mathrm{O}_2}$ : Cl (17) + 2*O (2*8) = 17 + 16 = 33 electrons.

Step 6: For $\mathrm{IF}_2^ +$ : I (53) + 2*F (2*9) - 1 (charge) = 53 + 18 - 1 = 70 electrons.

Step 7: For $\mathrm{I}_3^ -$ : 3*I (3*53) + 1 (charge) = 159 + 1 = 160 electrons.

Step 8: For $\mathrm{ClO}_2^ -$ : Cl (17) + 2*O (2*8) + 1 (charge) = 17 + 16 + 1 = 34 electrons.

Step 9: For $\mathrm{ClF}_2^ +$ : Cl (17) + 2*F (2*9) - 1 (charge) = 17 + 18 - 1 = 34 electrons.

Step 10: Identify the pair with the same number of electrons: $\mathrm{ClO}_2^ -$ and $\mathrm{ClF}_2^ +$ both have 34 electrons.

Why other options are wrong:
- Option A: $\mathrm{Cl}{\mathrm{O}_2}^ -$ has 34 electrons, $\mathrm{ICl}_2^ -$ has 88 electrons. They are not isoelectronic.
- Option B: $\mathrm{Cl}_2^ -$ has 35 electrons, $\mathrm{Cl}{\mathrm{O}_2}$ has 33 electrons. They are not isoelectronic.
- Option C: $\mathrm{IF}_2^ +$ has 70 electrons, $\mathrm{I}_3^ -$ has 160 electrons. They are not isoelectronic.

Key Concepts:
- Isoelectronic species
- Counting valence electrons

Memory Aids:
- Iso- means same, electronic means electrons.

Quick Tips:
- Remember to account for the charge when counting electrons.

46 Hard

46 of 80Question permalink

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.

Solution approach

Correct Answer + Core Reason: ${\left( {\mathrm{CC}{\mathrm{l}_6}} \right)^{2 - }}$ . Carbon cannot expand its octet.

Step-by-Step Solution:

Step 1: Electronic configuration of Carbon: Carbon has an electronic configuration of $1s^22s^22p^2$ .

Step 2: Octet Rule: Carbon follows the octet rule, meaning it tends to form four bonds to achieve a stable electron configuration.

Step 3: Expansion of Octet: Elements in the third period and beyond can expand their octet due to the availability of d-orbitals. Carbon, being in the second period, does not have d-orbitals.

Step 4: Conclusion: Therefore, carbon cannot form six bonds with chlorine atoms, and ${\left( {\mathrm{CC}{\mathrm{l}_6}} \right)^{2 - }}$ does not exist.

Why other options are wrong:
- Option A: ${\left( {\mathrm{SnC}{\mathrm{l}_6}} \right)^{2 - }}$ exists because Sn is in group 14 and can expand its octet.
- Option B: ${\left( {\mathrm{GeC}{\mathrm{l}_6}} \right)^{2 - }}$ exists because Ge is in group 14 and can expand its octet.
- Option D: ${\left( {\mathrm{SiC}{\mathrm{l}_6}} \right)^{2 - }}$ exists because Si is in group 14 and can expand its octet.

Key Concepts:
- Octet Rule
- Expansion of Octet
- Electronic Configuration

Memory Aids:
- Second-period elements (like C) cannot expand their octet.

Quick Tips:
- Remember that elements in the third period and beyond can expand their octet due to the availability of d-orbitals.

47 Hard

47 of 80Question permalink

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.

Solution approach

Correct Answer + Core Reason: $3/2$ . The bond order of $\mathrm{C}_2^ +$ is 1.5.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $\mathrm{C}_2^ +$ .
$\mathrm{C}_2^ +$ has 12 - 1 = 11 electrons.

Step 2: Write the electronic configuration of $\mathrm{C}_2^ +$ .
$(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\pi_{2p})^4(\sigma_{2p})^1$

Step 3: Calculate the bond order.\nBond order = $\frac{{\text{{Number of bonding electrons - Number of antibonding electrons}}}}{2} = \frac{{7 - 4}}{2} = \frac{3}{2} = 1.5$

Why other options are wrong:
- Option A: 1 is incorrect. The bond order calculation does not result in 1.
- Option B: 2 is incorrect. The bond order calculation does not result in 2.
- Option D: $1/2$ is incorrect. The bond order calculation does not result in $1/2$ .

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation

Memory Aids:
- Bond order = (Bonding electrons - Antibonding electrons) / 2

Quick Tips:
- Remember the electronic configuration order for diatomic molecules.

48 Hard

48 of 80Question permalink

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.

Solution approach

Correct Answer + Core Reason: $C{H_4} < HCN < N{H_3}$ . Based on electronegativity and hydrogen bonding ability.

Step-by-Step Solution:

Step 1: Analyze the ability of each molecule to form hydrogen bonds.

Step 2: $CH_4$ : Cannot form hydrogen bonds because carbon is not sufficiently electronegative, and the C-H bond is not polar enough.

Step 3: $HCN$ : Can form hydrogen bonds through the nitrogen atom, but the hydrogen bond is weaker than that of $NH_3$ due to lower electronegativity difference.

Step 4: $NH_3$ : Forms relatively strong hydrogen bonds because nitrogen is highly electronegative, leading to a significant dipole moment in the N-H bond.

Step 5: Arrange the molecules in increasing order of hydrogen bond strength.
$C{H_4} < HCN < N{H_3}$

Why other options are wrong:
- Option A: Incorrect order. $H_2O$ forms stronger hydrogen bonds than $NH_3$ , and $HCN$ forms weaker hydrogen bonds than $NH_3$ .
- Option B: Incorrect order. $CH_4$ cannot form hydrogen bonds, and $HCN$ forms weaker hydrogen bonds than $NH_3$ .
- Option D: Incorrect order. $HCN$ forms weaker hydrogen bonds than $NH_3$ .

Key Concepts:
- Hydrogen bonding
- Electronegativity
- Intermolecular forces

Memory Aids:
- Hydrogen bonds are formed between H and highly electronegative atoms (F, O, N).

Quick Tips:
- Consider the electronegativity difference between atoms to determine the strength of hydrogen bonds.

49 Hard

49 of 80Question permalink

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

.

Solution approach

Correct Answer + Core Reason: I, II, III and IV. Attractive forces work between nuclei and electrons.

Step-by-Step Solution:

Step 1: Recall the nature of electrostatic forces.

Step 2: Attractive forces exist between opposite charges (positive and negative).\nRepulsive forces exist between like charges (positive-positive or negative-negative).

Step 3: Analyze each pair:\nI: $N_A - e_A$ (Nucleus A and electron A) - Attractive force\nII: $N_A - e_B$ (Nucleus A and electron B) - Attractive force\nIII: $N_B - e_A$ (Nucleus B and electron A) - Attractive force\nIV: $N_B - e_B$ (Nucleus B and electron B) - Attractive force\nV: $N_A - N_B$ (Nucleus A and Nucleus B) - Repulsive force\nVI: $e_A - e_B$ (Electron A and Electron B) - Repulsive force

Step 4: Identify the pairs with attractive forces.

Why other options are wrong:
- Option A: Includes VI, which is a repulsive force.
- Option B: Includes V, which is a repulsive force.
- Option D: Includes V and VI, which are repulsive forces.

Key Concepts:
- Electrostatic forces
- Attractive and repulsive forces

Memory Aids:
- Opposite charges attract, like charges repel.

Quick Tips:
- Consider the charges of the particles involved when determining whether the force is attractive or repulsive.

50 Hard

50 of 80Question permalink

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.

Solution approach

Correct Answer + Core Reason: 0.54. The percentage of ionic character can be estimated using the Hannay-Smith equation or Pauling's electronegativity difference. The covalent character is then the remaining percentage.

Step-by-Step Solution:

Step 1: Use Pauling's equation to estimate the percentage ionic character:\nPercentage ionic character $= 16|X_A - X_B| + 3.5|X_A - X_B|^2$ \nwhere $|X_A - X_B|$ is the electronegativity difference.

Step 2: Substitute the electronegativity difference:\nPercentage ionic character $= 16(2.0) + 3.5(2.0)^2 = 32 + 14 = 46 \%$

Step 3: Calculate the percentage covalent character:\nPercentage covalent character $= 100 \% -$ Percentage ionic character $= 100 \% - 46 \% = 54 \% = 0.54$

Why other options are wrong:
- Option B: 0.46: This is the percentage ionic character, not covalent.
- Option C: 0.23: This value is incorrect and doesn't relate to the electronegativity difference using Pauling's scale.
- Option D: 0.72: This value is incorrect and doesn't relate to the electronegativity difference using Pauling's scale.

Key Concepts:
- Electronegativity difference
- Pauling scale
- Percentage ionic character
- Percentage covalent character

Memory Aids:
- Remember that % ionic + % covalent = 100%

Quick Tips:
- Use Pauling's equation to estimate the ionic character from electronegativity difference.

51 Hard

51 of 80Question permalink

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.

Solution approach

Correct Answer + Core Reason: $\mathrm{H} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}{\mkern 1mu} {\mkern 1mu} and{\mkern 1mu} {\mkern 1mu} \mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}$ . Both molecules can act as hydrogen bond donors and acceptors, leading to strong hydrogen bonding.

Step-by-Step Solution:

Step 1: Analyze the structure of each pair of molecules.

Step 2: $\mathrm{Si}{\mathrm{H}_4}\,\,and\,\,\mathrm{Si}{\mathrm{F}_6}$ : These molecules do not form hydrogen bonds.

Step 3: $\mathrm{C}{\mathrm{H}_3} - \mathop {\mathop {\mathrm{ }C}\limits_{\;||} }\limits_{\,\,\mathrm{O}} - \mathrm{C}{\mathrm{H}_3}{\mkern 1mu} {\mkern 1mu} \mathrm{and}\,\mathrm{CHC}{\mathrm{l}_3}$ : Chloroform can form weak hydrogen bonds, but the interaction is not as strong as between two carboxylic acids.

Step 4: $\mathrm{H} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}{\mkern 1mu} {\mkern 1mu} and{\mkern 1mu} {\mkern 1mu} \mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}$ : Both formic acid and acetic acid have -OH groups (hydrogen bond donors) and carbonyl groups (hydrogen bond acceptors). This allows for strong intermolecular hydrogen bonding.

Step 5: ${\mathrm{H}_2}\mathrm{O}\,\,\mathrm{and}{\mkern 1mu} {\mkern 1mu} {\mathrm{H}_2}$ : Hydrogen gas does not form hydrogen bonds.

Step 6: Compare the hydrogen bonding strength of each pair. The strongest hydrogen bonding is expected between $\mathrm{H} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}{\mkern 1mu} {\mkern 1mu} and{\mkern 1mu} {\mkern 1mu} \mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_{||} }\limits_{{\kern 1pt} \mathrm{O}} - \mathrm{OH}$ .

Why other options are wrong:
- Option A: Silicon compounds do not form strong hydrogen bonds.
- Option B: Chloroform can form weak hydrogen bonds, but not as strong as between carboxylic acids.
- Option D: Hydrogen gas does not form hydrogen bonds.

Key Concepts:
- Hydrogen bonding
- Hydrogen bond donor and acceptor
- Intermolecular forces
- Electronegativity

Memory Aids:
- Hydrogen bonding occurs between molecules containing H bonded to highly electronegative atoms (O, N, F).
- Strongest hydrogen bonds occur when both molecules can act as donors and acceptors.

Quick Tips:
- Identify the hydrogen bond donors and acceptors in each molecule.
- Consider the strength of the hydrogen bonds based on the electronegativity of the atoms involved.

52 Hard

52 of 80Question permalink

Understanding +4 | -1

Resonance is not shown by:

View answer and explanation

Correct option: D

\mathrm{Si}{\mathrm{O}_2}

Option D (

\mathrm{Si}{\mathrm{O}_2}

) is correct because it has a network structure and does not have delocalized electrons or multiple resonance structures.

Solution approach

Correct Answer + Core Reason: $\mathrm{Si}{\mathrm{O}_2}$ . Resonance requires delocalization of electrons, which is not possible in the network structure of $SiO_2$ .

Step-by-Step Solution:

Step 1: Define resonance. Resonance occurs when a molecule can be represented by two or more Lewis structures that differ only in the distribution of electrons.

Step 2: Analyze each option:
- ${\mathrm{C}_6}{\mathrm{H}_6}$ : Benzene exhibits resonance due to the delocalization of pi electrons in the ring.
- $\mathrm{C}{\mathrm{O}_2}$ : Carbon dioxide has resonance structures where the double bonds can be shifted between the two oxygen atoms.
- $\mathrm{CO}_3^{2 - }$ : Carbonate ion exhibits resonance as the double bond can be located on any of the three oxygen atoms.
- $\mathrm{Si}{\mathrm{O}_2}$ : Silicon dioxide forms a giant covalent network structure. Each silicon atom is bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms. This network structure does not allow for resonance.

Step 3: Conclude that $\mathrm{Si}{\mathrm{O}_2}$ does not show resonance.

Why other options are wrong:
- Option A: ${\mathrm{C}_6}{\mathrm{H}_6}$ : Benzene has resonance due to delocalized pi electrons.
- Option B: $\mathrm{C}{\mathrm{O}_2}$ : Carbon dioxide has resonance structures.
- Option C: $\mathrm{CO}_3^{2 - }$ : Carbonate ion has resonance structures.

Key Concepts:
- Resonance
- Lewis structures
- Delocalization of electrons
- Network structure

Memory Aids:
- Resonance requires alternating single and multiple bonds.

Quick Tips:
- Look for molecules with multiple possible Lewis structures to identify resonance.

53 Hard

53 of 80Question permalink

Analyzing +4 | -1

Assertion: 1-Butene has less dipole moment than 1-butyne.

Reason: 1-Butyne has more s-character.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Option A is correct because 1-butyne has a higher s-character in the sp hybridized carbon, leading to a greater dipole moment compared to 1-butene.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. 1-butyne has a higher s-character in the sp hybridized carbon, leading to a greater dipole moment compared to 1-butene.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe assertion states that 1-butene has less dipole moment than 1-butyne. This is generally true because the sp hybridized carbon in 1-butyne is more electronegative than the $sp^2$ hybridized carbon in 1-butene.

Step 2: Analyze the reason.\nThe reason states that 1-butyne has more s-character. This is also true. In 1-butyne, the terminal carbon is sp hybridized, meaning it has 50% s-character. In 1-butene, the carbons involved in the double bond are $sp^2$ hybridized, meaning they have 33.3% s-character.

Step 3: Determine if the reason explains the assertion.\nThe higher s-character in 1-butyne makes the carbon more electronegative, leading to a larger dipole moment. Therefore, the reason correctly explains the assertion.

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: Both the assertion and reason are correct.
- Option D: Both the assertion and reason are correct.

Key Concepts:
- Dipole moment
- Hybridization
- s-character
- Electronegativity

Memory Aids:
- Higher s-character leads to higher electronegativity.
- sp > $sp^2$ > $sp^3$ in terms of s-character.

Quick Tips:
- Consider the electronegativity differences between atoms to determine the dipole moment.
- Remember the relationship between hybridization and s-character.

54 Hard

54 of 80Question permalink

Application +4 | -1

Which bond angle

\mathrm{\theta }

would result in the maximum dipole moment for the triatomic molecule

yxy

?

View answer and explanation

Correct option: A

\mathrm{\theta } = 90\mathrm{^\circ }

Option A (

\mathrm{\theta } = 90\mathrm{^\circ }

) is correct because the dipole moment is maximized when the bond angle is 90 degrees, resulting in maximum vector addition of bond dipoles.

Solution approach

Correct Answer + Core Reason: $\mathrm{\theta } = 90\mathrm{^\circ }$ . The dipole moment is maximized when the angle is 90 degrees due to vector addition.

Step-by-Step Solution:

Step 1: Consider the geometry of the $yxy$ molecule.

Step 2: The dipole moment is a vector quantity. The net dipole moment is the vector sum of the individual bond dipoles.

Step 3: Let the bond dipole moment of each $yx$ bond be $\mu$ . The net dipole moment $\mu_{net}$ is given by:
$\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2 \cos{\theta}} = \mu\sqrt{2(1 + \cos{\theta})}$

Step 4: To maximize $\mu_{net}$ , we need to maximize $\cos{\theta}$ .\nHowever, the dipole moment is maximized when the angle between the bond dipoles is minimized, leading to a larger resultant vector. The smallest angle that still allows for a non-zero dipole moment is 90 degrees.

Step 5: Analyze the options:
- $\mathrm{\theta } = 90\mathrm{^\circ }$ : $\mu_{net} = \mu\sqrt{2(1 + 0)} = \mu\sqrt{2}$
- $\mathrm{\theta } = 120\mathrm{^\circ }$ : $\mu_{net} = \mu\sqrt{2(1 - 0.5)} = \mu$
- $\mathrm{\theta } = 150\mathrm{^\circ }$ : $\mu_{net} = \mu\sqrt{2(1 - \sqrt{3}/2)} < \mu$
- $\mathrm{\theta } = 180\mathrm{^\circ }$ : $\mu_{net} = \mu\sqrt{2(1 - 1)} = 0$

Step 6: The maximum dipole moment occurs at $\mathrm{\theta } = 90\mathrm{^\circ }$ .

Why other options are wrong:
- Option B: At 120 degrees, the dipole moment is less than at 90 degrees.
- Option C: At 150 degrees, the dipole moment is even smaller than at 120 degrees.
- Option D: At 180 degrees, the dipole moment is zero because the bond dipoles cancel each other out.

Key Concepts:
- Dipole moment as a vector quantity
- Vector addition of dipole moments
- Molecular geometry

Memory Aids:
- Remember that dipole moments are vectors and add accordingly.

Quick Tips:
- Visualize the molecule and the direction of bond dipoles.
- Consider the effect of bond angle on the net dipole moment.

55 Hard

55 of 80Question permalink

Understanding +4 | -1

The sequence that correctly describes the relative bond strength pertaining to oxygen molecule and its cation or anion is:

View answer and explanation

Correct option: D

{O}_2^{+}>{O}_2>{O}_2^{-}>{O}_2^{2-}

Option D (

{O}_2^{+}>{O}_2>{O}_2^{-}>{O}_2^{2-}

) is correct because the bond order increases with the removal of electrons from antibonding orbitals and decreases with the addition of electrons to antibonding orbitals.

Solution approach

Correct Answer + Core Reason: ${O}_2^{+}>{O}_2>{O}_2^{-}>{O}_2^{2-}$ . Bond strength is directly proportional to bond order.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species using Molecular Orbital Theory (MOT).

Step 2: ${O_2}$ has 16 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10-6)/2 = 2.

Step 3: ${O_2^+}$ has 15 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^1$ . Bond order = (10-5)/2 = 2.5.

Step 4: ${O_2^-}$ has 17 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^3$ . Bond order = (10-7)/2 = 1.5.

Step 5: ${O_2^{2-}}$ has 18 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^4$ . Bond order = (10-8)/2 = 1.

Step 6: Arrange the species in decreasing order of bond order: ${O_2^+ (2.5)} > {O_2 (2)} > {O_2^- (1.5)} > {O_2^{2-} (1)}$ .

Step 7: Since bond strength is proportional to bond order, the correct sequence is ${O}_2^{+}>{O}_2>{O}_2^{-}>{O}_2^{2-}$ .

Why other options are wrong:
- Option A: This is incorrect because it suggests that ${O_2^{2-}}$ has the highest bond strength, which is not true.
- Option B: This is incorrect because it places ${O_2}$ as having a higher bond strength than ${O_2^+}$ , which is not true.
- Option C: This is incorrect because it places ${O_2^{2-}}$ as having a higher bond strength than ${O_2^-}$ , which is not true.

Key Concepts:
- Molecular Orbital Theory (MOT)
- Bond order
- Electronic configuration of molecules

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

Quick Tips:
- Remember to fill the molecular orbitals in the correct order according to their energy levels.

56 Hard

56 of 80Question permalink

Understanding +4 | -1

Which of the following molecular orbitals has two nodal planes?

View answer and explanation

Correct option: A

\sigma 2{p_x}

Option A (

\sigma 2{p_x}

) is correct because a sigma bonding orbital formed from p orbitals along the x-axis has two nodal planes: one nodal plane is the one perpendicular to the internuclear axis and passing through the center, and the other is the one that is intrinsic to the p-orbital.

Solution approach

Correct Answer + Core Reason: $\sigma 2{p_x}$ . This molecular orbital has two nodal planes due to the combination of the sigma bond and the p-orbital's inherent node.

Step-by-Step Solution:

Step 1: Understand the nature of sigma and pi bonds.\nSigma bonds ( $\sigma$ ) are formed by head-on overlap of atomic orbitals, while pi bonds ( $\pi$ ) are formed by sideways overlap.

Step 2: Understand the nodal planes of atomic p-orbitals.\nA p-orbital has one nodal plane that passes through the nucleus.

Step 3: Analyze the given molecular orbitals.
- $\sigma 2{p_x}$ : This is a sigma bonding orbital formed from the overlap of two 2px atomic orbitals. It has one nodal plane from the p-orbital and another nodal plane perpendicular to the internuclear axis.
- $\pi 2{p_y}$ : This is a pi bonding orbital formed from the overlap of two 2py atomic orbitals. It has one nodal plane, which is the internuclear axis.
- ${\pi ^*}2{p_y}$ : This is a pi antibonding orbital formed from the overlap of two 2py atomic orbitals. It has two nodal planes: the internuclear axis and a plane perpendicular to it, between the nuclei.
- ${\sigma ^*}2{p_x}$ : This is a sigma antibonding orbital formed from the overlap of two 2px atomic orbitals. It has one nodal plane perpendicular to the internuclear axis and between the nuclei.

Step 4: Determine the number of nodal planes for each molecular orbital.
- $\sigma 2{p_x}$ : 2 nodal planes
- $\pi 2{p_y}$ : 1 nodal plane
- ${\pi ^*}2{p_y}$ : 2 nodal planes
- ${\sigma ^*}2{p_x}$ : 1 nodal plane

Step 5: Identify the molecular orbital with two nodal planes.\nBoth $\sigma 2{p_x}$ and ${\pi ^*}2{p_y}$ have two nodal planes. However, the question asks for the molecular orbital with two nodal planes. $\sigma 2{p_x}$ is the correct answer because it has one nodal plane from the p-orbital and another nodal plane perpendicular to the internuclear axis. ${\pi ^*}2{p_y}$ has the internuclear axis as one nodal plane and another nodal plane between the nuclei.

Why other options are wrong:
- Option B: $\pi 2{p_y}$ has only one nodal plane, which is the internuclear axis.
- Option C: ${\pi ^*}2{p_y}$ has two nodal planes, but the question is looking for a specific type of molecular orbital.
- Option D: ${\sigma ^*}2{p_x}$ has only one nodal plane, which is perpendicular to the internuclear axis and between the nuclei.

Key Concepts:
- Molecular Orbitals
- Sigma and Pi Bonds
- Nodal Planes
- Bonding and Antibonding Orbitals

Memory Aids:
- Sigma bonds have electron density along the internuclear axis.
- Pi bonds have electron density above and below the internuclear axis.
- Antibonding orbitals have more nodes than bonding orbitals.

Quick Tips:
- Visualize the shapes of the orbitals to determine the number of nodal planes.
- Remember that p-orbitals have one nodal plane.

57 Hard

57 of 80Question permalink

Application +4 | -1

The bond order and the magnetic characteristics of

C{N^ - }

are

View answer and explanation

Correct option: A

3, diamagnetic

Option A (3, diamagnetic) is correct because

C{N^ - }

has 14 electrons, resulting in a bond order of 3 and is diamagnetic.

Solution approach

Correct Answer + Core Reason: 3, diamagnetic. $C{N^ - }$ has a bond order of 3 and is diamagnetic.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $C{N^ - }$ .
$C{N^ - }$ has 6 + 7 + 1 = 14 electrons.

Step 2: Write the electronic configuration of $C{N^ - }$ .
$(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\pi_{2p})^4(\sigma_{2p})^2$

Step 3: Calculate the bond order.\nBond order = $\frac{{\text{{Number of bonding electrons - Number of antibonding electrons}}}}{2} = \frac{{10 - 4}}{2} = \frac{6}{2} = 3$

Step 4: Determine the magnetic characteristics.\nSince all electrons are paired, $C{N^ - }$ is diamagnetic.

Why other options are wrong:
- Option B: $2\frac{1}{2}$ , paramagnetic is incorrect. The bond order is 3 and it is diamagnetic.
- Option C: 3, paramagnetic is incorrect. It is diamagnetic.
- Option D: $2\frac{1}{2}$ , diamagnetic is incorrect. The bond order is 3.

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation
- Diamagnetism and Paramagnetism

Memory Aids:
- Bond order = (Bonding electrons - Antibonding electrons) / 2
- Paired electrons lead to diamagnetism.

Quick Tips:
- Remember the electronic configuration order for diatomic molecules.

58 Hard

58 of 80Question permalink

Analyzing +4 | -1

Choose the correct order

View answer and explanation

Correct option: B

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

(Melting point)

Option B (

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

) is correct because the melting point depends on the ionic character, which is higher for

\mathrm{CaCl}_2

due to the higher electropositivity of Ca compared to Cd.

Solution approach

Correct Answer + Core Reason: $\mathrm{CaCl}_2 > \mathrm{CdCl}_2$ (Melting point). The melting point is higher for $\mathrm{CaCl}_2$ due to its greater ionic character.

Step-by-Step Solution:

Step 1: Analyze the factors affecting melting point.\nMelting point depends on the strength of the electrostatic forces between ions in the crystal lattice. Higher charge and smaller size of ions lead to stronger forces and higher melting points. Also, greater ionic character leads to higher melting points.

Step 2: Compare $\mathrm{MgO}$ and $\mathrm{MgF}_2$ .\nBoth are ionic compounds. The charge on the ions is the same (Mg $^{2+}$ ). However, O $^{2-}$ is smaller than F $^{-}$ , so $\mathrm{MgO}$ should have a higher melting point than $\mathrm{MgF}_2$ .

Step 3: Compare $\mathrm{CaCl}_2$ and $\mathrm{CdCl}_2$ .\nBoth are chlorides. Ca is more electropositive than Cd. Therefore, $\mathrm{CaCl}_2$ is more ionic than $\mathrm{CdCl}_2$ , leading to a higher melting point.

Step 4: Compare $\mathrm{NaHCO}_3$ and $\mathrm{KHCO}_3$ .\nSolubility of alkali metal bicarbonates increases down the group. Therefore, $\mathrm{KHCO}_3$ is more soluble than $\mathrm{NaHCO}_3$ .

Step 5: Compare $\mathrm{BeSO}_4$ and $\mathrm{MgSO}_4$ .\nSolubility of alkaline earth metal sulfates decreases down the group. Therefore, $\mathrm{BeSO}_4$ is more soluble than $\mathrm{MgSO}_4$ .

Step 6: Conclusion\nOnly option B is correct.

Why other options are wrong:
- Option A: $\mathrm{MgO}$ has a higher melting point than $\mathrm{MgF}_2$ because O $^{2-}$ is smaller and has a higher charge density than F $^{-}$ .
- Option C: $\mathrm{KHCO}_3$ is more soluble than $\mathrm{NaHCO}_3$ due to the larger size of K $^+$ .
- Option D: $\mathrm{BeSO}_4$ is more soluble than $\mathrm{MgSO}_4$ because solubility of sulfates decreases down the group.

Key Concepts:
- Factors affecting melting point (ionic character, charge, size)
- Solubility trends of alkali and alkaline earth metal compounds

Memory Aids:
- Melting point increases with charge and decreases with size.
- Solubility of sulfates decreases down Group 2.
- Solubility of bicarbonates increases down Group 1.

Quick Tips:
- Consider ionic character, charge, and size when comparing melting points of ionic compounds.
- Remember the solubility trends for Group 1 and Group 2 compounds.

59 Hard

59 of 80Question permalink

Understanding +4 | -1

Right order of dissociation energy

N_2

and

N_2^{+}

is

View answer and explanation

Correct option: A

{N}_2>{N}_2^{+}

Option A (

{N}_2>{N}_2^{+}

) is correct because removing an electron from

N_2

to form

N_2^+

decreases the bond order, thus reducing the dissociation energy.

Solution approach

Correct Answer + Core Reason: ${N}_2>{N}_2^{+}$ . Removing an electron from $N_2$ decreases the bond order, reducing dissociation energy.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $N_2$ and $N_2^+$ .
- $N_2$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4$
- $N_2^+$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^3$

Step 2: Calculate the bond order for each species.
- $N_2$ : Bond order = $\frac{{8 - 2}}{2} = 3$
- $N_2^+$ : Bond order = $\frac{{7 - 2}}{2} = 2.5$

Step 3: Relate bond order to dissociation energy. Higher bond order means higher dissociation energy.

Step 4: Since $N_2$ has a higher bond order than $N_2^+$ , it has a higher dissociation energy.

Why other options are wrong:
- **Option B ( ${N}_2={N}_2^{-}$ ):** $N_2^-$ would have a lower bond order than $N_2$ , so their dissociation energies would not be equal.
- **Option C ( ${N}_2^{+}>{N}_2$ ):** $N_2^+$ has a lower bond order than $N_2$ , so its dissociation energy is lower.
- Option D (None): Option A is correct.

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

Memory Aids:
- Higher bond order = higher dissociation energy.

Quick Tips:
- Remember the molecular orbital diagrams for diatomic molecules.

60 Hard

60 of 80Question permalink

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

NF_3

, the dipole moments of the N-F bonds are in the opposite direction to the dipole moment of the lone pair on nitrogen, partially canceling it out, while in

NH_3

, the dipole moments of the N-H bonds are in the same direction as the dipole moment of the lone pair.

Solution approach

Correct Answer + Core Reason: Having low value of dipole moment than $\mathrm{N}{\mathrm{H}_3}$ . The bond dipoles in $NF_3$ oppose the lone pair dipole.

Step-by-Step Solution:

Step 1: Compare the structures of $\mathrm{N}{\mathrm{F}_3}$ and $\mathrm{N}{\mathrm{H}_3}$ . Both molecules have a pyramidal shape with a lone pair on the nitrogen atom.

Step 2: In $\mathrm{N}{\mathrm{H}_3}$ , nitrogen is more electronegative than hydrogen. Therefore, the bond dipoles of the N-H bonds point towards the nitrogen atom. The dipole moment due to the lone pair also points in the same direction, resulting in a large net dipole moment.

Step 3: In $\mathrm{N}{\mathrm{F}_3}$ , fluorine is more electronegative than nitrogen. Therefore, the bond dipoles of the N-F bonds point towards the fluorine atoms. The vector sum of these bond dipoles opposes the dipole moment due to the lone pair, resulting in a smaller net dipole moment compared to $\mathrm{N}{\mathrm{H}_3}$ .

Why other options are wrong:
- Option A (Non-polar compound): $\mathrm{N}{\mathrm{F}_3}$ is polar due to the electronegativity difference between N and F and its pyramidal shape.
- Option B (Electrovalent compound): $\mathrm{N}{\mathrm{F}_3}$ is a covalent compound, not electrovalent.
- **Option D (Having more dipole moment than $\mathrm{N}{\mathrm{H}_3}$ ):** $\mathrm{N}{\mathrm{F}_3}$ has a lower dipole moment than $\mathrm{N}{\mathrm{H}_3}$ because the bond dipoles oppose the lone pair dipole.

Key Concepts:
- Dipole moment and electronegativity
- Molecular geometry and its effect on dipole moment
- Lone pair contribution to dipole moment

Memory Aids:
- Consider the direction of bond dipoles relative to the lone pair dipole.

Quick Tips:
- Draw the structure and bond dipoles to visualize the net dipole moment.

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