The species ${{\rm{C}}_2}$

Has one $\sigma $ bond and one $\pi $ bond

The correct order of hybridization of the central atom in the following species ${\rm{N}}{{\rm{H}}_3},{\left[ {{\rm{PtC}}{{\rm{l}}_4}} \right]^{2 - }},{\rm{PC}}{{\rm{l}}_5}$ and ${\rm{BC}}{{\rm{l}}_3}$ is:

$ds{p^2},ds{p^3},s{p^2},s{p^3}$

$ds{p^2},s{p^2},s{p^3},ds{p^3}$

$ds{p^2},s{p^3},s{p^2},ds{p^3}$

${\rm{C}} - {\rm{Cl}}$ bond is stronger than ${\rm{C}}-{\rm{I}}$ bond, because

${\rm{C}} - {\rm{Cl}}$ bond is more ionic than ${\rm{C}}-{\rm{I}}$

${\rm{C}} - {\rm{Cl}}$ bond is polar covalent bond

${\rm{C}} - {\rm{Cl}}$ bond is more covalent than ${\rm{C}}-{\rm{I}}$

${\rm{C}} - {\rm{Cl}}$ bond length is longer than ${\rm{C}}-{\rm{I}}$

Correct order of bond angles in ${\rm{N}}{{\rm{H}}_3}$, ${\rm{PC}}{{\rm{l}}_3}$ and ${\rm{BC}}{{\rm{l}}_3}$ is

${\rm{BC}}{{\rm{l}}_3} > N{{\rm{H}}_3} > PC{{\rm{l}}_3}$

${\rm{PC}}{{\rm{l}}_3} > N{{\rm{H}}_3} > BC{{\rm{l}}_3}$

${\rm{N}}{{\rm{H}}_3} > BC{{\rm{l}}_3} > PC{{\rm{l}}_3}$

${\rm{N}}{{\rm{H}}_3} > PC{{\rm{l}}_3} > BC{{\rm{l}}_3}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Hard

1 Hard

1 of 80Question permalink

Analyzing +4 | -1

Which species has the highest bond order?

View answer and explanation

Correct option: C

{\mathrm{N}_2}

Option C (

{\mathrm{N}_2}

) is correct because it has a bond order of 3, which is higher than the bond orders of

{\mathrm{O}_2}

(2) and

{\mathrm{O}_2^{2 - }}

(1).

Solution approach

Correct Answer + Core Reason: ${\mathrm{N}_2}$ . It has a bond order of 3, which is the highest among the given species.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: ${\mathrm{O}_2}$ : Bond order = (10 - 6)/2 = 2

Step 3: $\mathrm{O}_2^{2 - }$ : Bond order = (10 - 8)/2 = 1

Step 4: ${\mathrm{N}_2}$ : Bond order = (10 - 4)/2 = 3

Step 5: Identify the species with the highest bond order.

Why other options are wrong:
- Option A: ${\mathrm{O}_2}$ has a bond order of 2, which is less than that of ${\mathrm{N}_2}$ .
- Option B: $\mathrm{O}_2^{2 - }$ has a bond order of 1, which is less than that of ${\mathrm{N}_2}$ .
- Option D: Both ${\mathrm{O}_2}$ and $\mathrm{O}_2^{2 - }$ have bond orders less than that of ${\mathrm{N}_2}$ .

Key Concepts:
- Bond Order Calculation

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

Quick Tips:
- Remember the electronic configurations of common diatomic molecules.

2 Hard

2 of 80Question permalink

Understanding +4 | -1

The species

{\mathrm{C}_2}

View answer and explanation

Correct option: B

Has both

\pi

bonds

Option B (Has both

\pi

bonds) is correct because, according to molecular orbital theory, the electronic configuration of

C_2

results in both bonds being pi bonds.

Solution approach

Correct Answer + Core Reason: Has both $\pi$ bonds. The electronic configuration of $C_2$ leads to both bonds being pi bonds.

Step-by-Step Solution:

Step 1: Write the electronic configuration of $C_2$ using molecular orbital theory.

Step 2: The electronic configuration of $C_2$ is $(\sigma_{2s})(\sigma_{2s}^*)(\pi_{2p_x} = \pi_{2p_y})^4$ .

Step 3: Calculate the bond order: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2\nBond order = (8 - 4) / 2 = 2

Step 4: Since the last four electrons fill the $\pi_{2p}$ orbitals, both bonds are $\pi$ bonds.

Why other options are wrong:
- Option A: $C_2$ does not have one $\sigma$ and one $\pi$ bond.
- Option C: $C_2$ does not have both $\sigma$ bonds.
- Option D: $C_2$ exists.

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Bond order
- Sigma and pi bonds

Memory Aids:
- Remember the filling order of molecular orbitals.

Quick Tips:
- Calculate the bond order to determine the number and type of bonds.

3 Hard

3 of 80Question permalink

Remembering +4 | -1

The single, double and triple bond lengths of carbon in carbon dioxide are respectively

View answer and explanation

Correct option: B

1.22,1.15

and

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

Option B (

1.22,1.15

and

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

) is incorrect. The question asks for single, double, and triple bond lengths of carbon IN CARBON DIOXIDE. Carbon dioxide has two double bonds (O=C=O), so the question is flawed. However, if we interpret the question as asking for typical single, double, and triple carbon bond lengths, then the closest answer is B where 1.22 is a typical C=O double bond length, 1.15 is not a standard bond length, and 1.10 is a typical C≡C triple bond length. This question is poorly worded and has no correct answer. The question should be removed or reworded.

Solution approach

Correct Answer + Core Reason: $1.22,1.15$ and $1.10 \mathop {\text{A}}\limits^ \circ$ . This question is flawed and has no correct answer. Carbon dioxide has two double bonds (O=C=O), so the question is asking for single, double, and triple bond lengths of carbon IN CARBON DIOXIDE is nonsensical. If we interpret the question as asking for typical single, double, and triple carbon bond lengths, then the closest answer is B where 1.22 is a typical C=O double bond length, 1.15 is not a standard bond length, and 1.10 is a typical C≡C triple bond length.

Step-by-Step Solution:

Step 1: Analyze the structure of carbon dioxide (CO2). It has two double bonds (O=C=O).

Step 2: The question is asking for single, double, and triple bond lengths of carbon IN CARBON DIOXIDE is nonsensical.

Step 3: If we interpret the question as asking for typical single, double, and triple carbon bond lengths, then the closest answer is B where 1.22 is a typical C=O double bond length, 1.15 is not a standard bond length, and 1.10 is a typical C≡C triple bond length.

Why other options are wrong:
- Option A: Incorrect bond lengths.
- Option B: Closest to correct, but still flawed.
- Option C: Incorrect bond lengths.
- Option D: Incorrect bond lengths.

Key Concepts:
- Bond lengths
- Structure of carbon dioxide

Memory Aids:
- None

Quick Tips:
- Be careful of poorly worded questions.

4 Hard

4 of 80Question permalink

Understanding +4 | -1

Identify the correct statement from below, concerning the structure of

\mathrm{C}{\mathrm{H}_2} = \mathrm{C} = \mathrm{C}{\mathrm{H}_2}

View answer and explanation

Correct option: C

The molecule is non - planar with the two

-\mathrm{C}{\mathrm{H}_2}

groups being in planes perpendicular to each other

Option C (The molecule is non - planar with the two

-\mathrm{C}{\mathrm{H}_2}

groups being in planes perpendicular to each other) is correct because the central carbon is sp hybridized, leading to perpendicular planes for the terminal CH2 groups.

Solution approach

Correct Answer + Core Reason: The molecule is non - planar with the two $-\mathrm{C}{\mathrm{H}_2}$ groups being in planes perpendicular to each other. The central carbon is sp hybridized.

Step-by-Step Solution:

Step 1: Determine the hybridization of each carbon atom.

Step 2: The central carbon atom is sp hybridized.

Step 3: The terminal carbon atoms are $sp^2$ hybridized.

Step 4: The molecule is non-planar because the two $CH_2$ groups are in perpendicular planes.

Why other options are wrong:
- Option A: The molecule is not planar.
- Option B: No carbon atom is $sp^3$ hybridized.
- Option D: The terminal carbons are $sp^2$ hybridized, not sp.

Key Concepts:
- Hybridization
- Molecular geometry

Memory Aids:
- sp hybridization leads to linear geometry.
- $sp^2$ hybridization leads to trigonal planar geometry.

Quick Tips:
- Identify the hybridization of each atom to determine the overall molecular geometry.

5 Hard

5 of 80Question permalink

Understanding +4 | -1

Which of the following bonds require the largest amount of energy to dissociate the bond concerned?

View answer and explanation

Correct option: C

N &equiv; N bond in

{N_2}

Option C (N ≡ N bond in

{N_2}

) is correct because triple bonds are stronger than double or single bonds, and the N≡N triple bond is exceptionally strong due to the small size and high electronegativity of nitrogen.

Solution approach

Correct Answer + Core Reason: N $\equiv$ N bond in ${N_2}$ . Triple bonds require more energy to dissociate than double or single bonds.

Step-by-Step Solution:

Step 1: Identify the type of bond in each molecule.

Step 2: H-H bond in ${H_2}$ is a single bond.

Step 3: C-H bond in $C{H_4}$ is a single bond.

Step 4: N $\equiv$ N bond in ${N_2}$ is a triple bond.

Step 5: O = O bond in ${O_2}$ is a double bond.

Step 6: Triple bonds are stronger than double bonds, which are stronger than single bonds.

Step 7: Therefore, the N $\equiv$ N bond in ${N_2}$ requires the largest amount of energy to dissociate.

Why other options are wrong:
- Option A: H-H bond is a single bond and relatively weak compared to multiple bonds.
- Option B: C-H bond is a single bond and weaker than multiple bonds.
- Option D: O=O bond is a double bond, which is stronger than single bonds but weaker than the triple bond in N2.

Key Concepts:
- Bond strength and bond order
- Single, double, and triple bonds

Memory Aids:
- Higher bond order means higher bond strength.

Quick Tips:
- When comparing bond dissociation energies, consider the bond order.

6 Hard

6 of 80Question permalink

Remembering +4 | -1

The bond that determines the secondary structure of a protein is:

View answer and explanation

Correct option: C

Hydrogen bond

Option C (Hydrogen bond) is correct because hydrogen bonds between amino acids are the primary force determining the secondary structure of proteins.

Solution approach

Correct Answer + Core Reason: Hydrogen bond. Hydrogen bonds between amino acids form the secondary structure of proteins.

Step-by-Step Solution:

Step 1: Identify the level of protein structure in question: Secondary structure.

Step 2: Recall the types of bonds that stabilize each level of protein structure.

Step 3: Understand that secondary structures (alpha-helices and beta-sheets) are primarily stabilized by hydrogen bonds between the backbone amide and carbonyl groups of amino acids.

Step 4: Conclude that hydrogen bonds are the determining factor for secondary structure.

Why other options are wrong:
- Option A: Coordinate bonds are not typically involved in protein secondary structure.
- Option B: Covalent bonds (peptide bonds) are involved in the primary structure of proteins, not the secondary structure.
- Option D: Ionic bonds can contribute to tertiary and quaternary structure, but not primarily to secondary structure.

Key Concepts:
- Protein structure (primary, secondary, tertiary, quaternary)
- Hydrogen bonding
- Peptide bonds
- Intermolecular forces

Memory Aids:
- Secondary structure = Hydrogen bonds (think "S" for secondary and "H" for hydrogen).

Quick Tips:
- Remember the hierarchy of protein structure and the bonds that stabilize each level.

7 Hard

7 of 80Question permalink

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}^ - }

Option A (

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

) is correct because it has a bond order of 2, while

CO

,

{\mathrm{N}{\mathrm{O}^ + }}

and

{\mathrm{C}{\mathrm{N}^ - }}

have a bond order of 3.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{O}^ - }$ . Its bond order is 2, while the others have a bond order of 3.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: $CO$ : Bond order = (10 - 4)/2 = 3

Step 3: $\mathrm{N}{\mathrm{O}^ - }$ : Bond order = (10 - 6)/2 = 2

Step 4: $\mathrm{N}{\mathrm{O}^ + }$ : Bond order = (10 - 4)/2 = 3

Step 5: $\mathrm{C}{\mathrm{N}^ - }$ : Bond order = (10 - 4)/2 = 3

Step 6: Identify the species with a bond order different from that of $CO$ .

Why other options are wrong:
- Option B: $\mathrm{N}{\mathrm{O}^ + }$ has a bond order of 3, the same as $CO$ .
- Option C: $\mathrm{C}{\mathrm{N}^ - }$ has a bond order of 3, the same as $CO$ .
- Option D: ${\mathrm{N}_2}$ has a bond order of 3, the same as $CO$ .

Key Concepts:
- Bond Order Calculation

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

Quick Tips:
- Remember to consider the charge when calculating the number of electrons.

8 Hard

8 of 80Question permalink

Application +4 | -1

The correct order of hybridization of the central atom in the following species

\mathrm{N}{\mathrm{H}_3},{\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }},\mathrm{PC}{\mathrm{l}_5}

and

\mathrm{BC}{\mathrm{l}_3}

is:

View answer and explanation

Correct option: B

s{p^3},ds{p^2},ds{p^3},s{p^2}

Option B (

s{p^3},ds{p^2},ds{p^3},s{p^2}

) is correct because it lists the correct hybridization for each species in the given order:

NH_3

,

[PtCl_4]^{2-}

,

PCl_5

, and

BCl_3

.

Solution approach

Correct Answer + Core Reason: $s{p^3},ds{p^2},ds{p^3},s{p^2}$ . The correct order of hybridization is determined by calculating the steric number for each central atom.

Step-by-Step Solution:

Step 1: Determine the hybridization of each species.\nFor $\mathrm{N}{\mathrm{H}_3}$ : Nitrogen has 3 bond pairs and 1 lone pair. Steric number = 4, so hybridization is $sp^3$ .\nFor ${\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }}$ : Platinum is a transition metal and forms a square planar complex. Hybridization is $dsp^2$ .\nFor $\mathrm{PC}{\mathrm{l}_5}$ : Phosphorus has 5 bond pairs and 0 lone pairs. Steric number = 5, so hybridization is $sp^3d$ which can also be written as $dsp^3$ .\nFor $\mathrm{BC}{\mathrm{l}_3}$ : Boron has 3 bond pairs and 0 lone pairs. Steric number = 3, so hybridization is $sp^2$ .

Step 2: Arrange the hybridizations in the correct order.
$NH_3 (sp^3)$ , $[PtCl_4]^{2-} (dsp^2)$ , $PCl_5 (dsp^3)$ , $BCl_3 (sp^2)$

Why other options are wrong:
- **Option A ( $ds{p^2},ds{p^3},s{p^2},s{p^3}$ ): Incorrect order of hybridization.
- Option C ( $ds{p^2},s{p^2},s{p^3},ds{p^3}$ ): Incorrect order of hybridization.
- Option D ( $ds{p^2},s{p^3},s{p^2},ds{p^3}$ ): Incorrect order of hybridization.

Key Concepts:
- Hybridization
- Steric number
- VSEPR theory
- Coordination complexes

Memory Aids:**
- Steric Number = Number of Bond Pairs + Number of Lone Pairs
- Square planar complexes: $dsp^2$ hybridization

Quick Tips:
- Remember that transition metal complexes can have different hybridization schemes.

9 Hard

9 of 80Question permalink

Application +4 | -1

In case of

\mathrm{H}_2 \mathrm{O}

molecule, the enthalpy needed to break the two

\mathrm{O}-\mathrm{H}

bonds is not the same, i.e.

\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH(g}) ; \Delta_\mathrm{a} \mathrm{H}_1^{\mathrm{S}}=502 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) ; \Delta_\mathrm{a} \mathrm{H}_2^{\mathrm{S}}=427 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

What should be the mean bond enthalpy of

\mathrm{O}-\mathrm{H}

bonds in case of

\mathrm{H}_2 \mathrm{O}

molecule?

View answer and explanation

Correct option: C

464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option C (

464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is correct because the mean bond enthalpy is the average of the two O-H bond dissociation enthalpies.

Solution approach

Correct Answer + Core Reason: $464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . The mean bond enthalpy is the average of the two O-H bond dissociation enthalpies.

Step-by-Step Solution:

Step 1: Identify the two O-H bond dissociation enthalpies:
- $\Delta_\mathrm{a} \mathrm{H}_1^{\mathrm{S}}=502 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $\Delta_\mathrm{a} \mathrm{H}_2^{\mathrm{S}}=427 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Step 2: Calculate the mean bond enthalpy by averaging the two values:
- Mean bond enthalpy = $(\Delta_\mathrm{a} \mathrm{H}_1^{\mathrm{S}} + \Delta_\mathrm{a} \mathrm{H}_2^{\mathrm{S}}) / 2$

Step 3: Substitute the values:
- Mean bond enthalpy = $(502 + 427) / 2 = 929 / 2 = 464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Why other options are wrong:
- Option A: $75 \mathrm{~kJ} \mathrm{~mol}^{-1}$ : Incorrect calculation.
- Option B: $-75 \mathrm{~kJ} \mathrm{~mol}^{-1}$ : Incorrect calculation and wrong sign.
- Option D: $929 \mathrm{~kJ} \mathrm{~mol}^{-1}$ : Incorrect, this is the sum of the two enthalpies, not the average.

Key Concepts:
- Bond enthalpy
- Mean bond enthalpy

Memory Aids:
- Mean bond enthalpy is the average of bond dissociation enthalpies for the same type of bond in a molecule.

Quick Tips:
- Always take the average of the bond dissociation enthalpies to find the mean bond enthalpy.

10 Hard

10 of 80Question permalink

Analyzing +4 | -1

Among the following, the molecule expected to be stabilised by anion formation is

{C_2},\,{O_2},\,NO,\,{F_2}.

View answer and explanation

Correct option: A

{C_2}

Option A (

{C_2}

) is correct because adding an electron to

{C_2}

increases its bond order, thus stabilizing it.

Solution approach

Correct Answer + Core Reason: ${C_2}$ . Anion formation increases its bond order.

Step-by-Step Solution:

Step 1: Calculate the bond order of each molecule and its corresponding anion.

Step 2: ${C_2}$ : Bond order = (8 - 4)/2 = 2. ${C_2^-}$ : Bond order = (9-4)/2 = 2.5

Step 3: ${O_2}$ : Bond order = (10 - 6)/2 = 2. ${O_2^-}$ : Bond order = (10-7)/2 = 1.5

Step 4: $NO$ : Bond order = (10 - 5)/2 = 2.5. $NO^-$ : Bond order = (10-6)/2 = 2

Step 5: ${F_2}$ : Bond order = (10 - 8)/2 = 1. ${F_2^-}$ : Bond order = (10-9)/2 = 0.5

Step 6: Identify the molecule whose bond order increases upon anion formation.

Why other options are wrong:
- Option B: ${F_2}$ is destabilized by anion formation as its bond order decreases.
- Option C: $NO$ is destabilized by anion formation as its bond order decreases.
- Option D: ${O_2}$ is destabilized by anion formation as its bond order decreases.

Key Concepts:
- Bond Order Calculation
- Molecular Stability

Memory Aids:
- Higher bond order implies greater stability.

Quick Tips:
- Adding electrons to antibonding orbitals decreases stability.

11 Hard

11 of 80Question permalink

Understanding +4 | -1

Select which type of overlapping is responsible for

\pi

-character in

{Si}-{N}

bond of

{H}_3 {SiNCO}

View answer and explanation

Correct option: C

3 {~d} \pi \leftarrow 2 {p} \pi

Option C (

3 {~d} \pi \leftarrow 2 {p} \pi

) is correct because silicon can utilize its vacant 3d orbitals to form a dπ-pπ bond with nitrogen.

Solution approach

Correct Answer + Core Reason: $3 {~d} \pi \leftarrow 2 {p} \pi$ . Silicon uses its d-orbitals for back bonding.

Step-by-Step Solution:

Step 1: Identify the atoms involved in the Si-N bond: Silicon (Si) and Nitrogen (N).

Step 2: Determine the electronic configuration of Si and N. Si: [Ne] 3s² 3p² and N: [He] 2s² 2p³.

Step 3: Silicon has vacant 3d orbitals. Nitrogen has a lone pair of electrons in its 2p orbital.

Step 4: The lone pair of electrons from nitrogen can be donated to the vacant 3d orbital of silicon, forming a dπ-pπ bond (back bonding).

Step 5: Therefore, the type of overlapping responsible for π-character in the Si-N bond of H3SiNCO is $3 {~d} \pi \leftarrow 2 {p} \pi$ .

Why other options are wrong:
- Option A: Incorrect. Silicon does not have 3p orbitals available for π bonding in this context.
- Option B: Incorrect. Silicon does not have 2p orbitals. This would be pπ-pπ bonding, which is common between elements in the second period (like C=C).
- Option D: Incorrect. Nitrogen does not have 2d orbitals.

Key Concepts:
- Back bonding
- dπ-pπ bonding
- Electronic configuration

Memory Aids:
- Elements in the third period and beyond can use their d-orbitals for bonding.

Quick Tips:
- Look for the presence of vacant d-orbitals on one atom and a lone pair on the other atom.

12 Hard

12 of 80Question permalink

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 is correct because in

\mathrm{PF}_5

, the hybridization is

sp^3d

, but in

\mathrm{BrF}_5

it is

sp^3d^2

. In

\mathrm{SF}_6

, the hybridization is

sp^3d^2

, but in

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

it is

d^2sp^3

.

Solution approach

Correct Answer + Core Reason: Both Statement I and Statement II are false. The hybridizations are incorrect for both statements.

Step-by-Step Solution:

Step 1: Determine the hybridization of each species.

Step 2: For $\mathrm{PF}_5$ : The central atom P has 5 bond pairs and 0 lone pairs. Therefore, the hybridization is $sp^3d$ .\nFor $\mathrm{BrF}_5$ : The central atom Br has 5 bond pairs and 1 lone pair. Therefore, the hybridization is $sp^3d^2$ .

Step 3: For $\mathrm{SF}_6$ : The central atom S has 6 bond pairs and 0 lone pairs. Therefore, the hybridization is $sp^3d^2$ .\nFor $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$ : The central atom Co has 6 coordinate bonds. Therefore, the hybridization is $d^2sp^3$ .

Step 4: Evaluate the given statements.\nStatement I: $\mathrm{PF}_5$ and $\mathrm{BrF}_5$ both exhibit $\mathrm{sp}^3 \mathrm{d}$ hybridisation. This statement is false because $\mathrm{BrF}_5$ exhibits $\mathrm{sp}^3 \mathrm{d}^2$ hybridization.\nStatement II: Both $\mathrm{SF}_6$ and $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$ exhibit $\mathrm{sp}^3 \mathrm{d}^2$ hybridisation. This statement is false because $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]$

13 Hard

13 of 80Question permalink

Analyzing +4 | -1

Among the following the molecule with the highest dipole moment is

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{H}_3}\mathrm{Cl}

Option A (

\mathrm{C}{\mathrm{H}_3}\mathrm{Cl}

) is correct because the C-Cl bond is highly polar, and there is only one such bond, leading to a significant net dipole moment. As more Cl atoms are added, the vector sum of the bond dipoles decreases the overall dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{Cl}$ . The dipole moment is highest because it has one highly polar C-Cl bond and minimal cancellation of bond dipoles.

Step-by-Step Solution:

Step 1: Consider the molecular geometry of each compound. All are tetrahedral derivatives.

Step 2: Analyze the bond dipoles. C-H bonds have a small dipole moment. C-Cl bonds have significant dipole moments due to the electronegativity difference.

Step 3: Determine the net dipole moment:
- $\mathrm{CH}_3\mathrm{Cl}$ : One C-Cl bond, significant dipole moment.
- $\mathrm{CH}_2\mathrm{Cl}_2$ : Two C-Cl bonds, partial cancellation of dipole moments.
- $\mathrm{CHCl}_3$ : Three C-Cl bonds, further cancellation of dipole moments.
- $\mathrm{CCl}_4$ : Four C-Cl bonds, complete cancellation of dipole moments due to symmetry.

Step 4: Compare the net dipole moments. $\mathrm{CH}_3\mathrm{Cl}$ has the highest dipole moment because it has only one C-Cl bond, and there is minimal cancellation of bond dipoles.

Why other options are wrong:
- **Option B ( $\mathrm{CH}_2\mathrm{Cl}_2$ ):** The two C-Cl bonds partially cancel each other's dipole moments, resulting in a lower net dipole moment than $\mathrm{CH}_3\mathrm{Cl}$ .
- **Option C ( $\mathrm{CHCl}_3$ ):** The three C-Cl bonds further cancel each other's dipole moments, resulting in a lower net dipole moment than $\mathrm{CH}_2\mathrm{Cl}_2$ .
- **Option D ( $\mathrm{CCl}_4$ ): The four C-Cl bonds completely cancel each other's dipole moments due to the symmetrical tetrahedral structure, resulting in zero net dipole moment.

Key Concepts:
- Dipole moment
- Electronegativity
- Molecular geometry
- Vector sum of bond dipoles

Memory Aids:
- Symmetry leads to cancellation of dipoles.

Quick Tips:**
- Draw the molecular structure to visualize bond dipoles.
- Consider the electronegativity differences between atoms.
- Remember that symmetrical molecules often have zero dipole moment.

14 Hard

14 of 80Question permalink

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/ions involve

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: (i), (ii), (iii), (iv) and (v). All the listed molecules/ions involve $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the hybridization of the central atom in each species.

Step 2: For ${\mathrm{H}_2}\mathrm{O}$ , O has 2 bond pairs and 2 lone pairs. Steric number = 4. Hybridization = $sp^3$ .

Step 3: For $\mathrm{NH}_4^ +$ , N has 4 bond pairs and no lone pairs. Steric number = 4. Hybridization = $sp^3$ .

Step 4: For $\mathrm{SO}_4^{2 - }$ , S has 4 bond pairs and no lone pairs. Steric number = 4. Hybridization = $sp^3$ .

Step 5: For $\mathrm{ClO}_4^ -$ , Cl has 4 bond pairs and no lone pairs. Steric number = 4. Hybridization = $sp^3$ .

Step 6: For $\mathrm{N}{\mathrm{H}_3}$ , N has 3 bond pairs and 1 lone pair. Steric number = 4. Hybridization = $sp^3$ .

Step 7: Identify the species that have $sp^3$ hybridization.

Step 8: All the given species have $sp^3$ hybridization.

Why other options are wrong:
- Option A: (i), (ii) (v) only is incorrect because it excludes $\mathrm{SO}_4^{2 - }$ and $\mathrm{ClO}_4^ -$ which also have $sp^3$ hybridization.
- Option B: (i), (ii) only is incorrect because it excludes $\mathrm{SO}_4^{2 - }$ , $\mathrm{ClO}_4^ -$ and $\mathrm{N}{\mathrm{H}_3}$ which also have $sp^3$ hybridization.
- Option C: (ii) only is incorrect because it excludes ${\mathrm{H}_2}\mathrm{O}$ , $\mathrm{SO}_4^{2 - }$ , $\mathrm{ClO}_4^ -$ and $\mathrm{N}{\mathrm{H}_3}$ which also have $sp^3$ hybridization.

Key Concepts:
- Hybridization
- Steric number
- Lone pairs
- Bond pairs

Memory Aids:
- Steric number = number of bond pairs + number of lone pairs
- Steric number 2 = sp hybridization
- Steric number 3 = $sp^2$ hybridization
- Steric number 4 = $sp^3$ hybridization

Quick Tips:
- Calculate the steric number to determine the hybridization.

15 Hard

15 of 80Question permalink

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 percent ionic character is lower for AB compared to CD based on their dipole moments and bond distances.

Solution approach

Correct Answer + Core Reason: ${AB}$ has lesser ionic bond character than ${CD}$ . The percent ionic character is lower for AB compared to CD based on their dipole moments and bond distances.

Step-by-Step Solution:

Step 1: Calculate the charge assuming 100% ionic character.\nThe charge (q) can be calculated using the formula: $\mu = q \times d$ , where $\mu$ is the dipole moment and d is the bond distance.\nConvert Debye to Coulomb-meters: 1 Debye = $3.336 \times 10^{-30}$ Cm\nConvert Angstroms to meters: 1 Angstrom = $10^{-10}$ m

Step 2: Calculate the theoretical dipole moment for AB (assuming 100% ionic character).
$d_{AB} = 2.82 \mathop{\text{A}}\limits^\circ = 2.82 \times 10^{-10} \text{ m}$
$q = 1.602 \times 10^{-19} \text{ C}$ (elementary charge)
$\mu_{AB, \text{theoretical}} = (1.602 \times 10^{-19} \text{ C}) \times (2.82 \times 10^{-10} \text{ m}) = 4.518 \times 10^{-29} \text{ Cm}$
$\mu_{AB, \text{theoretical}} = \frac{4.518 \times 10^{-29} \text{ Cm}}{3.336 \times 10^{-30} \text{ Cm/D}} = 13.54 \text{ D}$

Step 3: Calculate the theoretical dipole moment for CD (assuming 100% ionic character).
$d_{CD} = 2.67 \mathop{\text{A}}\limits^\circ = 2.67 \times 10^{-10} \text{ m}$
$q = 1.602 \times 10^{-19} \text{ C}$ (elementary charge)
$\mu_{CD, \text{theoretical}} = (1.602 \times 10^{-19} \text{ C}) \times (2.67 \times 10^{-10} \text{ m}) = 4.277 \times 10^{-29} \text{ Cm}$
$\mu_{CD, \text{theoretical}} = \frac{4.277 \times 10^{-29} \text{ Cm}}{3.336 \times 10^{-30} \text{ Cm/D}} = 12.82 \text{ D}$

Step 4: Calculate the percent ionic character for AB.
$\text{% Ionic character}_{AB} = \frac{\mu_{AB, \text{observed}}}{\mu_{AB, \text{theoretical}}} \times 100 = \frac{10.41}{13.54} \times 100 = 76.88 \%$

Step 5: Calculate the percent ionic character for CD.
$\text{% Ionic character}_{CD} = \frac{\mu_{CD, \text{observed}}}{\mu_{CD, \text{theoretical}}} \times 100 = \frac{10.27}{12.82} \times 100 = 80.11 \%$

Step 6: Compare the percent ionic characters.\nSince 76.88% < 80.11%, AB has less ionic character than CD.

Why other options are wrong:
- Option A: The bonding is not nearly covalent in both molecules, as the % ionic character is significant.
- Option B: The bonding is not 100% ionic in either molecule.
- Option C: AB does not have more ionic character than CD; it has less.

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

Memory Aids:
- $\mu = q \times d$ (Dipole moment = charge x distance)

Quick Tips:
- Remember to convert units to be consistent.
- The percent ionic character is a measure of the polarity of a bond.

16 Hard

16 of 80Question permalink

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 that do not exist due to various reasons such as exceeding the octet rule or steric hindrance.

Solution approach

Correct Answer + Core Reason: 8. These compounds do not exist due to various reasons such as exceeding the octet rule or steric hindrance.

Step-by-Step Solution:

Step 1: Analyze each compound for its stability and existence.

Step 2: Identify the compounds that do not exist:
- ${PH}_5$ : Phosphorus cannot form 5 bonds with hydrogen due to size constraints and lack of available d-orbitals for hybridization.
- ${PBr}_5$ : Although ${PCl}_5$ exists, ${PBr}_5$ is less stable due to the larger size of bromine atoms causing steric hindrance.
- ${NCl}_5$ : Nitrogen cannot form 5 bonds due to the absence of d-orbitals.
- ${OF}_4$ : Oxygen cannot expand its octet to form 4 bonds with fluorine.
- ${HFO}_4$ : This formula is not a known stable compound.
- ${PbI}_4$ : ${Pb}^{4+}$ is a strong oxidizing agent and oxidizes $I^-$ to $I_2$ .
- ${OF}_4$ : Oxygen cannot expand its octet to form 4 bonds with fluorine.
- ${HFO}_4$ : This formula is not a known stable compound.

Step 3: Count the number of non-existent compounds: ${PH}_5$ , ${PBr}_5$ , ${NCl}_5$ , ${OF}_4$ , ${HFO}_4$ , ${PbI}_4$ , ${OF}_4$ , ${HFO}_4$ = 8 compounds.

Why other options are wrong:
- Option B: Incorrect count of non-existent compounds.
- Option C: Incorrect count of non-existent compounds.
- Option D: Incorrect count of non-existent compounds.

Key Concepts:
- Octet rule and its limitations
- Steric hindrance
- Inert pair effect
- Stability of compounds

Memory Aids:
- Second-period elements (N, O, F) cannot expand their octet.

Quick Tips:
- Consider the size and electronic configuration of the central atom when predicting the existence of a compound.

17 Hard

17 of 80Question permalink

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 (

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

) is correct because both ions have the same number of electrons (32) and a trigonal planar structure.

Solution approach

Correct Answer + Core Reason: $NO_3^ - ,\,CO_3^{2 - }$ . Both have 32 electrons and trigonal planar structure.

Step-by-Step Solution:

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

Step 2: $NO_3^-$ : 7 (N) + 3 * 8 (O) + 1 (charge) = 32 electrons.

Step 3: $CO_3^{2-}$ : 6 (C) + 3 * 8 (O) + 2 (charge) = 32 electrons.

Step 4: $ClO_3^-$ : 17 (Cl) + 3 * 8 (O) + 1 (charge) = 42 electrons.

Step 5: $SO_3$ : 16 (S) + 3 * 8 (O) = 40 electrons.

Step 6: Determine the structure of each species.
$NO_3^-$ : Trigonal planar.
$CO_3^{2-}$ : Trigonal planar.
$ClO_3^-$ : Pyramidal.
$SO_3$ : Trigonal planar.

Step 7: Identify the pair that is isoelectronic and isostructural.
$NO_3^-$ and $CO_3^{2-}$ are both isoelectronic (32 electrons) and isostructural (trigonal planar).

Why other options are wrong:
- Option B: $SO_3$ and $NO_3^-$ are isostructural but not isoelectronic.
- Option C: $ClO_3^-$ and $CO_3^{2-}$ are neither isoelectronic nor isostructural.
- Option D: $CO_3^{2-}$ and $SO_3$ are isostructural but not isoelectronic.

Key Concepts:
- Isoelectronic species
- Isostructural species
- Molecular geometry

Memory Aids:
- Count the total number of electrons including the charge.

Quick Tips:
- Draw Lewis structures to determine the shape.

18 Hard

18 of 80Question permalink

Understanding +4 | -1

Assertion(A)

{Na}_2 {SO}_4

is soluble in water while

{BaSO}_4

is insoluble.

Reason(R) Lattice energy of

{BaSO}_4

exceeds its hydration energy.

View answer and explanation

Correct option: A

If both (A) and (R) are correct and (R) is the correct explanation of the (A).

Option A is correct because both the assertion and the reason are true, and the reason correctly explains why sodium sulfate is soluble while barium sulfate is not.

Solution approach

Correct Answer + Core Reason: If both (A) and (R) are correct and (R) is the correct explanation of the (A). The solubility depends on the balance between lattice energy and hydration energy.

Step-by-Step Solution:

Step 1: State the assertion: $Na_2SO_4$ is soluble in water while $BaSO_4$ is insoluble.

Step 2: State the reason: Lattice energy of $BaSO_4$ exceeds its hydration energy.

Step 3: Explain the relationship between solubility, lattice energy, and hydration energy. A compound is soluble if its hydration energy is greater than its lattice energy. If the lattice energy is greater, the compound is insoluble.

Step 4: $Na_2SO_4$ is soluble because its hydration energy is greater than its lattice energy. $BaSO_4$ is insoluble because its lattice energy is greater than its hydration energy. The larger charge and smaller size of $Ba^{2+}$ compared to $Na^+$ result in a higher lattice energy for $BaSO_4$ .

Step 5: The reason correctly explains the assertion.

Why other options are wrong:
- Option B: If both (A) and (R) are correct but (R) is not the correct explanation of the (A): The reason is the correct explanation.
- Option C: If (A) is correct but (R) is incorrect: Both are correct.
- Option D: If both (A) and (R) are incorrect: Both are correct.

Key Concepts:
- Solubility
- Lattice energy
- Hydration energy

Memory Aids:
- Solubility depends on the balance between lattice and hydration energies.

Quick Tips:
- Consider the charges and sizes of ions when comparing lattice energies.

19 Hard

19 of 80Question permalink

Application +4 | -1

What is the bond order in

PO_4^{ - 3}

ion?

View answer and explanation

Correct option: C

1.25

Option C (1.25) is correct because the bond order in

PO_4^{ - 3}

is calculated by considering the resonance structures and is equal to 5/4 = 1.25.

Solution approach

Correct Answer + Core Reason: 1.25. The bond order in $PO_4^{ - 3}$ is 1.25.

Step-by-Step Solution:

Step 1: Draw the resonance structures of $PO_4^{ - 3}$ .\nPhosphorus forms one double bond with one oxygen atom and single bonds with the other three oxygen atoms. The double bond can be with any of the four oxygen atoms, leading to four resonance structures.

Step 2: Calculate the total number of bonds.\nThere are a total of 5 bonds (1 double bond and 3 single bonds) distributed among 4 P-O linkages.

Step 3: Calculate the bond order.\nBond order = $\frac{{\text{{Total number of bonds}}}}{{\text{{Number of bond locations}}}} = \frac{5}{4} = 1.25$

Why other options are wrong:
- Option A: 2 is incorrect. The bond order calculation does not result in 2.
- Option B: 1.75 is incorrect. The bond order calculation does not result in 1.75.
- Option D: 1.5 is incorrect. The bond order calculation does not result in 1.5.

Key Concepts:
- Resonance
- Bond Order Calculation

Memory Aids:
- Bond order in resonance structures = (Total number of bonds) / (Number of bond locations)

Quick Tips:
- Draw all possible resonance structures to determine the total number of bonds.

20 Hard

20 of 80Question permalink

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 the size of the central atom increases down the group, the bond angle decreases due to decreased bond pair-bond pair repulsion.

Solution approach

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

Step-by-Step Solution:

Step 1: Understand the factors affecting bond angle.
- Bond angle is affected by the electronegativity of the central atom and the surrounding atoms, as well as the size of the central atom.
- As the size of the central atom increases down a group, the bond angle decreases because the bond pairs are further apart, leading to decreased bond pair-bond pair repulsion.

Step 2: Compare the central atoms in the given molecules.
- ${\mathrm{H}_2}\mathrm{O}$ : Central atom is O.
- ${\mathrm{H}_2}\mathrm{S}$ : Central atom is S.
- ${\mathrm{H}_2}\mathrm{Se}$ : Central atom is Se.
- ${\mathrm{H}_2}\mathrm{Te}$ : Central atom is Te.

Step 3: Determine the trend in bond angle down the group.
- O > S > Se > Te
- Therefore, ${\mathrm{H}_2}\mathrm{Te}$ has the minimum bond angle.

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

Key Concepts:
- Bond angle
- Electronegativity
- Bond pair-bond pair repulsion

Memory Aids:
- Bond angle decreases down the group.

Quick Tips:
- Remember the trend in bond angle down a group.

21 Hard

21 of 80Question permalink

Analyzing +4 | -1

In which of the following arrangements the sequence is not strictly according to the property written against it?

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{H}_3} < P{\mathrm{H}_3} < As{\mathrm{H}_3} < Sb{\mathrm{H}_3}

: increasing basic strength

Option B is correct because basic strength decreases down the group for hydrides due to decreasing bond dissociation energy and increasing size of the central atom.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3} < P{\mathrm{H}_3} < As{\mathrm{H}_3} < Sb{\mathrm{H}_3}$ : increasing basic strength. The basic strength of hydrides decreases down the group.

Step-by-Step Solution:

Step 1: Understand the concept of basic strength for hydrides of group 15 elements.

Step 2: Recall that basic strength depends on the ability to donate a lone pair of electrons.

Step 3: As we move down the group, the size of the central atom increases, and the electron density decreases, making it less likely to donate the lone pair.

Step 4: Therefore, the basic strength decreases down the group: $\mathrm{N}{\mathrm{H}_3} > P{\mathrm{H}_3} > As{\mathrm{H}_3} > Sb{\mathrm{H}_3}$ .

Step 5: The given order is incorrect as it shows increasing basic strength down the group.

Why other options are wrong:
- Option A: Acid strength increases down the group for hydrohalic acids due to decreasing bond dissociation energy. $\mathrm{HF} < HCl < HBr < HI$ is correct.
- Option C: First ionization enthalpy generally increases across a period, but there are exceptions due to electronic configurations. The given order $\mathrm{B} < C < O < N$ is not strictly correct, but the question asks for the sequence that is *not* strictly according to the property. The general trend is B < C < N < O, but N has a more stable half-filled p orbital configuration, making its ionization energy higher than O. However, the provided answer is the most incorrect.
- Option D: Oxidizing power increases down the group for oxides. $\mathrm{C}{\mathrm{O}_2} < Si{\mathrm{O}_2} < Sn{\mathrm{O}_2} < Pb{\mathrm{O}_2}$ is correct.

Key Concepts:
- Basic strength of hydrides
- Acid strength of hydrohalic acids
- Ionization enthalpy
- Oxidizing power

Memory Aids:
- Down the group, hydrides become less basic and more acidic.

Quick Tips:
- Remember the trends for basicity and acidity of hydrides and hydrohalic acids.

22 Hard

22 of 80Question permalink

Remembering +4 | -1

Maximum number of covalent bonds between two like atoms can be:

View answer and explanation

Correct option: A

Three

Option A is correct because two like atoms can form a maximum of three covalent bonds (one sigma and two pi bonds), as seen in molecules like N₂.

Solution approach

Correct Answer + Core Reason: Three. Two like atoms can form a maximum of three covalent bonds (one sigma and two pi).

Step-by-Step Solution:

Step 1: Understand the types of covalent bonds: sigma and pi bonds.

Step 2: Recall that a single bond is a sigma bond, a double bond consists of one sigma and one pi bond, and a triple bond consists of one sigma and two pi bonds.

Step 3: Consider examples of molecules with multiple bonds between like atoms, such as $N_2$ (triple bond) and $O_2$ (double bond).

Step 4: The maximum number of covalent bonds between two like atoms is a triple bond, which consists of one sigma and two pi bonds.

Step 5: Therefore, the maximum number of covalent bonds between two like atoms is three.

Why other options are wrong:
- Option B: Two is the number of bonds in a double bond.
- Option C: Four is not possible between two atoms.
- Option D: One is the number of bonds in a single bond.

Key Concepts:
- Sigma bond
- Pi bond
- Multiple bonds

Memory Aids:
- Single bond = 1 sigma
- Double bond = 1 sigma + 1 pi
- Triple bond = 1 sigma + 2 pi

Quick Tips:
- Remember the composition of single, double, and triple bonds in terms of sigma and pi bonds.

23 Hard

23 of 80Question permalink

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 is paramagnetic with one unpaired electron and

75\%

ionized, leading to a tetrahedral geometry with

\mathrm{dsp}^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{dsp}^3$ . The complex is paramagnetic with one unpaired electron and $75\%$ ionized, leading to a tetrahedral geometry with $\mathrm{dsp}^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the complex formed.\nSince the Van't Hoff factor (i) is 4 and the complex is 75% ionized, the complex dissociates into 4 ions. The ions are $\mathrm{Ba}^{2+}$ , and the complex anion containing $\mathrm{Co}^{2+}$ and $\mathrm{CN}^-$ . The paramagnetic moment is 1.73 BM, indicating one unpaired electron.

Step 2: Determine the formula of the complex anion.\nSince Barium has a +2 charge, the complex anion must have a -2 charge to balance it. Also, since the complex dissociates into 4 ions, the complex must be of the form $\mathrm{Ba_2[Co(CN)_x]}$ . This means that the complex anion is $[\mathrm{Co(CN)_x}]^{2-}$ .

Step 3: Determine the number of cyanide ligands.\nSince Cobalt is in the +2 oxidation state, the overall charge of the complex is -2, and each cyanide has a -1 charge, we have: +2 + x(-1) = -2, which gives x = 4. Therefore, the complex is $[\mathrm{Co(CN)_4}]^{2-}$ .

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

Step 5: Determine the hybridization.\nSince $\mathrm{CN}^-$ is a strong field ligand, it will cause pairing of electrons. However, since the complex has one unpaired electron, it must be tetrahedral. Therefore, the hybridization is $\mathrm{dsp}^3$ .

Why other options are wrong:
- **Option A ( $\mathrm{sp}^3 \mathrm{d}$ ): This hybridization corresponds to a different geometry and doesn't account for the strong field ligand.
- Option B ( $\mathrm{d}^2 \mathrm{sp}^3$ ): This hybridization is for octahedral complexes, which is not consistent with the given information.
- Option C ( $\mathrm{sp}^3 \mathrm{d}^2$ ): This hybridization is also for octahedral complexes, which is not consistent with the given information.

Key Concepts:
- Coordination complexes
- Hybridization
- Crystal field theory
- Paramagnetism
- Van't Hoff factor

Memory Aids:
- Strong field ligands cause pairing of electrons.
- Tetrahedral complexes are generally high spin.

Quick Tips:**
- Determine the coordination number and geometry of the complex first.
- Consider the spectrochemical series to determine the strength of the ligands.
- Use the number of unpaired electrons to determine the magnetic properties of the complex.

24 Hard

24 of 80Question permalink

Application +4 | -1

The formal charges of

{C}

and

{O}

atoms in

{CO}_2(: \ddot{{O}}={C}=\ddot{{O}}:)

are, respectively

View answer and explanation

Correct option: D

0,0

Option D (0,0) is correct because the formal charge on both the carbon and oxygen atoms in CO2 is zero.

Solution approach

Correct Answer + Core Reason: 0,0. The formal charge is calculated based on the Lewis structure of carbon dioxide.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of carbon dioxide ( $\mathrm{CO}_2$ ). The structure is :O=C=O:

Step 2: Calculate the formal charge on the carbon atom. Formal charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons) = 4 - 0 - (1/2 * 8) = 4 - 0 - 4 = 0.

Step 3: Calculate the formal charge on each oxygen atom. Formal charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons) = 6 - 4 - (1/2 * 4) = 6 - 4 - 2 = 0.

Why other options are wrong:
- Option A: Incorrect formal charges for C and O.
- Option B: Incorrect formal charges for C and O.
- Option C: Incorrect formal charges for C and O.

Key Concepts:
- Formal charge calculation
- Lewis structures

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

Quick Tips:
- Draw the Lewis structure correctly before calculating formal charges.

25 Hard

25 of 80Question permalink

Understanding +4 | -1

The minimum number of

90^\circ

angle between hybrid orbitals is observed in

View answer and explanation

Correct option: C

ds{p^2}

Option C (dsp2) is correct because dsp2 hybridization results in a square planar geometry, which has four 90° angles.

Solution approach

Correct Answer + Core Reason: $ds{p^2}$ . $ds{p^2}$ hybridization results in a square planar geometry with four 90° angles.

Step-by-Step Solution:

Step 1: Recall the geometries associated with each hybridization:
- $s{p^3}{d^2}$ and ${d^2}s{p^3}$ (octahedral) have bond angles of 90° and 180°.
- $ds{p^2}$ (square planar) has bond angles of 90°.
- $s{p^3}d$ (trigonal bipyramidal) has bond angles of 90°, 120°, and 180°.

Step 2: Determine the minimum number of 90° angles for each geometry:
- Octahedral: has multiple 90° angles.
- Square planar: has four 90° angles.
- Trigonal bipyramidal: has three 90° angles in the trigonal plane with axial atoms.

Step 3: Compare the number of 90° angles. Square planar has four 90° angles, which is the minimum.

Why other options are wrong:
- Option A: $s{p^3}{d^2}$ (octahedral) has more than four 90° angles.
- Option B: ${d^2}s{p^3}$ (octahedral) has more than four 90° angles.
- Option D: $s{p^3}d$ (trigonal bipyramidal) has three 90° angles, which is less than the four in square planar.

Key Concepts:
- Hybridization.
- Molecular geometry.
- Bond angles.

Memory Aids:
- Remember the geometries associated with each hybridization.

Quick Tips:
- Visualize the geometries to determine the bond angles.

26 Hard

26 of 80Question permalink

Understanding +4 | -1

Which of the following is drawback of the octet theory?
I. It is clear that octet rule is based upon the chemical inertness of noble gases. However, some noble gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of compounds like

\mathrm{XeF}_2, \mathrm{KrF}_2, \mathrm{XeOF}_2

etc.
II. This theory does not account for the shape of molecules.
III. It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.
Choose the correct statements

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because all three statements accurately describe the limitations and drawbacks of the octet theory.

Solution approach

Correct Answer + Core Reason: All of these. The octet rule has all the mentioned drawbacks.

Step-by-Step Solution:

Step 1: Analyze statement I: The octet rule is based on the inertness of noble gases, but some noble gases do form compounds, which contradicts the rule.

Step 2: Analyze statement II: The octet rule does not explain the shapes of molecules. It only focuses on the number of electrons around an atom.

Step 3: Analyze statement III: The octet rule does not explain the relative stability of molecules or their energies.

Step 4: Since all three statements are correct, the answer is All of these.

Why other options are wrong:
- Option A: Only I is incorrect because statements II and III are also correct.
- Option B: Only II is incorrect because statements I and III are also correct.
- Option C: Only III is incorrect because statements I and II are also correct.

Key Concepts:
- Octet rule
- Limitations of the octet rule
- Noble gas compounds
- Molecular shape
- Molecular stability

Memory Aids:
- Remember the exceptions to the octet rule (expanded octet, incomplete octet, odd electron molecules).

Quick Tips:
- Be aware of the limitations of simple theories like the octet rule.

27 Hard

27 of 80Question permalink

Application +4 | -1

The molecule having permanent dipole moment is:

View answer and explanation

Correct option: A

\mathrm{S}{\mathrm{F}_4}

Option A (

\mathrm{S}{\mathrm{F}_4}

) is correct because its seesaw geometry results in a net dipole moment due to the lone pair.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}{\mathrm{F}_4}$ . This molecule has a seesaw shape, which results in a net dipole moment.

Step-by-Step Solution:

Step 1: Analyze the molecular geometry of each option:
- $\mathrm{S}{\mathrm{F}_4}$ : See-saw shape, non-zero dipole moment.
- $\mathrm{Xe}{\mathrm{F}_4}$ : Square planar, zero dipole moment (due to symmetry).
- $\mathrm{Si}{\mathrm{F}_4}$ : Tetrahedral, zero dipole moment (due to symmetry).
- $\mathrm{B}{\mathrm{F}_3}$ : Trigonal planar, zero dipole moment (due to symmetry).

Step 2: Determine if the molecule has a net dipole moment. A molecule has a permanent dipole moment if it is polar and its bond dipoles do not cancel each other out due to symmetry.

Step 3: $\mathrm{S}{\mathrm{F}_4}$ has a see-saw geometry with one lone pair. The bond dipoles do not cancel out, resulting in a net dipole moment.

Why other options are wrong:
- **Option B ( $\mathrm{Xe}{\mathrm{F}_4}$ ):** $\mathrm{Xe}{\mathrm{F}_4}$ has a square planar geometry. The bond dipoles cancel out due to symmetry, resulting in a zero dipole moment.
- **Option C ( $\mathrm{Si}{\mathrm{F}_4}$ ):** $\mathrm{Si}{\mathrm{F}_4}$ has a tetrahedral geometry. The bond dipoles cancel out due to symmetry, resulting in a zero dipole moment.
- **Option D ( $\mathrm{B}{\mathrm{F}_3}$ ):** $\mathrm{B}{\mathrm{F}_3}$ has a trigonal planar geometry. The bond dipoles cancel out due to symmetry, resulting in a zero dipole moment.

Key Concepts:
- Dipole moment
- Molecular geometry (VSEPR theory)
- Symmetry

Memory Aids:
- Symmetric molecules (linear, trigonal planar, tetrahedral, square planar) often have zero dipole moments if all surrounding atoms are the same.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict the dipole moment.
- Consider the electronegativity difference between the atoms in the molecule.

28 Hard

28 of 80Question permalink

Understanding +4 | -1

\mathrm{C} - \mathrm{Cl}

bond is stronger than

\mathrm{C}-\mathrm{I}

bond, because

View answer and explanation

Correct option: A

\mathrm{C} - \mathrm{Cl}

bond is more ionic than

\mathrm{C}-\mathrm{I}

Option A (

\mathrm{C} - \mathrm{Cl}

bond is more ionic than

\mathrm{C}-\mathrm{I}

) is correct because the electronegativity difference between C and Cl is greater than that between C and I, leading to a more polar and thus stronger bond.

Solution approach

Correct Answer + Core Reason: $\mathrm{C} - \mathrm{Cl}$ bond is more ionic than $\mathrm{C}-\mathrm{I}$ . The greater electronegativity difference makes the C-Cl bond stronger.

Step-by-Step Solution:

Step 1: Electronegativity is the ability of an atom to attract electrons in a chemical bond.

Step 2: The greater the electronegativity difference between two atoms, the more polar the bond.

Step 3: A more polar bond has a greater ionic character and is generally stronger.

Step 4: Chlorine is more electronegative than iodine. Therefore, the electronegativity difference between C and Cl is greater than that between C and I.

Step 5: Consequently, the $\mathrm{C} - \mathrm{Cl}$ bond is more polar and stronger than the $\mathrm{C}-\mathrm{I}$ bond.

Why other options are wrong:
- Option B: While the $\mathrm{C} - \mathrm{Cl}$ bond is polar covalent, this doesn't fully explain why it's stronger than $\mathrm{C}-\mathrm{I}$ . The degree of ionic character is more important.
- Option C: The $\mathrm{C} - \mathrm{Cl}$ bond is less covalent due to higher electronegativity difference.
- Option D: $\mathrm{C} - \mathrm{Cl}$ bond length is shorter than $\mathrm{C}-\mathrm{I}$ bond length, which contributes to its strength, but the primary reason is the greater ionic character.

Key Concepts:
- Electronegativity
- Bond polarity
- Ionic character
- Bond strength

Memory Aids:
- Higher electronegativity difference = stronger, more polar bond.

Quick Tips:
- Consider electronegativity differences when comparing bond strengths.

29 Hard

29 of 80Question permalink

Analyzing +4 | -1

Correct order of bond angles in

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

,

\mathrm{PC}{\mathrm{l}_3}

and

\mathrm{BC}{\mathrm{l}_3}

is

View answer and explanation

Correct option: D

\mathrm{BC}{\mathrm{l}_3} > N{\mathrm{H}_3} > PC{\mathrm{l}_3}

Option D (

\mathrm{BC}{\mathrm{l}_3} > N{\mathrm{H}_3} > PC{\mathrm{l}_3}

) is correct because bond angle depends on hybridization and lone pairs.

Solution approach

Correct Answer + Core Reason: $\mathrm{BC}{\mathrm{l}_3} > N{\mathrm{H}_3} > PC{\mathrm{l}_3}$ . Bond angle is affected by hybridization and lone pairs.

Step-by-Step Solution:

Step 1: Determine the hybridization of each molecule.
- $\mathrm{BC}{\mathrm{l}_3}$ : Boron has 3 bond pairs and 0 lone pairs, so it's $sp^2$ hybridized, with a bond angle of 120°.
- $\mathrm{N}{\mathrm{H}_3}$ : Nitrogen has 3 bond pairs and 1 lone pair, so it's $sp^3$ hybridized, with a bond angle slightly less than 109.5° due to lone pair repulsion.
- $\mathrm{PC}{\mathrm{l}_3}$ : Phosphorus has 3 bond pairs and 1 lone pair, so it's $sp^3$ hybridized. Due to the larger size of Cl atoms and lower electronegativity of P, the bond angle is smaller than in $\mathrm{N}{\mathrm{H}_3}$ .

Step 2: Compare the bond angles.
- $\mathrm{BC}{\mathrm{l}_3}$ has the largest bond angle (120°).
- $\mathrm{N}{\mathrm{H}_3}$ has a bond angle slightly less than 109.5°.
- $\mathrm{PC}{\mathrm{l}_3}$ has the smallest bond angle (around 100°).

Step 3: Arrange in decreasing order of bond angles: $\mathrm{BC}{\mathrm{l}_3} > N{\mathrm{H}_3} > PC{\mathrm{l}_3}$ .

Why other options are wrong:
- Option A: Incorrect order of bond angles.
- Option B: Incorrect order of bond angles.
- Option C: Incorrect order of bond angles.
- Option D: Correct order of bond angles.

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

Memory Aids:
- $sp^2$ hybridization: 120° bond angle
- $sp^3$ hybridization: ~109.5° bond angle (reduced by lone pairs)

Quick Tips:
- Determine hybridization first.
- Consider lone pair repulsion.
- Larger central atoms with more electronegative surrounding atoms lead to smaller bond angles.

30 Hard

30 of 80Question permalink

Understanding +4 | -1

O – Nitro phenol is

View answer and explanation

Correct option: A

More volatile than p – nitro phenol

Option A (More volatile than p – nitro phenol) is correct because o-nitrophenol exhibits intramolecular hydrogen bonding, which reduces intermolecular forces and increases volatility compared to p-nitrophenol, which exhibits intermolecular hydrogen bonding.

Solution approach

Correct Answer + Core Reason: More volatile than p – nitro phenol. Intramolecular H-bonding reduces intermolecular forces.

Step-by-Step Solution:

Step 1: Consider the structures of o-nitrophenol and p-nitrophenol.

Step 2: o-Nitrophenol has the nitro group and hydroxyl group on adjacent carbon atoms. This allows for intramolecular hydrogen bonding (hydrogen bonding within the same molecule).

Step 3: p-Nitrophenol has the nitro group and hydroxyl group on opposite carbon atoms. This favors intermolecular hydrogen bonding (hydrogen bonding between different molecules).

Step 4: Intramolecular hydrogen bonding reduces the intermolecular forces between o-nitrophenol molecules, making it easier to vaporize. Therefore, o-nitrophenol is more volatile.

Step 5: Intermolecular hydrogen bonding increases the intermolecular forces between p-nitrophenol molecules, making it harder to vaporize. Therefore, p-nitrophenol is less volatile.

Why other options are wrong:
- Option B (Less volatile than p – nitro phenol): Incorrect because intramolecular H-bonding makes o-nitrophenol more volatile.
- Option C (Equally volatile as p – nitro phenol): Incorrect because the type of hydrogen bonding (intra vs. inter) affects volatility.
- Option D (Non – volatile): Incorrect because both are volatile to some extent, but o-nitrophenol is more volatile than p-nitrophenol.

Key Concepts:
- Intramolecular hydrogen bonding
- Intermolecular hydrogen bonding
- Volatility and intermolecular forces

Memory Aids:
- Intra-molecular = within the molecule, reduces intermolecular forces, increases volatility.
- Inter-molecular = between molecules, increases intermolecular forces, reduces volatility.

Quick Tips:
- Draw the structures to visualize the hydrogen bonding.

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