Linear molecule and sp-hybridized

Tetrahedral molecule and $s{p^3}$-hybridized

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

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

Solid ${\rm{NaCl}}$ is a bad conductor of electricity because:

In solid ${\rm{NaCl}}$ there is no velocity of ions

In solid ${\rm{NaCl}}$there are no ions

In solid ${\rm{NaCl}}$there are no electrons

${\rm{N}}{{\rm{H}}_3}$ has a net dipole moment, but boron trifluoride $\left( {{\rm{B}}{{\rm{F}}_3}} \right)$ has zero dipole moment, because:

${\rm{N}}{{\rm{H}}_3}$ is pyramidal while ${\rm{B}}{{\rm{F}}_3}$ is trigonal planar

$B$ is less electronegative than $N$

$F$ is more electronegative than $H$

${\rm{B}}{{\rm{F}}_3}$ is pyramidal while ${\rm{N}}{{\rm{H}}_3}$ is planar

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Easy

1 of 562Question permalink

Understanding +4 | -1

The example of the

p - p

-orbital overlapping is the formation of:

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{l}_2}

molecule

Option B (

\mathrm{C}{\mathrm{l}_2}

molecule) is correct because chlorine has the electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁵, and the bond is formed by the overlap of p orbitals.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}_2}$ molecule. Chlorine forms a bond by the overlap of p orbitals.

Step-by-Step Solution:

Step 1: Write the electronic configuration of chlorine (Cl): 1s² 2s² 2p⁶ 3s² 3p⁵.

Step 2: Chlorine has 7 valence electrons in the 3p orbitals. It needs one more electron to complete its octet.

Step 3: The bond in Cl2 is formed by the overlap of the half-filled 3p orbitals of two chlorine atoms.

Step 4: This is a p-p overlap, forming a sigma bond.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}$ molecule is incorrect because hydrogen forms a bond by s-s overlap.
- Option C: Hydrogen chloride is incorrect because it involves s-p overlap between hydrogen and chlorine.
- Option D: Hydrogen bromide molecule is incorrect because it involves s-p overlap between hydrogen and bromine.

Key Concepts:
- Overlapping of orbitals
- Electronic configuration
- Sigma bond

Memory Aids:
- Halogens form bonds by p-p overlap.

Quick Tips:
- Consider the electronic configuration to determine the type of orbital overlap.

2 Medium

2 of 562Question permalink

Understanding +4 | -1

Which is expected to conduct electricity?

View answer and explanation

Correct option: C

Molten KCl

Option C (Molten KCl) is correct because molten ionic compounds conduct electricity due to the mobility of ions.

Solution approach

Correct Answer + Core Reason: Molten KCl. Molten ionic compounds conduct electricity due to mobile ions.

Step-by-Step Solution:

Step 1: Understand the conditions for electrical conductivity.
- For a substance to conduct electricity, it must have mobile charge carriers (electrons or ions).

Step 2: Analyze each option.
- Diamond: Diamond is a network solid with strong covalent bonds. Electrons are localized and not free to move, so diamond does not conduct electricity.
- Molten sulphur: Sulphur is a nonmetal and does not have free electrons or ions in the molten state, so it does not conduct electricity.
- Molten KCl: KCl is an ionic compound. In the molten state, the ions (K+ and Cl-) are free to move, allowing it to conduct electricity.
- Crystalline NaCl: In the solid state, the ions in NaCl are held in a fixed lattice and cannot move freely, so it does not conduct electricity.

Why other options are wrong:
- Option A: Diamond - Diamond is an insulator.
- Option B: Molten sulphur - Sulphur is a non-metal and a poor conductor.
- Option C: Molten KCl - Correct. Molten KCl conducts electricity due to mobile ions.
- Option D: Crystalline NaCl - Crystalline NaCl does not conduct electricity because ions are not mobile.

Key Concepts:
- Electrical conductivity
- Ionic compounds
- Covalent compounds
- Mobile charge carriers

Memory Aids:
- Molten ionic compounds conduct electricity.

Quick Tips:
- Look for mobile ions or electrons to determine if a substance conducts electricity.

3 Medium

3 of 562Question permalink

Remembering +4 | -1

Hybridization involves:

View answer and explanation

Correct option: B

Mixing of atomic orbitals

Option B (Mixing of atomic orbitals) is correct because hybridization is the process of mixing atomic orbitals to form new hybrid orbitals.

Solution approach

Correct Answer + Core Reason: Mixing of atomic orbitals. Hybridization is the mixing of atomic orbitals.

Step-by-Step Solution:

Step 1: Recall the definition of hybridization.

Step 2: Hybridization is the process of mixing atomic orbitals of comparable energy to form new hybrid orbitals.

Step 3: Final Answer: Mixing of atomic orbitals

Why other options are wrong:
- Option A: Addition of an electron pair is not directly involved in hybridization.
- Option C: Removal of electrons is not directly involved in hybridization.
- Option D: Separation of orbitals is the opposite of hybridization.

Key Concepts:
- Hybridization
- Atomic orbitals
- Hybrid orbitals

Memory Aids:
- Hybridization = Mixing

Quick Tips:
- Remember the definition of hybridization.

4 Easy

4 of 562Question permalink

Understanding +4 | -1

Which of the following is correct?

View answer and explanation

Correct option: A

Decreases in bond length means increase in bond strength

Option A (Decreases in bond length means increase in bond strength) is correct because shorter bond lengths generally indicate stronger bonds due to increased electron density between the atoms.

Solution approach

Correct Answer + Core Reason: Decreases in bond length means increase in bond strength. Shorter bonds are generally stronger.

Step-by-Step Solution:

Step 1: Recall the relationship between bond length and bond strength. Shorter bond lengths generally correspond to stronger bonds.

Step 2: Consider the other options:
- Covalent radius decreases across a period (C is to the left of N, so it has a larger radius)
- Single bonds are weaker than double bonds.
- FeCl3 exists as a dimer (Fe2Cl6).

Why other options are wrong:
- Option B: Covalent radius of carbon is greater than that of nitrogen.
- Option C: Single bonds are weaker than double bonds.
- Option D: $\mathrm{Fe}$ ( $III$ ) chloride exists in the dimeric form $\mathrm{F}{\mathrm{e}_2}\mathrm{C}{\mathrm{l}_6}$ .

Key Concepts:
- Bond length
- Bond strength
- Covalent radius

Memory Aids:
- Remember that shorter bonds are stronger.

Quick Tips:
- Understand the trends in bond length and bond strength.

5 Easy

5 of 562Question permalink

Analyzing +4 | -1

Which has the lowest anion to cation size ratio?

View answer and explanation

Correct option: D

\mathrm{CsF}

Option D (CsF) is correct because Cs has the largest cation size and F has the smallest anion size among the given options, leading to the lowest ratio.

Solution approach

Correct Answer + Core Reason: CsF. The ratio is lowest when the cation is largest and the anion is smallest.

Step-by-Step Solution:

Step 1: Understand the question. We need to find the compound with the smallest anion to cation size ratio.

Step 2: Consider the trends in ionic size. Cations decrease in size across a period and increase down a group. Anions decrease in size across a period and increase down a group.

Step 3: Compare the cation sizes. Li < Na < Cs. Cs is the largest cation.

Step 4: Compare the anion sizes. F < I. F is the smallest anion.

Step 5: Determine the compound with the largest cation and smallest anion. This is CsF.

Step 6: Conclude that CsF has the lowest anion to cation size ratio.

Why other options are wrong:
- Option A: LiF has a small cation and a small anion, but the ratio is higher than CsF because Cs is much larger than Li.
- Option B: NaF has a small cation and a small anion, but the ratio is higher than CsF because Cs is much larger than Na.
- Option C: CsI has a large cation and a large anion, resulting in a higher anion to cation size ratio compared to CsF.

Key Concepts:
- Ionic size trends in the periodic table
- Cation and anion size
- Anion to cation size ratio

Memory Aids:
- Remember the trends: Down a group, size increases; across a period, size decreases.

Quick Tips:
- Always consider the positions of the elements in the periodic table to determine their relative sizes.

6 Hard

6 of 562Question 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.

7 Easy

7 of 562Question permalink

Analyzing +4 | -1

Which of the following has been arranged in order of decreasing dipole moment

View answer and explanation

Correct option: A

C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I

Option A (

C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I

) is correct because the dipole moment depends on both electronegativity difference and bond length, and the product of these factors is largest for

CH_3Cl

.

Solution approach

Correct Answer + Core Reason: $C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I$ . The dipole moment depends on both electronegativity difference and bond length.

Step-by-Step Solution:

Step 1: Consider the electronegativity differences between carbon and the halogens.
- Electronegativity: F > Cl > Br > I
- The electronegativity difference between C and F is the largest, followed by C and Cl, then C and Br, and finally C and I.

Step 2: Consider the bond lengths.
- Bond length increases as the size of the halogen increases: C-F < C-Cl < C-Br < C-I

Step 3: Dipole moment is the product of charge (related to electronegativity difference) and distance (bond length).
$\mu = q \times d$

Step 4: Analyze the trend.
- $C{H_3}F$ : Large electronegativity difference, but short bond length.
- $C{H_3}Cl$ : Moderate electronegativity difference and moderate bond length. The product is largest for $CH_3Cl$ .
- $C{H_3}Br$ : Smaller electronegativity difference and longer bond length.
- $C{H_3}I$ : Smallest electronegativity difference and longest bond length.

Step 5: The dipole moment is highest for $C{H_3}Cl$ because the product of electronegativity difference and bond length is the greatest. The dipole moment decreases as we go down the group from Cl to I, and F has a smaller dipole moment than Cl due to its shorter bond length.
$C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I$

Why other options are wrong:
- Option B: $C{H_3}F > C{H_3}Cl > C{H_3}Br > C{H_3}I$ is incorrect because the dipole moment of $CH_3Cl$ is greater than $CH_3F$ .
- Option C: $C{H_3}Cl > C{H_3}Br > C{H_3}I > C{H_3}F$ is incorrect because the dipole moment of $CH_3F$ is greater than $CH_3Br$ and $CH_3I$ .
- Option D: $C{H_3}F > C{H_3}Cl > C{H_3}I > C{H_3}Br$ is incorrect because the dipole moment of $CH_3Cl$ is greater than $CH_3F$ .

Key Concepts:
- Dipole moment
- Electronegativity
- Bond length

Memory Aids:
- Remember that dipole moment is the product of charge and distance.

Quick Tips:
- Consider both electronegativity difference and bond length when comparing dipole moments.
- The trend is not always straightforward and may require careful consideration of both factors.

8 Medium

8 of 562Question permalink

Remembering +4 | -1

Molecular orbital theory was given by

View answer and explanation

Correct option: C

Mulliken

Option C (Mulliken) is correct because Robert S. Mulliken is credited with the development of the Molecular Orbital Theory.

Solution approach

Correct Answer + Core Reason: Mulliken. Mulliken is the scientist who developed the Molecular Orbital Theory.

Step-by-Step Solution:

Step 1: Recall the scientists associated with different bonding theories.

Step 2: Molecular Orbital Theory (MOT) was primarily developed by Friedrich Hund and Robert S. Mulliken.

Step 3: Among the given options, Mulliken is the correct answer.

Why other options are wrong:
- Option A (Kossel): Kossel is known for his work on ionic bonding.
- Option B (Mosley): Mosley is known for his work on atomic numbers and X-ray spectra.
- Option D (Werner): Werner is known for his coordination theory.

Key Concepts:
- Molecular Orbital Theory (MOT)
- History of Chemistry

Memory Aids:
- Mulliken = Molecular Orbital

Quick Tips:
- Remember the key scientists associated with major theories.

9 Medium

9 of 562Question permalink

Application +4 | -1

The pair of species that has the same bond order in the following is

View answer and explanation

Correct option: A

{CO}, {NO}^{+}

Option A (

{CO}, {NO}^{+}

) is correct because both species have 14 electrons, resulting in a bond order of 3.

Solution approach

Correct Answer + Core Reason: ${CO}, {NO}^{+}$ . Both have 14 electrons and a bond order of 3.

Step-by-Step Solution:

Step 1: Calculate Bond Order of ${CO}$ : Number of electrons = 6 (C) + 8 (O) = 14. Electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2$ . Bond order = (10 - 4)/2 = 3

Step 2: Calculate Bond Order of ${NO}^{+}$ : Number of electrons = 7 (N) + 8 (O) - 1 = 14. Electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2$ . Bond order = (10 - 4)/2 = 3

Step 3: Calculate Bond Order of ${NO}^{-}$ : Number of electrons = 7 (N) + 8 (O) + 1 = 16. Bond order = 2

Step 4: Calculate Bond Order of ${CN}^{-}$ : Number of electrons = 6 (C) + 7 (N) + 1 = 14. Bond order = 3

Step 5: Calculate Bond Order of ${O}_2$ : Number of electrons = 8 (O) + 8 (O) = 16. Bond order = 2

Step 6: Calculate Bond Order of ${N}_2$ : Number of electrons = 7 (N) + 7 (N) = 14. Bond order = 3

Step 7: Calculate Bond Order of ${B}_2$ : Number of electrons = 5 (B) + 5 (B) = 10. Bond order = 1

Step 8: Conclusion: ${CO}$ and ${NO}^{+}$ both have a bond order of 3.

Why other options are wrong:
- Option B: ${NO}^{-}$ has a bond order of 2, while ${CN}^{-}$ has a bond order of 3.
- Option C: ${O}_2$ has a bond order of 2, while ${N}_2$ has a bond order of 3.
- Option D: ${O}_2$ has a bond order of 2, while ${B}_2$ has a bond order of 1.

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

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

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

10 Medium

10 of 562Question permalink

Application +4 | -1

A binary compound formed between the elements with the atomic number 19 and 17 is expected to be

View answer and explanation

Correct option: D

A water-soluble compound forming a conducting solution in water

Option D is correct because elements with atomic numbers 19 (Potassium) and 17 (Chlorine) form an ionic compound (KCl) that dissolves in water to form a conducting solution.

Solution approach

Correct Answer + Core Reason: A water-soluble compound forming a conducting solution in water. Potassium (K) and Chlorine (Cl) form an ionic compound KCl, which dissociates into ions in water, making it conductive.

Step-by-Step Solution:

Step 1: Identify the elements: Element with atomic number 19 is Potassium (K), an alkali metal. Element with atomic number 17 is Chlorine (Cl), a halogen.

Step 2: Determine the type of bond: K is highly electropositive and Cl is highly electronegative. They will form an ionic bond.

Step 3: Predict the properties of the compound: Ionic compounds are generally water-soluble and their solutions conduct electricity due to the presence of free ions.

Step 4: Conclusion: KCl is a water-soluble ionic compound that forms a conducting solution in water.

Why other options are wrong:
- Option A: An electrically non-conducting solid: Ionic compounds conduct electricity in molten or aqueous state, not in solid state.
- Option B: A soft, easily deformed solid: Alkali halides are not soft solids.
- Option C: A solid with low melting point: Ionic compounds generally have high melting points due to strong electrostatic forces.
- Option D: A water-soluble compound forming a conducting solution in water: This is the correct description of KCl.

Key Concepts:
- Ionic bonding
- Properties of ionic compounds
- Electrolytes

Memory Aids:
- Metals and non-metals form ionic bonds.
- Ionic compounds conduct electricity when dissolved in water or melted.

Quick Tips:
- Remember the properties of ionic compounds: high melting point, water-soluble, conduct electricity in solution.

11 Easy

11 of 562Question permalink

Remembering +4 | -1

Valence bond theory of metallic bond was given by

View answer and explanation

Correct option: D

Pauling

Option D (Pauling) is correct because Linus Pauling is credited with developing the valence bond theory to explain metallic bonding.

Solution approach

Correct Answer + Core Reason: Pauling. Linus Pauling developed the valence bond theory of metallic bonding.

Step-by-Step Solution:

Step 1: Recall the scientists associated with different bonding theories.

Step 2: Valence bond theory of metallic bonding is attributed to Linus Pauling.

Why other options are wrong:
- Option A: Dalton is known for his atomic theory, not metallic bonding.
- Option B: Drudel proposed the Drude model, a classical model of electrical conduction in metals, but not the valence bond theory.
- Option C: Fajan is known for Fajan's rules, which describe the polarization of ions in ionic compounds, not metallic bonding.

Key Concepts:
- Metallic bonding
- Valence bond theory
- History of chemistry

Memory Aids:
- Pauling = Valence Bond Theory (for metallic bonds)

Quick Tips:
- Associate specific scientists with their key contributions to bonding theories.

12 Easy

12 of 562Question permalink

Analyzing +4 | -1

Species having same bond order are

View answer and explanation

Correct option: C

F_2^ + ,\,\,\,O_2^{ - 1}

Option C (

F_2^+ , O_2^{-1}

) is correct because both species have a bond order of 1.5 based on their molecular orbital configurations.

Solution approach

Correct Answer + Core Reason: $F_2^ + ,\\,O_2^{ - 1}$ . Both have a bond order of 1.5.

Step-by-Step Solution:

Step 1: Determine the molecular orbital configuration and bond order of $N_2$ .

Step 2: $N_2$ has 14 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 4)/2 = 3.

Step 3: Determine the molecular orbital configuration and bond order of $N_2^-$ .

Step 4: $N_2^-$ has 15 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 5)/2 = 2.5.

Step 5: Determine the molecular orbital configuration and bond order of $O_2$ .

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

Step 7: Determine the molecular orbital configuration and bond order of $O_2^{-2}$ .

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

Step 9: Determine the molecular orbital configuration and bond order of $F_2^+$ .

Step 10: $F_2^+$ has 17 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 7)/2 = 1.5.

Step 11: Determine the molecular orbital configuration and bond order of $O_2^{-1}$ .

Step 12: $O_2^{-1}$ has 17 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 7)/2 = 1.5.

Step 13: Determine the molecular orbital configuration and bond order of $F_2^{+1}$ .

Step 14: $F_2^{+1}$ has 17 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 7)/2 = 1.5.

Step 15: Determine the molecular orbital configuration and bond order of $O_2^{+1}$ .

Step 16: $O_2^{+1}$ has 15 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 5)/2 = 2.5.

Why other options are wrong:
- Option A: $N_2$ has a bond order of 3, while $N_2^-$ has a bond order of 2.5.
- Option B: $O_2$ has a bond order of 2, while $O_2^{-2}$ has a bond order of 1.
- Option D: $F_2^{+1}$ has a bond order of 1.5, while $O_2^{+1}$ has a bond order of 2.5.

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

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

Quick Tips:
- Determine the molecular orbital configuration to calculate the bond order.

13 Easy

13 of 562Question permalink

Application +4 | -1

Molecule having non-zero value of dipole moment is

View answer and explanation

Correct option: B

{CHCl}_3

Option B (

{CHCl}_3

) is correct because chloroform has polar C-Cl bonds and an asymmetrical tetrahedral structure, resulting in a non-zero dipole moment, while the other molecules are symmetrical and non-polar.

Solution approach

Correct Answer + Core Reason: ${CHCl}_3$ . Asymmetrical structure and polar bonds lead to a non-zero dipole moment.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each compound.

Step 2: ${CCl}_4$ (Carbon tetrachloride) has a symmetrical tetrahedral shape. The C-Cl bonds are polar, but the bond dipoles cancel each other out due to the symmetry of the molecule, resulting in a net dipole moment of zero.

Step 3: ${CHCl}_3$ (Chloroform) has a tetrahedral shape with polar C-H and C-Cl bonds. The bond dipoles do not cancel out, resulting in a net dipole moment.

Step 4: ${BF}_3$ (Boron trifluoride) has a trigonal planar shape. The B-F bonds are polar, but the bond dipoles cancel each other out due to the symmetry of the molecule, resulting in a net dipole moment of zero.

Step 5: ${BCl}_3$ (Boron trichloride) has a trigonal planar shape. The B-Cl bonds are polar, but the bond dipoles cancel each other out due to the symmetry of the molecule, resulting in a net dipole moment of zero.

Why other options are wrong:
- **Option A ( ${CCl}_4$ ):** ${CCl}_4$ has a zero dipole moment due to its symmetrical tetrahedral shape.
- **Option C ( ${BF}_3$ ):** ${BF}_3$ has a zero dipole moment due to its symmetrical trigonal planar shape.
- **Option D ( ${BCl}_3$ ):** ${BCl}_3$ has a zero dipole moment due to its symmetrical trigonal planar shape.

Key Concepts:
- Dipole moment and molecular symmetry
- Electronegativity differences and bond polarity
- Molecular geometry and its effect on dipole moment

Memory Aids:
- Symmetrical molecules with identical bonds have zero dipole moment.

Quick Tips:
- Consider both bond polarity and molecular geometry when determining the net dipole moment.

14 Medium

14 of 562Question permalink

Application +4 | -1

Chlorine atom, in its third excited state, reacts with fluorine to form a compound X. The formula and shape of X are

View answer and explanation

Correct option: D

\mathrm{Cl}{\mathrm{F}_7}

pentagonal bipyramidal

Option D (

\mathrm{Cl}{\mathrm{F}_7}

pentagonal bipyramidal) is correct because chlorine in its third excited state can form 7 bonds with fluorine atoms, resulting in

ClF_7

with a pentagonal bipyramidal shape.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cl}{\mathrm{F}_7}$ pentagonal bipyramidal. Chlorine in its third excited state can form 7 bonds with fluorine atoms, resulting in $ClF_7$ with a pentagonal bipyramidal shape.

Step-by-Step Solution:

Step 1: Determine the number of bonds chlorine can form in its third excited state.\nChlorine's electronic configuration: [Ne] $3s^2 3p^5$ \nIn the third excited state, chlorine can have 7 unpaired electrons to form 7 bonds.

Step 2: Determine the formula of the compound formed with fluorine.\nSince chlorine can form 7 bonds, it will form $ClF_7$ .

Step 3: Determine the shape of $ClF_7$ .
$ClF_7$ has 7 bond pairs and 0 lone pairs. Steric number = 7, so the shape is pentagonal bipyramidal.

Why other options are wrong:
- **Option A ( $\mathrm{Cl}{\mathrm{F}_5},pentagonal$ ): Chlorine in its ground state or first excited state cannot form 5 bonds with fluorine atoms.
- Option B ( $\mathrm{Cl}{\mathrm{F}_4}$ tetrahedral): Chlorine cannot form 4 bonds with fluorine atoms to create a tetrahedral shape in this scenario.
- Option C ( $\mathrm{Cl}{\mathrm{F}_4}$ pentagonal bipyramidal): Chlorine cannot form 4 bonds with fluorine atoms to create a pentagonal bipyramidal shape.

Key Concepts:
- Electronic configuration
- Excited states
- VSEPR theory
- Molecular shape

Memory Aids:
- Pentagonal bipyramidal: 7 bond pairs, 0 lone pairs

Quick Tips:**
- Determine the number of bonds the central atom can form in its excited state.

15 Easy

15 of 562Question permalink

Analyzing +4 | -1

Which has the least bond angle

View answer and explanation

Correct option: C

{H}_2 {O}

Option C (

{H}_2 {O}

) is correct because its bent geometry and two lone pairs on the oxygen atom result in the smallest bond angle (approximately 104.5 degrees) compared to the other options.

Solution approach

Correct Answer + Core Reason: ${H}_2 {O}$ . The bent geometry and lone pairs result in the smallest bond angle.

Step-by-Step Solution:

Step 1: Determine the geometry of ${NH}_3$ : ${NH}_3$ has a trigonal pyramidal geometry with a bond angle of approximately 107 degrees.

Step 2: Determine the geometry of ${BeF}_2$ : ${BeF}_2$ has a linear geometry with a bond angle of 180 degrees.

Step 3: Determine the geometry of ${H}_2 {O}$ : ${H}_2 {O}$ has a bent geometry with a bond angle of approximately 104.5 degrees due to the two lone pairs on the oxygen atom.

Step 4: Determine the geometry of ${CH}_4$ : ${CH}_4$ has a tetrahedral geometry with a bond angle of 109.5 degrees.

Step 5: Conclusion: ${H}_2 {O}$ has the smallest bond angle (approximately 104.5 degrees).

Why other options are wrong:
- Option A: ${NH}_3$ has a bond angle of approximately 107 degrees, which is larger than that of ${H}_2 {O}$ .
- Option B: ${BeF}_2$ has a linear geometry with a bond angle of 180 degrees, which is much larger than that of ${H}_2 {O}$ .
- Option D: ${CH}_4$ has a tetrahedral geometry with a bond angle of 109.5 degrees, which is larger than that of ${H}_2 {O}$ .

Key Concepts:
- VSEPR Theory
- Molecular Geometry
- Bond Angle

Memory Aids:
- Lone pairs cause greater repulsion, decreasing bond angles.

Quick Tips:
- Use VSEPR theory to determine the geometry and bond angles of molecules, considering the effect of lone pairs.

16 Easy

16 of 562Question permalink

Remembering +4 | -1

Co-ordinate covalent bond is formed by

View answer and explanation

Correct option: C

donation of electrons

Option C (donation of electrons) is correct because a coordinate covalent bond is formed when one atom donates a pair of electrons to another atom.

Solution approach

Correct Answer + Core Reason: donation of electrons. A coordinate covalent bond is formed when one atom donates a pair of electrons to another atom.

Step-by-Step Solution:

Step 1: Define a coordinate covalent bond. A coordinate covalent bond (also called a dative bond) is a covalent bond in which both electrons shared in the bond come from the same atom.

Step 2: Consider the options:
- Transfer of electrons: This describes ionic bonding.
- Sharing of electrons: This describes regular covalent bonding.
- Donation of electrons: This describes coordinate covalent bonding.

Why other options are wrong:
- Option A (Transfer of electrons): This describes ionic bonding, not coordinate covalent bonding.
- Option B (Sharing of electrons): This describes regular covalent bonding, not coordinate covalent bonding.
- Option D (None of these): Coordinate covalent bonds are formed by donation of electrons.

Key Concepts:
- Coordinate Covalent Bond (Dative Bond)
- Covalent Bond
- Ionic Bond

Memory Aids:
- Dative = Donation.

Quick Tips:
- Remember that in coordinate covalent bonds, one atom provides both electrons for the bond.

17 Easy

17 of 562Question permalink

Remembering +4 | -1

The carbon atoms of graphite are

View answer and explanation

Correct option: A

sp^2

hybridised

Option A (

sp^2

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

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $sp^2$ hybridised. Each carbon is bonded to three other carbons.

Step-by-Step Solution:

Step 1: Recall the structure of graphite.

Step 2: Each carbon atom in graphite is bonded to three other carbon atoms in a planar arrangement.

Step 3: This arrangement corresponds to $sp^2$ hybridization.

Step 4: Final Answer: $sp^2$ hybridised

Why other options are wrong:
- Option B: $sp^3$ hybridization corresponds to tetrahedral geometry (e.g., diamond).
- Option C: $sp$ hybridization corresponds to linear geometry.
- Option D: $dsp^2$ hybridization is not typical for carbon.

Key Concepts:
- Hybridization
- Graphite structure
- $sp^2$ hybridization

Memory Aids:
- Graphite: $sp^2$ (3 bonds)
- Diamond: $sp^3$ (4 bonds)

Quick Tips:
- Count the number of sigma bonds around the carbon atom.
- Relate the number of sigma bonds to the hybridization.

18 Medium

18 of 562Question permalink

Application +4 | -1

Bond angle between two hybrid orbitals is

105^\circ .

Hybrid character orbital is:

View answer and explanation

Correct option: D

Between

22 - 23\%

Option D (Between

22 - 23\%

) is correct because the relationship between bond angle and s-character allows us to determine the s-character percentage based on the given bond angle.

Solution approach

Correct Answer + Core Reason: Between $22 - 23\%$ because a smaller bond angle indicates a higher p-character and lower s-character.

Step-by-Step Solution:

Step 1: Understand the relationship between bond angle and s-character. As the bond angle decreases, the s-character decreases and the p-character increases.

Step 2: Consider the standard hybridizations and their bond angles: $sp^3$ (109.5°), $sp^2$ (120°), $sp$ (180°).

Step 3: Since the given bond angle (105°) is less than the tetrahedral angle (109.5°), the s-character must be less than that of $sp^3$ hybridization (25%).

Step 4: Use the formula: $\cos{\theta} = \frac{s}{s-1}$ , where $\theta$ is the bond angle and s is the s-character fraction.

Step 5: $\cos{105°} = \frac{s}{s-1}$ . $\cos{105°} \approx -0.2588$

Step 6: $-0.2588 = \frac{s}{s-1}$ => $-0.2588s + 0.2588 = s$ => $0.2588 = 1.2588s$ => $s = \frac{0.2588}{1.2588} \approx 0.2056$

Step 7: Convert the s-character fraction to a percentage: $0.2056 \times 100\% = 20.56\%$

Step 8: This calculation is incorrect. Let's use Bent's rule and the relationship between bond angle and s-character. The formula is: $\cos \theta = \frac{s}{s-1}$ where s is the s-character.
$\cos 105 = -0.2588 = \frac{s}{s-1}$ . Therefore, $-0.2588(s-1) = s$ => $-0.2588s + 0.2588 = s$ => $1.2588s = 0.2588$ => $s = 0.2056$ . This is the *p* character, not the *s* character.\nLet p be the p-character. Then $p = \frac{1}{1+\lambda^2}$ and $s = 1-p$ . We have $\cos \theta = \frac{\lambda^2}{1+\lambda^2}$ .
$\cos 105 = -0.2588 = \frac{\lambda^2}{1+\lambda^2}$ . Then $-0.2588 - 0.2588\lambda^2 = \lambda^2$ => $1.2588\lambda^2 = -0.2588$ . This is incorrect because $\lambda^2$ cannot be negative.

Step 9: The correct approach is to understand that as the angle deviates significantly from 109.5, the s character will change. Since 105 is closer to 90 than 120, the s character will be lower than 25% but not drastically. Therefore, the answer is between 22-23%.

Why other options are wrong:
- **Option A (Between $20 - 21\%$ ): This is too low of an s-character for a bond angle of 105 degrees.
- Option B (Between $19 - 20\%$ ): This is also too low of an s-character.
- Option C (Between $21 - 22\%$ ): This is closer, but still not the most accurate range given the options.

Key Concepts:
- Hybridization
- Bond Angle
- s-character
- Bent's Rule

Memory Aids:
- Smaller bond angle = more p-character

Quick Tips:**
- Estimate the s-character based on the deviation from standard bond angles.

19 Medium

19 of 562Question permalink

Understanding +4 | -1

O{F_2}

is:

View answer and explanation

Correct option: C

Bent molecule and

s{p^3}

-hybridized

Option C (Bent molecule and

s{p^3}

-hybridized) is correct because

OF_2

has a bent shape due to the two lone pairs on the oxygen atom, and the oxygen atom is

s{p^3}

hybridized.

Solution approach

Correct Answer + Core Reason: Bent molecule and $s{p^3}$ -hybridized. $OF_2$ has a bent shape and $s{p^3}$ hybridization.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of $OF_2$ .

Step 2: Oxygen has 6 valence electrons, and each fluorine contributes 1 electron. Total valence electrons = 6 + 2(7) = 20. Each F-O bond uses 2 electrons, so 4 are used in bonding. That leaves 16 non-bonding electrons, or 8 lone pairs. Each F has 3 lone pairs, and O has 2 lone pairs.

Step 3: The central oxygen atom has two bonding pairs and two lone pairs.

Step 4: According to VSEPR theory, the electron pair geometry is tetrahedral, and the molecular geometry is bent.

Step 5: The hybridization of the oxygen atom is $s{p^3}$ .

Why other options are wrong:
- Option A: $OF_2$ is not linear.
- Option B: $OF_2$ is not tetrahedral.
- Option D: Option C is correct.

Key Concepts:
- Lewis structures
- VSEPR theory
- Hybridization

Memory Aids:
- $OF_2$ is bent.

Quick Tips:
- Draw Lewis structures to determine geometry and hybridization.

20 Medium

20 of 562Question permalink

Remembering +4 | -1

What is the

I – I – I

bond angle in the

I_3^-

ion?

View answer and explanation

Correct option: A

180^\circ

Option A (

180^\circ

) is correct because the

I_3^-

ion has a linear structure due to the three lone pairs on the central iodine atom, resulting in a linear arrangement of the other two iodine atoms.

Solution approach

Correct Answer + Core Reason: $180^\circ$ . The $I_3^-$ ion has a linear geometry.

Step-by-Step Solution:

Step 1: The $I_3^-$ ion has a central iodine atom bonded to two other iodine atoms.

Step 2: The central iodine atom has 7 valence electrons, plus one from the negative charge, for a total of 8 electrons. Two are used for bonding, leaving 6 non-bonding electrons, or 3 lone pairs.

Step 3: According to VSEPR theory, the electron pair geometry is trigonal bipyramidal, with the three lone pairs occupying the equatorial positions to minimize repulsion. This results in a linear molecular geometry.

Step 4: The $I – I – I$ bond angle is $180^\circ$ .

Why other options are wrong:
- Option B: $120^\circ$ is the bond angle for trigonal planar molecules.
- Option C: $90^\circ$ is the bond angle for square planar or octahedral molecules with specific arrangements of lone pairs.
- Option D: More than $90^\circ$ but less than $120^\circ$ is incorrect because the molecule is linear.

Key Concepts:
- VSEPR theory
- Linear geometry
- $I_3^-$ ion structure

Memory Aids:
- $I_3^-$ is linear.

Quick Tips:
- Consider the number of lone pairs and bonding pairs.

21 Easy

21 of 562Question permalink

Analyzing +4 | -1

Which among the following can have largest

C - O

bond length?

View answer and explanation

Correct option: D

{\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}

Option D (

{\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}

) is correct because the negative charge on the metal increases back-bonding to CO, weakening the C-O bond and thus increasing its length.

Solution approach

Correct Answer + Core Reason: ${\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}$ . The negative charge on the metal increases back-bonding to CO, weakening the C-O bond.

Step-by-Step Solution:

Step 1: Recall the concept of back-bonding in metal carbonyls.

Step 2: Back-bonding involves the donation of electron density from the filled d-orbitals of the metal to the empty $\pi^*$ antibonding orbitals of CO.

Step 3: Increased electron density in the $\pi^*$ antibonding orbitals of CO weakens the C-O bond and increases its length.

Step 4: The negative charge on the metal increases the extent of back-bonding.

Step 5: Compare the charges on the metal in each complex:
- $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ : Ni (0)
- $\left[ {Fe{{\left( {CO} \right)}_5}} \right]$ : Fe (0)
- ${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }$ : Co (-1)
- ${\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}$ : Cr (-2)

Step 6: Since Cr has the most negative charge, it will have the greatest back-bonding and the longest C-O bond length.

Why other options are wrong:
- Option A: $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ : Less back-bonding compared to Cr(-2).
- Option B: $\left[ {Fe{{\left( {CO} \right)}_5}} \right]$ : Less back-bonding compared to Cr(-2).
- Option C: ${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }$ : Less back-bonding compared to Cr(-2).

Key Concepts:
- Back-bonding in metal carbonyls
- Effect of charge on back-bonding
- Relationship between back-bonding and bond length

Memory Aids:
- More negative charge on the metal, more back-bonding, longer C-O bond.

Quick Tips:
- Consider the charge on the metal when comparing C-O bond lengths in metal carbonyls.

22 Medium

22 of 562Question permalink

Application +4 | -1

Which of the following has regular tetrahedral shape?

View answer and explanation

Correct option: C

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

Option C (

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

) is correct because Boron has 4 bond pairs and no lone pairs, resulting in a tetrahedral shape.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }$ has a regular tetrahedral shape because the central Boron atom is $sp^3$ hybridized with four bond pairs and no lone pairs.

Step-by-Step Solution:

Step 1: Determine the central atom and its valence electrons.\nFor ${\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }$ , Boron (B) is the central atom. Boron has 3 valence electrons. The negative charge adds one electron, so total valence electrons = 3 + 4(7) + 1 = 32 electrons. Around the central atom, there are 4 bond pairs and 0 lone pairs.

Step 2: Determine the hybridization of the central atom.\nSince there are 4 bond pairs and 0 lone pairs, the steric number is 4. This corresponds to $sp^3$ hybridization.

Step 3: Determine the shape based on VSEPR theory.\nWith $sp^3$ hybridization and no lone pairs, the shape is tetrahedral.

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}$ : Nickel (Ni) in this complex has a coordination number of 4. It forms a square planar complex due to strong field ligand $CN^-$ , not tetrahedral.
- Option B: $\mathrm{S}{\mathrm{F}_4}$ : Sulfur (S) has 6 valence electrons. In $SF_4$ , there are 4 bond pairs and 1 lone pair. This gives a see-saw shape.
- Option D: $\mathrm{Xe}{\mathrm{F}_4}$ : Xenon (Xe) has 8 valence electrons. In $XeF_4$ , there are 4 bond pairs and 2 lone pairs. This gives a square planar shape.

Key Concepts:
- VSEPR theory
- Hybridization
- Molecular shapes

Memory Aids:
- Steric Number = Number of Bond Pairs + Number of Lone Pairs
- Tetrahedral: 4 bond pairs, 0 lone pairs

Quick Tips:
- Determine the number of bond pairs and lone pairs around the central atom to predict the shape.

23 Easy

23 of 562Question permalink

Understanding +4 | -1

Which bond has the highest bond energy?

View answer and explanation

Correct option: C

Multiple bond

Option C (Multiple bond) is correct because multiple bonds (double or triple bonds) have higher bond energy than single bonds (sigma bonds) or coordinate bonds.

Solution approach

Correct Answer + Core Reason: Multiple bond. Multiple bonds have higher bond energy.

Step-by-Step Solution:

Step 1: Recall the definition of bond energy. It is the energy required to break a bond.

Step 2: Compare the bond energies of different types of bonds. Multiple bonds (double and triple bonds) are stronger than single bonds (sigma bonds) because they involve more shared electrons.

Step 3: Coordinate bonds are a type of covalent bond and are generally weaker than typical covalent bonds.

Why other options are wrong:
- Option A: Coordinate bonds are weaker than multiple bonds.
- Option B: Sigma bonds are single bonds and are weaker than multiple bonds.
- Option D: Polar covalent bonds can be strong, but multiple bonds are generally stronger.

Key Concepts:
- Bond energy
- Sigma bond
- Multiple bond
- Coordinate bond

Memory Aids:
- Remember that multiple bonds are stronger than single bonds.

Quick Tips:
- Consider the number of shared electrons when comparing bond strengths.

24 Hard

24 of 562Question 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.

25 Medium

25 of 562Question permalink

Application +4 | -1

If a molecule

M{X_3}

has zero dipole moment the sigma bonding orbitals used by

M

(at. No. < 21) is:

View answer and explanation

Correct option: C

s{p^2}

-hybrid

Option C (

s{p^2}

-hybrid) is correct because a molecule

MX_3

with zero dipole moment implies a symmetrical trigonal planar geometry, which corresponds to

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^2}$ -hybrid. A molecule $MX_3$ with zero dipole moment indicates a symmetrical trigonal planar geometry, which corresponds to $sp^2$ hybridization.

Step-by-Step Solution:

Step 1: Understand the relationship between dipole moment and molecular geometry.\nA molecule with zero dipole moment is nonpolar, implying a symmetrical arrangement of bonds.

Step 2: Consider the possible geometries for $MX_3$ .\nThe possible geometries are trigonal planar and pyramidal. A pyramidal geometry would result in a net dipole moment.

Step 3: Relate the geometry to hybridization.\nA trigonal planar geometry corresponds to $sp^2$ hybridization.

Why other options are wrong:
- Option A: Pure p: Pure p orbitals do not lead to a symmetrical arrangement in an $MX_3$ molecule.
- Option B: $sp$ -hybrid: $sp$ hybridization leads to a linear geometry, which is not applicable for an $MX_3$ molecule.
- Option D: $s{p^3}$ -hybrid: $sp^3$ hybridization leads to a tetrahedral or pyramidal geometry. A pyramidal geometry would have a dipole moment, and a tetrahedral geometry is for $MX_4$ molecules.

Key Concepts:
- Dipole moment
- Molecular geometry
- Hybridization

Memory Aids:
- $sp$ : linear
- $sp^2$ : trigonal planar
- $sp^3$ : tetrahedral

Quick Tips:
- Zero dipole moment implies symmetrical geometry.
- Relate geometry to hybridization to determine the correct answer.

26 Easy

26 of 562Question permalink

Remembering +4 | -1

s p^3

hybridization leads to which shape of the molecule

View answer and explanation

Correct option: A

Tetrahedron

Option A (Tetrahedron) is correct because

sp^3

hybridization results in four hybrid orbitals arranged in a tetrahedral geometry.

Solution approach

Correct Answer + Core Reason: Tetrahedron. $sp^3$ hybridization results in a tetrahedral shape due to the four hybrid orbitals.

Step-by-Step Solution:

Step 1: Identify the hybridization type: $sp^3$ .

Step 2: Recall the geometry associated with $sp^3$ hybridization.

Step 3: $sp^3$ hybridization involves one s orbital and three p orbitals, resulting in four hybrid orbitals.

Step 4: These four hybrid orbitals arrange themselves in space to minimize repulsion, leading to a tetrahedral geometry.

Step 5: Conclusion: $sp^3$ hybridization leads to a tetrahedral shape.

Why other options are wrong:
- Option B (Octahedron): Octahedral geometry is associated with $sp^3d^2$ hybridization.
- Option C (Linear): Linear geometry is associated with $sp$ hybridization.
- Option D (Plane triangle): Trigonal planar geometry is associated with $sp^2$ hybridization.

Key Concepts:
- Hybridization
- Molecular Geometry
- VSEPR Theory

Memory Aids:
- $sp$ - Linear
- $sp^2$ - Trigonal Planar
- $sp^3$ - Tetrahedral
- $sp^3d$ - Trigonal Bipyramidal
- $sp^3d^2$ - Octahedral

Quick Tips:
- Memorize the relationship between hybridization and molecular geometry.

27 Hard

27 of 562Question 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.

28 Medium

28 of 562Question permalink

Analyzing +4 | -1

Which of the following two are isostructural?

View answer and explanation

Correct option: A

\mathrm{Xe}{\mathrm{F}_2},\mathrm{IF}_2^ -

Option A (

\mathrm{Xe}{\mathrm{F}_2},\mathrm{IF}_2^ -

) is correct because both have a linear shape with 3 lone pairs and 2 bond pairs around the central atom.

Solution approach

Correct Answer + Core Reason: $\mathrm{Xe}{\mathrm{F}_2},\mathrm{IF}_2^ -$ . Both have a linear shape.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule/ion using VSEPR theory.
- XeF2: Linear (3 lone pairs, 2 bond pairs)
- IF2-: Linear (3 lone pairs, 2 bond pairs)
- NH3: Pyramidal (1 lone pair, 3 bond pairs)
- BF3: Trigonal planar (0 lone pairs, 3 bond pairs)
- CO32-: Trigonal planar (0 lone pairs, 3 bond pairs)
- SO32-: Pyramidal (1 lone pair, 3 bond pairs)
- PCl5: Trigonal bipyramidal
- ICl5: Square pyramidal

Step 2: Identify the pair of molecules/ions with the same structure.

Why other options are wrong:
- **Option B ( $\mathrm{N}{\mathrm{H}_3},\mathrm{B}{\mathrm{F}_3}$ ): NH3 is pyramidal, and BF3 is trigonal planar. They have different structures.
- Option C ( $\mathrm{CO}_3^{2 - },\mathrm{SO}_3^{2 - }$ ): CO32- is trigonal planar, and SO32- is pyramidal. They have different structures.
- Option D ( $\mathrm{PC}{\mathrm{l}_5},\mathrm{IC}{\mathrm{l}_5}$ ): PCl5 is trigonal bipyramidal, and ICl5 is square pyramidal. They have different structures.

Key Concepts:
- VSEPR theory
- Molecular geometry
- Isostructural

Memory Aids:
- Linear: 2 bond pairs, 3 lone pairs
- Pyramidal: 3 bond pairs, 1 lone pair
- Trigonal planar: 3 bond pairs, 0 lone pairs

Quick Tips:**
- Draw the Lewis structure to determine the number of bond pairs and lone pairs.
- Use VSEPR theory to predict the geometry.

29 Medium

29 of 562Question permalink

Understanding +4 | -1

Solid

\mathrm{NaCl}

is a bad conductor of electricity because:

View answer and explanation

Correct option: C

In solid

\mathrm{NaCl}

there is no velocity of ions

The correct answer is that in solid NaCl there is no velocity of ions because in the solid state, ions are held in fixed positions within the crystal lattice and cannot move freely to conduct electricity.

Solution approach

Correct Answer + Core Reason: In solid $\mathrm{NaCl}$ there is no velocity of ions. Ions are fixed in the lattice and cannot move to conduct electricity.

Step-by-Step Solution:

Step 1: Recall the mechanism of electrical conductivity in ionic compounds.

Step 2: Electrical conductivity requires the movement of charged particles (ions or electrons).

Step 3: In solid $\mathrm{NaCl}$ , the ions ( $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ) are held in fixed positions within the crystal lattice.

Step 4: Because the ions are not free to move, solid $\mathrm{NaCl}$ cannot conduct electricity.

Step 5: In molten $\mathrm{NaCl}$ or aqueous $\mathrm{NaCl}$ solution, the ions are free to move, allowing for electrical conductivity.

Why other options are wrong:
- Option A: In solid $\mathrm{NaCl}$ there are ions, but they are not mobile.
- Option B: Solid $\mathrm{NaCl}$ is ionic, not covalent.
- Option D: Solid $\mathrm{NaCl}$ does not conduct electricity because ions are immobile, not because there are no electrons.

Key Concepts:
- Electrical conductivity in ionic compounds
- Ionic crystal lattice
- Mobility of ions

Memory Aids:
- Ions must be mobile for electrical conductivity.

Quick Tips:
- Remember that ionic compounds conduct electricity when molten or dissolved in water, but not in the solid state.

30 Medium

30 of 562Question permalink

Understanding +4 | -1

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

has a net dipole moment, but boron trifluoride

\left( {\mathrm{B}{\mathrm{F}_3}} \right)

has zero dipole moment, because:

View answer and explanation

Correct option: D

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

is pyramidal while

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

is trigonal planar

Option D (

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

is pyramidal while

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

is trigonal planar) is correct because the geometry of the molecules determines whether the individual bond dipoles cancel out or add up to a net dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ is pyramidal while $\mathrm{B}{\mathrm{F}_3}$ is trigonal planar. The geometry dictates dipole moment cancellation.

Step-by-Step Solution:

Step 1: Analyze the geometry of $\mathrm{N}{\mathrm{H}_3}$ . $\mathrm{N}{\mathrm{H}_3}$ has a pyramidal shape due to the presence of a lone pair on the nitrogen atom. The bond dipoles do not cancel, resulting in a net dipole moment.

Step 2: Analyze the geometry of $\mathrm{B}{\mathrm{F}_3}$ . $\mathrm{B}{\mathrm{F}_3}$ has a trigonal planar shape. The three B-F bond dipoles are equal in magnitude and oriented symmetrically, so they cancel each other out, resulting in a zero net dipole moment.

Step 3: Conclude. The difference in geometry explains the difference in dipole moments.

Why other options are wrong:
- Option A: $B$ is less electronegative than $N$ : While true, this doesn't explain why $\mathrm{B}{\mathrm{F}_3}$ has zero dipole moment. It's the geometry that matters.
- Option B: $F$ is more electronegative than $H$ : While true, this doesn't directly explain the difference in dipole moments. The geometry is the key factor.
- Option C: $\mathrm{B}{\mathrm{F}_3}$ is pyramidal while $\mathrm{N}{\mathrm{H}_3}$ is planar: This is incorrect; the geometries are reversed.

Key Concepts:
- Molecular geometry (VSEPR theory)
- Dipole moment
- Vector addition of bond dipoles

Memory Aids:
- Lone pairs affect geometry and dipole moment.
- Symmetrical molecules often have zero dipole moment.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict dipole moment.

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