CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 8

211 Easy

Remembering +4 | -1

Which type of compounds show high melting and boiling points

View answer and explanation

Correct option: A

Electrovalent compounds

Option A (Electrovalent compounds) is correct because electrovalent compounds have strong electrostatic forces of attraction between ions, leading to high melting and boiling points.

Solution approach

Correct Answer + Core Reason: Electrovalent compounds have high melting and boiling points due to strong electrostatic forces.

Step-by-Step Solution:

Step 1: Electrovalent compounds (also known as ionic compounds) are formed by the complete transfer of electrons from one atom to another, resulting in the formation of ions.

Step 2: These ions are held together by strong electrostatic forces of attraction.

Step 3: Breaking these strong electrostatic forces requires a large amount of energy, hence the high melting and boiling points.

Why other options are wrong:
- Option B: Covalent compounds have relatively weaker intermolecular forces (e.g., van der Waals forces) compared to the strong electrostatic forces in electrovalent compounds. Therefore, they have lower melting and boiling points.
- Option C: Coordinate compounds generally have properties intermediate between ionic and covalent compounds, and their melting and boiling points are usually lower than those of electrovalent compounds.
- Option D: This is incorrect because the melting and boiling points of these three types of compounds are significantly different.

Key Concepts:
- Electrovalent (Ionic) bonding
- Covalent bonding
- Intermolecular forces
- Melting and boiling points

Memory Aids:
- Ionic = Ions = Strong attraction = High MP/BP

Quick Tips:
- Remember that ionic compounds generally have much higher melting and boiling points than covalent compounds.

212 Easy

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Application +4 | -1

Which of the following is diamagnetic?

View answer and explanation

Correct option: C

\mathrm{L}{\mathrm{i}_2}

Option C (

\mathrm{L}{\mathrm{i}_2}

) is correct because it has all its electrons paired in its molecular orbital configuration, resulting in no net magnetic moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{L}{\mathrm{i}_2}$ . Diamagnetic species have all paired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species using molecular orbital theory.

Step 2: The electronic configurations are:
$\mathrm{H}_2^ +$ : $(\sigma_{1s})^1$
${\mathrm{O}_2}$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$
$\mathrm{L}{\mathrm{i}_2}$ : $(\sigma_{2s})^2$
$\mathrm{Fe}_2^ +$ : This is complex and not easily determined by simple MO theory. However, Fe typically has unpaired electrons.

Step 3: Identify the species with all paired electrons.
$\mathrm{H}_2^ +$ has one unpaired electron.
${\mathrm{O}_2}$ has two unpaired electrons.
$\mathrm{L}{\mathrm{i}_2}$ has all paired electrons.
$\mathrm{Fe}_2^ +$ is likely to have unpaired electrons.

Step 4: Diamagnetic species have all paired electrons.

Why other options are wrong:
- Option A: $\mathrm{H}_2^ +$ is paramagnetic.
- Option B: ${\mathrm{O}_2}$ is paramagnetic.
- Option D: $\mathrm{Fe}_2^ +$ is paramagnetic.

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Paramagnetism and diamagnetism

Memory Aids:
- Paramagnetic: unpaired electrons; Diamagnetic: all paired electrons.

Quick Tips:
- Determine the electronic configuration and check for unpaired electrons.

213 Easy

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Application +4 | -1

The bond order in

{O}_2^{+}

is

View answer and explanation

Correct option: B

2.5

Option B (2.5) is correct because the bond order is calculated using molecular orbital theory, and for

O_2^+

, it is 2.5.

Solution approach

Correct Answer + Core Reason: 2.5. The bond order is calculated using molecular orbital theory.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $O_2^+$ .
- $O_2$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2$
- $O_2^+$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^1$

Step 2: Calculate the bond order.
- Bond order = $\frac{{Number of bonding electrons - Number of antibonding electrons}}{2}$
- Bond order = $\frac{{10 - 5}}{2} = 2.5$

Why other options are wrong:
- Option A (2): This would be the bond order of $O_2^{2+}$ .
- Option C (1.5): This is not a possible bond order for an oxygen species.
- Option D (3): This is not a possible bond order for an oxygen species.

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

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

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

214 Easy

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Application +4 | -1

The hybridization of phosphorus in

\mathrm{POC}{\mathrm{l}_3}

is same as in:

View answer and explanation

Correct option: A

\mathrm{P in PC}{\mathrm{l}_3}

Option A (

\mathrm{P in PCl_3}

) is correct because both

POCl_3

and

PCl_3

have phosphorus as the central atom with three bonding pairs and one lone pair (or a double bond and two single bonds in

POCl_3

), resulting in

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{P in PC}{\mathrm{l}_3}$ . Both have $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the hybridization of P in $POCl_3$ .

Step 2: P in $POCl_3$ has 3 single bonds to Cl and one double bond to O. This is equivalent to 4 sigma bonds. Therefore, the hybridization is $sp^3$ .

Step 3: Determine the hybridization of P in $PCl_3$ .

Step 4: P in $PCl_3$ has 3 single bonds to Cl and one lone pair. Therefore, the hybridization is $sp^3$ .

Step 5: Determine the hybridization of S in $SF_6$ .

Step 6: S in $SF_6$ has 6 single bonds to F. Therefore, the hybridization is $sp^3d^2$ .

Step 7: Determine the hybridization of Cl in $ClF_3$ .

Step 8: Cl in $ClF_3$ has 3 single bonds to F and 2 lone pairs. Therefore, the hybridization is $sp^3d$ .

Step 9: Determine the hybridization of B in $BCl_3$ .

Step 10: B in $BCl_3$ has 3 single bonds to Cl. Therefore, the hybridization is $sp^2$ .

Why other options are wrong:
- **Option B ( $\mathrm{S in S}{\mathrm{F}_6}$ ):** The hybridization is $sp^3d^2$ .
- **Option C ( $\mathrm{Cl in Cl}{\mathrm{F}_3}$ ):** The hybridization is $sp^3d$ .
- **Option D ( $\mathrm{B in BC}{\mathrm{l}_3}$ ):** The hybridization is $sp^2$ .

Key Concepts:
- Hybridization
- VSEPR theory

Memory Aids:
- Steric number = number of sigma bonds + number of lone pairs

Quick Tips:
- Draw the Lewis structures to determine the number of sigma bonds and lone pairs.

215 Easy

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Application +4 | -1

Which of the following species would be expected paramagnetic

View answer and explanation

Correct option: C

{Cu}^{++}

Option C (

{Cu}^{++}

) is correct because

Cu^{++}

has an unpaired electron in its electronic configuration, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: ${Cu}^{++}$ . Paramagnetic species have unpaired electrons.

Step-by-Step Solution:

Step 1: Define paramagnetism: Paramagnetism is the property of a substance being attracted to a magnetic field due to the presence of unpaired electrons.

Step 2: Determine the electronic configuration of each species.

Step 3: Copper (Cu) has an electronic configuration of [Ar] 3d¹⁰ 4s¹.

Step 4: ${Cu}^{+}$ has an electronic configuration of [Ar] 3d¹⁰. All electrons are paired, so it is diamagnetic.

Step 5: ${Cu}^{++}$ has an electronic configuration of [Ar] 3d⁹. There is one unpaired electron, so it is paramagnetic.

Step 6: ${H}_2$ has a configuration of $\sigma_{1s}^2$ . All electrons are paired, so it is diamagnetic.

Step 7: Copper crystals are metallic and have complex electronic band structures, but the presence of unpaired electrons at the Fermi level makes them paramagnetic.

Step 8: However, the question asks for a *species*, implying an ion or molecule, not a bulk metal.

Why other options are wrong:
- Option A: Copper crystals are paramagnetic, but the question is looking for a *species* (ion or molecule).
- Option B: ${Cu}^{+}$ is diamagnetic because it has a filled d-shell (3d¹⁰) with no unpaired electrons.
- Option D: ${H}_2$ is diamagnetic because all its electrons are paired.

Key Concepts:
- Paramagnetism
- Diamagnetism
- Electronic configuration

Memory Aids:
- Unpaired electrons = paramagnetic
- Paired electrons = diamagnetic

Quick Tips:
- Determine the number of unpaired electrons in each species.

216 Easy

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Debye is

View answer and explanation

Correct option: C

Unit of dipole momentum

Option C is correct because Debye is the unit used to measure dipole moment.

Solution approach

Correct Answer + Core Reason: Unit of dipole momentum. Debye is the standard unit for measuring dipole moment.

Step-by-Step Solution:

Step 1: Recall the definition of Debye. Debye is a unit used to quantify the dipole moment of a molecule.

Step 2: Conclusion: Debye is the unit of dipole momentum.

Why other options are wrong:
- Option A: Unit of radius: This is incorrect because radius is measured in units of length (e.g., meters, angstroms).
- Option B: Unit of bond order: This is incorrect because bond order is a dimensionless quantity.
- Option D: Unit of bond length: This is incorrect because bond length is measured in units of length (e.g., meters, angstroms).

Key Concepts:
- Dipole Moment
- Debye Unit

Memory Aids:
- Associate Debye with dipole moment.

Quick Tips:
- Remember that dipole moment is a measure of the polarity of a molecule.

217 Easy

217 of 220Question permalink

Application +4 | -1

Bond order of

\mathrm{N}_2^ -

anion is :

View answer and explanation

Correct option: C

2.5

Option C (2.5) is correct because the bond order is calculated as (bonding electrons - antibonding electrons)/2, and for

N_2^-

it is (10-5)/2 = 2.5.

Solution approach

Correct Answer + Core Reason: 2.5. The bond order is calculated using the formula (bonding electrons - antibonding electrons) / 2.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $\mathrm{N}_2^ -$ using molecular orbital theory.

Step 2: The electronic configuration of $N_2$ is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2$ .\nSince $\mathrm{N}_2^ -$ has one extra electron, the electronic configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p}^*)^1$ .

Step 3: Count the number of bonding and antibonding electrons.\nBonding electrons: 2 ( $\sigma_{2s}$ ) + 2 ( $\sigma_{2p_z}$ ) + 4 ( $\pi_{2p_x}$ and $\pi_{2p_y}$ ) = 10\nAntibonding electrons: 2 ( $\sigma_{2s}^*$ ) + 1 ( $\pi_{2p}^*$ ) = 5

Step 4: Calculate the bond order.\nBond order = (Number of bonding electrons - Number of antibonding electrons) / 2\nBond order = (10 - 5) / 2 = 2.5

Why other options are wrong:
- Option A: 3 is the bond order of $N_2$ .
- Option B: 2 is not the correct bond order for $\mathrm{N}_2^ -$ .
- Option D: 1.5 is not the correct bond order for $\mathrm{N}_2^ -$ .

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

Memory Aids:
- Remember the formula for calculating bond order.

Quick Tips:
- Determine the electronic configuration first, then calculate the bond order.

218 Easy

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Remembering +4 | -1

Electrolytes are compound containing

View answer and explanation

Correct option: A

Electrovalent bond

Option A (Electrovalent bond) is correct because electrolytes are compounds that dissociate into ions in solution, which is characteristic of electrovalent or ionic bonds.

Solution approach

Correct Answer + Core Reason: Electrovalent bond. Electrolytes dissociate into ions due to the presence of electrovalent bonds.

Step-by-Step Solution:

Step 1: Define Electrolytes: Electrolytes are substances that produce ions when dissolved in a polar solvent like water, making the solution electrically conductive.

Step 2: Identify Bond Types: Consider the different types of chemical bonds: electrovalent (ionic), covalent, coordinate, and hydrogen bonds.

Step 3: Relate Bond Type to Electrolytic Behavior: Electrovalent bonds involve the complete transfer of electrons, resulting in ions. These ions are responsible for the electrolytic behavior.

Step 4: Conclusion: Electrolytes contain electrovalent bonds because these bonds lead to the formation of ions when dissolved.

Why other options are wrong:
- Option B: Covalent bonds involve sharing of electrons and do not typically result in the formation of ions in solution.
- Option C: Coordinate bonds are a type of covalent bond where one atom provides both electrons, but they don't necessarily lead to ion formation in solution.
- Option D: Hydrogen bonds are intermolecular forces, not intramolecular bonds, and do not lead to ion formation.

Key Concepts:
- Electrolytes
- Electrovalent (Ionic) Bonds
- Covalent Bonds
- Ionization

Memory Aids:
- Electrolytes = Ions (think of \"electro\" relating to ions)

Quick Tips:
- Remember that electrolytes conduct electricity due to the presence of ions, which are formed by ionic compounds.

219 Easy

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The value of one debye unit is equal to :

View answer and explanation

Correct option: C

3.33 \times 10^{-30} \mathrm{~Cm}

Option C (

3.33 \times 10^{-30} \mathrm{~Cm}

) is correct because one debye unit is defined as

3.33 \times 10^{-30}

Coulomb-meters.

Solution approach

Correct Answer + Core Reason: $3.33 \times 10^{-30} \mathrm{~Cm}$ . This is the definition of a Debye unit.

Step-by-Step Solution:

Step 1: Recall the definition of a Debye unit.
1 Debye (D) = $3.33564 \times 10^{-30}$ Coulomb-meters (Cm).

Step 2: Identify the closest value among the options.
$3.33 \times 10^{-30} \mathrm{~Cm}$ is the closest value.

Why other options are wrong:
- Option A: Incorrect value for Debye unit.
- Option B: Incorrect value for Debye unit.
- Option D: Incorrect value for Debye unit.

Key Concepts:
- Debye unit
- Dipole moment

Memory Aids:
- Remember the value of 1 Debye in Coulomb-meters.

Quick Tips:
- Know the definition of common units.

220 Easy

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Remembering +4 | -1

Molecular orbitals theory was proposed by:

View answer and explanation

Correct option: D

Mullikan

Option D (Mullikan) is correct because Robert S. Mulliken is one of the primary developers of the molecular orbital theory.

Solution approach

Correct Answer + Core Reason: Mullikan. Robert S. Mulliken is known for his contributions to the molecular orbital theory.

Step-by-Step Solution:

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

Step 2: Molecular orbital theory is attributed to Robert S. Mulliken and Friedrich Hund.

Why other options are wrong:
- Option A: Werner is known for coordination chemistry.
- Option B: Kossel is known for his theory of ionic bonding.
- Option C: Moseley is known for his work on atomic numbers and X-ray spectra.

Key Concepts:
- Molecular orbital theory
- Chemical bonding
- History of chemistry

Memory Aids:
- Mullikan = Molecular Orbital Theory

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

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