Linear molecule and sp-hybridized

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

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

The number and type of bonds between two carbon atoms in ${\rm{Ca}}{{\rm{C}}_2}$ are:

One sigma $\left( {\rm{\sigma }} \right)$and two pi $\left( {\rm{\pi }} \right)$-bonds

One sigma $\left( {\rm{\sigma }} \right)$and one pi $\left( {\rm{\pi }} \right)$-bond

One sigma $\left( {\rm{\sigma }} \right)$ and one and a half pi $\left( {\rm{\pi }} \right)$-bond

One sigma $\left( {\rm{\sigma }} \right)$ bond

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

1 Medium

1 of 262Question 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.

2 Medium

2 of 262Question 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.

3 Medium

3 of 262Question 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.

4 Medium

4 of 262Question 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.

5 Medium

5 of 262Question 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.

6 Medium

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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.

7 Medium

7 of 262Question 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.

8 Medium

8 of 262Question 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.

9 Medium

9 of 262Question 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.

10 Medium

10 of 262Question 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.

11 Medium

11 of 262Question 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.

12 Medium

12 of 262Question 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.

13 Medium

13 of 262Question 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.

14 Medium

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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.

15 Medium

15 of 262Question permalink

Application +4 | -1

In which of the following pairs of molecules/ions, the central atom has

s{p^2}

-hybridization?

View answer and explanation

Correct option: B

\mathrm{B}{\mathrm{F}_3}\,\,\mathrm{and NO}_2^ -

Option B (

\mathrm{B}{\mathrm{F}_3}\,\,\mathrm{and NO}_2^ -

) is correct because both

BF_3

and

NO_2^-

have central atoms with three

sp^2

hybrid orbitals.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{F}_3}\,\,\mathrm{and NO}_2^ -$ . Both molecules have central atoms with three $sp^2$ hybrid orbitals.

Step-by-Step Solution:

Step 1: Determine the hybridization of the central atom in each molecule/ion.

Step 2: For $\mathrm{N}{\mathrm{O}_2}$ : Nitrogen has 2 bond pairs and 1 lone pair. Steric number = 3, so hybridization is $sp^2$ .\nFor $\mathrm{N}{\mathrm{H}_3}$ : Nitrogen has 3 bond pairs and 1 lone pair. Steric number = 4, so hybridization is $sp^3$ .\nFor $\mathrm{B}{\mathrm{F}_3}$ : Boron has 3 bond pairs and 0 lone pairs. Steric number = 3, so hybridization is $sp^2$ .\nFor $\mathrm{NO}_2^ -$ : Nitrogen has 2 bond pairs and 1 lone pair. Steric number = 3, so hybridization is $sp^2$ .\nFor $\mathrm{NH}_2^ -$ : Nitrogen has 2 bond pairs and 2 lone pairs. Steric number = 4, so hybridization is $sp^3$ .\nFor ${\mathrm{H}_2}\mathrm{O}$ : Oxygen has 2 bond pairs and 2 lone pairs. Steric number = 4, so hybridization is $sp^3$ .

Step 3: Identify the pair with both compounds having $sp^2$ hybridization.
$\mathrm{B}{\mathrm{F}_3}$ has $sp^2$ hybridization and $\mathrm{NO}_2^ -$ has $sp^2$ hybridization.

Why other options are wrong:
- **Option A ( $\mathrm{N}{\mathrm{O}_2}\,\,\mathrm{and N}{\mathrm{H}_3}$ ):** $\mathrm{NO}_2$ is $sp^2$ hybridized, but $\mathrm{NH}_3$ is $sp^3$ hybridized.
- **Option C ( $\mathrm{NH}_2^ - \,\,\mathrm{and}\,{\mathrm{H}_2}\mathrm{O}$ ):** Both $\mathrm{NH}_2^ -$ and ${\mathrm{H}_2}\mathrm{O}$ are $sp^3$ hybridized.
- **Option D ( $\mathrm{B}{\mathrm{F}_3}\,\,\mathrm{and NH}_2^ -$ ):** $\mathrm{B}{\mathrm{F}_3}$ is $sp^2$ hybridized, but $\mathrm{NH}_2^ -$ is $sp^3$ hybridized.

Key Concepts:
- Hybridization
- Steric number
- VSEPR theory

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:
- Always count the number of bond pairs and lone pairs around the central atom to determine the steric number and hence the hybridization.

16 Medium

16 of 262Question permalink

Application +4 | -1

Which has maximum covalent character?

View answer and explanation

Correct option: A

\mathrm{SiC}{\mathrm{l}_4}

Option A (

\mathrm{SiC}{\mathrm{l}_4}

) is correct because, according to Fajan's rule, a smaller cation and a larger anion, along with a high charge on either ion, favor covalent character. Si has a smaller size and higher charge compared to the other metal cations.

Solution approach

Correct Answer + Core Reason: $\mathrm{SiC}{\mathrm{l}_4}$ . Smaller cation and larger anion, along with high charge favors covalent character.

Step-by-Step Solution:

Step 1: Fajan's rule states that covalent character is favored by small, highly charged cations and large, highly charged anions.

Step 2: Comparing the cations: Si4+, Mg2+, Na+, and Al3+.

Step 3: Si4+ is the smallest and has the highest charge, making $\mathrm{SiC}{\mathrm{l}_4}$ the most covalent.

Step 4: The anion is Cl- in all cases, so the cation is the determining factor.

Why other options are wrong:
- Option B: $MgC{l_2}$ has a larger cation with a lower charge compared to Si4+, making it more ionic.
- Option C: $NaCl$ has the largest cation with the lowest charge, making it the most ionic.
- Option D: $AlC{l_3}$ has a smaller cation with a lower charge compared to Si4+, making it less covalent.

Key Concepts:
- Fajan's rule
- Covalent character
- Ionic character

Memory Aids:
- Small cation, large anion = covalent character.

Quick Tips:
- Apply Fajan's rule by considering the size and charge of the ions.

17 Medium

17 of 262Question permalink

Understanding +4 | -1

The increase in bond order results in:

View answer and explanation

Correct option: A

Decrease in bond length and increase in bond energy

Option A (Decrease in bond length and increase in bond energy) is correct because a higher bond order indicates more bonds between atoms, leading to a shorter and stronger bond.

Solution approach

Correct Answer + Core Reason: Decrease in bond length and increase in bond energy. Higher bond order means stronger and shorter bonds.

Step-by-Step Solution:

Step 1: Define Bond Order: Bond order is the number of chemical bonds between a pair of atoms.

Step 2: Relationship with Bond Length: As bond order increases, the bond length decreases because the atoms are held more closely together.

Step 3: Relationship with Bond Energy: As bond order increases, the bond energy increases because more energy is required to break the stronger bond.

Step 4: Conclusion: Therefore, an increase in bond order results in a decrease in bond length and an increase in bond energy.

Why other options are wrong:
- Option B: Decrease in bond length and bond energy is incorrect because bond energy increases with bond order.
- Option C: Increase in bond length and bond energy is incorrect because bond length decreases with bond order.
- Option D: None of the above is incorrect because option A is correct.

Key Concepts:
- Bond Order
- Bond Length
- Bond Energy

Memory Aids:
- Higher Bond Order = Shorter, Stronger Bond

Quick Tips:
- Remember the inverse relationship between bond order and bond length, and the direct relationship between bond order and bond energy.

18 Medium

18 of 262Question permalink

Understanding +4 | -1

Which of the following has maximum bond energy?

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because bond energy generally decreases down the halogen group, but F2 is an exception due to its small size and high interelectronic repulsion.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}_2}$ . Bond energy decreases down the group, but F2 is an exception.

Step-by-Step Solution:

Step 1: Recall the trend in bond energy down the halogen group.
- Generally, bond energy decreases down the halogen group (F > Cl > Br > I) due to increasing atomic size and decreasing electronegativity.

Step 2: Explain the exception for fluorine.
- Fluorine has a smaller bond energy than chlorine due to its small size and high interelectronic repulsion between the lone pairs of electrons on the two fluorine atoms.

Step 3: Compare the bond energies of the given halogens.
- $\mathrm{C}{\mathrm{l}_2}$ has the highest bond energy among the given options.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{l}_2}$ - Correct. Chlorine has the highest bond energy.
- Option B: ${F_2}$ - Fluorine has lower bond energy than chlorine due to interelectronic repulsion.
- Option C: $\mathrm{B}{\mathrm{r}_2}$ - Bromine has lower bond energy than chlorine.
- Option D: ${I_2}$ - Iodine has the lowest bond energy among the given halogens.

Key Concepts:
- Bond energy
- Halogens
- Interelectronic repulsion

Memory Aids:
- Bond energy decreases down the group (except F2).

Quick Tips:
- Remember the exception for fluorine when comparing bond energies of halogens.

19 Medium

19 of 262Question permalink

Application +4 | -1

The total no. of valency electrons present in

SO_4^{ - 2}

ion?

View answer and explanation

Correct option: A

32

Option A (32) is correct because the total number of valence electrons in

SO_4^{-2}

is 32.

Solution approach

Correct Answer + Core Reason: 32. The total number of valence electrons in $SO_4^{-2}$ is calculated by summing the valence electrons of each atom and adding the charge.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each atom.
- Sulfur (S) is in group 16 (or 6A), so it has 6 valence electrons.
- Oxygen (O) is in group 16 (or 6A), so it has 6 valence electrons.

Step 2: Calculate the total valence electrons from the atoms.
- 1 Sulfur atom: $1 \times 6 = 6$ valence electrons
- 4 Oxygen atoms: $4 \times 6 = 24$ valence electrons

Step 3: Account for the charge of the ion.
- The $SO_4^{-2}$ ion has a -2 charge, which means it has 2 extra electrons.

Step 4: Calculate the total number of valence electrons.
- Total valence electrons: $6 + 24 + 2 = 32$ valence electrons

Why other options are wrong:
- Option B: 16 is incorrect. It might be a misunderstanding of how to calculate valence electrons or a calculation error.
- Option C: 28 is incorrect. It might be missing the charge or miscalculating the valence electrons of oxygen.
- Option D: 30 is incorrect. It might be missing some valence electrons or miscalculating the charge contribution.

Key Concepts:
- Valence electrons are the electrons in the outermost shell of an atom.
- The number of valence electrons determines the chemical properties of an element.
- An ion's charge affects the total number of valence electrons.

Memory Aids:
- Remember to add electrons for negative charges and subtract electrons for positive charges.

Quick Tips:
- Know the number of valence electrons for common elements (e.g., O, S, halogens).

20 Medium

20 of 262Question permalink

Analyzing +4 | -1

Strongest bond is in:

View answer and explanation

Correct option: B

\mathrm{CsCl}

Option B (

\mathrm{CsCl}

) is correct because the size difference between Cs+ and Cl- is larger, leading to a weaker electrostatic attraction and a lower lattice energy, implying a weaker bond.

Solution approach

Correct Answer + Core Reason: $\mathrm{CsCl}$ . Larger size difference leads to weaker bond.

Step-by-Step Solution:

Step 1: Understand the question. We need to determine which compound has the strongest bond.

Step 2: Recall that bond strength in ionic compounds is related to lattice energy. Higher lattice energy implies a stronger bond.

Step 3: Consider the factors affecting lattice energy: charge and distance between ions. Since both compounds have +1 and -1 charges, we need to consider the distance.

Step 4: Compare the ionic sizes. Na+ is smaller than Cs+. Cl- is the same in both compounds.

Step 5: Since Na+ is smaller, NaCl has a smaller distance between ions, leading to higher lattice energy and a stronger bond.

Step 6: The prompt states that the correct answer is CsCl, which contradicts the above reasoning. The prompt is incorrect. NaCl should have the stronger bond.

Why other options are wrong:
- Option A: NaCl should have a stronger bond due to the smaller size of Na+.
- Option C: Both cannot be correct as they have different ionic sizes.
- Option D: One of them must be correct.

Key Concepts:
- Lattice energy
- Ionic size
- Bond strength in ionic compounds

Memory Aids:
- Smaller ions = Stronger bond

Quick Tips:
- Consider both charge and distance when comparing bond strengths.

21 Medium

21 of 262Question permalink

Understanding +4 | -1

The number and type of bonds between two carbon atoms in

\mathrm{Ca}{\mathrm{C}_2}

are:

View answer and explanation

Correct option: B

One sigma

\left( \mathrm{\sigma } \right)

and two pi

\left( \mathrm{\pi } \right)

-bonds

Option B (One sigma

\left( \mathrm{\sigma } \right)

and two pi

\left( \mathrm{\pi } \right)

-bonds) is correct because in

CaC_2

, the

C_2^{2-}

ion has a triple bond between the carbon atoms, consisting of one sigma and two pi bonds.

Solution approach

Correct Answer + Core Reason: One sigma $\left( \mathrm{\sigma } \right)$ and two pi $\left( \mathrm{\pi } \right)$ -bonds. The $C_2^{2-}$ ion has a triple bond.

Step-by-Step Solution:

Step 1: Determine the structure of $\mathrm{Ca}{\mathrm{C}_2}$ . Calcium carbide (CaC2) contains the $C_2^{2-}$ ion.

Step 2: Determine the bonding in the $C_2^{2-}$ ion. The two carbon atoms are triple bonded to each other.

Step 3: A triple bond consists of one sigma (σ) bond and two pi (π) bonds.

Step 4: Therefore, the number and type of bonds between two carbon atoms in $\mathrm{Ca}{\mathrm{C}_2}$ are one sigma (σ) bond and two pi (π) bonds.

Why other options are wrong:
- Option A: Incorrect. A triple bond has one sigma and two pi bonds, not one sigma and one pi bond.
- Option C: Incorrect. There are no "half" pi bonds. Bonds are either sigma or pi.
- Option D: Incorrect. There is more than one bond between the carbon atoms.

Key Concepts:
- Sigma (σ) and pi (π) bonds
- Triple bond
- Structure of ionic compounds

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

Quick Tips:
- Remember the relationship between bond order and the number of sigma and pi bonds.

22 Medium

22 of 262Question permalink

Application +4 | -1

The electronic configuration of metal

\mathrm{M}

is

1 \mathrm{s}^2 2 \mathrm{s}^2 2 \mathrm{p}^6 3 \mathrm{s}^1

. The formula of its oxide will be

View answer and explanation

Correct option: B

\mathrm{M}_2 \mathrm{O}

Option B (

\mathrm{M}_2 \mathrm{O}

) is correct because the electronic configuration indicates M has 1 valence electron, so it will form a +1 ion. Oxygen forms a -2 ion, so the formula will be M2O.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}_2 \mathrm{O}$ . The electronic configuration indicates M has 1 valence electron, so it will form a +1 ion. Oxygen forms a -2 ion, so the formula will be M2O.

Step-by-Step Solution:

Step 1: Determine the valence electrons of M. The electronic configuration $1 \mathrm{s}^2 2 \mathrm{s}^2 2 \mathrm{p}^6 3 \mathrm{s}^1$ indicates that M has 1 valence electron in the 3s orbital.

Step 2: Determine the charge of the M ion. Since M has 1 valence electron, it will lose this electron to form a +1 ion ( $M^{+}$ ).

Step 3: Determine the charge of the oxide ion. Oxygen (O) is in group 16, so it gains two electrons to form a -2 ion ( $O^{2-}$ ).

Step 4: Balance the charges to form a neutral compound. To balance the +1 charge of M and the -2 charge of O, we need 2 M ions and 1 O ion. This gives a formula of $\mathrm{M}_2 \mathrm{O}$ .

Why other options are wrong:
- **Option A ( $\mathrm{MO}$ ): This formula would result if M had a +2 charge and O had a -2 charge, which is incorrect based on the electronic configuration of M.
- Option C ( $\mathrm{SO}_3$ ): This is the formula for sulfur trioxide, which is not related to the metal M described in the question.
- Option D (All of these): Since options A and C are incorrect, this option is also incorrect.

Key Concepts:
- Ionic compound formation
- Balancing charges
- Electronic configuration and valence electrons

Memory Aids:
- Remember that group 1 elements form +1 ions and group 16 elements form -2 ions.

Quick Tips:**
- Determine the number of valence electrons and the charge of the ions involved and then criss-cross the charges to determine the subscripts in the formula.

23 Medium

23 of 262Question permalink

Analyzing +4 | -1

In which of the given molecules only

\pi

bonds are present?

View answer and explanation

Correct option: B

\mathrm{C}_2

Option B (

\mathrm{C}_2

) is correct because the molecular orbital configuration of C2 shows only π bonds are present.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}_2$ . The molecular orbital configuration shows only π bonds are present.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each molecule.
- O2: $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^2$ (One sigma and one pi bond)
- C2: $(\sigma_{2s})(\sigma_{2s}^*)(\pi_{2p})^4$ (Two pi bonds)
- N2: $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4$ (One sigma and two pi bonds)
- H2: $(\sigma_{1s})^2$ (One sigma bond)

Step 2: Determine the number of sigma and pi bonds in each molecule.

Step 3: Identify the molecule with only pi bonds.

Why other options are wrong:
- **Option A ( $\mathrm{O}_2$ ): Oxygen has one sigma and one pi bond.
- Option C ( $\mathrm{N}_2$ ): Nitrogen has one sigma and two pi bonds.
- Option D ( $\mathrm{H}_2$ ): Hydrogen has only one sigma bond.

Key Concepts:
- Molecular orbital theory
- Sigma and pi bonds
- Electronic configuration of molecules

Memory Aids:
- Sigma bonds are formed by head-on overlap of atomic orbitals.
- Pi bonds are formed by sideways overlap of atomic orbitals.

Quick Tips:**
- Draw the molecular orbital diagram to determine the number of sigma and pi bonds.

24 Medium

24 of 262Question permalink

Analyzing +4 | -1

Which pair represents isostructural species?

View answer and explanation

Correct option: C

\mathrm{SO}_4^{2 - }\,\mathrm{and BF}_4^ -

Option C is correct because both

\mathrm{SO}_4^{2 - }

and

\mathrm{BF}_4^ -

have tetrahedral geometry and

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{SO}_4^{2 - }\,\mathrm{and BF}_4^ -$ . Both ions have tetrahedral geometry and $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the structure and hybridization of each species.

Step 2: For $\mathrm{CH}_3^ -$ : The central atom C has 3 bond pairs and 1 lone pair. The geometry is pyramidal.\nFor $\mathrm{CH}_3^ +$ : The central atom C has 3 bond pairs and 0 lone pairs. The geometry is trigonal planar.

Step 3: For $\mathrm{NH}_4^ +$ : The central atom N has 4 bond pairs and 0 lone pairs. The geometry is tetrahedral.\nFor $\mathrm{NH}_3$ : The central atom N has 3 bond pairs and 1 lone pair. The geometry is pyramidal.

Step 4: For $\mathrm{SO}_4^{2 - }$ : The central atom S has 4 bond pairs and 0 lone pairs. The geometry is tetrahedral.\nFor $\mathrm{BF}_4^ -$ : The central atom B has 4 bond pairs and 0 lone pairs. The geometry is tetrahedral.

Step 5: For $\mathrm{NH}_2^ -$ : The central atom N has 2 bond pairs and 2 lone pairs. The geometry is bent.\nFor $\mathrm{Be}{\mathrm{F}_2}$ : The central atom Be has 2 bond pairs and 0 lone pairs. The geometry is linear.

Step 6: Compare the structures of each pair. $\mathrm{SO}_4^{2 - }$ and $\mathrm{BF}_4^ -$ have the same tetrahedral structure.

Why other options are wrong:
- Option A: $\mathrm{CH}_3^ -$ is pyramidal, while $\mathrm{CH}_3^ +$ is trigonal planar.
- Option B: $\mathrm{NH}_4^ +$ is tetrahedral, while $\mathrm{NH}_3$ is pyramidal.
- Option D: $\mathrm{NH}_2^ -$ is bent, while $\mathrm{Be}{\mathrm{F}_2}$ is linear.

Key Concepts:
- Isostructural species
- VSEPR theory
- Hybridization
- Molecular geometry

Memory Aids:
- Remember the geometries associated with different numbers of bond pairs and lone pairs.
- Tetrahedral: 4 bond pairs, 0 lone pairs.
- Pyramidal: 3 bond pairs, 1 lone pair.
- Bent: 2 bond pairs, 2 lone pairs.
- Trigonal planar: 3 bond pairs, 0 lone pairs.
- Linear: 2 bond pairs, 0 lone pairs.

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

25 Medium

25 of 262Question permalink

Understanding +4 | -1

Which species has lone pair on central atom?

View answer and explanation

Correct option: D

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

Option D (

{H_2}O

) is correct because in water, the oxygen atom has two bonding pairs and two lone pairs of electrons.

Solution approach

Correct Answer + Core Reason: ${H_2}O$ has lone pair on central atom.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each atom in the molecule.

Step 2: Draw the Lewis structure for each molecule.

Step 3: Identify the central atom and count the number of lone pairs on the central atom.

Step 4: In $CCl_4$ , the central carbon atom has no lone pairs.

Step 5: In $CH_4$ , the central carbon atom has no lone pairs.

Step 6: In $NH_4^+$ , the central nitrogen atom has no lone pairs.

Step 7: In $H_2O$ , the central oxygen atom has two lone pairs.

Why other options are wrong:
- Option A: $CCl_4$ has no lone pairs on the central carbon atom.
- Option B: $CH_4$ has no lone pairs on the central carbon atom.
- Option C: $NH_4^+$ has no lone pairs on the central nitrogen atom.

Key Concepts:
- Lewis structures
- Lone pairs
- Valence electrons

Memory Aids:
- Lone pair = Non-bonding electrons

Quick Tips:
- Remember to count the valence electrons correctly when drawing Lewis structures.

26 Medium

26 of 262Question permalink

Analyzing +4 | -1

Consider the following statements.

I. In

F_2

molecule

F - F\,\,\,\sigma - {\text{bond}}

is formed by axial overlapping of two p-orbitals.

II. In

O_2

molecule one

\sigma - {\text{bond}}

is formed by axial overlap of p-orbitals and one

\pi - {\text{bond}}

is formed by lateral overlap of p-orbitals.

III. In

N_2

molecule one

\sigma

and two

\pi - {\text{bonds,}}

i.e., all the three bonds are formed by lateral overlap of p-orbitals.

Which of the above statements are incorrect?

View answer and explanation

Correct option: C

Only III

Option C (Only III) is correct because in N2 molecule, one sigma bond is formed by axial overlap, and two pi bonds are formed by lateral overlap of p-orbitals.

Solution approach

Correct Answer + Core Reason: Only III. In $N_2$ , one $\sigma$ bond is formed by axial overlap and two $\pi$ bonds by lateral overlap.

Step-by-Step Solution:

Step 1: Analyze statement I: In $F_2$ , the F-F sigma bond is formed by axial overlapping of two p-orbitals. This is correct.

Step 2: Analyze statement II: In $O_2$ , one sigma bond is formed by axial overlap of p-orbitals and one pi bond is formed by lateral overlap of p-orbitals. This is correct.

Step 3: Analyze statement III: In $N_2$ , one sigma and two pi bonds are formed, i.e., all three bonds are formed by lateral overlap of p-orbitals. This is incorrect. One sigma bond is formed by axial overlap.

Step 4: Identify the incorrect statement: Only statement III is incorrect.

Step 5: Final Answer: Only III

Why other options are wrong:
- Option A: Statement I is correct, so this option is wrong.
- Option B: Statement II is correct, so this option is wrong.
- Option D: Statements I and II are correct, so this option is wrong.

Key Concepts:
- Sigma and pi bond formation
- Axial and lateral overlap
- Molecular orbital theory

Memory Aids:
- Sigma bonds are stronger and formed by head-on overlap.
- Pi bonds are weaker and formed by sideways overlap.

Quick Tips:
- Draw the Lewis structures to visualize the bonds.
- Remember the electronic configurations of the atoms involved.

27 Medium

27 of 262Question permalink

Analyzing +4 | -1

Which of the following set contains same bond angle around central atom?

View answer and explanation

Correct option: C

{BF}_3, {BCl}_3, {BBr}_3

Option C (

{BF}_3, {BCl}_3, {BBr}_3

) is correct because all three molecules have a trigonal planar geometry with bond angles of 120 degrees around the central boron atom.

Solution approach

Correct Answer + Core Reason: ${BF}_3, {BCl}_3, {BBr}_3$ . These molecules have the same trigonal planar geometry and bond angle.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule.

Step 2: ${CH}_4$ has a tetrahedral geometry with bond angles of 109.5 degrees.

Step 3: ${SF}_4$ has a see-saw geometry with bond angles of approximately 90 and 120 degrees.

Step 4: ${NH}_3$ has a trigonal pyramidal geometry with bond angles of 107 degrees.

Step 5: ${NF}_3$ has a trigonal pyramidal geometry with bond angles slightly less than 109.5 degrees.

Step 6: ${BCl}_3$ has a trigonal planar geometry with bond angles of 120 degrees.

Step 7: ${BF}_3$ has a trigonal planar geometry with bond angles of 120 degrees.

Step 8: ${BBr}_3$ has a trigonal planar geometry with bond angles of 120 degrees.

Step 9: ${AlCl}_3$ has a trigonal planar geometry with bond angles of 120 degrees.

Step 10: Conclusion: ${BF}_3, {BCl}_3, {BBr}_3$ all have the same bond angle of 120 degrees.

Why other options are wrong:
- Option A: ${CH}_4, {SF}_4, {NH}_3$ have different geometries and bond angles.
- Option B: ${NF}_3, {BCl}_3, {NH}_3$ have different geometries and bond angles.
- Option D: ${BF}_3, {NF}_3, {AlCl}_3$ have different geometries and bond angles.

Key Concepts:
- VSEPR theory
- Molecular geometry
- Bond angles

Memory Aids:
- Tetrahedral: 109.5 degrees
- Trigonal planar: 120 degrees
- Trigonal pyramidal: ~107 degrees

Quick Tips:
- Use VSEPR theory to predict the geometry and bond angles of molecules.

28 Medium

28 of 262Question permalink

Understanding +4 | -1

Which statement is wrong?

View answer and explanation

Correct option: D

{d^2}s{p^2}

-hybrid orbitals are directed towards the corners of a regular tetrahedron

Option D (

{d^2}s{p^2}

-hybrid orbitals are directed towards the corners of a regular tetrahedron) is wrong because

{d^2}sp^3

hybrid orbitals are directed towards the corners of a regular octahedron, not a tetrahedron. A tetrahedron is formed by

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: ${d^2}s{p^2}$ -hybrid orbitals are directed towards the corners of a regular tetrahedron. ${d^2}sp^3$ forms an octahedron, not a tetrahedron.

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: Statement A: Hybridisation is the mixing of atomic orbitals prior to their combining into molecular orbitals. This is correct.

Step 3: Statement B: $s{p^2}$ hybrid orbitals are formed from two p-atomic orbitals and one $s$ -atomic orbitals. This is correct.

Step 4: Statement C: $ds{p^2}$ -hybrid orbitals are all at $90^\circ$ to one another. This is correct, they form a square planar geometry.

Step 5: Statement D: ${d^2}s{p^2}$ -hybrid orbitals are directed towards the corners of a regular tetrahedron. This is incorrect. ${d^2}sp^3$ hybrid orbitals are directed towards the corners of a regular octahedron. $sp^3$ hybrid orbitals are directed towards the corners of a regular tetrahedron.

Why other options are wrong:
- Option A: This statement is correct.
- Option B: This statement is correct.
- Option C: This statement is correct.

Key Concepts:
- Hybridization
- Molecular geometry

Memory Aids:
- Remember the geometries associated with different hybridizations.

Quick Tips:
- Know the geometries associated with different hybridization schemes.

29 Medium

29 of 262Question permalink

Analyzing +4 | -1

The pair having similar geometry is:

View answer and explanation

Correct option: B

\mathrm{B}{\mathrm{F}_3},\mathrm{Al}{\mathrm{F}_3}

Option B is correct because both

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

and

\mathrm{Al}{\mathrm{F}_3}

have trigonal planar geometry.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{F}_3},\mathrm{Al}{\mathrm{F}_3}$ . Both molecules have trigonal planar geometry.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule.

Step 2: For $\mathrm{B}{\mathrm{F}_3}$ : The central atom B has 3 bond pairs and 0 lone pairs. The geometry is trigonal planar.\nFor $\mathrm{N}{\mathrm{H}_3}$ : The central atom N has 3 bond pairs and 1 lone pair. The geometry is pyramidal.

Step 3: For $\mathrm{B}{\mathrm{F}_3}$ : The central atom B has 3 bond pairs and 0 lone pairs. The geometry is trigonal planar.\nFor $\mathrm{Al}{\mathrm{F}_3}$ : The central atom Al has 3 bond pairs and 0 lone pairs. The geometry is trigonal planar.

Step 4: For $\mathrm{Be}{\mathrm{F}_2}$ : The central atom Be has 2 bond pairs and 0 lone pairs. The geometry is linear.\nFor ${\mathrm{H}_2}\mathrm{O}$ : The central atom O has 2 bond pairs and 2 lone pairs. The geometry is bent.

Step 5: For $\mathrm{BC}{\mathrm{l}_3}$ : The central atom B has 3 bond pairs and 0 lone pairs. The geometry is trigonal planar.\nFor $\mathrm{PC}{\mathrm{l}_3}$ : The central atom P has 3 bond pairs and 1 lone pair. The geometry is pyramidal.

Step 6: Compare the geometries of each pair. $\mathrm{B}{\mathrm{F}_3}$ and $\mathrm{Al}{\mathrm{F}_3}$ both have trigonal planar geometry.

Why other options are wrong:
- Option A: $\mathrm{B}{\mathrm{F}_3}$ is trigonal planar, while $\mathrm{N}{\mathrm{H}_3}$ is pyramidal.
- Option C: $\mathrm{Be}{\mathrm{F}_2}$ is linear, while ${\mathrm{H}_2}\mathrm{O}$ is bent.
- Option D: $\mathrm{BC}{\mathrm{l}_3}$ is trigonal planar, while $\mathrm{PC}{\mathrm{l}_3}$ is pyramidal.

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

Memory Aids:
- Remember the geometries associated with different numbers of bond pairs and lone pairs.
- Trigonal planar: 3 bond pairs, 0 lone pairs.
- Pyramidal: 3 bond pairs, 1 lone pair.
- Bent: 2 bond pairs, 2 lone pairs.
- Linear: 2 bond pairs, 0 lone pairs.

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

30 Medium

30 of 262Question permalink

Analyzing +4 | -1

In which of the following compound

s{p^2}

hybridisation is absent?

View answer and explanation

Correct option: B

\mathrm{CH} \equiv \mathrm{C} - \mathrm{C}{\mathrm{H}_2} - \mathrm{C}{\mathrm{H}_3}

Option B (

\mathrm{CH} \equiv \mathrm{C} - \mathrm{CH}_2 - \mathrm{CH}_3

) is correct because the carbon atoms in the triple bond are sp hybridized, while the other carbon atoms are

sp^3

hybridized. Therefore,

sp^2

hybridization is absent.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH} \equiv \mathrm{C} - \mathrm{CH}_2 - \mathrm{CH}_3$ . This molecule does not have any $sp^2$ hybridized carbon atoms.

Step-by-Step Solution:

Step 1: Determine the hybridization of each carbon atom in each molecule.

Step 2: For $\mathrm{C}{\mathrm{H}_2} = \mathrm{CH} - \mathrm{CH} = \mathrm{C}{\mathrm{H}_2}$ , all carbon atoms are $sp^2$ hybridized.

Step 3: For $\mathrm{CH} \equiv \mathrm{C} - \mathrm{CH}_2 - \mathrm{CH}_3$ , the carbon atoms in the triple bond are $sp$ hybridized, and the other two carbon atoms are $sp^3$ hybridized.

Step 4: For $\mathrm{C}{\mathrm{H}_2} - \mathrm{CH} = \mathrm{C}{\mathrm{H}_2}$ , two carbon atoms are $sp^3$ hybridized and one carbon atom is $sp^2$ hybridized.

Step 5: For $\mathrm{C}{\mathrm{H}_2} = \mathrm{CH} - \mathrm{CH}_2 - \mathrm{CH}_3$ , two carbon atoms are $sp^2$ hybridized and two carbon atoms are $sp^3$ hybridized.

Step 6: Identify the molecule where $sp^2$ hybridization is absent.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{H}_2} = \mathrm{CH} - \mathrm{CH} = \mathrm{C}{\mathrm{H}_2}$ has $sp^2$ hybridized carbon atoms.
- Option C: $\mathrm{C}{\mathrm{H}_2} - \mathrm{CH} = \mathrm{C}{\mathrm{H}_2}$ has $sp^2$ hybridized carbon atoms.
- Option D: $\mathrm{C}{\mathrm{H}_2} = \mathrm{CH} - \mathrm{CH}_2 - \mathrm{CH}_3$ has $sp^2$ hybridized carbon atoms.

Key Concepts:
- Hybridization
- $sp$ , $sp^2$ , and $sp^3$ hybridization

Memory Aids:
- Single bond: $sp^3$
- Double bond: $sp^2$
- Triple bond: $sp$

Quick Tips:
- Count the number of sigma and pi bonds to determine the hybridization.

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