A number of ionic compounds, e.g., ${\rm{AgCl}},{\rm{Ca}}{{\rm{F}}_2},{\rm{BaS}}{{\rm{O}}_4}$ are insoluble in water. This is because:

Ionic compounds do not dissolve in water

Water has a high dielectric constant

Water is not a good ionizing solvent

${{\rm{H}}_2}{\rm{O}}$ has a net dipole moment, while ${\rm{Be}}{{\rm{F}}_2}$ has zero dipole moment, because:

${\rm{Be}}{{\rm{F}}_2}$ molecule is linear, while ${{\rm{H}}_2}{\rm{O}}$ is bent

${{\rm{H}}_2}{\rm{O}}$ molecule as linear while ${\rm{Be}}{{\rm{F}}_2}$ is bent

Fluorine is more electronegative than oxygen

${\rm{Be}}$ is more electronegative than oxygen

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 4

91 Easy

91 of 220Question permalink

Understanding +4 | -1

\mathrm{Xe}{\mathrm{F}_6}

is:

View answer and explanation

Correct option: B

Pentagonal pyramidal

Option B (Pentagonal pyramidal) is correct because XeF6 has a distorted octahedral structure due to the lone pair, resulting in a pentagonal pyramidal shape.

Solution approach

Correct Answer + Core Reason: Pentagonal pyramidal. XeF6 has one lone pair and six bond pairs, leading to a distorted octahedral geometry.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for Xe and F. Xe has 8 valence electrons, and each F has 7.

Step 2: Calculate the total number of valence electrons in XeF6: 8 + 6(7) = 50.

Step 3: Determine the number of electron pairs: 50/2 = 25.

Step 4: Determine the number of bond pairs: 6 (since there are 6 F atoms bonded to Xe).

Step 5: Determine the number of lone pairs: (50 - 12)/2 = 1 (one lone pair on Xe).

Step 6: Apply VSEPR theory. With 6 bond pairs and 1 lone pair, the electron geometry is distorted octahedral. The molecular geometry is pentagonal pyramidal.

Why other options are wrong:
- Option A: Octahedral would require 6 bond pairs and 0 lone pairs.
- Option C: Planar structures are typically seen with sp2 hybridization and do not fit the electron count of XeF6.
- Option D: Tetrahedral structures are typically seen with sp3 hybridization and do not fit the electron count of XeF6.

Key Concepts:
- VSEPR theory.
- Lone pairs and bond pairs.
- Molecular geometry.
- Electron geometry.

Memory Aids:
- VSEPR: Valence Shell Electron Pair Repulsion.

Quick Tips:
- Always determine the number of lone pairs and bond pairs before predicting the shape of a molecule.

92 Easy

92 of 220Question permalink

Remembering +4 | -1

A

\pi

-bond is formed by sideways overlapping of:

View answer and explanation

Correct option: B

p-p orbitals

Option B (p-p orbitals) is correct because pi bonds are formed by the sideways or lateral overlapping of p-orbitals.

Solution approach

Correct Answer + Core Reason: p-p orbitals. $\pi$ -bonds are formed by sideways overlapping of p-orbitals.

Step-by-Step Solution:

Step 1: Recall the definition of a $\pi$ -bond.

Step 2: $\pi$ -bonds are formed by the lateral or sideways overlap of p-orbitals.

Step 3: s-orbitals are spherical and form $\sigma$ -bonds by head-on overlap.

Step 4: Final Answer: p-p orbitals

Why other options are wrong:
- Option A: s-s orbitals form sigma bonds.
- Option C: s-p orbitals can form sigma bonds.
- Option D: s-p-s is not a standard orbital interaction.

Key Concepts:
- Sigma and pi bond formation
- s and p orbitals
- Lateral and axial overlap

Memory Aids:
- $\sigma$ bonds are stronger and formed by head-on overlap.
- $\pi$ bonds are weaker and formed by sideways overlap.

Quick Tips:
- Visualize the shapes of s and p orbitals.
- Remember that $\pi$ bonds require p-orbitals.

93 Easy

93 of 220Question permalink

Understanding +4 | -1

Two electrons of one atom

A

and two electrons of another atom

B

are utilized to form a compound

A B

. This is an example of

View answer and explanation

Correct option: A

Polar covalent bond

Option A is correct because if atoms A and B have different electronegativities, the bond formed will be a polar covalent bond.

Solution approach

Correct Answer + Core Reason: Polar covalent bond. If A and B have different electronegativities, the bond formed will be a polar covalent bond.

Step-by-Step Solution:

Step 1: Understand the difference between polar and non-polar covalent bonds.

Step 2: Recall that a covalent bond is formed by sharing of electrons between atoms.

Step 3: If the electronegativity difference between the atoms is significant, the electrons are not shared equally, resulting in a polar covalent bond.

Step 4: In a polar covalent bond, one atom has a partial positive charge and the other has a partial negative charge.

Step 5: If the electronegativity difference is negligible, the electrons are shared equally, resulting in a non-polar covalent bond.

Step 6: Since the question does not specify that A and B have the same electronegativity, we assume they are different, leading to a polar covalent bond.

Why other options are wrong:
- Option B: Non-polar covalent bond is incorrect because it assumes equal electronegativity.
- Option C: Polar bond is a general term, but polar covalent bond is more specific and accurate.
- Option D: Dative bond (coordinate covalent bond) involves one atom donating both electrons, which is not described in the question.

Key Concepts:
- Polar covalent bond
- Non-polar covalent bond
- Electronegativity
- Dative bond

Memory Aids:
- Unequal sharing = Polar
- Equal sharing = Non-polar

Quick Tips:
- Remember that electronegativity difference determines the polarity of a covalent bond.

94 Easy

94 of 220Question permalink

Remembering +4 | -1

Polar covalent compounds are soluble in:

View answer and explanation

Correct option: A

Polar solvents

Option A (Polar solvents) is correct because polar covalent compounds dissolve best in solvents with similar polarity due to intermolecular forces.

Solution approach

Correct Answer + Core Reason: Polar solvents. \"Like dissolves like,\" meaning polar compounds dissolve in polar solvents.

Step-by-Step Solution:

Step 1: Identify the type of solute and solvent.\nWe have a polar covalent compound as the solute and we are considering different types of solvents.

Step 2: Apply the \"like dissolves like\" principle.\nPolar solutes dissolve in polar solvents due to favorable interactions between the solute and solvent molecules (dipole-dipole interactions, hydrogen bonding). Non-polar solutes dissolve in non-polar solvents (London dispersion forces).

Step 3: Conclude the solubility.\nTherefore, polar covalent compounds are soluble in polar solvents.

Why other options are wrong:
- Option B: Non-polar solvents are incorrect because polar compounds do not interact favorably with non-polar solvents.
- Option C: Concentrated acids are incorrect because while some polar compounds might react with acids, solubility is a general property related to intermolecular forces, not necessarily chemical reactions.
- Option D: All solvents is incorrect because the \"like dissolves like\" principle dictates that solubility depends on the similarity in polarity between solute and solvent.

Key Concepts:
- Polarity of molecules
- Intermolecular forces (dipole-dipole, hydrogen bonding, London dispersion forces)
- \"Like dissolves like\" principle
- Solubility

Memory Aids:
- \"Like dissolves like\" - helps remember the relationship between solute and solvent polarity.

Quick Tips:
- Remember that polarity is a key factor in determining solubility. Consider the intermolecular forces between solute and solvent.

95 Easy

95 of 220Question permalink

Analyzing +4 | -1

The increasing order of bond angle among

\mathrm{BeCl}_2(\mathrm{I})

,

\mathrm{NH}_3(\mathrm{II}), \mathrm{CH}_4(\mathrm{III}), \mathrm{BF}_3 \mathrm{(IV})

will be:

View answer and explanation

Correct option: B

II

<

III

<

IV

<

I

Option B (II

<

III

<

IV

<

I) is correct because it represents the increasing order of bond angles based on molecular geometry and VSEPR theory.

Solution approach

Correct Answer + Core Reason: II $<$ III $<$ IV $<$ I. The bond angles increase in the order: $\mathrm{NH}_3 < \mathrm{CH}_4 < \mathrm{BF}_3 < \mathrm{BeCl}_2$ based on VSEPR theory and molecular geometry.

Step-by-Step Solution:

Step 1: Determine the molecular geometry of each molecule using VSEPR theory.
- $\mathrm{BeCl}_2$ (I): Linear, bond angle = 180°
- $\mathrm{NH}_3$ (II): Trigonal pyramidal, bond angle < 109.5° (due to lone pair repulsion)
- $\mathrm{CH}_4$ (III): Tetrahedral, bond angle = 109.5°
- $\mathrm{BF}_3$ (IV): Trigonal planar, bond angle = 120°

Step 2: Arrange the bond angles in increasing order.
$\mathrm{NH}_3$ (II) < $\mathrm{CH}_4$ (III) < $\mathrm{BF}_3$ (IV) < $\mathrm{BeCl}_2$ (I)

Step 3: Therefore, the increasing order of bond angle is II < III < IV < I.

Why other options are wrong:
- **Option A (I $<$ II $<$ III $<$ IV):** This is incorrect because it places the linear $\mathrm{BeCl}_2$ with the largest bond angle at the beginning of the order.
- **Option C (II $<$ IV $<$ I $<$ III):** This is incorrect because it places the tetrahedral $\mathrm{CH}_4$ with a bond angle of 109.5° at the end of the order.
- **Option D (I $<$ IV $<$ II $<$ III):** This is incorrect because it places the linear $\mathrm{BeCl}_2$ with the largest bond angle at the beginning of the order and incorrectly orders $\mathrm{NH}_3$ and $\mathrm{CH}_4$ .

Key Concepts:
- VSEPR theory
- Molecular geometry
- Bond angle
- Lone pair repulsion

Memory Aids:
- Linear: 180°
- Trigonal planar: 120°
- Tetrahedral: 109.5°
- Trigonal pyramidal: < 109.5°

Quick Tips:
- Determine the molecular geometry first using VSEPR theory.
- Consider the effect of lone pairs on bond angles.
- Remember the standard bond angles for common geometries.

96 Easy

96 of 220Question permalink

Application +4 | -1

Formula of a metallic oxide is

M O

. The formula of its phosphate will be

View answer and explanation

Correct option: D

{M}_3\left({PO}_4\right)_2

Option D (

{M}_3\left({PO}_4\right)_2

) is correct because if MO has a +2 and -2 charge, then M has a +2 charge. Phosphate has a -3 charge, so the formula will be M3(PO4)2.

Solution approach

Correct Answer + Core Reason: ${M}_3\left({PO}_4\right)_2$ . If MO has a +2 and -2 charge, then M has a +2 charge. Phosphate has a -3 charge, so the formula will be M3(PO4)2.

Step-by-Step Solution:

Step 1: Determine the charge of the metal M. Since the oxide is MO, and oxygen has a -2 charge, the metal M must have a +2 charge.

Step 2: Determine the charge of the phosphate ion. The phosphate ion is $PO_4^{3-}$ , so it has a -3 charge.

Step 3: Balance the charges to form a neutral compound. To balance the +2 charge of M and the -3 charge of phosphate, we need 3 M ions and 2 phosphate ions. This gives a formula of ${M}_3\left({PO}_4\right)_2$ .

Why other options are wrong:
- **Option A ( ${M}_2\left({PO}_4\right)_2$ ): This formula would result if M had a +1 charge and phosphate had a -1 charge, which is incorrect.
- Option B ( ${M}\left({PO}_4\right)$ ): This formula would result if M had a +3 charge and phosphate had a -3 charge, which is incorrect.
- Option C ( ${M}_2 {PO}_4$ ): This formula would result if M had a +1 charge and phosphate had a -2 charge, which is incorrect.

Key Concepts:
- Ionic compound formation
- Balancing charges

Memory Aids:
- Remember the charge of the phosphate ion is -3.

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

97 Easy

97 of 220Question permalink

Remembering +4 | -1

Who provide explanation of valence based on intertness of noble gases?

View answer and explanation

Correct option: B

Kössel-Lewis

Option B (Kössel-Lewis) is correct because Kössel and Lewis independently developed theories explaining valence based on the inertness of noble gases and the tendency of atoms to achieve a noble gas configuration.

Solution approach

Correct Answer + Core Reason: Kössel-Lewis. They independently explained valence based on the inertness of noble gases.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify who provided the explanation of valence based on the inertness of noble gases.

Step 2: Recall the historical development of valence theory.
- Kössel and Lewis independently developed theories based on the octet rule and the tendency of atoms to achieve a noble gas configuration.

Step 3: Identify Kössel-Lewis as the correct answer.

Why other options are wrong:
- Option A: Lewis contributed significantly, but Kössel also independently developed similar ideas.
- Option C: Langmuir made contributions to chemical bonding but is not primarily associated with the initial explanation based on noble gas inertness.
- Option D: Sidgwick & Powell developed VSEPR theory, which is a later development.

Key Concepts:
- Valence theory
- Octet rule
- Noble gas configuration

Memory Aids:
- Kössel and Lewis are the key names associated with the early valence theories.

Quick Tips:
- Remember the historical context of the development of valence theory.

98 Easy

98 of 220Question permalink

Application +4 | -1

Chloride of metal is

{MCl}_2

. The formula of its phosphate will be

View answer and explanation

Correct option: B

{M}_3\left({PO}_4\right)_2

Option B (

{M}_3\left({PO}_4\right)_2

) is correct because if the metal chloride is

{MCl}_2

, then the metal has a +2 charge (

M^{2+}

). To balance the -3 charge of two phosphate ions (

2PO_4^{3-}

), three metal ions are required.

Solution approach

Correct Answer + Core Reason: ${M}_3\left({PO}_4\right)_2$ . The metal has a +2 charge, and phosphate has a -3 charge; therefore, the formula is ${M}_3\left({PO}_4\right)_2$ .

Step-by-Step Solution:

Step 1: Determine the charge of the metal ion from the formula of its chloride.

Step 2: The formula of the metal chloride is ${MCl}_2$ . Since chlorine has a -1 charge, the metal M must have a +2 charge ( $M^{2+}$ ).

Step 3: Determine the charge of the phosphate ion.

Step 4: The phosphate ion is ${PO_4^{3-}}$ , which has a -3 charge.

Step 5: Write the formula of the metal phosphate by balancing the charges.

Step 6: To balance the +2 charge of the metal and the -3 charge of the phosphate, we need three metal ions and two phosphate ions.

Step 7: Therefore, the formula of the metal phosphate is ${M}_3\left({PO}_4\right)_2$ .

Why other options are wrong:
- Option A: ${M}_2 {PO}_4$ would imply that the metal has a +1 charge, which is incorrect based on the given chloride formula.
- Option C: ${M}_2\left({PO}_4\right)_3$ would imply that the metal has a +3 charge, which is incorrect based on the given chloride formula.
- Option D: ${MPO}_4$ would imply that the metal has a +3 charge, which is incorrect based on the given chloride formula.

Key Concepts:
- Chemical formula
- Charge balance
- Ionic compounds

Memory Aids:
- Remember the charges of common ions (e.g., chloride, phosphate).

Quick Tips:
- Use the criss-cross method to balance charges when writing chemical formulas.

99 Easy

99 of 220Question permalink

Application +4 | -1

Which of the following is not

sp^3

hybridise

View answer and explanation

Correct option: A

BH_3

Option A (

BH_3

) is correct because Boron in

BH_3

has only 6 electrons in its valence shell, resulting in

sp^2

hybridization, not

sp^3

.

Solution approach

Correct Answer + Core Reason: $BH_3$ is not $sp^3$ hybridized because Boron has only 6 electrons around it, leading to $sp^2$ hybridization.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for Boron in $BH_3$ . Boron has 3 valence electrons, and each Hydrogen contributes 1, totaling 6 electrons around Boron.

Step 2: Calculate the steric number (number of sigma bonds + lone pairs). In $BH_3$ , Boron forms 3 sigma bonds with Hydrogen and has 0 lone pairs. Therefore, the steric number is 3.

Step 3: Determine the hybridization based on the steric number. A steric number of 3 corresponds to $sp^2$ hybridization.

Step 4: For $BH_4^-$ , Boron has 3 valence electrons + 1 from the negative charge + 4 from the hydrogens = 8 electrons. Steric number = 4, so $sp^3$ hybridization.

Step 5: For $NH_4^+$ , Nitrogen has 5 valence electrons - 1 from the positive charge + 4 from the hydrogens = 8 electrons. Steric number = 4, so $sp^3$ hybridization.

Step 6: For $NH_3$ , Nitrogen has 5 valence electrons + 3 from the hydrogens = 8 electrons. Steric number = 4 (3 sigma bonds + 1 lone pair), so $sp^3$ hybridization.

Why other options are wrong:
- **Option B ( $BH_4^-$ ):** Boron in $BH_4^-$ has 4 bond pairs and 0 lone pairs, resulting in $sp^3$ hybridization.
- **Option C ( $NH_4^+$ ):** Nitrogen in $NH_4^+$ has 4 bond pairs and 0 lone pairs, resulting in $sp^3$ hybridization.
- **Option D ( $NH_3$ ):** Nitrogen in $NH_3$ has 3 bond pairs and 1 lone pair, resulting in $sp^3$ hybridization.

Key Concepts:
- Hybridization
- Steric Number
- Valence Electrons

Memory Aids:
- Steric Number = Number of Sigma Bonds + Number of Lone Pairs
- sp = 2, sp2 = 3, sp3 = 4

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

100 Easy

100 of 220Question permalink

Understanding +4 | -1

The double bonds between the two carbon atoms in ethylene consists of:

View answer and explanation

Correct option: B

One sigma-bond and one pi-bond

Option B (One sigma-bond and one pi-bond) is correct because a double bond consists of one sigma bond formed by head-on overlap and one pi bond formed by sideways overlap.

Solution approach

Correct Answer + Core Reason: One sigma-bond and one pi-bond. A double bond consists of one sigma and one pi bond.

Step-by-Step Solution:

Step 1: Recall the definition of sigma and pi bonds.

Step 2: A sigma bond is formed by head-on overlap of atomic orbitals.

Step 3: A pi bond is formed by sideways overlap of atomic orbitals.

Step 4: A single bond consists of one sigma bond.

Step 5: A double bond consists of one sigma bond and one pi bond.

Step 6: A triple bond consists of one sigma bond and two pi bonds.

Why other options are wrong:
- Option A (Two sigma-bonds at right angles to each other): Incorrect, a double bond has only one sigma bond.
- Option C (Two pi-bonds at right angles to each other): Incorrect, a double bond has only one pi bond.
- **Option D (Two pi-bonds at an angle of $60^\circ$ to each other): Incorrect, a double bond has only one pi bond.

Key Concepts:
- Sigma bond
- Pi bond
- Double bond

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

Quick Tips:**
- Remember the composition of single, double, and triple bonds.

101 Easy

101 of 220Question permalink

Application +4 | -1

Hybridization present in

{ClF}_3

is

View answer and explanation

Correct option: D

{sp}^3 {~d}

Option D (

{sp}^3 {~d}

) is correct because

ClF_3

has 3 bond pairs and 2 lone pairs, resulting in a steric number of 5, which corresponds to

sp^3d

hybridization.

Solution approach

Correct Answer + Core Reason: ${sp}^3 {~d}$ because $ClF_3$ has 3 bond pairs and 2 lone pairs, leading to a steric number of 5.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for Chlorine. Chlorine has 7 valence electrons.

Step 2: Determine the number of bond pairs and lone pairs in $ClF_3$ . Chlorine forms 3 bonds with Fluorine, so there are 3 bond pairs. The remaining electrons form 2 lone pairs (7 - 3 = 4 electrons, which is 2 lone pairs).

Step 3: Calculate the steric number (number of sigma bonds + lone pairs). In $ClF_3$ , the steric number is 3 (bond pairs) + 2 (lone pairs) = 5.

Step 4: Determine the hybridization based on the steric number. A steric number of 5 corresponds to $sp^3d$ hybridization.

Why other options are wrong:
- **Option A ( ${sp}^2$ ):** $sp^2$ hybridization corresponds to a steric number of 3, which is incorrect for $ClF_3$ .
- **Option B ( ${sp}^3$ ):** $sp^3$ hybridization corresponds to a steric number of 4, which is incorrect for $ClF_3$ .
- **Option C ( ${dsp}^2$ ):** $dsp^2$ hybridization corresponds to a square planar geometry and is not applicable to $ClF_3$ .

Key Concepts:
- Hybridization
- Steric Number
- Valence Electrons
- Lone Pairs
- Bond Pairs

Memory Aids:
- Steric Number = Number of Sigma Bonds + Number of Lone Pairs
- sp = 2, sp2 = 3, sp3 = 4, sp3d = 5, sp3d2 = 6

Quick Tips:
- Draw the Lewis structure to determine the number of bond pairs and lone pairs.

102 Easy

102 of 220Question permalink

Application +4 | -1

Which of the following molecules will have polar bonds but zero dipole moment

View answer and explanation

Correct option: C

C{F_4}

Option C (

C{F_4}

) is correct because carbon tetrafluoride has polar C-F bonds, but due to its symmetrical tetrahedral geometry, the bond dipoles cancel each other out, resulting in a zero net dipole moment.

Solution approach

Correct Answer + Core Reason: $C{F_4}$ . Symmetrical tetrahedral structure cancels bond dipoles.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each compound.

Step 2: ${O_2}$ (Oxygen) is a diatomic molecule with a nonpolar covalent bond. Therefore, it has no dipole moment.

Step 3: $CHC{l_3}$ (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: $C{F_4}$ (Carbon tetrafluoride) has a symmetrical tetrahedral shape. The C-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.

Why other options are wrong:
- **Option A ( ${O_2}$ ):** ${O_2}$ has a nonpolar bond, so it has no dipole moment.
- **Option B ( $CHC{l_3}$ ):** $CHC{l_3}$ has a net dipole moment because its bond dipoles do not cancel.
- Option D (None): $C{F_4}$ fits the description.

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.

103 Easy

103 of 220Question permalink

Understanding +4 | -1

Which statement is correct?

View answer and explanation

Correct option: A

Pi-bond always exists with sigma-bond

Option A is correct because a

\pi

-bond always exists in conjunction with a

\sigma

-bond; it cannot exist independently.

Solution approach

Correct Answer + Core Reason: Pi-bond always exists with sigma-bond. A pi-bond is always formed after a sigma-bond.

Step-by-Step Solution:

Step 1: Understand the formation of sigma and pi bonds.

Step 2: Recall that sigma bonds are formed by head-on overlapping of atomic orbitals, and pi bonds are formed by sideways overlapping.

Step 3: A sigma bond must be formed first to provide the necessary framework for the sideways overlapping that forms a pi bond.

Step 4: Therefore, a pi bond always exists with a sigma bond.

Why other options are wrong:
- Option B: Pi-bond can exist independently is incorrect. It always needs a sigma bond.
- Option C: Sigma-bond is weaker than pi-bond is incorrect. Sigma bonds are generally stronger.
- Option D: Pi-bond is less reactive than sigma-bond is incorrect. Pi bonds are more reactive due to their exposed electrons.

Key Concepts:
- Sigma bond
- Pi bond
- Bond formation

Memory Aids:
- Sigma before pi.

Quick Tips:
- Remember that sigma bonds are the foundation for pi bonds.

104 Easy

104 of 220Question permalink

Remembering +4 | -1

The directional nature of covalent bond was introduced by

View answer and explanation

Correct option: A

VB theory

Option A (VB theory) is correct because Valence Bond (VB) theory introduced the concept of directional nature of covalent bonds based on the overlap of atomic orbitals.

Solution approach

Correct Answer + Core Reason: VB theory. Valence Bond (VB) theory introduced the concept of directional nature of covalent bonds.

Step-by-Step Solution:

Step 1: Recall the different theories of chemical bonding.

Step 2: Kossel and Lewis theory focused on the formation of ionic and covalent bonds through the transfer or sharing of electrons to achieve a stable octet.

Step 3: VSEPR theory focuses on predicting the shapes of molecules based on the repulsion between electron pairs around the central atom.

Step 4: Hybridization theory explains the mixing of atomic orbitals to form new hybrid orbitals with different shapes and energies.

Step 5: Valence Bond (VB) theory explains the formation of covalent bonds through the overlap of atomic orbitals. The directional nature of covalent bonds arises from the specific orientation of the overlapping atomic orbitals.

Step 6: Therefore, the directional nature of covalent bonds was introduced by VB theory.

Why other options are wrong:
- Option B: VSEPR theory explains the shapes of molecules but does not introduce the directional nature of covalent bonds.
- Option C: Hybridisation theory explains the mixing of atomic orbitals but does not directly introduce the directional nature of covalent bonds; it complements VB theory.
- Option D: Kossel and Lewis theory explains the formation of chemical bonds but does not focus on the directional nature of covalent bonds.

Key Concepts:
- Valence Bond (VB) theory
- VSEPR theory
- Hybridization theory
- Kossel and Lewis theory
- Covalent bond
- Directional nature of covalent bond

Memory Aids:
- VB theory: Overlap of atomic orbitals leads to bond formation.

Quick Tips:
- Understand the basic postulates of each bonding theory to answer such questions.

105 Easy

105 of 220Question permalink

Understanding +4 | -1

Elements have electronegativities

1.2\,\,and\,3.0

, bond formed between them would be

View answer and explanation

Correct option: B

Covalent

Option B (Covalent) is correct because an electronegativity difference between 0.4 and 1.7 typically indicates a polar covalent bond.

Solution approach

Correct Answer + Core Reason: Covalent. An electronegativity difference between 0.4 and 1.7 typically indicates a polar covalent bond.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference: $3.0 - 1.2 = 1.8$

Step 2: An electronegativity difference of less than 0.4 indicates a nonpolar covalent bond.

Step 3: An electronegativity difference between 0.4 and 1.7 indicates a polar covalent bond.

Step 4: An electronegativity difference greater than 1.7 indicates an ionic bond.

Step 5: Since the electronegativity difference is 1.8, it is borderline, but closer to covalent character. Considering the options, covalent is the most appropriate answer.

Why other options are wrong:
- Option A: Ionic bonds typically form with a larger electronegativity difference (greater than 1.7).
- Option C: Coordinate bonds are a type of covalent bond where one atom donates both electrons, but the electronegativity difference doesn't directly determine this.
- Option D: Metallic bonds occur between metal atoms, not between atoms with a significant electronegativity difference.

Key Concepts:
- Electronegativity
- Covalent bond
- Ionic bond

Memory Aids:
- Small difference = Covalent, Large difference = Ionic

Quick Tips:
- Remember the electronegativity difference ranges for different types of bonds.

106 Easy

106 of 220Question permalink

Understanding +4 | -1

Select the compound of highest covalent character.

View answer and explanation

Correct option: D

NaI

Option D (NaI) is correct because, according to Fajan's rule, larger anions increase covalent character, and iodide (I-) is the largest anion among the given options.

Solution approach

Correct Answer + Core Reason: NaI. Larger anion size leads to higher covalent character according to Fajan's rule.

Step-by-Step Solution:

Step 1: State Fajan's Rule: Fajan's rule describes the conditions that favor covalent character in ionic compounds.

Step 2: Identify the Anions: The anions in the given compounds are $Cl^-$ , $F^-$ , $Br^-$ , and $I^-$ .

Step 3: Compare Anion Sizes: The size of the anions increases in the order: $F^- < Cl^- < Br^- < I^-$ .

Step 4: Apply Fajan's Rule: According to Fajan's rule, larger anions favor covalent character. Therefore, NaI has the highest covalent character.

Why other options are wrong:
- Option A: NaCl: Chloride is smaller than iodide, so less covalent character.
- Option B: NaF: Fluoride is the smallest anion, so least covalent character.
- Option C: NaBr: Bromide is smaller than iodide, so less covalent character.

Key Concepts:
- Fajan's Rule
- Covalent Character
- Anion Size

Memory Aids:
- Larger anion = more covalent character

Quick Tips:
- Remember the trend of increasing anion size down the halogen group in the periodic table.

107 Easy

107 of 220Question permalink

Application +4 | -1

The bond order of

N O

molecule is

View answer and explanation

Correct option: C

2.5

Option C (2.5) is correct because the bond order of NO is calculated using Molecular Orbital Theory, resulting in a bond order of 2.5.

Solution approach

Correct Answer + Core Reason: 2.5. The bond order of NO is calculated using Molecular Orbital Theory.

Step-by-Step Solution:

Step 1: Determine the number of electrons in the NO molecule.

Step 2: Nitrogen (N) has 7 electrons, and Oxygen (O) has 8 electrons. Therefore, NO has 7 + 8 = 15 electrons.

Step 3: Write the electronic configuration of NO using Molecular Orbital Theory (MOT).

Step 4: The electronic configuration of NO is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^1$ .

Step 5: Calculate the bond order using the formula: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2.

Step 6: Number of bonding electrons = 2 ( $\sigma_{2s}$ ) + 2 ( $\sigma_{2p}$ ) + 4 ( $\pi_{2p}$ ) = 8.

Step 7: Number of antibonding electrons = 2 ( $\sigma_{2s}^*$ ) + 1 ( $\pi_{2p}^*$ ) = 3.

Step 8: Bond order = (8 - 3) / 2 = 5 / 2 = 2.5.

Why other options are wrong:
- Option A: A bond order of 1 is too low for NO, which has a significant amount of multiple bonding character.
- Option B: A bond order of 2 is also too low for NO, which has a significant amount of multiple bonding character.
- Option D: A bond order of 3 is too high for NO, as it does not have a full triple bond.

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

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

Quick Tips:
- Use MOT to determine the electronic configuration and calculate the bond order.

108 Easy

108 of 220Question permalink

Understanding +4 | -1

A number of ionic compounds, e.g.,

\mathrm{AgCl},\mathrm{Ca}{\mathrm{F}_2},\mathrm{BaS}{\mathrm{O}_4}

are insoluble in water. This is because:

View answer and explanation

Correct option: D

These molecules have exceptionally high attractive forces in their lattice

Option D (These molecules have exceptionally high attractive forces in their lattice) is correct because the high lattice energy (strong attractive forces) prevents the ions from being easily separated and solvated by water molecules.

Solution approach

Correct Answer + Core Reason: High attractive forces in their lattice makes them insoluble.

Step-by-Step Solution:

Step 1: The solubility of an ionic compound in water depends on the balance between the lattice energy (energy required to separate the ions) and the hydration energy (energy released when ions are solvated by water molecules).

Step 2: If the lattice energy is much greater than the hydration energy, the compound will be insoluble.

Step 3: $AgCl$ , $CaF_2$ , and $BaSO_4$ have exceptionally high attractive forces in their lattice, resulting in high lattice energies.

Step 4: This high lattice energy makes it difficult for water molecules to overcome the attractive forces between the ions and dissolve the compound.

Why other options are wrong:
- Option A: Ionic compounds can dissolve in water if the hydration energy is greater than the lattice energy.
- Option B: Water's high dielectric constant helps to reduce the electrostatic attraction between ions, but it cannot overcome exceptionally high lattice energies.
- Option C: Water is a good ionizing solvent for many ionic compounds, but not for those with very high lattice energies.

Key Concepts:
- Solubility of ionic compounds
- Lattice energy
- Hydration energy
- Dielectric constant

Memory Aids:
- High lattice energy = Insoluble

Quick Tips:
- Remember that the balance between lattice energy and hydration energy determines the solubility of ionic compounds.

109 Easy

109 of 220Question permalink

Remembering +4 | -1

Which is present in peroxides?

View answer and explanation

Correct option: C

\mathrm{O}_2^{2 - }

Option C (

\mathrm{O}_2^{2 - }

) is correct because peroxides contain the peroxide ion, which is

\mathrm{O}_2^{2 - }

.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^{2 - }$ . Peroxides contain the peroxide ion, $\mathrm{O}_2^{2 - }$ .

Step-by-Step Solution:

Step 1: Define peroxides.

Step 2: Peroxides are compounds containing the peroxide ion, which is $\mathrm{O}_2^{2 - }$ . Each oxygen atom has an oxidation state of -1.

Step 3: Examples of peroxides include hydrogen peroxide ( $\mathrm{H}_2\mathrm{O}_2$ ) and sodium peroxide ( $\mathrm{Na}_2\mathrm{O}_2$ ).

Why other options are wrong:
- Option A: ${\mathrm{O}_2}$ is molecular oxygen.
- Option B: ${\mathrm{O}^{2 - }}$ is the oxide ion.
- Option D: $\mathrm{O}_2^ -$ is the superoxide ion.

Key Concepts:
- Peroxides
- Peroxide ion ( $\mathrm{O}_2^{2 - }$ )
- Oxidation states of oxygen

Memory Aids:
- Peroxide: $\mathrm{O}_2^{2 - }$
- Superoxide: $\mathrm{O}_2^ -$
- Oxide: ${\mathrm{O}^{2 - }}$

Quick Tips:
- Know the different oxidation states of oxygen in various compounds.

110 Easy

110 of 220Question permalink

Remembering +4 | -1

According to Fajan’s rule formation of a covalent bond is favoured by

View answer and explanation

Correct option: C

Small cation & Large anion

Option C (Small cation & Large anion) is correct because according to Fajan's rule, small, highly charged cations and large, highly charged anions favor covalent bond formation due to increased polarization.

Solution approach

Correct Answer + Core Reason: Small cation & Large anion. Fajan's rule states that small cations and large anions favor covalent bond formation.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify the conditions that favor the formation of a covalent bond according to Fajan's rule.

Step 2: Recall Fajan's rule. Covalent character is favored by:
- Small, highly charged cation
- Large, highly charged anion

Step 3: Choose the option that matches Fajan's rule: Small cation & Large anion.

Why other options are wrong:
- Option A: Large cation & Large anion favors ionic character.
- Option B: Small cation & small anion doesn't maximize polarization.
- Option D: Large cation & Small anion favors ionic character.

Key Concepts:
- Fajan's rule
- Polarization of ions
- Covalent character

Memory Aids:
- Small cation, Large anion = Covalent

Quick Tips:
- Remember Fajan's rule to predict covalent character.

111 Easy

111 of 220Question permalink

Understanding +4 | -1

In

{H_2}^ -

ion, the bond order is

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because

{H_2}^-

has a bond order of 0.5, calculated from its electronic configuration.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}$ . The bond order of H2- is 0.5.

Step-by-Step Solution:

Step 1: Write the electronic configuration of ${H_2}^-$ .

Step 2: ${H_2}^-$ has 3 electrons. Its electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^1$ .

Step 3: Calculate the bond order using the formula: Bond order = (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals) / 2.

Step 4: Bond order = (2 - 1) / 2 = 1/2 = 0.5.

Why other options are wrong:
- Option A (Zero): This is incorrect. The bond order calculation does not result in 0.
- **Option C ( $\frac{{ - 1}}{2}$ ): Bond order cannot be negative.
- Option D (1): This is incorrect. The bond order calculation does not result in 1.

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

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

Quick Tips:**
- Write the correct electronic configuration before calculating the bond order.

112 Easy

112 of 220Question permalink

Understanding +4 | -1

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

has a net dipole moment, while

\mathrm{Be}{\mathrm{F}_2}

has zero dipole moment, because:

View answer and explanation

Correct option: B

\mathrm{Be}{\mathrm{F}_2}

molecule is linear, while

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

is bent

Option B is correct because BeF2 is linear, causing the bond dipoles to cancel out, while H2O is bent, resulting in a net dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be}{\mathrm{F}_2}$ molecule is linear, while ${\mathrm{H}_2}\mathrm{O}$ is bent. The shape of the molecule determines whether the bond dipoles cancel out or not.

Step-by-Step Solution:

Step 1: Determine the molecular geometry of $BeF_2$ . Beryllium has 2 bond pairs and 0 lone pairs, resulting in a linear geometry. The bond dipoles cancel out, making it nonpolar.

Step 2: Determine the molecular geometry of $H_2O$ . Oxygen has 2 bond pairs and 2 lone pairs, resulting in a bent geometry. The bond dipoles do not cancel out, making it polar.

Step 3: Conclusion: The difference in polarity is due to the difference in molecular geometry.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\mathrm{O}$ molecule as linear while $\mathrm{Be}{\mathrm{F}_2}$ is bent: This is incorrect because the geometries are reversed.
- Option C: Fluorine is more electronegative than oxygen: This is incorrect because while electronegativity is important, the geometry is the deciding factor in this case.
- Option D: $\mathrm{Be}$ is more electronegative than oxygen: This is incorrect because Be is less electronegative than O.

Key Concepts:
- Molecular Geometry
- Polarity of Molecules
- Dipole Moment

Memory Aids:
- Remember that linear molecules with identical bonds are nonpolar.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict the polarity of a molecule.

113 Easy

113 of 220Question permalink

Analyzing +4 | -1

Which of the following statements are not correct?
A. The electron gain enthalpy of

F

is more negative than that of

Cl.

B. Ionization enthalpy decreases in a group of periodic table.
C. The electronegativity of an atom depends upon the atoms bonded to it.
D.

Al_2O_3

and

NO

are examples of amphoteric oxides.
Choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: D

A, C and D only

Option D is the correct answer because statements A, C, and D are incorrect regarding electron gain enthalpy, electronegativity, and amphoteric oxides, respectively.

Solution approach

Correct Answer + Core Reason: A, C and D only. These statements are incorrect regarding periodic trends and properties of oxides.

Step-by-Step Solution:

Step 1: Analyze statement A: The electron gain enthalpy of F is less negative than that of Cl. This is due to the small size of F, which leads to interelectronic repulsions, making it less favorable to add an electron.

Step 2: Analyze statement B: Ionization enthalpy decreases down a group due to increasing atomic size and shielding effect. This statement is correct.

Step 3: Analyze statement C: Electronegativity is an intrinsic property of an atom and does not depend on the atoms bonded to it. It influences the polarity of bonds formed, but the electronegativity value itself is a characteristic of the atom.

Step 4: Analyze statement D: $Al_2O_3$ is an amphoteric oxide, but $NO$ is a neutral oxide. Amphoteric oxides can react with both acids and bases.

Step 5: Identify the incorrect statements: A, C, and D are incorrect.

Why other options are wrong:
- Option A: Includes B, which is a correct statement.
- Option B: Includes B, which is a correct statement.
- Option C: Includes only B and D, but A and C are also incorrect.
- Option D: Correctly identifies A, C, and D as incorrect.

Key Concepts:
- Electron gain enthalpy
- Ionization enthalpy
- Electronegativity
- Amphoteric oxides

Memory Aids:
- Down a group, ionization energy decreases.
- Electronegativity is an intrinsic property of an element.

Quick Tips:
- Remember the exceptions to periodic trends.
- Know the common amphoteric and neutral oxides.

114 Easy

114 of 220Question permalink

Application +4 | -1

Elements whose electronegativities are

1.2

and

3.0

form:

View answer and explanation

Correct option: A

Ionic bond

Option A (Ionic bond) is correct because a large electronegativity difference (greater than 1.7) between two elements typically results in the formation of an ionic bond.

Solution approach

Correct Answer + Core Reason: Ionic bond. A large electronegativity difference leads to ionic bond formation.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference between the two elements.

Step 2: Electronegativity difference = |3.0 - 1.2| = 1.8.

Step 3: Determine the type of bond based on the electronegativity difference.

Step 4: If the electronegativity difference is greater than 1.7, the bond is considered ionic.

Step 5: If the electronegativity difference is between 0.4 and 1.7, the bond is considered polar covalent.

Step 6: If the electronegativity difference is less than 0.4, the bond is considered non-polar covalent.

Step 7: Since the electronegativity difference is 1.8, the bond is ionic.

Why other options are wrong:
- Option B: Covalent bonds form when the electronegativity difference is small.
- Option C: Coordinate bonds are a type of covalent bond where both electrons are donated by one atom, but the electronegativity difference still plays a role in bond formation.
- Option D: Metallic bonds occur between metal atoms.

Key Concepts:
- Electronegativity
- Ionic bond
- Covalent bond
- Electronegativity difference and bond type

Memory Aids:
- Large electronegativity difference = Ionic bond.

Quick Tips:
- Remember the electronegativity difference ranges for different bond types.

115 Easy

115 of 220Question permalink

Understanding +4 | -1

Which of the following has covalent bond?

View answer and explanation

Correct option: B

\mathrm{AlC}{\mathrm{l}_3}

Option B (

\mathrm{AlCl}_3

) is correct because aluminum chloride has a significant degree of covalent character due to the small size and high charge of the aluminum ion, leading to polarization of the chloride ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{AlCl}_3$ . Due to the polarizing power of $Al^{3+}$ ion, it exhibits covalent character.

Step-by-Step Solution:

Step 1: Analyze the Compounds: Consider the electronegativity differences between the elements in each compound.

Step 2: Apply Electronegativity Concepts: Large electronegativity differences typically indicate ionic bonds, while smaller differences suggest covalent bonds.

Step 3: Evaluate Each Compound:
- $\mathrm{N}{\mathrm{a}_2}\mathrm{S}$ : Sodium sulfide is primarily ionic due to the large electronegativity difference between Na and S.
- $\mathrm{AlCl}_3$ : Aluminum chloride has some covalent character due to the polarizing power of the $Al^{3+}$ ion.
- $NaH$ : Sodium hydride is primarily ionic.
- $\mathrm{MgCl}_2$ : Magnesium chloride is primarily ionic.

Step 4: Conclusion: $\mathrm{AlCl}_3$ has the most covalent character among the given options.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{a}_2}\mathrm{S}$ : Primarily ionic due to large electronegativity difference.
- Option C: $NaH$ : Primarily ionic due to large electronegativity difference.
- Option D: $\mathrm{MgC}{\mathrm{l}_2}$ : Primarily ionic due to large electronegativity difference.

Key Concepts:
- Electronegativity
- Ionic Bonds
- Covalent Bonds
- Fajan's Rule

Memory Aids:
- Large electronegativity difference = ionic bond

Quick Tips:
- Consider the polarizing power of cations when determining covalent character.

116 Easy

116 of 220Question permalink

Analyzing +4 | -1

In which of the process the bond order increases and magnetic behaviour changes

View answer and explanation

Correct option: D

{O_2} \to {O_2}^{ + 2}

Option D (

{O_2} \to {O_2}^{ + 2}

) is correct because the bond order increases from 2 to 3 and the magnetic behavior changes from paramagnetic to diamagnetic.

Solution approach

Correct Answer + Core Reason: ${O_2} \to {O_2}^{ + 2}$ . Bond order increases and magnetic behavior changes.

Step-by-Step Solution:

Step 1: Determine the bond order and magnetic behavior of each species.
${N_2}$ : Bond order = 3, Diamagnetic
${N_2}^ +$ : Bond order = 2.5, Paramagnetic
${O_2}$ : Bond order = 2, Paramagnetic
${O_2}^ -$ : Bond order = 1.5, Paramagnetic
${C_2}$ : Bond order = 2, Diamagnetic
${C_2}^ +$ : Bond order = 1.5, Paramagnetic
${O_2}^{ + 2}$ : Bond order = 3, Diamagnetic

Step 2: Analyze each process.
${N_2} \to {N_2}^ +$ : Bond order decreases, magnetic behavior changes.
${O_2} \to {O_2}^ -$ : Bond order decreases, magnetic behavior remains paramagnetic.
${C_2} \to {C_2}^ +$ : Bond order decreases, magnetic behavior changes.
${O_2} \to {O_2}^{ + 2}$ : Bond order increases, magnetic behavior changes.

Step 3: Identify the process where bond order increases and magnetic behavior changes.
${O_2} \to {O_2}^{ + 2}$ is the correct process.

Why other options are wrong:
- Option A: ${N_2} \to {N_2}^ +$ bond order decreases.
- Option B: ${O_2} \to {O_2}^ -$ bond order decreases.
- Option C: ${C_2} \to {C_2}^ +$ bond order decreases.

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

Memory Aids:
- Remember the effect of adding or removing electrons on bond order and magnetic behavior.

Quick Tips:
- Calculate the bond order and determine the magnetic behavior for each species.

117 Easy

117 of 220Question permalink

Analyzing +4 | -1

In which of the process, the bond order increases and magnetic behaviour changes?

View answer and explanation

Correct option: C

\mathrm{NO} \to \mathrm{N}{\mathrm{O}^ + }

Option C (

\mathrm{NO} \to \mathrm{N}{\mathrm{O}^ + }

) is correct because the bond order increases from 2.5 to 3 and the magnetic behavior changes from paramagnetic to diamagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{NO} \to \mathrm{N}{\mathrm{O}^ + }$ . The bond order increases and magnetic behaviour changes.

Step-by-Step Solution:

Step 1: Determine the bond order and magnetic behavior of each species.
${\mathrm{N}_2}$ : Bond order = 3, Diamagnetic
$\mathrm{N}_2^ +$ : Bond order = 2.5, Paramagnetic
${\mathrm{C}_2}$ : Bond order = 2, Diamagnetic
$\mathrm{C}_2^ +$ : Bond order = 1.5, Paramagnetic
$\mathrm{NO}$ : Bond order = 2.5, Paramagnetic
$\mathrm{N}{\mathrm{O}^ + }$ : Bond order = 3, Diamagnetic
${\mathrm{O}_2}$ : Bond order = 2, Paramagnetic
$\mathrm{O}_2^ +$ : Bond order = 2.5, Paramagnetic

Step 2: Analyze each process.
${\mathrm{N}_2} \to \mathrm{N}_2^ +$ : Bond order decreases, magnetic behavior changes.
${\mathrm{C}_2} \to \mathrm{C}_2^ +$ : Bond order decreases, magnetic behavior changes.
$\mathrm{NO} \to \mathrm{N}{\mathrm{O}^ + }$ : Bond order increases, magnetic behavior changes.
${\mathrm{O}_2} \to \mathrm{O}_2^ +$ : Bond order increases, magnetic behavior remains paramagnetic.

Step 3: Identify the process where bond order increases and magnetic behavior changes.
$\mathrm{NO} \to \mathrm{N}{\mathrm{O}^ + }$ is the correct process.

Why other options are wrong:
- Option A: ${\mathrm{N}_2} \to \mathrm{N}_2^ +$ bond order decreases.
- Option B: ${\mathrm{C}_2} \to \mathrm{C}_2^ +$ bond order decreases.
- Option D: ${\mathrm{O}_2} \to \mathrm{O}_2^ +$ magnetic behavior remains paramagnetic.

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

Memory Aids:
- Remember the effect of adding or removing electrons on bond order and magnetic behavior.

Quick Tips:
- Calculate the bond order and determine the magnetic behavior for each species.

118 Easy

118 of 220Question permalink

Understanding +4 | -1

Both

B{F_3}\,and\,N{F_3}

are covalent compounds.

B{F_3}\,

is non polar compound but

N{F_3}

is polar, due to

View answer and explanation

Correct option: B

B{F_3}\,

is planar but

\,N{F_3}

is pyramidal is shape

Option B is correct because BF3 is planar with symmetrical bond dipoles that cancel out, while NF3 is pyramidal with asymmetrical bond dipoles that do not cancel out, resulting in a net dipole moment.

Solution approach

Correct Answer + Core Reason: $B{F_3}\,\,$ is planar but $\,N{F_3}$ is pyramidal is shape. The difference in molecular geometry leads to different dipole moments.

Step-by-Step Solution:

Step 1: Determine the molecular geometry of $BF_3$ . Boron has 3 bond pairs and 0 lone pairs, resulting in a trigonal planar geometry. The bond dipoles cancel out, making it nonpolar.

Step 2: Determine the molecular geometry of $NF_3$ . Nitrogen has 3 bond pairs and 1 lone pair, resulting in a pyramidal geometry. The bond dipoles do not cancel out, making it polar.

Step 3: Conclusion: The difference in polarity is due to the difference in molecular geometry.

Why other options are wrong:
- Option A: Boron is a solid and Nitrogen is a gas in free state: This is incorrect because the physical state of the elements does not directly determine the polarity of the compounds.
- Option C: Boron is a metalloid while nitrogen is a non-metal: This is incorrect because the metallic character of the elements does not directly determine the polarity of the compounds.
- Option D: Atomic size of boron is small than Nitrogen: This is incorrect because the atomic size difference does not directly determine the polarity of the compounds.

Key Concepts:
- Molecular Geometry
- Polarity of Molecules
- Dipole Moment

Memory Aids:
- Remember that symmetrical molecules are generally nonpolar, while asymmetrical molecules can be polar.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict the polarity of a molecule.

119 Easy

119 of 220Question permalink

Understanding +4 | -1

The total number of bonds in acetylene molecule is:

View answer and explanation

Correct option: D

Five

Option D (Five) is correct because acetylene (C₂H₂) has a triple bond between the carbon atoms (one sigma and two pi bonds) and a single bond between each carbon and hydrogen atom (two sigma bonds), totaling five bonds.

Solution approach

Correct Answer + Core Reason: Five. Acetylene (C₂H₂) has a triple bond between the carbon atoms (one sigma and two pi bonds) and a single bond between each carbon and hydrogen atom (two sigma bonds), totaling five bonds.

Step-by-Step Solution:

Step 1: The formula for acetylene is C₂H₂.

Step 2: The structure of acetylene is H-C≡C-H.

Step 3: The triple bond between the carbon atoms consists of one sigma bond and two pi bonds.

Step 4: Each carbon atom is also bonded to one hydrogen atom with a single (sigma) bond.

Step 5: Therefore, there are 3 bonds in the triple bond + 1 bond to each H atom = 3 + 1 + 1 = 5 bonds.

Why other options are wrong:
- Option A: One is incorrect; there are more than one bond in acetylene.
- Option B: Two is incorrect; there are more than two bonds in acetylene.
- Option C: Three is incorrect; there are more than three bonds in acetylene.

Key Concepts:
- Chemical structure
- Sigma bond
- Pi bond

Memory Aids:
- Acetylene = Triple Bond + 2 Single Bonds

Quick Tips:
- Draw the Lewis structure to visualize the bonds.

120 Easy

120 of 220Question permalink

Application +4 | -1

An element

X

with the electronic configuration

1 s^2, 2 s^2 2 p^6, 3 s^2

would be expected to form the chloride with the formula

View answer and explanation

Correct option: B

{XCl}_2

Option B (

{XCl}_2

) is correct because the electronic configuration indicates X has 2 valence electrons, so it will form a +2 ion. Chlorine forms a -1 ion, so the formula will be XCl2.

Solution approach

Correct Answer + Core Reason: ${XCl}_2$ . The electronic configuration indicates X has 2 valence electrons, so it will form a +2 ion. Chlorine forms a -1 ion, so the formula will be XCl2.

Step-by-Step Solution:

Step 1: Determine the valence electrons of X. The electronic configuration $1 s^2, 2 s^2 2 p^6, 3 s^2$ indicates that X has 2 valence electrons in the 3s orbital.

Step 2: Determine the charge of the X ion. Since X has 2 valence electrons, it will lose these 2 electrons to form a +2 ion ( $X^{2+}$ ).

Step 3: Determine the charge of the chloride ion. Chlorine (Cl) is in group 17, so it gains one electron to form a -1 ion ( $Cl^-$ ).

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

Why other options are wrong:
- **Option A ( ${XCl}_3$ ): This formula would result if X had a +3 charge and Cl had a -1 charge, which is incorrect based on the electronic configuration of X.
- Option C ( ${XCl}$ ): This formula would result if X had a +1 charge and Cl had a -1 charge, which is incorrect based on the electronic configuration of X.
- Option D ( ${X}_2 {Cl}$ ): This formula would result if X had a +1/2 charge and Cl had a -1 charge, which is incorrect.

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

Memory Aids:
- Remember that group 2 elements form +2 ions and group 17 elements form -1 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.

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