CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 6

151 Easy

Understanding +4 | -1

Which of the following statement is not true

View answer and explanation

Correct option: C

HF

is less polar than

HBr

Option C (

HF

is less polar than

HBr

) is not true because HF is more polar than HBr due to the higher electronegativity of fluorine compared to bromine.

Solution approach

Correct Answer + Core Reason: $HF$ is less polar than $HBr$ is not true.

Step-by-Step Solution:

Step 1: Polarity of a bond depends on the electronegativity difference between the two atoms forming the bond.

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

Step 3: Fluorine (F) is more electronegative than Bromine (Br).

Step 4: Therefore, the $H-F$ bond is more polar than the $H-Br$ bond.

Why other options are wrong:
- Option A: $CuCl$ is more covalent than $NaCl$ is true because Cu has a higher polarizing power due to its smaller size and higher charge density compared to Na.
- Option B: $HF$ is more polar than $HBr$ is true because F is more electronegative than Br.
- Option D: Chemical bond formation takes place when forces of attraction is true because bond formation releases energy and stabilizes the system.

Key Concepts:
- Polarity of bonds
- Electronegativity
- Polarizing power

Memory Aids:
- Higher electronegativity difference = More polar

Quick Tips:
- Remember that electronegativity increases across a period and decreases down a group.

152 Easy

152 of 220Question permalink

Application +4 | -1

For the dot structure shown, the most likely elements

X=

and

Y=

are

View answer and explanation

Correct option: A

carbon, fluorine

Option A (carbon, fluorine) is correct because carbon typically forms four bonds, and fluorine typically forms one bond, matching the dot structure.

Solution approach

Correct Answer + Core Reason: carbon, fluorine. Carbon forms 4 bonds and Fluorine forms 1.

Step-by-Step Solution:

Step 1: Analyze the dot structure. X has four bonds, and Y has one bond.

Step 2: Consider the valency of each element in the options.

Step 3: Carbon has a valency of 4, and fluorine has a valency of 1.

Step 4: Therefore, X is carbon, and Y is fluorine.

Why other options are wrong:
- Option B: carbon, hydrogen. Hydrogen only forms one bond, but the structure shows Y forming only one bond, which is consistent. However, the structure implies a neutral molecule, and carbon bonding to four hydrogens would be CH4. The dot structure does not represent CH4 correctly.
- Option C: carbon, oxygen. Oxygen typically forms two bonds, not one.
- Option D: oxygen, carbon. Oxygen typically forms two bonds, and carbon forms four, but the structure has X forming four and Y forming one, so this is reversed.

Key Concepts:
- Valency
- Dot structure

Memory Aids:
- Carbon always forms four bonds.
- Fluorine always forms one bond.
- Oxygen usually forms two bonds.
- Hydrogen always forms one bond.

Quick Tips:
- Remember the common valencies of elements.

153 Easy

153 of 220Question permalink

Analyzing +4 | -1

The bond angle in

\mathrm{As}{\mathrm{H}_3}

is greater than that in

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the bond angle in

AsH_3

is smaller than in

NH_3

and

H_2O

, and

BCl_3

has a bond angle of 120 degrees.

Solution approach

Correct Answer + Core Reason: None of these. The bond angle in $AsH_3$ is smaller than that in $NH_3$ and $H_2O$ , and $BCl_3$ has a bond angle of 120 degrees.

Step-by-Step Solution:

Step 1: Consider the factors affecting bond angles.\nElectronegativity of the central atom, lone pairs, and size of the surrounding atoms affect bond angles.

Step 2: Compare the bond angles in $NH_3$ and $AsH_3$ .\nNitrogen (N) is more electronegative than Arsenic (As). As the electronegativity of the central atom decreases, the bond angle decreases. Therefore, the bond angle in $NH_3$ is greater than in $AsH_3$ .

Step 3: Compare the bond angles in $H_2O$ and $AsH_3$ .\nOxygen (O) has two lone pairs, which cause greater repulsion and a smaller bond angle than in $NH_3$ , but still larger than $AsH_3$ .

Step 4: Consider the bond angle in $BCl_3$ .
$BCl_3$ has a trigonal planar geometry with a bond angle of 120 degrees. This is larger than the bond angle in $AsH_3$ .

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{H}_3}$ : $NH_3$ has a larger bond angle than $AsH_3$ .
- Option B: ${\mathrm{H}_2}\mathrm{O}$ : $H_2O$ has a larger bond angle than $AsH_3$ .
- Option C: $\mathrm{BC}{\mathrm{l}_3}$ : $BCl_3$ has a larger bond angle than $AsH_3$ .

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

Memory Aids:
- As electronegativity of central atom increases, bond angle increases.
- Lone pairs decrease bond angle.

Quick Tips:
- Consider electronegativity and lone pairs when comparing bond angles.

154 Easy

154 of 220Question permalink

Analyzing +4 | -1

In which of the following molecules, the type of hybridisation changes when

View answer and explanation

Correct option: B

Al{H_3}

combines with

{H^ - }

Option B (

AlH_3

combines with

H^−

) is correct because the hybridization of Al changes from

sp^2

to

sp^3

upon accepting a hydride ion.

Solution approach

Correct Answer + Core Reason: $AlH_3$ combines with $H^-$ because the hybridization of Al changes from $sp^2$ to $sp^3$ .

Step-by-Step Solution:

Step 1: Determine the hybridization of $NH_3$ .

Step 2: $NH_3$ has 3 bond pairs and 1 lone pair, so it is $sp^3$ hybridized.

Step 3: Determine the hybridization of $NH_4^+$ .

Step 4: $NH_4^+$ has 4 bond pairs and 0 lone pairs, so it is $sp^3$ hybridized.

Step 5: The hybridization of N does not change when $NH_3$ combines with $H^+$ .

Step 6: Determine the hybridization of $AlH_3$ .

Step 7: $AlH_3$ has 3 bond pairs and 0 lone pairs, so it is $sp^2$ hybridized.

Step 8: Determine the hybridization of $AlH_4^-$ .

Step 9: $AlH_4^-$ has 4 bond pairs and 0 lone pairs, so it is $sp^3$ hybridized.

Step 10: The hybridization of Al changes when $AlH_3$ combines with $H^-$ .

Why other options are wrong:
- Option A: The hybridization of nitrogen does not change when $NH_3$ combines with $H^+$ .
- Option C: Only in the case of $AlH_3$ combining with $H^-$ does the hybridization change.
- Option D: The hybridization changes in one of the cases.

Key Concepts:
- Hybridization
- Lewis acid-base reactions

Memory Aids:
- Remember the common hybridizations and their corresponding geometries.

Quick Tips:
- Determine the number of bond pairs and lone pairs to predict the hybridization.

155 Easy

155 of 220Question permalink

Understanding +4 | -1

B.P of

{H_2}O\left( {{{100}^0}C} \right)

and

{H_2}S\left( { - {{42}^0}C} \right)

is explained by

View answer and explanation

Correct option: C

hydrogen bond

Option C (hydrogen bond) is correct because the high boiling point of water compared to hydrogen sulfide is due to the presence of hydrogen bonding in water.

Solution approach

Correct Answer + Core Reason: hydrogen bond. The high boiling point of water is due to hydrogen bonding.

Step-by-Step Solution:

Step 1: Compare the molecular structures of H2O and H2S.

Step 2: H2O has O-H bonds, allowing for hydrogen bonding. H2S has S-H bonds, but sulfur is not electronegative enough to form strong hydrogen bonds.

Step 3: The stronger intermolecular forces (hydrogen bonds) in H2O lead to a higher boiling point compared to H2S, which only has weaker dipole-dipole interactions and London dispersion forces.

Step 4: Conclude that hydrogen bonding explains the difference in boiling points.

Why other options are wrong:
- Option A: Van der Waals forces are present in both molecules, but they don't explain the significant difference in boiling points.
- Option B: Covalent bonds are intramolecular forces and do not directly affect boiling points.
- Option D: Ionic bonds are not present in either molecule.

Key Concepts:
- Hydrogen bonding.
- Intermolecular forces.
- Boiling point.

Memory Aids:
- Hydrogen bonding increases boiling point.

Quick Tips:
- Look for H bonded to F, O, or N to identify potential hydrogen bonding.

156 Easy

156 of 220Question permalink

Remembering +4 | -1

The electrons in an incomplete outershell are known as:

View answer and explanation

Correct option: B

Valency electrons

Option B (Valency electrons) is correct because valency electrons are the electrons in the outermost shell of an atom, which are involved in chemical bonding.

Solution approach

Correct Answer + Core Reason: Valency electrons. These are the electrons in the outermost shell involved in chemical bonding.

Step-by-Step Solution:

Step 1: Define Kernel electrons: Kernel electrons are the inner electrons, not in the outermost shell.

Step 2: Define Valency electrons: Valency electrons are the electrons in the incomplete outermost shell, which participate in chemical bonding.

Step 3: Define Shell electrons: This is a general term and not specific to incomplete outershells.

Why other options are wrong:
- Option A: Kernel electrons are inner electrons, not the ones in the incomplete outershell.
- Option C: Shell electrons is a general term and doesn't specifically refer to the outermost, bonding electrons.
- Option D: Valency electrons are the correct answer.

Key Concepts:
- Valency electrons
- Kernel electrons
- Electronic configuration

Memory Aids:
- Valency = Valence = Outermost shell electrons

Quick Tips:
- Remember that valency electrons are responsible for chemical bonding.

157 Easy

157 of 220Question permalink

Understanding +4 | -1

The bond in the formation of fluorine molecule is

View answer and explanation

Correct option: C

Due to p - p overlapping

Option C (Due to p - p overlapping) is correct because fluorine atoms have the electronic configuration 1s² 2s² 2p⁵, and the bond is formed by the overlap of p orbitals.

Solution approach

Correct Answer + Core Reason: Due to p - p overlapping. Fluorine forms a bond by the overlap of p orbitals.

Step-by-Step Solution:

Step 1: Write the electronic configuration of fluorine (F): 1s² 2s² 2p⁵.

Step 2: Fluorine has 7 valence electrons. It needs one more electron to complete its octet.

Step 3: The bond in F2 is formed by the overlap of the half-filled 2p orbitals of two fluorine atoms.

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

Why other options are wrong:
- Option A: Due to s - s overlapping is incorrect because s orbitals are already filled and do not participate in bonding in F2.
- Option B: Due to s - p overlapping is incorrect because s orbitals are filled and do not participate in bonding in F2.
- Option D: Due to hybridisation is incorrect because fluorine does not undergo hybridization in F2.

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

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

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

158 Easy

158 of 220Question permalink

Application +4 | -1

Which of the following exhibits diamagnetic behavior:

View answer and explanation

Correct option: B

\mathrm{O}_2^{2 - }

Option B (

\mathrm{O}_2^{2 - }

) is correct because it has all its electrons paired, making it diamagnetic, while the other options have unpaired electrons, making them paramagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^{2 - }$ . Diamagnetic substances have all electrons paired.

Step-by-Step Solution:

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

Step 2: NO has 15 electrons.

Step 3: $\mathrm{O}_2^{2 - }$ has 18 electrons.

Step 4: $\mathrm{O}_2^ +$ has 15 electrons.

Step 5: ${\mathrm{O}_2}$ has 16 electrons.

Step 6: Write the electronic configuration for each species using Molecular Orbital Theory (MOT).

Step 7: NO: $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^1$ . It has one unpaired electron, so it is paramagnetic.

Step 8: $\mathrm{O}_2^{2 - }$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^4$ . It has all electrons paired, so it is diamagnetic.

Step 9: $\mathrm{O}_2^ +$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^1$ . It has one unpaired electron, so it is paramagnetic.

Step 10: ${\mathrm{O}_2}$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^2$ . It has two unpaired electrons, so it is paramagnetic.

Step 11: Therefore, $\mathrm{O}_2^{2 - }$ exhibits diamagnetic behavior.

Why other options are wrong:
- Option A: NO has an unpaired electron, making it paramagnetic.
- Option C: $\mathrm{O}_2^ +$ has an unpaired electron, making it paramagnetic.
- Option D: ${\mathrm{O}_2}$ has two unpaired electrons, making it paramagnetic.

Key Concepts:
- Diamagnetism and paramagnetism
- Molecular Orbital Theory (MOT)
- Unpaired electrons

Memory Aids:
- Diamagnetic: all electrons paired; Paramagnetic: at least one unpaired electron.

Quick Tips:
- Use MOT to determine the electronic configuration and identify unpaired electrons.

159 Easy

159 of 220Question permalink

Understanding +4 | -1

As the

s

-character of hybridization orbitals increases, the bond angle:

View answer and explanation

Correct option: A

Increases

Option A (Increases) is correct because as the s-character increases, the bond angle also increases due to the increased directionality of the hybrid orbitals.

Solution approach

Correct Answer + Core Reason: Increases. Higher s-character leads to larger bond angles.

Step-by-Step Solution:

Step 1: Recall the relationship between s-character and bond angle.

Step 2: Understand that higher s-character means the hybrid orbital is more like an s orbital, which is spherically symmetrical.

Step 3: Recognize that this increased s-character leads to greater electron density closer to the nucleus, resulting in increased repulsion between electron pairs.

Step 4: This increased repulsion forces the bonding pairs further apart, increasing the bond angle.

Why other options are wrong:
- Option B (Decreases): The bond angle increases with s-character.
- Option C (Does not change): The bond angle changes with s-character.
- Option D (Becomes zero): The bond angle does not become zero.

Key Concepts:
- Hybridization
- s-character
- Bond angle
- Electron repulsion

Memory Aids:
- More s-character, more angle.

Quick Tips:
- Remember the trend: $sp > sp^2 > sp^3$ in terms of bond angle.

160 Easy

160 of 220Question permalink

Remembering +4 | -1

Kossel and Lewis approach was based on the

View answer and explanation

Correct option: B

Inertness of noble gases

Option B (Inertness of noble gases) is correct because Kossel and Lewis based their theory on the stable electronic configurations of noble gases.

Solution approach

Correct Answer + Core Reason: Inertness of noble gases. Kossel and Lewis based their theory on the stable electronic configurations of noble gases.

Step-by-Step Solution:

Step 1: Kossel and Lewis observed that noble gases are chemically inert.

Step 2: They attributed this inertness to their stable electronic configurations (octet rule).

Step 3: They proposed that atoms achieve stability by gaining, losing, or sharing electrons to attain the same electronic configuration as the nearest noble gas.

Step 4: Therefore, their approach was based on the inertness of noble gases.

Why other options are wrong:
- Option A: Reactivity of elements is a consequence of their tendency to achieve noble gas configurations, not the basis of the Kossel and Lewis approach.
- Option C: Reactivity of metals is related to their tendency to lose electrons, but it's not the fundamental basis of the theory.
- Option D: Inertness of non-metals is not the primary focus; the theory explains how elements achieve stability, including metals and non-metals.

Key Concepts:
- Octet rule
- Electronic configuration
- Inertness of noble gases

Memory Aids:
- Noble gases are stable because they have a full outer shell of electrons.

Quick Tips:
- Remember the electronic configurations of noble gases.

161 Easy

161 of 220Question permalink

Application +4 | -1

Which of the following molecule is linear

View answer and explanation

Correct option: B

{NO}_2^{+}

Option B (

{NO}_2^{+}

) is correct because the central nitrogen atom in

NO_2^+

has two bond pairs and no lone pairs, resulting in a linear shape.

Solution approach

Correct Answer + Core Reason: ${NO}_2^{+}$ . The central nitrogen atom has two bond pairs and no lone pairs, resulting in a linear shape.

Step-by-Step Solution:

Step 1: Draw the Lewis structure for each molecule/ion.

Step 2: Determine the number of bond pairs and lone pairs around the central atom.\nFor ${SO}_2$ : Sulfur has 2 bond pairs and 1 lone pair. Bent shape.\nFor ${NO}_2^{+}$ : Nitrogen has 2 bond pairs and 0 lone pairs. Linear shape.\nFor ${NO}_2^{-}$ : Nitrogen has 2 bond pairs and 1 lone pair. Bent shape.\nFor ${SCl}_2$ : Sulfur has 2 bond pairs and 2 lone pairs. Bent shape.

Step 3: Determine the shape based on the number of bond pairs and lone pairs.

Why other options are wrong:
- **Option A ( ${SO}_2$ ):** ${SO}_2$ has a bent shape due to one lone pair on the sulfur atom.
- **Option C ( ${NO}_2^{-}$ ):** ${NO}_2^{-}$ has a bent shape due to one lone pair on the nitrogen atom.
- **Option D ( ${SCl}_2$ ):** ${SCl}_2$ has a bent shape due to two lone pairs on the sulfur atom.

Key Concepts:
- Lewis structure
- VSEPR theory
- Molecular shape

Memory Aids:
- Linear: 2 bond pairs, 0 lone pairs
- Bent: 2 bond pairs, 1 or 2 lone pairs

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

162 Easy

162 of 220Question permalink

Analyzing +4 | -1

Which of the following compound has highest
lattice energy?

View answer and explanation

Correct option: B

A{l_2}{O_3}

Option B (

Al_2O_3

) is correct because lattice energy is directly proportional to the charge of the ions and inversely proportional to the size of the ions.

Al_2O_3

has the highest charges (+3 and -2) compared to the other options.

Solution approach

Correct Answer + Core Reason: $A{l_2}{O_3}$ . Lattice energy is proportional to the product of the charges of the ions and inversely proportional to the distance between them. $Al_2O_3$ has the highest charge product.

Step-by-Step Solution:

Step 1: Recall the factors affecting lattice energy: Charge and size of ions.

Step 2: Consider the charges of the ions in each compound:
- $CaF_2$ : $Ca^{2+}$ and $F^-$
- $Al_2O_3$ : $Al^{3+}$ and $O^{2-}$
- $AlF_3$ : $Al^{3+}$ and $F^-$
- $Na_2S$ : $Na^+$ and $S^{2-}$

Step 3: Compare the charge products:
- $CaF_2$ : $(+2)(-1) = -2$ , product is 2
- $Al_2O_3$ : $(+3)(-2) = -6$ , but we have 2 Al and 3 O, so $(2 \times 3) \times (3 \times 2) = 36$
- $AlF_3$ : $(+3)(-1) = -3$ , product is 3
- $Na_2S$ : $(+1)(-2) = -2$ , product is 2

Step 4: $Al_2O_3$ has the highest charge product, therefore the highest lattice energy.

Why other options are wrong:
- Option A: $Ca{F_2}$ : Lower charges than $Al_2O_3$ .
- Option C: $Al{F_3}$ : Lower charges than $Al_2O_3$ .
- Option D: $N{a_2}S$ : Lower charges than $Al_2O_3$ .

Key Concepts:
- Lattice energy
- Charge of ions
- Size of ions

Memory Aids:
- Higher charge, higher lattice energy.
- Smaller size, higher lattice energy.

Quick Tips:
- Compare the charges of the ions first.
- If charges are similar, consider the size of the ions.

163 Easy

163 of 220Question permalink

Application +4 | -1

Which has

s p^3

hybridization of central atom

View answer and explanation

Correct option: A

{PCl}_3

Option A (

{PCl}_3

) is correct because the central phosphorus atom in

PCl_3

has three bond pairs and one lone pair, resulting in

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: ${PCl}_3$ . The central phosphorus atom has three bond pairs and one lone pair, resulting in $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the number of bond pairs and lone pairs around the central atom in each molecule.

Step 2: For ${PCl}_3$ : Phosphorus has 3 bond pairs (with chlorine atoms) and 1 lone pair. Steric number = 4, so hybridization is $sp^3$ .\nFor ${SO}_3$ : Sulfur has 3 bond pairs (with oxygen atoms) and 0 lone pairs. Steric number = 3, so hybridization is $sp^2$ .\nFor ${BF}_3$ : Boron has 3 bond pairs (with fluorine atoms) and 0 lone pairs. Steric number = 3, so hybridization is $sp^2$ .\nFor ${NO}_3^{-}$ : Nitrogen has 3 bond pairs (with oxygen atoms) and 0 lone pairs. Steric number = 3, so hybridization is $sp^2$ .

Step 3: Identify the molecule with $sp^3$ hybridization.
${PCl}_3$ has $sp^3$ hybridization.

Why other options are wrong:
- **Option B ( ${SO}_3$ ):** ${SO}_3$ has $sp^2$ hybridization.
- **Option C ( ${BF}_3$ ):** ${BF}_3$ has $sp^2$ hybridization.
- **Option D ( ${NO}_3^{-}$ ):** ${NO}_3^{-}$ has $sp^2$ hybridization.

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.

164 Easy

164 of 220Question permalink

Understanding +4 | -1

A covalent bond is formed between the atoms by the overlapping of orbitals containing:

View answer and explanation

Correct option: D

Single electron with opposite spin

Option D (Single electron with opposite spin) is correct because covalent bonds are formed by the sharing of electrons with opposite spins to achieve a stable electron configuration.

Solution approach

Correct Answer + Core Reason: Single electron with opposite spin. Covalent bonds require electrons with opposite spins to pair up.

Step-by-Step Solution:

Step 1: Define a covalent bond.

Step 2: A covalent bond is formed by the sharing of electrons between two atoms.

Step 3: For a stable covalent bond to form, the overlapping orbitals must contain single electrons with opposite spins (Pauli Exclusion Principle).

Step 4: This allows the electrons to pair up and form a stable bond.

Why other options are wrong:
- Option A: Single electron is incomplete because it doesn't specify the spin.
- Option B: Paired electron is incorrect because paired electrons are already stable and do not form covalent bonds.
- Option C: Single electron with parallel spin is incorrect because electrons with parallel spins cannot form a stable bond (Hund's rule).

Key Concepts:
- Covalent bonding
- Spin pairing
- Pauli Exclusion Principle

Memory Aids:
- Opposite spins pair up to form a bond.

Quick Tips:
- Remember that covalent bonds involve sharing of electrons with opposite spins.

165 Easy

165 of 220Question permalink

Understanding +4 | -1

Which is not true according to VBT

View answer and explanation

Correct option: B

A covalent bond is formed by the overlapping of orbitals with unpaired electrons of same spins

Option B (A covalent bond is formed by the overlapping of orbitals with unpaired electrons of same spins) is correct because VBT states that covalent bonds are formed by overlapping orbitals with unpaired electrons of *opposite* spins.

Solution approach

Correct Answer + Core Reason: A covalent bond is formed by the overlapping of orbitals with unpaired electrons of same spins. This statement contradicts the VBT.

Step-by-Step Solution:

Step 1: Review the postulates of Valence Bond Theory (VBT).

Step 2: VBT states that a covalent bond is formed by the overlapping of atomic orbitals with unpaired electrons of *opposite* spins.

Step 3: The greater the extent of overlapping, the stronger the bond.

Step 4: Overlapping takes place in the direction of maximum electron density of the orbital.

Why other options are wrong:
- Option A: A covalent bond is formed by the overlapping of orbitals with unpaired electrons of opposite spins is a correct statement according to VBT.
- Option C: The greater the extent of overlapping the stronger is the bond is a correct statement according to VBT.
- Option D: Overlapping takes place only in the direction of maximum electron density of the orbital is a correct statement according to VBT.

Key Concepts:
- Valence Bond Theory (VBT)
- Overlapping of orbitals
- Spin pairing

Memory Aids:
- Opposite spins attract to form a bond.

Quick Tips:
- Remember the key postulates of VBT.

166 Easy

166 of 220Question permalink

Remembering +4 | -1

Electrovalence of calcium and chlorine respectively is

View answer and explanation

Correct option: A

+2,-1

Option A (+2,-1) is correct because calcium loses 2 electrons to achieve a stable configuration, resulting in a +2 charge, while chlorine gains 1 electron, resulting in a -1 charge.

Solution approach

Correct Answer + Core Reason: $+2,-1$ . Calcium loses 2 electrons, and chlorine gains 1 electron.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of calcium and chlorine.

Step 2: Calcium (Ca) has 2 valence electrons and tends to lose 2 electrons to form a $\mathrm{Ca}^{2+}$ ion.

Step 3: Chlorine (Cl) has 7 valence electrons and tends to gain 1 electron to form a $\mathrm{Cl}^ -$ ion.

Step 4: The electrovalence of calcium is +2, and the electrovalence of chlorine is -1.

Why other options are wrong:
- Option B: Incorrect electrovalence for calcium.
- Option C: Incorrect electrovalence for chlorine.
- Option D: Incorrect electrovalence for chlorine.

Key Concepts:
- Electrovalence
- Electronic configuration
- Ion formation

Memory Aids:
- Metals tend to lose electrons (positive charge).
- Nonmetals tend to gain electrons (negative charge).

Quick Tips:
- Relate electrovalence to the number of electrons gained or lost.

167 Easy

167 of 220Question permalink

Understanding +4 | -1

Which element among the following cannot exhibit variable electrovalency?

View answer and explanation

Correct option: D

{11^{Na}}

Option D (

{11^{Na}}

) is correct because sodium (Na) is an alkali metal and typically exhibits only a +1 oxidation state, thus it cannot exhibit variable electrovalency.

Solution approach

Correct Answer + Core Reason: ${11^{Na}}$ . Sodium, being an alkali metal, has a fixed +1 oxidation state and cannot exhibit variable electrovalency.

Step-by-Step Solution:

Step 1: Understand the concept of variable electrovalency. Variable electrovalency means an element can exhibit more than one oxidation state.

Step 2: Consider the electronic configurations of the given elements.

Step 3: Analyze each element:
- Copper ( ${29^{Cu}}$ ) can exhibit +1 and +2 oxidation states.
- Tin ( ${50^{Sn}}$ ) can exhibit +2 and +4 oxidation states.
- Manganese ( ${25^{Mn}}$ ) can exhibit multiple oxidation states like +2, +3, +4, +6, and +7.
- Sodium ( ${11^{Na}}$ ) has an electronic configuration of $1s^2 2s^2 2p^6 3s^1$ . It readily loses one electron to form $Na^+$ , achieving a stable noble gas configuration. It does not exhibit other oxidation states under normal conditions.

Step 4: Conclusion: Sodium cannot exhibit variable electrovalency.

Why other options are wrong:
- **Option A ( ${29^{Cu}}$ ):** Copper can exhibit +1 and +2 oxidation states due to its electronic configuration and the stability of $Cu^+$ and $Cu^{2+}$ ions.
- **Option B ( ${50^{Sn}}$ ): Tin can exhibit +2 and +4 oxidation states. The +2 state is due to the inert pair effect.
- Option C ( ${25^{Mn}}$ ): Manganese exhibits a wide range of oxidation states from +2 to +7 due to the involvement of both 4s and 3d electrons in bonding.

Key Concepts:
- Variable electrovalency
- Oxidation states
- Electronic configuration
- Inert pair effect

Memory Aids:
- Alkali metals (Group 1) generally have a fixed +1 oxidation state.

Quick Tips:**
- Remember the common oxidation states of transition metals and main group elements.

168 Easy

168 of 220Question permalink

Remembering +4 | -1

Valency means:

View answer and explanation

Correct option: A

Combining capacity of an element

Option A (Combining capacity of an element) is correct because valency refers to the number of chemical bonds an atom can form.

Solution approach

Correct Answer + Core Reason: Combining capacity of an element. Valency is the number of chemical bonds an atom can form.

Step-by-Step Solution:

Step 1: Define valency.

Step 2: Valency is the measure of the combining power of an element with other atoms when it forms chemical compounds or molecules. It represents the number of chemical bonds that an atom can form.

Why other options are wrong:
- Option B: Atomicity is the number of atoms in a molecule.
- Option C: Oxidation number is the charge an atom would have if all bonds were ionic.
- Option D: None of the above is incorrect because option A is correct.

Key Concepts:
- Valency
- Chemical bonding

Memory Aids:
- Valency is the combining capacity.

Quick Tips:
- Valency is related to the number of valence electrons an atom has.

169 Easy

169 of 220Question permalink

Application +4 | -1

Valency of sulphur in

{Na}_2 {~S}_2 {O}_3

is

View answer and explanation

Correct option: A

Two

Option A (Two) is correct because in

Na_2S_2O_3

, each sulfur atom has an oxidation state of +2 on average, and it forms two bonds with other atoms.

Solution approach

Correct Answer + Core Reason: Two. The average oxidation state of S is +2, suggesting a valency of 2.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Na and O in $Na_2S_2O_3$ . Na has an oxidation state of +1, and O has an oxidation state of -2.

Step 2: Let the oxidation state of S be x. The sum of the oxidation states in a neutral compound is zero.

Step 3: Write the equation: 2(+1) + 2(x) + 3(-2) = 0

Step 4: Solve for x: 2 + 2x - 6 = 0 => 2x = 4 => x = +2

Step 5: The average oxidation state of sulfur is +2. This corresponds to a valency of 2.

Why other options are wrong:
- Option B: Three is incorrect. The oxidation state calculation does not support a valency of 3.
- Option C: Four is incorrect. The oxidation state calculation does not support a valency of 4.
- Option D: Six is incorrect. The oxidation state calculation does not support a valency of 6. While sulfur *can* have a valency of 6, it doesn't in this compound.

Key Concepts:
- Oxidation state
- Valency

Memory Aids:
- Sum of oxidation states in a neutral compound is zero.

Quick Tips:
- Calculate the oxidation state of the element in question.

170 Easy

170 of 220Question permalink

Understanding +4 | -1

Lattice energy of a solid increases if

View answer and explanation

Correct option: A

Size of ions is small

Option A (Size of ions is small) is correct because smaller ions lead to a higher charge density and stronger electrostatic attraction, thus increasing lattice energy.

Solution approach

Correct Answer + Core Reason: Lattice energy increases if the size of ions is small.

Step-by-Step Solution:

Step 1: Lattice energy is the energy required to separate one mole of a solid ionic compound into its gaseous ions.

Step 2: Lattice energy is directly proportional to the charge of the ions and inversely proportional to the size of the ions.

Step 3: Smaller ions have a higher charge density, leading to stronger electrostatic attraction and higher lattice energy.

Why other options are wrong:
- Option B: Charges of ions are small would decrease the lattice energy.
- Option C: Ions are neutral would mean there is no ionic bond and hence no lattice energy.
- Option D: None of the above is incorrect because option A is correct.

Key Concepts:
- Lattice energy
- Charge density
- Electrostatic attraction

Memory Aids:
- Small size = High charge density = High lattice energy

Quick Tips:
- Remember that smaller, highly charged ions have higher lattice energies.

171 Easy

171 of 220Question permalink

Application +4 | -1

In ozone, the formal charge on each oxygen atom is

View answer and explanation

Correct option: B

– 1, +1, 0

Option B (– 1, +1, 0) is correct because the formal charges on the oxygen atoms in ozone are -1, +1, and 0, respectively, based on the Lewis structure.

Solution approach

Correct Answer + Core Reason: – 1, +1, 0. The formal charges are calculated based on the Lewis structure of ozone.

Step-by-Step Solution:

Step 1: Draw the resonance structures of ozone ( $O_3$ ). Ozone has two resonance structures where one oxygen atom is double-bonded to the central oxygen, and the other is single-bonded.

Step 2: Assign formal charges:
- For the singly bonded oxygen: Formal charge = 6 (valence electrons) - 1 (non-bonding electrons) - 1/2 * 6 (bonding electrons) = 6 - 6 - 1 = -1
- For the doubly bonded oxygen: Formal charge = 6 (valence electrons) - 4 (non-bonding electrons) - 1/2 * 4 (bonding electrons) = 6 - 4 - 2 = 0
- For the central oxygen: Formal charge = 6 (valence electrons) - 2 (non-bonding electrons) - 1/2 * 6 (bonding electrons) = 6 - 2 - 3 = +1

Step 3: Determine the formal charges on each oxygen atom: -1, +1, 0.

Why other options are wrong:
- Option A: + 1, +1, +1: Incorrect formal charges.
- Option C: – 1, – 1, 0: Incorrect formal charges.
- Option D: 0, 0, +1: Incorrect formal charges.

Key Concepts:
- Formal charge
- Resonance structures
- Lewis structures

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

Quick Tips:
- Draw all possible resonance structures to accurately determine formal charges.

172 Easy

172 of 220Question permalink

Application +4 | -1

Which shows highest magnetic moment?

View answer and explanation

Correct option: C

\mathrm{Fe}^{3+}

Option C (

{\mathrm{Fe}^{3+}}

) is correct because it has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}^{3+}$ . It has the highest number of unpaired electrons.

Step-by-Step Solution:

Step 1: Electronic Configuration of $V^{3+}$ : $[Ar]3d^2$ . Number of unpaired electrons = 2.

Step 2: Electronic Configuration of $Cr^{3+}$ : $[Ar]3d^3$ . Number of unpaired electrons = 3.

Step 3: Electronic Configuration of $Fe^{3+}$ : $[Ar]3d^5$ . Number of unpaired electrons = 5.

Step 4: Electronic Configuration of $Co^{3+}$ : $[Ar]3d^6$ . Number of unpaired electrons = 4 (assuming high spin).

Step 5: Magnetic Moment Calculation: Magnetic moment is proportional to the number of unpaired electrons. The more unpaired electrons, the higher the magnetic moment.

Step 6: Conclusion: $Fe^{3+}$ has the highest number of unpaired electrons (5), therefore it has the highest magnetic moment.

Why other options are wrong:
- Option A: $V^{3+}$ has only 2 unpaired electrons, so its magnetic moment is lower than $Fe^{3+}$ .
- Option B: $Cr^{3+}$ has only 3 unpaired electrons, so its magnetic moment is lower than $Fe^{3+}$ .
- Option D: $Co^{3+}$ has 4 unpaired electrons (assuming high spin), so its magnetic moment is lower than $Fe^{3+}$ .

Key Concepts:
- Electronic Configuration
- Hund's Rule
- Magnetic Moment
- Unpaired Electrons

Memory Aids:
- More unpaired electrons = Higher magnetic moment

Quick Tips:
- Write out the electronic configurations to determine the number of unpaired electrons.

173 Easy

173 of 220Question permalink

Application +4 | -1

The order of increasing lattice energy of the following salts is

View answer and explanation

Correct option: B

NaBr < NaCl < BaO < CaO

Option B (

NaBr < NaCl < BaO < CaO

) is correct because lattice energy increases with increasing charge and decreasing size of the ions. CaO has the highest charges and relatively small sizes.

Solution approach

Correct Answer + Core Reason: $NaBr < NaCl < BaO < CaO$ . Lattice energy is proportional to charge and inversely proportional to size.

Step-by-Step Solution:

Step 1: Lattice energy is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

Step 2: According to the Born-Landé equation, lattice energy is directly proportional to the product of the charges of the ions and inversely proportional to the distance between them (which is related to the ionic radii).

Step 3: Comparing NaCl and NaBr, Cl- is smaller than Br-, so NaCl has a higher lattice energy than NaBr.

Step 4: Comparing BaO and CaO, Ca2+ is smaller than Ba2+, so CaO has a higher lattice energy than BaO.

Step 5: Comparing NaCl and BaO, BaO has higher charges (+2 and -2) compared to NaCl (+1 and -1), so BaO has a higher lattice energy than NaCl.

Step 6: CaO has the highest charges and smallest ionic radii among all the given compounds, so it has the highest lattice energy.

Why other options are wrong:
- Option A: Incorrect order. It places NaCl before NaBr, which is wrong because Cl is smaller than Br.
- Option C: Incorrect order. It places NaCl before NaBr, which is wrong because Cl is smaller than Br.
- Option D: Incorrect order. It places BaO before CaO, which is wrong because Ca is smaller than Ba.

Key Concepts:
- Lattice energy
- Born-Landé equation
- Ionic charge
- Ionic radius

Memory Aids:
- Higher charge, smaller size = higher lattice energy.

Quick Tips:
- Consider both charge and size when comparing lattice energies.

174 Easy

174 of 220Question permalink

Remembering +4 | -1

Which one is the weakest bond?

View answer and explanation

Correct option: A

Hydrogen

Option A (Hydrogen) is correct because hydrogen bonds are the weakest among the listed types of chemical bonds.

Solution approach

Correct Answer + Core Reason: Hydrogen. Hydrogen bonds are the weakest intermolecular force.

Step-by-Step Solution:

Step 1: Consider the strength of each type of bond.

Step 2: Hydrogen bonds are weak intermolecular forces (typically 5-30 kJ/mol).

Step 3: Ionic bonds are strong electrostatic attractions between ions (typically 400-1000 kJ/mol).

Step 4: Covalent bonds are strong bonds formed by sharing electrons (typically 200-800 kJ/mol).

Step 5: Metallic bonds are formed by the delocalization of electrons in a metal lattice (typically 100-1000 kJ/mol).

Step 6: Compare the bond strengths.\nHydrogen bonds are the weakest.

Why other options are wrong:
- Option B: Ionic bonds are much stronger than hydrogen bonds.
- Option C: Covalent bonds are much stronger than hydrogen bonds.
- Option D: Metallic bonds are much stronger than hydrogen bonds.

Key Concepts:
- Chemical bonds
- Intermolecular forces
- Bond strength

Memory Aids:
- Hydrogen bonds are weaker than chemical bonds.

Quick Tips:
- Remember the relative strengths of different types of bonds.

175 Easy

175 of 220Question permalink

Application +4 | -1

In

O{F_2}\,

oxygen has hybridization of

View answer and explanation

Correct option: C

s{p^3}

Option C (

s{p^3}

) is correct because oxygen in

OF_2

has two bond pairs and two lone pairs, resulting in tetrahedral electron geometry and

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^3}$ . Oxygen in $OF_2$ has two bond pairs and two lone pairs, leading to $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the central atom: Oxygen is the central atom in $OF_2$ .

Step 2: Count the number of bond pairs and lone pairs around the central atom.\nOxygen forms two single bonds with two fluorine atoms (2 bond pairs).\nOxygen has 6 valence electrons. Two are used for bonding, leaving 4 electrons or 2 lone pairs.

Step 3: Calculate the steric number (number of bond pairs + number of lone pairs):\nSteric number = 2 bond pairs + 2 lone pairs = 4

Step 4: Determine the hybridization based on the steric number:\nSteric number 4 corresponds to $sp^3$ hybridization.

Why other options are wrong:
- **Option A ( $sp$ ):** $sp$ hybridization corresponds to a steric number of 2, which is incorrect for oxygen in $OF_2$ .
- **Option B ( $s{p^2}$ ):** $s{p^2}$ hybridization corresponds to a steric number of 3, which is incorrect for oxygen in $OF_2$ .
- Option D (None of these): $sp^3$ is the correct hybridization.

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.

176 Easy

176 of 220Question permalink

Analyzing +4 | -1

Select the incorrect matching. Here E stands for lone pair of electrons.

View answer and explanation

Correct option: D

A{B_5} - \mathrm{ }\,\mathrm{Square \,\,pyramidal }

Option D (

A{B_5} - \mathrm{ }\,\mathrm{Square \,pyramidal }

) is the incorrect matching because

AB_5

has a trigonal bipyramidal geometry, not square pyramidal.

Solution approach

Correct Answer + Core Reason: $A{B_5} - \mathrm{ }\,\mathrm{Square \,pyramidal }$ is incorrect because $AB_5$ has a trigonal bipyramidal geometry.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule based on VSEPR theory.

Step 2: $A{B_3}{E_2}$ has 3 bond pairs and 2 lone pairs. This corresponds to a T-shape.

Step 3: $A{B_5}E$ has 5 bond pairs and 1 lone pair. This corresponds to a square pyramidal shape.

Step 4: $A{B_4}{E_2}$ has 4 bond pairs and 2 lone pairs. This corresponds to a square planar shape.

Step 5: $A{B_5}$ has 5 bond pairs and 0 lone pairs. This corresponds to a trigonal bipyramidal shape.

Step 6: Identify the incorrect matching. The incorrect matching is $A{B_5} - \mathrm{ }\,\mathrm{Square \,pyramidal }$ because $AB_5$ is trigonal bipyramidal.

Why other options are wrong:
- Option A: $A{B_3}{E_2} - T - \mathrm{ shape}$ is correct. The T-shape arises from the trigonal bipyramidal arrangement with two lone pairs occupying the equatorial positions.
- Option B: $A{B_5}E - \mathrm{\,Square\, pyramidal }$ is correct. One lone pair distorts the octahedral arrangement to a square pyramidal shape.
- Option C: $A{B_4}{E_2} - \mathrm{\,Square \,planar}$ is correct. Two lone pairs in an octahedral arrangement lead to a square planar shape.
- Option D: $A{B_5} - \mathrm{ }\,\mathrm{Square \,pyramidal }$ is incorrect. $AB_5$ has a trigonal bipyramidal shape.

Key Concepts:
- VSEPR theory
- Molecular geometry
- Lone pairs
- Bond pairs

Memory Aids:
- VSEPR: Valence Shell Electron Pair Repulsion
- AXE notation: A = central atom, X = bonding atoms, E = lone pairs

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

177 Easy

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Understanding +4 | -1

In compound X, all the bond angles are exactly

109^\circ 28,X

is:

View answer and explanation

Correct option: C

Carbon tetrachloride

Option C (Carbon tetrachloride) is correct because carbon tetrachloride (

CCl_4

) has four identical bonds and a tetrahedral geometry, resulting in all bond angles being exactly

109^\circ 28'

.

Solution approach

Correct Answer + Core Reason: Carbon tetrachloride. $CCl_4$ has perfect tetrahedral geometry.

Step-by-Step Solution:

Step 1: Consider the structure of each molecule.

Step 2: Chloromethane ( $CH_3Cl$ ) has slightly distorted tetrahedral geometry due to the different electronegativities of H and Cl.

Step 3: Iodoform ( $CHI_3$ ) has slightly distorted tetrahedral geometry due to the different electronegativities of H and I.

Step 4: Chloroform ( $CHCl_3$ ) has slightly distorted tetrahedral geometry due to the different electronegativities of H and Cl.

Step 5: Carbon tetrachloride ( $CCl_4$ ) has perfect tetrahedral geometry because all four substituents are identical.

Why other options are wrong:
- Option A: Chloromethane has a slightly distorted tetrahedral geometry.
- Option B: Iodoform has a slightly distorted tetrahedral geometry.
- Option D: Chloroform has a slightly distorted tetrahedral geometry.

Key Concepts:
- Tetrahedral geometry
- Molecular symmetry

Memory Aids:
- Identical substituents lead to perfect symmetry.

Quick Tips:
- Look for molecules with identical substituents.

178 Easy

178 of 220Question permalink

Remembering +4 | -1

In a triple bond there is a sharing of

View answer and explanation

Correct option: D

6 electrons

Option D (6 electrons) is correct because a triple bond involves the sharing of three pairs of electrons, totaling six electrons.

Solution approach

Correct Answer + Core Reason: 6 electrons. A triple bond involves the sharing of three pairs of electrons.

Step-by-Step Solution:

Step 1: Define a single bond: A single bond involves the sharing of 2 electrons (1 pair).

Step 2: Define a double bond: A double bond involves the sharing of 4 electrons (2 pairs).

Step 3: Define a triple bond: A triple bond involves the sharing of 6 electrons (3 pairs).

Why other options are wrong:
- Option A: 3 electrons would represent a 1.5 bond order, which is not a standard triple bond.
- Option B: 4 electrons represent a double bond.
- Option C: While several electrons are involved in bonding generally, a triple bond specifically involves 6.

Key Concepts:
- Chemical bonding
- Single, double, and triple bonds

Memory Aids:
- Single = 2 electrons, Double = 4 electrons, Triple = 6 electrons

Quick Tips:
- Remember the number of electron pairs shared in each type of bond.

179 Easy

179 of 220Question permalink

Remembering +4 | -1

The maximum number of hydrogen bonds formed by a water molecule in ice is

View answer and explanation

Correct option: A

4

Option A (4) is correct because each water molecule can act as both a hydrogen bond donor (through its two H atoms) and a hydrogen bond acceptor (through its two lone pairs), allowing it to form a maximum of four hydrogen bonds.

Solution approach

Correct Answer + Core Reason: 4. Each water molecule can form 4 hydrogen bonds.

Step-by-Step Solution:

Step 1: Understand the nature of hydrogen bonding in water.\nWater molecules can act as both hydrogen bond donors and acceptors.

Step 2: Explain the maximum number of hydrogen bonds a water molecule can form.\nEach water molecule has two hydrogen atoms that can form hydrogen bonds with oxygen atoms of other water molecules. Also, the oxygen atom has two lone pairs that can accept hydrogen bonds from hydrogen atoms of other water molecules. Therefore, each water molecule can form a maximum of four hydrogen bonds.

Why other options are wrong:
- Option B: Incorrect. A water molecule can form more than one hydrogen bond.
- Option C: Incorrect. A water molecule can form more than two hydrogen bonds.
- Option D: Incorrect. A water molecule can form more than three hydrogen bonds.

Key Concepts:
- Hydrogen bonding
- Hydrogen bond donor
- Hydrogen bond acceptor

Memory Aids:
- Water has two H atoms and two lone pairs, allowing it to form four hydrogen bonds.

Quick Tips:
- Visualize the hydrogen bonding network in water.

180 Easy

180 of 220Question permalink

Application +4 | -1

According to VSEPR theory, shapes of

S{F_4}

and

Br{F_5}

are respectively

View answer and explanation

Correct option: D

See – Saw and square pyramidal

Option D (See – Saw and square pyramidal) is correct because

SF_4

has a see-saw shape and

BrF_5

has a square pyramidal shape according to VSEPR theory.

Solution approach

Correct Answer + Core Reason: See – Saw and square pyramidal. These are the shapes according to VSEPR theory.

Step-by-Step Solution:

Step 1: Determine the shape of $S{F_4}$ .

Step 2: Sulfur has 4 bond pairs and 1 lone pair, so it has a see-saw shape.

Step 3: Determine the shape of $Br{F_5}$ .

Step 4: Bromine has 5 bond pairs and 1 lone pair, so it has a square pyramidal shape.

Why other options are wrong:
- Option A: $S{F_4}$ is see-saw, but $Br{F_5}$ is not octahedral.
- Option B: Neither molecule has these shapes.
- Option C: Neither molecule has these shapes.

Key Concepts:
- VSEPR theory
- Molecular geometry

Memory Aids:
- VSEPR: Valence Shell Electron Pair Repulsion

Quick Tips:
- Count bond pairs and lone pairs to predict the shape.

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