Molecular size of ${\rm{ICl}}$ and ${\rm{B}}{{\rm{r}}_2}$ is nearly same, but boiling point of ${\rm{ICl}}$ is about $40^\circ C$ higher than ${\rm{B}}{{\rm{r}}_2}$. This might be due to:

${\rm{ICl}}$ is polar where as ${\rm{B}}{{\rm{r}}_2}$ is non-polar

${\rm{I - Cl}}$ bond is stronger than ${\rm{Br}} - {\rm{Br}}$ bond

Ionisation energy of ${\rm{I}} < $ ionisation energy of Br

The size of ${\rm{I}} > $ size of ${\rm{Br}}$

The bond order in $NO$ is 2.5 while that in ${\rm{N}}{{\rm{O}}^ + }$ is 3. Which of the following statements is true for these two species?

Bond length in $NO$ is greater than in ${\rm{N}}{{\rm{O}}^ + }$

Bond length in ${\rm{N}}{{\rm{O}}^ + }$ is greater than in $NO$

Bond length in ${\rm{N}}{{\rm{O}}^ + }$ is equal to than in $NO$

The correct order of increasing bond length of ${\rm{C}} - {\rm{H}},{\rm{C}} - {\rm{O}},{\rm{C}} - {\rm{C}}$ and ${\rm{C}} = {\rm{C}}$ is:

${\rm{C}} - {\rm{H}} < C = C < C - O < C - C$

${\rm{C}} - {\rm{H}} < C - O < C - C < C = C$

${\rm{C}} - {\rm{C}} < C = C < C - O < C - H$

${\rm{C}} - {\rm{O}} < C - H < C - C < C = C$

Molecular shape of ${\rm{S}}{{\rm{F}}_4},{\rm{C}}{{\rm{F}}_4}$ and ${\rm{Xe}}{{\rm{F}}_4}$ are:

Different with $1,{\rm{ }}0{\rm{ }} and {\rm{ }}2$ lone pairs of electrons respectively.

The same with $2,{\rm{ }}0{\rm{ }} and {\rm{ }}1$ lone pair of electrons respectively

The same with $1,{\rm{ }}1{\rm{ }} and {\rm{ }}1$lone pair of electrons respectively.

Different with $0,{\rm{ }}1{\rm{ }} and {\rm{ }}2$ lone pairs of electrons respectively.

A compound contains $X,{\rm{ }}Y\,and\,Z$ atoms. The oxidation states of $X,{\rm{ }}Y\,and\,Z$ are $ + 2$, $ + 2$ and $ - 2$ respectively. The possible formula of the compound is

${Y_2}{\left( {X{Z_3}} \right)_2}$

${X_3}{\left( {{Y_4}Z} \right)_2}$

${X_3}{\left( {Y{Z_4}} \right)_3}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 9

241 Medium

241 of 262Question permalink

Remembering +4 | -1

According to Fajan's rule covalent bond is favoured by:

View answer and explanation

Correct option: A

Small cation and large anion

Option A (Small cation and large anion) is correct because, according to Fajan's rule, a small cation and a large anion favor covalent character due to increased polarization.

Solution approach

Correct Answer + Core Reason: Small cation and large anion. Fajan's rule dictates that covalent character is favored by small cations and large anions.

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: Define Polarization: Polarization is the distortion of the electron cloud of an anion by a cation, leading to increased covalent character.

Step 3: Identify Factors Favoring Covalent Character: According to Fajan's rule, covalent character is favored by:
- Small, highly charged cations (high polarizing power)
- Large, highly charged anions (high polarizability)

Step 4: Apply Fajan's Rule: Therefore, a small cation and a large anion favor covalent bond formation.

Why other options are wrong:
- Option B: Small cation and small anion: A small anion is less polarizable, reducing covalent character.
- Option C: Large cation and large anion: A large cation has less polarizing power, reducing covalent character.
- Option D: Large cation and small anion: A large cation has less polarizing power, and a small anion is less polarizable, both reducing covalent character.

Key Concepts:
- Fajan's Rule
- Covalent Character
- Polarization
- Polarizing Power
- Polarizability

Memory Aids:
- Fajan's Rule: Small cation, Large anion = Covalent character

Quick Tips:
- Remember that covalent character increases with increased polarization, which is favored by small cations and large anions.

242 Medium

242 of 262Question permalink

Understanding +4 | -1

proton plays an important role in…bonding.

View answer and explanation

Correct option: D

Coordinate

Option D (Coordinate) is correct because a proton (H+) accepts a lone pair of electrons from another atom to form a coordinate bond.

Solution approach

Correct Answer + Core Reason: Coordinate. A proton accepts a lone pair in coordinate bonding.

Step-by-Step Solution:

Step 1: Define each type of bonding.

Step 2: Electrovalent (ionic) bonding involves the transfer of electrons.

Step 3: Hydrogen bonding is an intermolecular force between a hydrogen atom bonded to a highly electronegative atom and another electronegative atom.

Step 4: Covalent bonding involves the sharing of electrons.

Step 5: Coordinate bonding involves one atom donating both electrons to form a bond. A proton (H+) can accept a lone pair of electrons from another atom, forming a coordinate bond.

Why other options are wrong:
- Option A: Electrovalent is incorrect because protons do not directly participate in electron transfer.
- Option B: Hydrogen is incorrect because while protons are hydrogen ions, hydrogen bonding is an intermolecular force, not a direct bonding mechanism involving protons in the same way as coordinate bonding.
- Option C: Covalent is incorrect because protons do not directly participate in electron sharing in a typical covalent bond.

Key Concepts:
- Coordinate bonding
- Lewis acids and bases

Memory Aids:
- Coordinate bonds involve donation of electron pairs.

Quick Tips:
- Remember that coordinate bonds involve one atom donating both electrons.

243 Medium

243 of 262Question permalink

Remembering +4 | -1

According to Fajan's rule polarization is more when:

View answer and explanation

Correct option: A

Small cation and large anion

Option A (Small cation and large anion) is correct because according to Fajan's rule, polarization is favored by a small, highly charged cation and a large, highly charged anion.

Solution approach

Correct Answer + Core Reason: Small cation and large anion. Fajan's rule states that small cations and large anions favor polarization.

Step-by-Step Solution:

Step 1: State Fajan's Rule: Fajan's rule describes the conditions that favor covalent character in ionic compounds, which is related to polarization.

Step 2: Define Polarization: Polarization is the distortion of the electron cloud of an anion by a cation.

Step 3: Identify Factors Affecting Polarization: The polarizing power of a cation is high when it is small and highly charged. The polarizability of an anion is high when it is large and highly charged.

Step 4: Apply Fajan's Rule: Therefore, polarization is more when the cation is small and the anion is large.

Why other options are wrong:
- Option B: Small cation and small anion: A small anion is less polarizable.
- Option C: Large cation and large anion: A large cation has less polarizing power.
- Option D: Large cation and small anion: A large cation has less polarizing power and a small anion is less polarizable.

Key Concepts:
- Fajan's Rule
- Polarization
- Polarizing Power
- Polarizability

Memory Aids:
- Fajan's Rule: Small cation, Large anion = Polarization

Quick Tips:
- Remember the trends: smaller cations and larger anions lead to greater polarization.

244 Medium

244 of 262Question permalink

Remembering +4 | -1

When hydrogen bond is formed between two molecules, it is called:

View answer and explanation

Correct option: B

intermolecular

{H}

-bond

Option B (intermolecular

{H}

-bond) is correct because hydrogen bonds formed between two separate molecules are called intermolecular hydrogen bonds.

Solution approach

Correct Answer + Core Reason: intermolecular ${H}$ -bond. Hydrogen bonds formed between different molecules are intermolecular.

Step-by-Step Solution:

Step 1: Define hydrogen bond.

Step 2: A hydrogen bond is an electrostatic attraction between a hydrogen atom bonded to a highly electronegative atom (such as oxygen, nitrogen, or fluorine) and another electronegative atom in a different molecule or part of the same molecule.

Step 3: Differentiate between intramolecular and intermolecular hydrogen bonds.

Step 4: Intramolecular hydrogen bonds occur within the same molecule.

Step 5: Intermolecular hydrogen bonds occur between different molecules.

Why other options are wrong:
- Option A (intra-molecular H-bond): Intramolecular hydrogen bonds occur within the same molecule, not between two molecules.
- **Option C (directional ${H}$ -bond): While hydrogen bonds are directional, this term doesn't define the type of bond formed between two molecules.
- Option D (Both (2) and (3)): While hydrogen bonds are directional, the primary distinction here is whether the bond is within a molecule or between molecules.

Key Concepts:
- Hydrogen Bonding
- Intermolecular Forces
- Intramolecular Forces

Memory Aids:
- Inter = Between
- Intra = Within

Quick Tips:**
- Understand the difference between intermolecular and intramolecular forces.

245 Medium

245 of 262Question permalink

Application +4 | -1

Atoms

\mathrm{X}

and

\mathrm{Y}

combine to create an ionic compound with the formula

\mathrm{XY}_2.

Among the following options, which pair of atomic numbers for

\mathrm{X}

and

\mathrm{Y}

is suitable for this compound?

View answer and explanation

Correct option: C

\mathrm{X}=12 ; \mathrm{Y}=17

Option C (

\mathrm{X}=12 ; \mathrm{Y}=17

) is correct because X with 12 electrons (Magnesium) will lose 2 electrons to form

X^{2+}

, and Y with 17 electrons (Chlorine) will gain 1 electron to form

Y^{-}

. Thus, the formula

XY_2

is formed.

Solution approach

Correct Answer + Core Reason: $\mathrm{X}=12 ; \mathrm{Y}=17$ . This combination leads to the formation of $X^{2+}$ and $Y^-$ ions, resulting in the formula $XY_2$ .

Step-by-Step Solution:

Step 1: Determine the electronic configuration and valency of each element in each option.
- Option A: X = 11 (Na), Y = 16 (S). Na forms $Na^+$ , and S forms $S^{2-}$ . The formula would be $X_2Y$ .
- Option B: X = 3 (Li), Y = 17 (Cl). Li forms $Li^+$ , and Cl forms $Cl^-$ . The formula would be $XY$ .
- Option C: X = 12 (Mg), Y = 17 (Cl). Mg forms $Mg^{2+}$ , and Cl forms $Cl^-$ . The formula would be $XY_2$ .
- Option D: X = 6 (C), Y = 16 (O). C can form various compounds with O, but not typically ionic $XY_2$ .

Step 2: Identify the combination that leads to the formation of $XY_2$ .
- Option C: Mg forms $Mg^{2+}$ and Cl forms $Cl^-$ , so the compound is $MgCl_2$ , which fits the $XY_2$ formula.

Why other options are wrong:
- Option A: $\mathrm{X}=11 ; \mathrm{Y}=16$ : Forms $X_2Y$ compound.
- Option B: $\mathrm{X}=3 ; \mathrm{Y}=17$ : Forms $XY$ compound.
- Option D: $\mathrm{X}=6 ; \mathrm{Y}=16$ : Does not typically form an ionic compound of the form $XY_2$ .

Key Concepts:
- Ionic compound formation
- Valency
- Electronic configuration

Memory Aids:
- Metals lose electrons to form cations, and nonmetals gain electrons to form anions.

Quick Tips:
- Determine the charges of the ions formed by each element to predict the formula of the ionic compound.

246 Medium

246 of 262Question permalink

Understanding +4 | -1

A coordinate bond is a dative covalent bond. Which of the below is true?

View answer and explanation

Correct option: C

Two atoms form bond and one of them provides both electrons

Option C (Two atoms form bond and one of them provides both electrons) is correct because that is the definition of a coordinate covalent bond.

Solution approach

Correct Answer + Core Reason: Two atoms form bond and one of them provides both electrons. This is the definition of a coordinate bond.

Step-by-Step Solution:

Step 1: Define a coordinate bond.

Step 2: A coordinate bond is formed when one atom donates both electrons to form the bond.

Step 3: Compare the definition to the options.

Why other options are wrong:
- Option A: Three atoms forming a bond is not a coordinate bond.
- Option B: Two atoms sharing electrons, where each contributes one electron, is a covalent bond, not a coordinate bond.
- Option D: Electrons obtained from a third atom doesn't describe a coordinate bond.

Key Concepts:
- Coordinate bond
- Covalent bond

Memory Aids:
- Coordinate = one donates both

Quick Tips:
- Remember the difference between covalent and coordinate bonds.

247 Medium

247 of 262Question permalink

Understanding +4 | -1

Molecular size of

\mathrm{ICl}

and

\mathrm{B}{\mathrm{r}_2}

is nearly same, but boiling point of

\mathrm{ICl}

is about

40^\circ C

higher than

\mathrm{B}{\mathrm{r}_2}

. This might be due to:

View answer and explanation

Correct option: C

\mathrm{ICl}

is polar where as

\mathrm{B}{\mathrm{r}_2}

is non-polar

Option C is correct because the higher boiling point of ICl compared to Br2 is due to the dipole-dipole interactions present in polar ICl molecules, which are absent in non-polar Br2 molecules.

Solution approach

Correct Answer + Core Reason: $\mathrm{ICl}$ is polar where as $\mathrm{B}{\mathrm{r}_2}$ is non-polar. The difference in boiling points is due to the presence of dipole-dipole interactions in ICl, which are stronger than the London dispersion forces in Br2.

Step-by-Step Solution:

Step 1: Identify the intermolecular forces present in each molecule.

Step 2: ICl is a polar molecule due to the difference in electronegativity between iodine and chlorine. This results in dipole-dipole interactions.

Step 3: Br2 is a non-polar molecule because it consists of two identical bromine atoms. The only intermolecular forces present are London dispersion forces.

Step 4: Dipole-dipole interactions are generally stronger than London dispersion forces when the molecular sizes are comparable.

Step 5: Therefore, ICl has a higher boiling point than Br2.

Why other options are wrong:
- Option A: Bond strength is related to bond dissociation energy, not boiling point directly. While it might influence intermolecular forces, the polarity difference is the primary factor here.
- Option B: Ionization energy is a property of individual atoms and doesn't directly explain the difference in boiling points.
- Option D: While the size of I is larger than Br, the molecular sizes are stated to be nearly the same in the question, making this less relevant than the polarity difference.

Key Concepts:
- Intermolecular forces (dipole-dipole, London dispersion forces)
- Polarity of molecules
- Boiling point and intermolecular forces

Memory Aids:
- "Like dissolves like" - polar molecules interact more strongly with other polar molecules.

Quick Tips:
- When comparing boiling points, consider the types and strengths of intermolecular forces present.

248 Medium

248 of 262Question permalink

Understanding +4 | -1

Shortest bond length among the following is of

View answer and explanation

Correct option: B

O – H

Option B (O – H) is correct because oxygen is more electronegative than carbon and nitrogen, leading to a smaller atomic radius and thus a shorter bond length.

Solution approach

Correct Answer + Core Reason: O – H. Oxygen is more electronegative, smaller atomic radius.

Step-by-Step Solution:

Step 1: Understand the relationship between bond length and atomic size.
- Bond length is the distance between the nuclei of two bonded atoms.
- Smaller atoms form shorter bonds.
- Higher electronegativity difference can also lead to shorter bond lengths due to increased ionic character.

Step 2: Compare the atomic sizes and electronegativities of C, N, and O.
- Atomic size decreases across a period (C > N > O).
- Electronegativity increases across a period (C < N < O).

Step 3: Analyze the bond lengths of the given bonds.
- C – H: Carbon is larger than oxygen, so the C-H bond is longer than the O-H bond.
- O – H: Oxygen is the smallest and most electronegative, so the O-H bond is the shortest.
- C – N: Both C and N are larger than O, so C-N bond is longer than O-H bond.
- N – O: N-O bond is longer than O-H bond because N is larger than H.

Why other options are wrong:
- Option A: C – H - Incorrect. C-H bond is longer than O-H bond.
- Option B: O – H - Correct. O-H bond is the shortest.
- Option C: C – N - Incorrect. C-N bond is longer than O-H bond.
- Option D: N – O - Incorrect. N-O bond is longer than O-H bond.

Key Concepts:
- Bond length
- Atomic size
- Electronegativity

Memory Aids:
- Smaller atoms = shorter bonds

Quick Tips:
- Consider both atomic size and electronegativity when comparing bond lengths.

249 Medium

249 of 262Question permalink

Understanding +4 | -1

The bond order in

NO

is 2.5 while that in

\mathrm{N}{\mathrm{O}^ + }

is 3. Which of the following statements is true for these two species?

View answer and explanation

Correct option: B

Bond length in

NO

is greater than in

\mathrm{N}{\mathrm{O}^ + }

Option B (Bond length in

NO

is greater than in

\mathrm{N}{\mathrm{O}^ + }

) is correct because a higher bond order indicates a stronger and shorter bond, so

\mathrm{N}{\mathrm{O}^ + }

with a bond order of 3 has a shorter bond length than

NO

with a bond order of 2.5.

Solution approach

Correct Answer + Core Reason: Bond length in $NO$ is greater than in $\mathrm{N}{\mathrm{O}^ + }$ . Higher bond order implies shorter bond length.

Step-by-Step Solution:

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

Step 2: Relate Bond Order to Bond Length: Higher bond order implies a stronger and shorter bond.

Step 3: Compare Bond Orders: The bond order in $NO$ is 2.5, while that in $\mathrm{N}{\mathrm{O}^ + }$ is 3.

Step 4: Apply the Relationship: Since $\mathrm{N}{\mathrm{O}^ + }$ has a higher bond order, it has a shorter bond length compared to $NO$ .

Step 5: Conclusion: Therefore, the bond length in $NO$ is greater than in $\mathrm{N}{\mathrm{O}^ + }$ .

Why other options are wrong:
- Option A: Bond length in $\mathrm{N}{\mathrm{O}^ + }$ is greater than in $NO$ : Incorrect; higher bond order means shorter bond length.
- Option C: Bond length in $\mathrm{N}{\mathrm{O}^ + }$ is equal to than in $NO$ : Incorrect; bond lengths are different due to different bond orders.
- Option D: Bond length is unpredictable: Incorrect; bond length is predictable based on bond order.

Key Concepts:
- Bond Order
- Bond Length
- Relationship between Bond Order and Bond Length

Memory Aids:
- Higher bond order = shorter bond length

Quick Tips:
- Remember that bond order and bond length are inversely related.

250 Medium

250 of 262Question permalink

Application +4 | -1

The structure of

\mathrm{ICl}_2^ -

is:

View answer and explanation

Correct option: D

Distorted trigonal bipyramid

Option D (Distorted trigonal bipyramid) is correct because

\mathrm{ICl}_2^ -

has 2 bond pairs and 3 lone pairs around the central iodine atom, leading to a linear shape derived from a trigonal bipyramidal arrangement.

Solution approach

Correct Answer + Core Reason: Distorted trigonal bipyramid. $\mathrm{ICl}_2^ -$ has 2 bond pairs and 3 lone pairs.

Step-by-Step Solution:

Step 1: Determine the number of bond pairs and lone pairs around the central iodine atom.

Step 2: Iodine has 7 valence electrons. In $\mathrm{ICl}_2^ -$ , it forms 2 bonds with chlorine atoms and has one extra electron from the negative charge. Therefore, it has 2 bond pairs and 3 lone pairs.

Step 3: According to VSEPR theory, 2 bond pairs and 3 lone pairs result in a linear shape derived from a trigonal bipyramidal arrangement.

Step 4: The lone pairs occupy the equatorial positions to minimize repulsion, resulting in a linear shape for the atoms.

Why other options are wrong:
- Option A: Trigonal is incorrect.
- Option B: Octahedral is incorrect.
- Option C: Square planar is incorrect.

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

Memory Aids:
- Lone pairs cause distortion in geometry.

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

251 Medium

251 of 262Question permalink

Understanding +4 | -1

The correct order of increasing bond length of

\mathrm{C} - \mathrm{H},\mathrm{C} - \mathrm{O},\mathrm{C} - \mathrm{C}

and

\mathrm{C} = \mathrm{C}

is:

View answer and explanation

Correct option: B

\mathrm{C} - \mathrm{H} < C = C < C - O < C - C

Option B (

\mathrm{C} - \mathrm{H} < C = C < C - O < C - C

) is correct because single bonds are longer than double bonds, and bond length increases with the size of the atoms involved.

Solution approach

Correct Answer + Core Reason: $\mathrm{C} - \mathrm{H} < C = C < C - O < C - C$ . This order reflects the relative bond lengths based on bond order and atomic size.

Step-by-Step Solution:

Step 1: Consider the factors affecting bond length: bond order and atomic size.

Step 2: Higher bond order (e.g., double bond) results in shorter bond length.

Step 3: Larger atomic size results in longer bond length.

Step 4: C-H bond length is the shortest because hydrogen is the smallest atom.

Step 5: C=C bond is shorter than C-O and C-C bonds due to its double bond character.

Step 6: C-O bond is shorter than C-C bond because oxygen is more electronegative than carbon, leading to a slightly shorter bond.

Step 7: Therefore, the correct order is C-H < C=C < C-O < C-C.

Why other options are wrong:
- Option A: $\mathrm{C} - \mathrm{H} < C - O < C - C < C = C$ is incorrect because C=C is shorter than C-O and C-C.
- Option C: $\mathrm{C} - \mathrm{C} < C = C < C - O < C - H$ is incorrect because C-H is the shortest and C-C is longer than C=C and C-O.
- Option D: $\mathrm{C} - \mathrm{O} < C - H < C - C < C = C$ is incorrect because C-H is the shortest and C=C is shorter than C-O and C-C.

Key Concepts:
- Bond length
- Bond order
- Atomic size
- Electronegativity

Memory Aids:
- Higher bond order, shorter bond length.
- Larger atoms, longer bond length.

Quick Tips:
- Consider both bond order and atomic size when comparing bond lengths.

252 Medium

252 of 262Question permalink

Application +4 | -1

Molecular shape of

\mathrm{S}{\mathrm{F}_4},\mathrm{C}{\mathrm{F}_4}

and

\mathrm{Xe}{\mathrm{F}_4}

are:

View answer and explanation

Correct option: D

Different with

1,\mathrm{ }0\mathrm{ } and \mathrm{ }2

lone pairs of electrons respectively.

Option D (Different with

1,\mathrm{ }0\mathrm{ } and \mathrm{ }2

lone pairs of electrons respectively) is correct because the shapes and lone pairs differ:

SF_4

(see-saw, 1 lone pair),

CF_4

(tetrahedral, 0 lone pairs), and

XeF_4

(square planar, 2 lone pairs).

Solution approach

Correct Answer + Core Reason: Different with $1,\mathrm{ }0\mathrm{ } and \mathrm{ }2$ lone pairs of electrons respectively. The shapes and lone pairs differ: $SF_4$ (see-saw, 1 lone pair), $CF_4$ (tetrahedral, 0 lone pairs), and $XeF_4$ (square planar, 2 lone pairs).

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each central atom.\nS in $SF_4$ : 6 valence electrons\nC in $CF_4$ : 4 valence electrons\nXe in $XeF_4$ : 8 valence electrons

Step 2: Determine the number of bond pairs and lone pairs for each molecule.
$SF_4$ : 4 bond pairs, 1 lone pair
$CF_4$ : 4 bond pairs, 0 lone pairs
$XeF_4$ : 4 bond pairs, 2 lone pairs

Step 3: Determine the shape of each molecule using VSEPR theory.
$SF_4$ : See-saw
$CF_4$ : Tetrahedral
$XeF_4$ : Square planar

Why other options are wrong:
- Option A: The same with $2,\mathrm{ }0\mathrm{ } and \mathrm{ }1$ lone pair of electrons respectively: Incorrect number of lone pairs for each molecule.
- Option B: The same with $1,\mathrm{ }1\mathrm{ } and \mathrm{ }1$ lone pair of electrons respectively: Incorrect number of lone pairs for each molecule and incorrect shapes.
- Option C: Different with $0,\mathrm{ }1\mathrm{ } and \mathrm{ }2$ lone pairs of electrons respectively: Incorrect number of lone pairs for $SF_4$ and $CF_4$ .

Key Concepts:
- VSEPR theory
- Lone pairs
- Molecular shapes

Memory Aids:
- VSEPR theory helps predict shapes based on electron pair repulsion.

Quick Tips:
- Count valence electrons, determine bond pairs and lone pairs, and then apply VSEPR theory.

253 Medium

253 of 262Question permalink

Application +4 | -1

The bond order of

\mathrm{N}_2^ +

is

View answer and explanation

Correct option: C

2.5

Option C (2.5) is correct because

\mathrm{N}_2^ +

has 13 electrons, resulting in a bond order of 2.5.

Solution approach

Correct Answer + Core Reason: 2.5. The bond order of $\mathrm{N}_2^ +$ is 2.5.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $\mathrm{N}_2^ +$ .
$\mathrm{N}_2^ +$ has 14 - 1 = 13 electrons.

Step 2: Write the electronic configuration of $\mathrm{N}_2^ +$ .
$(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^{\ast})^3$

Step 3: Calculate the bond order.\nBond order = $\frac{{\text{{Number of bonding electrons - Number of antibonding electrons}}}}{2} = \frac{{9 - 4}}{2} = \frac{5}{2} = 2.5$

Why other options are wrong:
- Option A: 1.5 is incorrect. The bond order calculation does not result in 1.5.
- Option B: 3.0 is incorrect. The bond order calculation does not result in 3.0.
- Option D: 2.0 is incorrect. The bond order calculation does not result in 2.0.

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation

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

Quick Tips:
- Remember the electronic configuration order for diatomic molecules.

254 Medium

254 of 262Question permalink

Remembering +4 | -1

Among the following, which is water insoluble?

View answer and explanation

Correct option: D

Magnesium fluoride

Option D (Magnesium fluoride) is correct because magnesium fluoride (

MgF_2

) has a relatively low solubility in water compared to the other alkali metal fluorides due to its higher lattice energy and lower hydration energy.

Solution approach

Correct Answer + Core Reason: Magnesium fluoride is water insoluble due to high lattice energy.

Step-by-Step Solution:

Step 1: Solubility of ionic compounds in water depends on the balance between lattice energy and hydration energy.

Step 2: Lattice energy is the energy required to separate the ions in the crystal lattice, while hydration energy is the energy released when ions are hydrated by water molecules.

Step 3: For a compound to be soluble, the hydration energy must be greater than the lattice energy.

Step 4: Among the given options, magnesium fluoride ( $MgF_2$ ) has a higher lattice energy compared to sodium fluoride ( $NaF$ ) and potassium fluoride ( $KF$ ). Beryllium fluoride ( $BeF_2$ ) is soluble due to the small size and high charge density of $Be^{2+}$ ion, which leads to high hydration energy.

Step 5: Due to the higher charge density and smaller size of $Mg^{2+}$ compared to $Na^+$ and $K^+$ , $MgF_2$ has a higher lattice energy and is less soluble.

Why other options are wrong:
- Option A: Sodium fluoride ( $NaF$ ) is soluble in water because the hydration energy is greater than the lattice energy.
- Option B: Potassium fluoride ( $KF$ ) is soluble in water because the hydration energy is greater than the lattice energy.
- Option C: Beryllium fluoride ( $BeF_2$ ) is soluble due to the small size and high charge density of $Be^{2+}$ ion, which leads to high hydration energy.

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

Memory Aids:
- High lattice energy = Low solubility

Quick Tips:
- Remember that smaller ions with higher charges tend to have higher lattice energies and lower solubility.

255 Medium

255 of 262Question permalink

Application +4 | -1

A compound contains

X,\mathrm{ }Y\,and\,Z

atoms. The oxidation states of

X,\mathrm{ }Y\,and\,Z

are

+ 2

,

+ 2

and

- 2

respectively. The possible formula of the compound is

View answer and explanation

Correct option: A

XY{Z_2}

Option A (

XY{Z_2}

) is correct because the sum of oxidation states in a neutral compound must be zero.

Solution approach

Correct Answer + Core Reason: $XY{Z_2}$ . The sum of oxidation states in a neutral compound must be zero.

Step-by-Step Solution:

Step 1: Define oxidation state: The oxidation state is the hypothetical charge an atom would have if all bonds were completely ionic.

Step 2: Apply the rule that the sum of oxidation states in a neutral compound is zero.

Step 3: Analyze each compound:
- $XY{Z_2}$ : (+2) + (+2) + 2*(-2) = 2 + 2 - 4 = 0
- ${Y_2}{\left( {X{Z_3}} \right)_2}$ : 2*(+2) + 2*((+2) + 3*(-2)) = 4 + 2*(2 - 6) = 4 + 2*(-4) = 4 - 8 = -4 ≠ 0
- ${X_3}{\left( {{Y_4}Z} \right)_2}$ : 3*(+2) + 2*(4*(+2) + (-2)) = 6 + 2*(8 - 2) = 6 + 2*(6) = 6 + 12 = 18 ≠ 0
- ${X_3}{\left( {Y{Z_4}} \right)_3}$ : 3*(+2) + 3*((+2) + 4*(-2)) = 6 + 3*(2 - 8) = 6 + 3*(-6) = 6 - 18 = -12 ≠ 0

Step 4: Identify the compound where the sum of oxidation states is zero: $XY{Z_2}$ .

Why other options are wrong:
- Option B: ${Y_2}{\left( {X{Z_3}} \right)_2}$ : The sum of oxidation states is not zero.
- Option C: ${X_3}{\left( {{Y_4}Z} \right)_2}$ : The sum of oxidation states is not zero.
- Option D: ${X_3}{\left( {Y{Z_4}} \right)_3}$ : The sum of oxidation states is not zero.

Key Concepts:
- Oxidation state
- Neutral compound

Memory Aids:
- The sum of oxidation states in a neutral compound is always zero.

Quick Tips:
- Remember the common oxidation states of elements to quickly determine the formula of a compound.

256 Medium

256 of 262Question permalink

Application +4 | -1

Two elements

X

and

Y

have following electronic configuration

X = 1{s^2}\,2{s^2}2{p^6}\,\,3{s^2}3{p^6}\,\,4{s^2}

Y = 1{s^2}\,2{s^2}2{p^6}\,\,3{s^2}3{p^5}\,

The expected compound formed by combination of

X

and

Y

will be expressed as

View answer and explanation

Correct option: A

X{Y_2}

Option A (

X{Y_2}

) is correct because X has a +2 charge and Y has a -1 charge, thus requiring two Y atoms to balance the charge of one X atom.

Solution approach

Correct Answer + Core Reason: $X{Y_2}$ . X forms a +2 cation and Y forms a -1 anion, leading to the compound $X{Y_2}$ .

Step-by-Step Solution:

Step 1: Determine the valence electrons for element X.\nElement X has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ . The outermost shell is the 4th shell, which has 2 electrons in the 4s orbital. Therefore, X has 2 valence electrons.

Step 2: Determine the valence electrons for element Y.\nElement Y has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^5$ . The outermost shell is the 3rd shell, which has 2 electrons in the 3s orbital and 5 electrons in the 3p orbital. Therefore, Y has 7 valence electrons.

Step 3: Determine the charge of the ions formed by X and Y.\nX tends to lose 2 electrons to achieve a stable noble gas configuration, forming a +2 cation ( $X^{2+}$ ).\nY tends to gain 1 electron to achieve a stable noble gas configuration, forming a -1 anion ( $Y^{-}$ ).

Step 4: Determine the formula of the compound formed by X and Y.\nTo form a neutral compound, the charges must balance. Therefore, one $X^{2+}$ ion requires two $Y^{-}$ ions. The formula of the compound is $X{Y_2}$ .

Why other options are wrong:
- **Option B ( ${X_5}{Y_2}$ ): This formula would imply that X has a charge of +2/5 and Y has a charge of -1/2, which is not consistent with their electronic configurations and the octet rule.
- Option C ( ${X_2}{Y_5}$ ): This formula would imply that X has a charge of +5/2 and Y has a charge of -2/5, which is not consistent with their electronic configurations and the octet rule.
- Option D ( $X{Y_5}$ ): This formula would imply that X has a charge of +5 and Y has a charge of -1, which is not consistent with the fact that X has only 2 valence electrons and tends to form a +2 ion.

Key Concepts:
- Electronic configuration
- Valence electrons
- Ion formation
- Octet rule
- Chemical formula

Memory Aids:
- Elements tend to gain or lose electrons to achieve a stable noble gas configuration (octet rule).
- Metals tend to lose electrons and form positive ions (cations).
- Nonmetals tend to gain electrons and form negative ions (anions).

Quick Tips:**
- Determine the number of valence electrons for each element.
- Determine the charge of the ions formed by each element based on their tendency to gain or lose electrons.
- Balance the charges to determine the formula of the compound.

257 Medium

257 of 262Question permalink

Analyzing +4 | -1

Which of the following is paramagnetic with bond order 0.5?

View answer and explanation

Correct option: B

\mathrm{H}_2^ +

Option B (

{\mathrm{H}_2^ + }

) is correct because it has one electron, resulting in a bond order of 0.5 and is paramagnetic due to the unpaired electron.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_2^ +$ . It has one electron, resulting in a bond order of 0.5 and is paramagnetic.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: ${\mathrm{F}_2}$ : Bond order = (10 - 8)/2 = 1 - Diamagnetic.

Step 3: $\mathrm{H}_2^ +$ : Bond order = (1 - 0)/2 = 0.5 - Paramagnetic (1 unpaired electron).

Step 4: ${\mathrm{N}_2}$ : Bond order = (10 - 4)/2 = 3 - Diamagnetic.

Step 5: $\mathrm{O}_2^ -$ : Bond order = (10 - 7)/2 = 1.5 - Paramagnetic.

Step 6: Identify the species with a bond order of 0.5 and is paramagnetic.

Why other options are wrong:
- Option A: ${\mathrm{F}_2}$ has a bond order of 1 and is diamagnetic.
- Option C: ${\mathrm{N}_2}$ has a bond order of 3 and is diamagnetic.
- Option D: $\mathrm{O}_2^ -$ has a bond order of 1.5 and is paramagnetic.

Key Concepts:
- Bond Order Calculation
- Paramagnetism and Diamagnetism

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

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

258 Medium

258 of 262Question permalink

Analyzing +4 | -1

Among the following which has the highest cation to anion size ratio?

View answer and explanation

Correct option: B

\mathrm{CsF}

Option B (CsF) is correct because Cs has the largest cation size and F has the smallest anion size among the given options, maximizing the ratio.

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

Why other options are wrong:
- Option A: CsI has a large cation and a large anion, but the ratio is lower than CsF because F is much smaller than I.
- Option C: LiF has a small cation and a small anion, but the ratio is lower than CsF because Cs is much larger than Li.
- Option D: NaF has a small cation and a small anion, but the ratio is lower than CsF because Cs is much larger than Na.

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

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

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

259 Medium

259 of 262Question permalink

Application +4 | -1

Which of the following species has four lone pairs in its outer shell?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{l}^ - }

Option C (

\mathrm{C}{\mathrm{l}^ - }

) is correct because a chloride ion has 7 valence electrons plus one from the negative charge, giving it 8 electrons, which is 4 lone pairs.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}^ - }$ . A chloride ion has 7 valence electrons plus one from the negative charge, giving it 8 electrons, which is 4 lone pairs.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each species.
- I (Iodine): 7 valence electrons
- ${\mathrm{O}^ - }$ (Oxygen ion): 6 + 1 = 7 valence electrons
- $\mathrm{C}{\mathrm{l}^ - }$ (Chloride ion): 7 + 1 = 8 valence electrons
- $\mathrm{He}$ (Helium): 2 valence electrons

Step 2: Determine the number of lone pairs for each species.
- I: 7 valence electrons = 3 lone pairs and 1 single electron
- ${\mathrm{O}^ - }$ : 7 valence electrons = 3 lone pairs and 1 single electron
- $\mathrm{C}{\mathrm{l}^ - }$ : 8 valence electrons = 4 lone pairs
- $\mathrm{He}$ : 2 valence electrons = 1 lone pair

Step 3: Identify the species with four lone pairs. $\mathrm{C}{\mathrm{l}^ - }$ has 4 lone pairs.

Why other options are wrong:
- **Option A ( $\mathrm{I}$ ): Iodine has 3 lone pairs.
- Option B ( ${\mathrm{O}^ - }$ ): Oxygen ion has 3 lone pairs.
- Option D ( $\mathrm{He}$ ): Helium has 1 lone pair.

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

Memory Aids:
- Remember that each lone pair consists of two electrons.

Quick Tips:**
- Draw the Lewis structure to visualize the lone pairs.

260 Medium

260 of 262Question permalink

Remembering +4 | -1

The energy of hydrogen bond is of the order of

View answer and explanation

Correct option: A

40 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option A (

40 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is correct because the energy of a hydrogen bond typically falls in the range of 5 to 40 kJ/mol.

Solution approach

Correct Answer + Core Reason: $40 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . This is within the typical range for hydrogen bond energy.

Step-by-Step Solution:

Step 1: Recall the typical energy range for hydrogen bonds.\nThe energy of a hydrogen bond is typically between 5 and 40 kJ/mol.

Step 2: Identify the option that falls within this range.
$40 \mathrm{~kJ} \mathrm{~mol}^{-1}$ falls within this range.

Why other options are wrong:
- Option B: Incorrect. This value is too high for a single hydrogen bond.
- Option C: Incorrect. This value is too high for a single hydrogen bond.
- Option D: Incorrect. This value is too low for a single hydrogen bond.

Key Concepts:
- Hydrogen bonding
- Bond energy

Memory Aids:
- Hydrogen bonds are relatively weak compared to covalent bonds.

Quick Tips:
- Remember the approximate energy range for hydrogen bonds.

261 Medium

261 of 262Question permalink

Understanding +4 | -1

The molecule having smallest bond angle is

View answer and explanation

Correct option: D

{\mathrm{H}_2}\mathrm{Te}

Option D (

{\mathrm{H}_2}\mathrm{Te}

) is correct because as we go down the group, the bond angle decreases due to the increasing size and decreasing electronegativity of the central atom.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{Te}$ . Down the group, bond angle decreases due to increasing size and decreasing electronegativity of the central atom.

Step-by-Step Solution:

Step 1: Identify the central atom in each molecule: O, S, N, and Te.

Step 2: Consider the position of these elements in the periodic table. O, S, and Te are in Group 16, and N is in Group 15.

Step 3: Analyze the factors affecting bond angle:
- Electronegativity of the central atom: Higher electronegativity leads to greater bond angle.
- Size of the central atom: Larger size leads to smaller bond angle.
- Lone pairs on the central atom: More lone pairs lead to smaller bond angle.

Step 4: Compare the bond angles:
- ${\mathrm{H}_2}\mathrm{O}$ : Bond angle is approximately 104.5°.
- ${\mathrm{H}_2}\mathrm{S}$ : Bond angle is approximately 92°.
- $\mathrm{N}{\mathrm{H}_3}$ : Bond angle is approximately 107°.
- ${\mathrm{H}_2}\mathrm{Te}$ : Bond angle is approximately 90°.

Step 5: Conclusion: ${\mathrm{H}_2}\mathrm{Te}$ has the smallest bond angle.

Why other options are wrong:
- **Option A ( ${\mathrm{H}_2}\mathrm{O}$ ): Oxygen is smaller and more electronegative than S and Te, leading to a larger bond angle.
- Option B ( ${\mathrm{H}_2}\mathrm{S}$ ): Sulfur is smaller and more electronegative than Te, leading to a larger bond angle.
- Option C ( ${\mathrm{N}{\mathrm{H}_3}}$ ): Nitrogen has a lone pair, but its smaller size and higher electronegativity compared to Te result in a larger bond angle.

Key Concepts:
- Bond angle
- Electronegativity
- Size of atom
- Lone pairs

Memory Aids:
- As you go down a group, bond angle decreases.

Quick Tips:**
- Consider the combined effects of electronegativity, size, and lone pairs when comparing bond angles.

262 Medium

262 of 262Question permalink

Remembering +4 | -1

Which of the following has the highest melting point

View answer and explanation

Correct option: B

Diamond

Option B (Diamond) is correct because diamond has a giant covalent network structure with strong covalent bonds throughout, requiring a large amount of energy to break.

Solution approach

Correct Answer + Core Reason: Diamond. Diamond has a giant covalent network structure with strong covalent bonds throughout, requiring a large amount of energy to break.

Step-by-Step Solution:

Step 1: Melting point is related to the strength of the bonds holding the substance together.

Step 2: Diamond has a giant covalent network structure where each carbon atom is covalently bonded to four other carbon atoms.

Step 3: These strong covalent bonds require a large amount of energy to break, resulting in a very high melting point.

Why other options are wrong:
- Option A: Lead (Pb) is a metal with metallic bonding, which is weaker than the covalent bonding in diamond.
- Option C: Iron (Fe) is a metal with metallic bonding, which is weaker than the covalent bonding in diamond.
- Option D: Sodium (Na) is an alkali metal with metallic bonding, which is weaker than the covalent bonding in diamond.

Key Concepts:
- Melting point
- Giant covalent structures
- Metallic bonding

Memory Aids:
- Diamond = Strongest Covalent Network = Highest Melting Point

Quick Tips:
- Remember the relationship between bond strength and melting point.

Your Progress

Chemical bonding & molecular structure practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice