CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 7

181 Medium

Application +4 | -1

The shape of

\mathrm{ClO}_3^ -

according to VSEPR model is:

View answer and explanation

Correct option: B

Pyramidal

Option B (Pyramidal) is correct because in

ClO_3^-

, Chlorine has 3 bond pairs and 1 lone pair, resulting in a pyramidal shape.

Solution approach

Correct Answer + Core Reason: Pyramidal. In $ClO_3^-$ , Chlorine has 3 bond pairs and 1 lone pair, resulting in a pyramidal shape.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for the central atom.\nChlorine (Cl) has 7 valence electrons. The negative charge adds one electron, so total valence electrons = 7 + 3(6) + 1 = 26 electrons.

Step 2: Determine the number of bond pairs and lone pairs.\nThere are 3 bond pairs (Cl-O bonds) and 1 lone pair on the Cl atom.

Step 3: Determine the shape using VSEPR theory.\nWith 3 bond pairs and 1 lone pair, the shape is pyramidal.

Why other options are wrong:
- Option A: Planar triangle: This shape would require 3 bond pairs and 0 lone pairs.
- Option C: Tetrahedral: This shape would require 4 bond pairs and 0 lone pairs.
- Option D: Square planar: This shape requires 4 bond pairs and 2 lone pairs.

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

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

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

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Remembering +4 | -1

Polarization of an anion is maximum by high charged cation

View answer and explanation

Correct option: A

Polarization of an anion is maximum by high charged cation

Option A (Polarization of an anion is maximum by high charged cation) is correct because a highly charged cation has a greater ability to distort the electron cloud of the anion, leading to maximum polarization.

Solution approach

Correct Answer + Core Reason: Polarization of an anion is maximum by high charged cation. Higher charge on cation leads to greater polarization.

Step-by-Step Solution:

Step 1: Define Polarization: Polarization is the distortion of the electron cloud of an anion by a cation.

Step 2: Identify Factors Affecting Polarization: The polarizing power of a cation is directly proportional to its charge. A higher charge means greater polarizing power.

Step 3: Relate Charge to Polarization: Therefore, a high charged cation will cause maximum polarization of an anion.

Why other options are wrong:
- Option B: Small sized cation minimises the polarization: A small cation *maximizes* polarization, not minimizes it.
- Option C: A small anion brings about a large degree of polarisation: Anions don't bring about polarization; they *undergo* polarization.
- Option D: A small anion undergoes a high degree of polarization: A *large* anion undergoes a higher degree of polarization.

Key Concepts:
- Polarization
- Cation Charge
- Anion Size

Memory Aids:
- High charge = High polarization

Quick Tips:
- Remember that higher charge on the cation leads to increased polarization of the anion.

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Remembering +4 | -1

Dipole-dipole interaction energy between stationary polar molecules is proportional to

View answer and explanation

Correct option: C

\frac{1}{{{r^3}}}

Option C is correct because the dipole-dipole interaction energy between stationary polar molecules is inversely proportional to the cube of the distance (r) between them, hence 1/r³.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{{r^3}}}$ . The dipole-dipole interaction energy is proportional to $\frac{1}{r^3}$ for stationary polar molecules.

Step-by-Step Solution:

Step 1: Recall the relationship between dipole-dipole interaction energy and distance. For stationary polar molecules, the dipole-dipole interaction energy is inversely proportional to the cube of the distance between the molecules.

Step 2: Express the relationship mathematically: Dipole-dipole interaction energy $\propto \frac{1}{r^3}$ .

Step 3: Conclusion: The dipole-dipole interaction energy is proportional to $\frac{1}{r^3}$ .

Why other options are wrong:
- Option A: $\frac{1}{r}$ : This is incorrect because the interaction energy is proportional to $\frac{1}{r^3}$ , not $\frac{1}{r}$ .
- Option B: $\frac{1}{{{r^2}}}$ : This is incorrect because the interaction energy is proportional to $\frac{1}{r^3}$ , not $\frac{1}{r^2}$ .
- Option D: $\frac{1}{{{r^6}}}$ : This is incorrect because the interaction energy is proportional to $\frac{1}{r^3}$ , not $\frac{1}{r^6}$ . This relationship is for London dispersion forces.

Key Concepts:
- Dipole-Dipole Interactions
- Intermolecular Forces

Memory Aids:
- Remember that dipole-dipole interaction energy is inversely proportional to $r^3$ .

Quick Tips:
- Be aware of the distance dependence of different types of intermolecular forces.

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Remembering +4 | -1

The structure of

P F_5

molecule is

View answer and explanation

Correct option: B

Trigonal bipyramidal

Option B (Trigonal bipyramidal) is correct because

PF_5

has five bonding pairs and no lone pairs, leading to a trigonal bipyramidal geometry.

Solution approach

Correct Answer + Core Reason: Trigonal bipyramidal. $PF_5$ has a trigonal bipyramidal structure due to $sp^3d$ hybridization.

Step-by-Step Solution:

Step 1: Determine the central atom: Phosphorus (P).

Step 2: Count the number of valence electrons for P: 5.

Step 3: Count the number of bonding pairs: 5 (one for each F atom).

Step 4: Calculate the steric number: 5 (5 bonding pairs + 0 lone pairs).

Step 5: Determine the hybridization: $sp^3d$ .

Step 6: Determine the geometry based on the hybridization: Trigonal bipyramidal.

Why other options are wrong:
- Option A (Tetrahedral): Tetrahedral geometry corresponds to $sp^3$ hybridization with 4 bonding pairs and 0 lone pairs.
- Option C (Square planar): Square planar geometry typically involves $dsp^2$ hybridization and lone pairs.
- Option D (Pentagonal bipyramidal): Pentagonal bipyramidal geometry is associated with $sp^3d^3$ hybridization, which is not the case for $PF_5$ .

Key Concepts:
- VSEPR theory
- Hybridization
- Molecular geometry

Memory Aids:
- Steric number = number of bonding pairs + number of lone pairs

Quick Tips:
- Use VSEPR theory to predict molecular geometry.

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Understanding +4 | -1

In which of the following, the bond length between hybridised carbon atom and other carbon atom is minimum?

View answer and explanation

Correct option: A

Propyne

Option A (Propyne) is correct because the carbon atom bonded to the other carbon atom is sp hybridized, resulting in the shortest bond length.

Solution approach

Correct Answer + Core Reason: Propyne. The sp hybridized carbon has the shortest bond length to the other carbon.

Step-by-Step Solution:

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

Step 2: Propyne (CH3C≡CH): The carbon atom bonded to the methyl group is sp hybridized.

Step 3: Propene (CH3CH=CH2): The carbon atom bonded to the methyl group is sp2 hybridized.

Step 4: Butane (CH3CH2CH2CH3): All carbon atoms are sp3 hybridized.

Step 5: Propane (CH3CH2CH3): All carbon atoms are sp3 hybridized.

Step 6: The order of bond length is: sp < sp2 < sp3.

Step 7: Therefore, the bond length between the hybridized carbon atom and the other carbon atom is minimum in propyne.

Why other options are wrong:
- Option B: Propene has sp2 hybridized carbon, which has a longer bond length than sp.
- Option C: Butane has sp3 hybridized carbon, which has a longer bond length than sp.
- Option D: Propane has sp3 hybridized carbon, which has a longer bond length than sp.

Key Concepts:
- Hybridization
- Bond length

Memory Aids:
- sp hybridization: shortest bond length
- sp2 hybridization: intermediate bond length
- sp3 hybridization: longest bond length

Quick Tips:
- Remember the relationship between hybridization and bond length.

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Application +4 | -1

Which of the following molecules or ions is not linear

View answer and explanation

Correct option: D

ICl_2^ +

Option D (

ICl_2^+

) is correct because it has a bent shape due to the presence of two bond pairs and two lone pairs around the central iodine atom, resulting in

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $ICl_2^+$ . It is bent due to two bond pairs and two lone pairs.

Step-by-Step Solution:

Step 1: Determine the structure of $BeCl_2$ .

Step 2: Be has 2 valence electrons. In $BeCl_2$ , both are used for bonding with Cl atoms. So, there are 0 lone pairs. The steric number is 2 (2 bond pairs + 0 lone pairs), leading to $sp$ hybridization. The shape is linear.

Step 3: Determine the structure of $ICl_2^-$ .

Step 4: I has 7 valence electrons. In $ICl_2^-$ , two are used for bonding with Cl atoms, and one negative charge adds one electron. So, there are 6 non-bonding electrons, i.e., 3 lone pairs. The steric number is 5 (2 bond pairs + 3 lone pairs), leading to $sp^3d$ hybridization. The shape is linear.

Step 5: Determine the structure of $CS_2$ .

Step 6: C has 4 valence electrons. In $CS_2$ , all four are used for bonding with S atoms (two double bonds). So, there are 0 lone pairs. The steric number is 2 (2 bond pairs + 0 lone pairs), leading to $sp$ hybridization. The shape is linear.

Step 7: Determine the structure of $ICl_2^+$ .

Step 8: I has 7 valence electrons. In $ICl_2^+$ , two are used for bonding with Cl atoms, and one positive charge removes one electron. So, there are 4 non-bonding electrons, i.e., 2 lone pairs. The steric number is 4 (2 bond pairs + 2 lone pairs), leading to $sp^3$ hybridization. The shape is bent.

Why other options are wrong:
- Option A: $BeCl_2$ is linear.
- Option B: $ICl_2^-$ is linear.
- Option C: $CS_2$ is linear.

Key Concepts:
- VSEPR theory
- Hybridization
- Molecular shapes

Memory Aids:
- Steric number = Number of bond pairs + Number of lone pairs

Quick Tips:
- Determine the number of bond pairs and lone pairs around the central atom to predict the shape.

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Application +4 | -1

\mathrm{CaO}

and

\mathrm{NaCl}

have the same crystal structure and approximately the same ionic radii. If

U

is the lattice energy of

NaCl

, the approximate lattice of

CaO

is

View answer and explanation

Correct option: C

4U

Option C (

4U

) is correct because lattice energy is proportional to the product of the charges of the ions. CaO has charges of +2 and -2, while NaCl has charges of +1 and -1. Therefore, the lattice energy of CaO is approximately 4 times that of NaCl.

Solution approach

Correct Answer + Core Reason: $4U$ . Lattice energy is proportional to the product of the charges.

Step-by-Step Solution:

Step 1: Lattice energy (U) is proportional to the product of the charges of the ions: $U \propto Q_1Q_2$ .

Step 2: For NaCl, the charges are +1 and -1, so $Q_1Q_2 = (1)(-1) = -1$ .

Step 3: For CaO, the charges are +2 and -2, so $Q_1Q_2 = (2)(-2) = -4$ .

Step 4: Since the ionic radii are approximately the same, the lattice energy is mainly determined by the product of the charges.

Step 5: Therefore, the lattice energy of CaO is approximately 4 times that of NaCl: $U_{CaO} \approx 4U_{NaCl}$ .

Why other options are wrong:
- Option A: Incorrect. It suggests the lattice energy of CaO is less than that of NaCl, which is incorrect.
- Option B: Incorrect. It suggests the lattice energy of CaO is the same as that of NaCl, which is incorrect.
- Option D: Incorrect. It suggests the lattice energy of CaO is twice that of NaCl, which is incorrect.

Key Concepts:
- Lattice energy
- Ionic charge

Memory Aids:
- Higher charges = higher lattice energy.

Quick Tips:
- Consider the charges of the ions when comparing lattice energies.

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Analyzing +4 | -1

Which one is the strongest bond?

View answer and explanation

Correct option: C

Br - \mathrm{F}

Option C is correct because bond strength increases with increasing electronegativity difference between the bonded atoms.

Solution approach

Correct Answer + Core Reason: $Br - \mathrm{F}$ . The greater the electronegativity difference, the stronger the bond.

Step-by-Step Solution:

Step 1: Determine the electronegativity values of the elements involved:
- Electronegativity of F: 3.98
- Electronegativity of Cl: 3.16
- Electronegativity of Br: 2.96

Step 2: Calculate the electronegativity differences for each bond:
- $\mathrm{Cl} - \mathrm{F}$ : |3.98 - 3.16| = 0.82
- $\mathrm{F} - \mathrm{F}$ : |3.98 - 3.98| = 0
- $Br - \mathrm{F}$ : |3.98 - 2.96| = 1.02
- $Br - Cl$ : |3.16 - 2.96| = 0.20

Step 3: Compare the electronegativity differences: The largest electronegativity difference corresponds to the strongest bond.
- $Br - \mathrm{F}$ has the largest electronegativity difference (1.02).

Step 4: Conclude: The strongest bond is $Br - \mathrm{F}$ .

Why other options are wrong:
- Option A: $\mathrm{Cl} - \mathrm{F}$ has a smaller electronegativity difference than $Br - \mathrm{F}$ .
- Option B: $\mathrm{F} - \mathrm{F}$ has no electronegativity difference, so it's a weaker bond.
- Option D: $Br - Cl$ has a smaller electronegativity difference than $Br - \mathrm{F}$ .

Key Concepts:
- Electronegativity
- Bond strength
- Electronegativity difference

Memory Aids:
- The greater the electronegativity difference, the more polar and stronger the bond.

Quick Tips:
- Use electronegativity values to predict bond strength.
- Remember that a larger electronegativity difference leads to a stronger bond.

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Analyzing +4 | -1

The correct order of decreasing polarity is

View answer and explanation

Correct option: B

\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}

Option B (

\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}

) is correct because it correctly orders the molecules based on their dipole moments, which reflect the polarity of the bonds and molecular geometry.

Solution approach

Correct Answer + Core Reason: $\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}$ . This order reflects the decreasing dipole moments of the molecules.

Step-by-Step Solution:

Step 1: Consider the electronegativity differences and molecular geometry of each molecule:
- HF: Large electronegativity difference between H and F, and a linear geometry, resulting in a high dipole moment.
- ${\mathrm{H}_2}\mathrm{O}$ : Significant electronegativity difference between H and O, and a bent geometry, resulting in a substantial dipole moment.
- $S{\mathrm{O}_2}$ : Electronegativity difference between S and O, and a bent geometry, resulting in a dipole moment.
- $N{\mathrm{H}_3}$ : Electronegativity difference between N and H, and a trigonal pyramidal geometry, resulting in a dipole moment.

Step 2: Compare the dipole moments: HF has the highest dipole moment due to the large electronegativity difference and simple linear structure. Water has a significant dipole moment due to its bent shape and electronegativity difference. Sulfur dioxide is polar, but less so than water. Ammonia is the least polar among these.

Step 3: Conclusion: The correct order of decreasing polarity is $\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}$ .

Why other options are wrong:
- Option A: $\mathrm{HF} > S{\mathrm{O}_2} > {\mathrm{H}_2}\mathrm{O} > N{\mathrm{H}_3}$ is incorrect because water is more polar than sulfur dioxide.
- Option C: $\mathrm{HF} > N{\mathrm{H}_3} > S{\mathrm{O}_2} > {\mathrm{H}_2}\mathrm{O}$ is incorrect because water is more polar than both ammonia and sulfur dioxide.
- Option D: ${\mathrm{H}_2}\mathrm{O} > N{\mathrm{H}_3} > S{\mathrm{O}_2} > H\mathrm{F}$ is incorrect because HF is the most polar molecule.

Key Concepts:
- Electronegativity
- Dipole Moment
- Molecular Geometry
- Bond Polarity

Memory Aids:
- Higher electronegativity difference and asymmetrical geometry = Higher polarity

Quick Tips:
- Consider both the electronegativity difference and the molecular geometry when determining the polarity of a molecule.

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Analyzing +4 | -1

Which of the following have identical bond order?

\begin{array}{cccccccccccccccccccc} {1.\,\,\,C{N^ - }}&{2.\,\,\,O_2^ - }&{3.\,\,\,N{O^ + }}&{4.\,\,\,C{N^ + }} \end{array}

View answer and explanation

Correct option: A

1,3

Option A (1,3) is correct because both

C{N^-}

and

N{O^+}

have 14 electrons, resulting in a bond order of 3.

Solution approach

Correct Answer + Core Reason: 1,3. Both $C{N^-}$ and $N{O^+}$ have 14 electrons, leading to the same bond order.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each species:
- $C{N^-}$ : 6 (C) + 7 (N) + 1 (charge) = 14 electrons
- $O_2^-$ : 16 (O2) + 1 (charge) = 17 electrons
- $N{O^+}$ : 7 (N) + 8 (O) - 1 (charge) = 14 electrons
- $C{N^+}$ : 6 (C) + 7 (N) - 1 (charge) = 12 electrons

Step 2: Determine the bond order for each species:
- $C{N^-}$ : Bond order = 3
- $O_2^-$ : Bond order = 1.5
- $N{O^+}$ : Bond order = 3
- $C{N^+}$ : Bond order = 2

Step 3: Identify the species with identical bond orders. $C{N^-}$ and $N{O^+}$ both have a bond order of 3.

Why other options are wrong:
- Option B (1,2): $C{N^-}$ and $O_2^-$ have different bond orders (3 and 1.5, respectively).
- Option C (2,4): $O_2^-$ and $C{N^+}$ have different bond orders (1.5 and 2, respectively).
- Option D (1,2,3): Only $C{N^-}$ and $N{O^+}$ have the same bond order.

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation

Memory Aids:
- Remember the electronic configurations for common diatomic molecules and ions.

Quick Tips:
- Calculate the number of electrons first, then determine the bond order.

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Analyzing +4 | -1

Which of the following order is incorrect?

View answer and explanation

Correct option: A

Ionic character

={MCl} < {MCl}_2 < {MCl}_3

Option A is correct because ionic character increases with increasing charge on the cation and decreasing size of the cation, which is not reflected in the given order.

Solution approach

Correct Answer + Core Reason: Ionic character $={MCl} < {MCl}_2 < {MCl}_3$ . Ionic character increases with increasing charge on the cation.

Step-by-Step Solution:

Step 1: Understand the concept of ionic character and its dependence on charge and size of ions.

Step 2: Recall Fajans' rules, which state that smaller cations with higher charges favor covalent character, while larger cations with smaller charges favor ionic character.

Step 3: Analyze the given order: ${MCl} < {MCl}_2 < {MCl}_3$ . This implies that as the charge on the metal ion increases, the ionic character increases.

Step 4: This is generally true because a higher charge on the cation leads to a stronger electrostatic attraction with the anion, thus increasing the ionic character.

Step 5: However, the increase in charge also increases polarization, which favors covalent character. The question is asking for the incorrect order, and this order is generally correct.

Why other options are wrong:
- Option B: Polarizability increases down the group due to increasing size. ${F}^{-}< {Cl}^{-} < {Br}^{-}< {I}^{-}$ is correct.
- Option C: Polarizing power increases with increasing charge and decreasing size. ${Na}^{+} < {Ca}^{+2} < {Mg}^{+2} < {Al}^{+3}$ is correct because ${Mg}^{+2}$ is smaller than ${Ca}^{+2}$ .
- Option D: Covalent character increases down the group due to increasing polarizability of the anion. ${LiF} < {LiCl} < {LiBr} < LiI$ is correct.

Key Concepts:
- Ionic character
- Fajans' rules
- Polarizability
- Polarizing power

Memory Aids:
- Fajans' Rules: Small, high charge = covalent; Large, low charge = ionic

Quick Tips:
- Remember that ionic character is favored by larger cations with smaller charges.

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Understanding +4 | -1

Van der Waals forces are maximum in:

View answer and explanation

Correct option: D

AgBr

Option D (AgBr) is correct because Van der Waals forces increase with increasing molecular size and polarizability, and AgBr has the largest and most polarizable atoms among the given options.

Solution approach

Correct Answer + Core Reason: AgBr. Larger size and higher polarizability lead to stronger Van der Waals forces.

Step-by-Step Solution:

Step 1: Understand Van der Waals Forces: Van der Waals forces are intermolecular forces that arise from temporary fluctuations in electron distribution.

Step 2: Identify Factors Affecting Van der Waals Forces: Van der Waals forces increase with:
- Increasing molecular size (number of electrons)
- Increasing polarizability (ease of distortion of the electron cloud)

Step 3: Analyze the Compounds:
- HBr: Small molecule, relatively low polarizability.
- LiBr: Smaller ions compared to AgBr.
- LiCl: Smallest ions among the options.
- AgBr: Silver and bromine are larger and more polarizable than the other elements.

Step 4: Conclusion: AgBr has the maximum Van der Waals forces due to the larger size and higher polarizability of its atoms.

Why other options are wrong:
- Option A: HBr: Smaller size and lower polarizability compared to AgBr.
- Option B: LiBr: Smaller ions compared to AgBr.
- Option C: LiCl: Smallest ions among the options, so weakest Van der Waals forces.

Key Concepts:
- Van der Waals Forces
- Molecular Size
- Polarizability

Memory Aids:
- Larger size = stronger Van der Waals forces

Quick Tips:
- Remember that Van der Waals forces are generally stronger for larger and more polarizable molecules.

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Analyzing +4 | -1

Among the following the pair in which the two species are NOT iso structural is

View answer and explanation

Correct option: A

Si{F_4}\mathrm{ and}\,\,S{F_4}

Option A (

Si{F_4}\mathrm{ and}\,S{F_4}

) is correct because

SiF_4

is tetrahedral and

SF_4

is see-saw shaped, hence they are not isostructural.

Solution approach

Correct Answer + Core Reason: $Si{F_4}\mathrm{ and}\,S{F_4}$ . $SiF_4$ is tetrahedral, $SF_4$ is see-saw.

Step-by-Step Solution:

Step 1: Determine the structure of each species.

Step 2: $SiF_4$ : Silicon has 4 bond pairs and 0 lone pairs. It is tetrahedral.

Step 3: $SF_4$ : Sulfur has 4 bond pairs and 1 lone pair. It is see-saw shaped.

Step 4: $IO_3^-$ : Iodine has 3 bond pairs and 1 lone pair. It is pyramidal.
$XeO_3$ : Xenon has 3 bond pairs and 1 lone pair. It is pyramidal.

Step 5: $BH_4^-$ : Boron has 4 bond pairs and 0 lone pairs. It is tetrahedral.
$NH_4^+$ : Nitrogen has 4 bond pairs and 0 lone pairs. It is tetrahedral.

Step 6: $PF_6^-$ : Phosphorus has 6 bond pairs and 0 lone pairs. It is octahedral.
$SF_6$ : Sulfur has 6 bond pairs and 0 lone pairs. It is octahedral.

Step 7: Identify the pair that is not isostructural.
$SiF_4$ and $SF_4$ are not isostructural.

Why other options are wrong:
- Option B: $IO_3^-$ and $XeO_3$ are both pyramidal and isostructural.
- Option C: $BH_4^-$ and $NH_4^+$ are both tetrahedral and isostructural.
- Option D: $PF_6^-$ and $SF_6$ are both octahedral and isostructural.

Key Concepts:
- Molecular structure
- VSEPR theory
- Isostructural species

Memory Aids:
- Lone pairs affect molecular shape.

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

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Analyzing +4 | -1

Which of the following represents the incorrect decreasing order of bond angles?

View answer and explanation

Correct option: D

\mathrm{IO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{ClO}_3^{-}

Option D (

\mathrm{IO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{ClO}_3^{-}

) is correct because the bond angle decreases down the group due to increasing size and decreasing electronegativity of the central atom, making the given order incorrect.

Solution approach

Correct Answer + Core Reason: $\mathrm{IO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{ClO}_3^{-}$ . Bond angle decreases down the group due to decreasing electronegativity of the central atom.

Step-by-Step Solution:

Step 1: Analyze Option A: $\mathrm{CO}_2 > \mathrm{BF}_3 > \mathrm{CH}_4 > \mathrm{H}_2 \mathrm{O}$ . Bond angles: 180° > 120° > 109.5° > 104.5°. This order is correct.

Step 2: Analyze Option B: $\mathrm{BCl}_3 > \mathrm{PCl}_3 > \mathrm{AsCl}_3 > \mathrm{BiCl}_3$ . Bond angles decrease down the group due to increasing lone pair repulsion. This order is correct.

Step 3: Analyze Option C: $\stackrel{+}{\mathrm{NO}_2} > \mathrm{NO}_2 > \mathrm{NO}_2^{-}$ . Bond angles: > 120° > 115°. This order is correct.

Step 4: Analyze Option D: $\mathrm{IO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{ClO}_3^{-}$ . Due to the increasing size and decreasing electronegativity of the central atom down the group, the bond angle decreases. Therefore, the correct order should be $\mathrm{ClO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{IO}_3^{-}$ . This order is incorrect.

Step 5: Final Answer: $\mathrm{IO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{ClO}_3^{-}$

Why other options are wrong:
- Option A: The decreasing order of bond angles is correct.
- Option B: The decreasing order of bond angles is correct.
- Option C: The decreasing order of bond angles is correct.

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

Memory Aids:
- Bond angle decreases with increasing lone pair repulsion.
- Electronegativity decreases down the group.

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

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Understanding +4 | -1

The number of types of bonds between two carbon atoms in calcium carbide is

View answer and explanation

Correct option: A

One sigma, two pi

Option A is correct because in calcium carbide (CaC₂), each carbon atom is bonded to one calcium atom (ionic) and the two carbon atoms are triple bonded (one sigma and two pi bonds).

Solution approach

Correct Answer + Core Reason: One sigma, two pi. In $CaC_2$ , the $C_2^{2-}$ ion has a triple bond between the carbon atoms (one sigma and two pi).

Step-by-Step Solution:

Step 1: Understand the structure of calcium carbide ( $CaC_2$ ).

Step 2: Recall that calcium carbide consists of $Ca^{2+}$ ions and $C_2^{2-}$ ions.

Step 3: The $C_2^{2-}$ ion is similar to $N_2$ and has a triple bond between the two carbon atoms.

Step 4: A triple bond consists of one sigma bond and two pi bonds.

Step 5: Therefore, the number of types of bonds between two carbon atoms in calcium carbide is one sigma and two pi.

Why other options are wrong:
- Option B: One sigma, one pi is incorrect. It's a double bond.
- Option C: Two sigma, one pi is incorrect.
- Option D: Two sigma, two pi is incorrect.

Key Concepts:
- Calcium carbide structure
- Sigma bond
- Pi bond
- Triple bond

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

Quick Tips:
- Remember the structure of $CaC_2$ and the nature of the bond between the carbon atoms.

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Understanding +4 | -1

Species having two unpaired electrons is

View answer and explanation

Correct option: B

{B_2}

Option B (

{B_2}

) is correct because

{B_2}

has two unpaired electrons according to molecular orbital theory.

Solution approach

Correct Answer + Core Reason: ${B_2}$ . B2 has two unpaired electrons according to MO theory.

Step-by-Step Solution:

Step 1: Write the electronic configuration of ${B_2}$ .

Step 2: Each B atom has 5 electrons, so ${B_2}$ has 10 electrons.

Step 3: The molecular orbital configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^2$ .

Step 4: According to Hund's rule, the two electrons in the $\pi_{2p}$ orbitals will occupy separate orbitals with parallel spins, resulting in two unpaired electrons.

Why other options are wrong:
- **Option A ( $He_2^ +$ ):** $He_2^+$ has one unpaired electron.
- **Option C ( $O_2^ -$ ):** $O_2^-$ has one unpaired electron.
- **Option D ( ${N_2}$ ):** ${N_2}$ has no unpaired electrons.

Key Concepts:
- Molecular Orbital Theory
- Hund's Rule
- Electronic Configuration
- Unpaired Electrons

Memory Aids:
- Hund's Rule: Electrons fill orbitals individually before pairing up.

Quick Tips:
- Remember Hund's rule when filling degenerate orbitals.

197 Medium

197 of 262Question permalink

Analyzing +4 | -1

For two ionic solids

CaO

and

KI

, identify the wrong statement among the following?

View answer and explanation

Correct option: D

KI

has high M.P

Option D (

KI

has high M.P) is wrong because

KI

has a lower melting point compared to

CaO

due to weaker electrostatic forces between the ions.

Solution approach

Correct Answer + Core Reason: $KI$ has high M.P. $CaO$ has a much higher melting point than $KI$ due to higher charges on the ions, leading to greater lattice energy. Therefore, the statement that KI has a high M.P is incorrect compared to CaO.

Step-by-Step Solution:

Step 1: Identify the ionic compounds: $CaO$ and $KI$ .

Step 2: Consider the charges of the ions: $Ca^{2+}$ and $O^{2-}$ in $CaO$ , $K^+$ and $I^-$ in $KI$ .

Step 3: Recall the factors affecting lattice energy: Charge and size of ions. Lattice energy is proportional to the product of the charges and inversely proportional to the distance between the ions.

Step 4: Compare the lattice energies: $CaO$ has higher charges (+2 and -2) compared to $KI$ (+1 and -1). Therefore, $CaO$ has a much higher lattice energy.

Step 5: Relate lattice energy to melting point: Higher lattice energy leads to higher melting point.

Step 6: KI is soluble in water, not benzene.

Step 7: Conclusion: $KI$ does not have a high M.P compared to $CaO$ .

Why other options are wrong:
- Option A: Lattice energy of $CaO$ is much higher that that of $KI$ : Correct statement.
- Option B: $KI$ is soluble in benzene: Incorrect, KI is soluble in water.
- Option C: $CaO$ has high M.P: Correct statement.

Key Concepts:
- Lattice energy
- Melting point of ionic compounds
- Charge of ions
- Solubility

Memory Aids:
- Higher charge, higher lattice energy, higher melting point.

Quick Tips:
- Compare the charges of the ions first when comparing lattice energies.

198 Medium

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Application +4 | -1

If the bond length and dipole moment of a diatomic molecule are 1.25 A and 1.0 D respectively, what is the per cent ionic character of the bond?

View answer and explanation

Correct option: C

16.66

Option C (16.66) is correct because it is the calculated percentage ionic character using the given bond length and dipole moment.

Solution approach

Correct Answer + Core Reason: 16.66. The percentage ionic character is calculated using the formula relating dipole moment, bond length, and electronic charge.

Step-by-Step Solution:

Step 1: Convert bond length from Angstroms to meters: $d = 1.25 \text{ A} = 1.25 \times 10^{-10} \text{ m}$

Step 2: Convert dipole moment from Debye to Coulomb-meters: $\mu = 1.0 \text{ D} = 1.0 \times 3.336 \times 10^{-30} \text{ C m}$

Step 3: Calculate the theoretical dipole moment assuming 100% ionic character: $\mu_{\text{theoretical}} = q \times d = (1.602 \times 10^{-19} \text{ C}) \times (1.25 \times 10^{-10} \text{ m}) = 2.0025 \times 10^{-29} \text{ C m}$

Step 4: Calculate the percentage ionic character: $\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \times 100 = \frac{3.336 \times 10^{-30} \text{ C m}}{2.0025 \times 10^{-29} \text{ C m}} \times 100 = 16.66 \%$

Why other options are wrong:
- Option A (10.66): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option B (12.33): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option D (19.33): This is incorrect due to a calculation error or using incorrect conversion factors.

Key Concepts:
- Dipole moment
- Percentage ionic character
- Conversion of units (Angstroms to meters, Debye to Coulomb-meters)

Memory Aids:
- Remember the formula: $\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \times 100$

Quick Tips:
- Always convert units to SI units before calculations.
- Remember the conversion factor for Debye to Coulomb-meters: 1 D = 3.336 × 10^-30 C m.

199 Medium

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Understanding +4 | -1

Greater the dipole moment:

View answer and explanation

Correct option: A

Greater is the ionic nature

Option A (Greater is the ionic nature) is correct because a larger dipole moment indicates a greater separation of charge, hence a greater ionic character.

Solution approach

Correct Answer + Core Reason: Greater is the ionic nature. A larger dipole moment signifies a greater degree of charge separation and thus, a higher ionic character.

Step-by-Step Solution:

Step 1: Define dipole moment. Dipole moment is a measure of the polarity of a chemical bond or molecule. It arises from differences in electronegativity between bonded atoms.

Step 2: Relate dipole moment to ionic character. A larger dipole moment indicates a greater separation of charge, which is characteristic of ionic compounds.

Step 3: Conclude. Therefore, a greater dipole moment implies a greater ionic nature.

Why other options are wrong:
- Option B: Lesser the polarity: This is incorrect because a greater dipole moment implies greater polarity, not lesser.
- Option C: Smaller the ionic nature: This is incorrect because a greater dipole moment implies greater ionic nature, not smaller.
- Option D: None of these: This is incorrect because option A is correct.

Key Concepts:
- Dipole moment
- Electronegativity
- Ionic character
- Polarity

Memory Aids:
- Dipole moment = charge x distance. Higher charge separation = higher dipole moment = higher ionic character.

Quick Tips:
- Remember that dipole moment is directly proportional to the magnitude of the partial charges and the distance between them.

200 Medium

200 of 262Question permalink

Analyzing +4 | -1

Which one of the following molecules has the smallest bond angle?

View answer and explanation

Correct option: D

{\mathrm{H}_2}\mathrm{Se}

Option D is correct because as you go down the group, bond angle decreases due to increased size and decreased electronegativity of the central atom.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{Se}$ . Bond angle decreases down the group due to decreasing bond pair-bond pair repulsion.

Step-by-Step Solution:

Step 1: Analyze $\mathrm{N}{\mathrm{H}_3}$ : N has 5 valence electrons. In $\mathrm{N}{\mathrm{H}_3}$ , there are three bond pairs and one lone pair around N. The hybridization is $sp^3$ , and the shape is pyramidal. The bond angle is approximately 107°.

Step 2: Analyze $\mathrm{P}{\mathrm{H}_3}$ : P has 5 valence electrons. In $\mathrm{P}{\mathrm{H}_3}$ , there are three bond pairs and one lone pair around P. The hybridization is approximately $sp^3$ , but the bond angle is smaller than in $\mathrm{N}{\mathrm{H}_3}$ due to the larger size of P and decreased electronegativity. The bond angle is approximately 93.5°.

Step 3: Analyze ${\mathrm{H}_2}\mathrm{O}$ : O has 6 valence electrons. In ${\mathrm{H}_2}\mathrm{O}$ , there are two bond pairs and two lone pairs around O. The hybridization is $sp^3$ , and the shape is bent. The bond angle is approximately 104.5°.

Step 4: Analyze ${\mathrm{H}_2}\mathrm{Se}$ : Se has 6 valence electrons. In ${\mathrm{H}_2}\mathrm{Se}$ , there are two bond pairs and two lone pairs around Se. The hybridization is approximately $sp^3$ , but the bond angle is smaller than in ${\mathrm{H}_2}\mathrm{O}$ due to the larger size of Se and decreased electronegativity. The bond angle is approximately 91°.

Step 5: Compare the bond angles: ${\mathrm{H}_2}\mathrm{Se}$ has the smallest bond angle.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{H}_3}$ has a larger bond angle than ${\mathrm{H}_2}\mathrm{Se}$ .
- Option B: $\mathrm{P}{\mathrm{H}_3}$ has a larger bond angle than ${\mathrm{H}_2}\mathrm{Se}$ .
- Option C: ${\mathrm{H}_2}\mathrm{O}$ has a larger bond angle than ${\mathrm{H}_2}\mathrm{Se}$ .

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

Memory Aids:
- Down a group, bond angles generally decrease.

Quick Tips:
- Consider the effect of lone pairs and electronegativity on bond angles.

201 Medium

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Analyzing +4 | -1

If

\mathrm{N}{\mathrm{a}^ + }

ion is larger than

\mathrm{M}{\mathrm{g}^{2 + }}

ion and

{\mathrm{S}^{2 - }}

is larger than

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

ion, which of the following will be stable soluble in water?

View answer and explanation

Correct option: D

Magnesium sulphide

Option D (Magnesium sulphide) is correct because larger cation and larger anion leads to lower lattice energy and hence more soluble.

Solution approach

Correct Answer + Core Reason: Magnesium sulphide. Larger cation and anion size leads to lower lattice energy and higher solubility.

Step-by-Step Solution:

Step 1: Understand the question. We need to determine which compound will be most stable and soluble in water, given the relative sizes of the ions.

Step 2: Consider the factors affecting solubility. Lattice energy is a major factor. Lower lattice energy generally leads to higher solubility.

Step 3: Recall that lattice energy is inversely proportional to the size of the ions. Larger ions result in lower lattice energy.

Step 4: Analyze the given information. Na+ < Mg2+ and Cl- < S2-. Therefore, MgS has the largest cation (Mg2+) and largest anion (S2-) among the options.

Step 5: Conclude that MgS will have the lowest lattice energy and be the most soluble.

Why other options are wrong:
- Option A: Sodium chloride has smaller ions compared to magnesium sulfide, leading to higher lattice energy and lower solubility.
- Option B: Sodium sulphide has a smaller cation compared to magnesium sulfide, leading to higher lattice energy and lower solubility.
- Option C: Magnesium chloride has a smaller anion compared to magnesium sulfide, leading to higher lattice energy and lower solubility.

Key Concepts:
- Solubility of ionic compounds
- Lattice energy
- Ionic size and its effect on lattice energy

Memory Aids:
- "Like dissolves Like" - compounds with lower lattice energy are more soluble in polar solvents like water.

Quick Tips:
- Relate ionic size to lattice energy and then to solubility.

202 Medium

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Remembering +4 | -1

What is

{X}, {Y}

and

{Z}

in the following expression of formal charge.
Formal charge (F.C) on an atom in a Lewis structure

=X-Y-\frac{1}{2}(Z)

View answer and explanation

Correct option: C

{X}=

Total number of valence electrons in the free atom

{Y}=

Total number of non bonding electrons

{Z}=

Total number of bonding electrons

Option C (

{X}=

Total number of valence electrons in the free atom,

{Y}=

Total number of non bonding electrons,

{Z}=

Total number of bonding electrons) is correct because it correctly represents the formula for calculating formal charge.

Solution approach

Correct Answer + Core Reason: ${X}=$ Total number of valence electrons in the free atom, ${Y}=$ Total number of non bonding electrons, ${Z}=$ Total number of bonding electrons. This is the correct formula for formal charge.

Step-by-Step Solution:

Step 1: Recall the Formal Charge Formula: The formal charge (FC) on an atom in a Lewis structure is calculated using the formula: FC = (Valence electrons) - (Non-bonding electrons) - (1/2 * Bonding electrons).

Step 2: Identify the Variables: In the given expression, $X$ , $Y$ , and $Z$ represent different components of the formula.

Step 3: Match Variables to Formula Components:
- $X$ corresponds to the total number of valence electrons in the free atom.
- $Y$ corresponds to the total number of non-bonding electrons.
- $Z$ corresponds to the total number of bonding electrons.

Step 4: Conclusion: Therefore, the correct assignment is: $X$ = Valence electrons, $Y$ = Non-bonding electrons, $Z$ = Bonding electrons.

Why other options are wrong:
- Option A: Incorrect assignment of variables to the components of the formal charge formula.
- Option B: Incorrect assignment of variables to the components of the formal charge formula.
- Option D: Incorrect assignment of variables to the components of the formal charge formula.

Key Concepts:
- Formal Charge
- Lewis Structures
- Valence Electrons
- Bonding Electrons
- Non-bonding Electrons

Memory Aids:
- Formal Charge = Valence - Non-bonding - 1/2 Bonding

Quick Tips:
- Always remember the correct formula for formal charge to avoid errors.

203 Medium

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Analyzing +4 | -1

The bond order in

\mathrm{O}_2^ +

is equal to bond order in:

View answer and explanation

Correct option: A

\mathrm{N}_2^ +

Option A (

\mathrm{N}_2^ +

) is correct because both

\mathrm{O}_2^ +

and

\mathrm{N}_2^ +

have the same number of electrons and therefore the same bond order.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}_2^ +$ . Both species have the same number of electrons and thus the same bond order.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $\mathrm{O}_2^ +$ . Oxygen has 8 electrons, so $\mathrm{O}_2$ has 16. The +1 charge means one electron is removed, so $\mathrm{O}_2^ +$ has 15 electrons.

Step 2: Calculate the bond order of $\mathrm{O}_2^ +$ using molecular orbital theory. The electronic configuration is $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^{\,*})^1$ . Bond order = (8 - 3)/2 = 2.5

Step 3: Determine the number of electrons in each option:
- $\mathrm{N}_2^ +$ : Nitrogen has 7 electrons, so $\mathrm{N}_2$ has 14. The +1 charge means one electron is removed, so $\mathrm{N}_2^ +$ has 13 electrons. Bond order = (7-2)/2 = 2.5
- $\mathrm{C}{\mathrm{N}^ - }$ : Carbon has 6 electrons, Nitrogen has 7, and the -1 charge adds one electron, so $\mathrm{C}{\mathrm{N}^ - }$ has 14 electrons. Bond order = (9-5)/2 = 3
- $\mathrm{CO}$ : Carbon has 6 electrons, Oxygen has 8, so $\mathrm{CO}$ has 14 electrons. Bond order = (9-5)/2 = 3
- $\mathrm{N}{\mathrm{O}^ + }$ : Nitrogen has 7 electrons, Oxygen has 8, and the +1 charge removes one electron, so $\mathrm{N}{\mathrm{O}^ + }$ has 14 electrons. Bond order = (9-5)/2 = 3

Step 4: Compare the bond orders. Only $\mathrm{N}_2^ +$ has the same bond order (2.5) as $\mathrm{O}_2^ +$ .

Why other options are wrong:
- **Option B ( $\mathrm{C}{\mathrm{N}^ - }$ ):** $\mathrm{C}{\mathrm{N}^ - }$ has 14 electrons, resulting in a bond order of 3, which is different from $\mathrm{O}_2^ +$ .
- **Option C ( $\mathrm{CO}$ ):** $\mathrm{CO}$ has 14 electrons, resulting in a bond order of 3, which is different from $\mathrm{O}_2^ +$ .
- **Option D ( $\mathrm{N}{\mathrm{O}^ + }$ ):** $\mathrm{N}{\mathrm{O}^ + }$ has 14 electrons, resulting in a bond order of 3, which is different from $\mathrm{O}_2^ +$ .

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation
- Number of Electrons in Ions

Memory Aids:
- Remember the electron configurations for common diatomic molecules and ions.

Quick Tips:
- Calculate the number of electrons first, then determine the electronic configuration and bond order.

204 Medium

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Analyzing +4 | -1

Which of the following has lowest bond angle?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the presence of two lone pairs on the oxygen atom causes greater repulsion, leading to a smaller bond angle compared to the other options.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{O}$ . Due to the presence of two lone pairs on the central oxygen atom, the bond angle is significantly reduced due to lone pair-lone pair repulsion.

Step-by-Step Solution:

Step 1: Determine the hybridization and geometry of each molecule:
- $\mathrm{Be}{\mathrm{F}_2}$ : Be has 2 valence electrons. The molecule is linear with a bond angle of 180°.
- ${\mathrm{H}_2}\mathrm{O}$ : O has 6 valence electrons. The molecule has a bent shape with two bond pairs and two lone pairs. The bond angle is approximately 104.5°.
- $\mathrm{N}{\mathrm{H}_3}$ : N has 5 valence electrons. The molecule has a trigonal pyramidal shape with three bond pairs and one lone pair. The bond angle is approximately 107°.
- $\mathrm{C}{\mathrm{H}_4}$ : C has 4 valence electrons. The molecule has a tetrahedral shape with four bond pairs and no lone pairs. The bond angle is 109.5°.

Step 2: Compare the bond angles. ${\mathrm{H}_2}\mathrm{O}$ has the smallest bond angle due to the greater repulsion from two lone pairs.

Why other options are wrong:
- **Option A ( $\mathrm{Be}{\mathrm{F}_2}$ ):** $\mathrm{Be}{\mathrm{F}_2}$ is a linear molecule with a bond angle of 180°, which is the largest among the given options.
- **Option C ( $\mathrm{N}{\mathrm{H}_3}$ ):** $\mathrm{N}{\mathrm{H}_3}$ has one lone pair, resulting in a bond angle of approximately 107°, which is larger than that of ${\mathrm{H}_2}\mathrm{O}$ .
- **Option D ( $\mathrm{C}{\mathrm{H}_4}$ ):** $\mathrm{C}{\mathrm{H}_4}$ has no lone pairs, resulting in a bond angle of 109.5°, which is larger than that of ${\mathrm{H}_2}\mathrm{O}$ .

Key Concepts:
- VSEPR Theory (Valence Shell Electron Pair Repulsion)
- Hybridization
- Lone Pair Repulsion
- Bond Angle

Memory Aids:
- Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair repulsion.

Quick Tips:
- Determine the number of lone pairs and bond pairs around the central atom to predict the bond angle.

205 Medium

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Remembering +4 | -1

Ionic compounds in the crystalline state consist of orderly three-dimensional arrangements of cations and anions held together by

View answer and explanation

Correct option: A

Coulombic interaction energies

Option A (Coulombic interaction energies) is correct because ionic compounds are held together by the electrostatic attraction between oppositely charged ions, which is described by Coulomb's law.

Solution approach

Correct Answer + Core Reason: Coulombic interaction energies. Ionic compounds are held together by electrostatic forces.

Step-by-Step Solution:

Step 1: Ionic compounds are formed by the transfer of electrons from one atom to another, resulting in the formation of ions (cations and anions).

Step 2: These ions are arranged in a crystal lattice structure, where each ion is surrounded by ions of opposite charge.

Step 3: The electrostatic attraction between the oppositely charged ions is the primary force holding the crystal lattice together. This attraction is described by Coulomb's law: $F = k \frac{{q_1 q_2}}{{r^2}}$ , where F is the force, k is Coulomb's constant, q1 and q2 are the charges of the ions, and r is the distance between them.

Step 4: The energy associated with this electrostatic interaction is the Coulombic interaction energy.

Why other options are wrong:
- Option B: Electron gain energies refer to the energy released when an atom gains an electron, which is related to the formation of anions, but not the force holding the crystal together.
- Option C: Ionization energies refer to the energy required to remove an electron from an atom, which is related to the formation of cations, but not the force holding the crystal together.
- Option D: None of the above is incorrect because Coulombic interaction energies are the primary force.

Key Concepts:
- Ionic bonding
- Coulomb's law
- Electrostatic forces

Memory Aids:
- Opposites attract.

Quick Tips:
- Remember that ionic compounds are held together by electrostatic forces between ions.

206 Medium

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Remembering +4 | -1

The bond angle between

\mathrm{H} - \mathrm{O} - \mathrm{H}

in ice is closest to:

View answer and explanation

Correct option: B

109\;^\circ 28\prime

Option B is correct because the bond angle in ice is close to the tetrahedral angle due to hydrogen bonding.

Solution approach

Correct Answer + Core Reason: $109\;^\circ 28\prime$ . The bond angle in ice is close to the tetrahedral angle due to the tetrahedral arrangement of water molecules in the ice lattice.

Step-by-Step Solution:

Step 1: Understand the structure of ice: In ice, each oxygen atom is tetrahedrally surrounded by four hydrogen atoms, two through covalent bonds and two through hydrogen bonds.

Step 2: Recall the ideal tetrahedral angle: The ideal tetrahedral angle is $109.5^\circ$ or $109^\circ 28\prime$ .

Step 3: Consider the effect of hydrogen bonding: Hydrogen bonding in ice causes a slight distortion from the ideal tetrahedral angle, but the angle remains close to $109.5^\circ$ .

Step 4: Conclude: The bond angle between $\mathrm{H} - \mathrm{O} - \mathrm{H}$ in ice is closest to $109\;^\circ 28\prime$ .

Why other options are wrong:
- Option A: $115\;^\circ$ is incorrect. This value is too high for the bond angle in ice.
- Option C: $110\;^\circ$ is incorrect. This value is slightly higher than the actual bond angle.
- Option D: $90\;^\circ$ is incorrect. This value is significantly lower than the actual bond angle.

Key Concepts:
- Hydrogen bonding
- Tetrahedral geometry
- Bond angle

Memory Aids:
- Tetrahedral angle: $109.5^\circ$

Quick Tips:
- Remember the structure of ice and the role of hydrogen bonding.
- Recall the ideal tetrahedral angle.

207 Medium

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Understanding +4 | -1

The correct order of decreasing polarisability of ion is:

View answer and explanation

Correct option: C

{\mathrm{I}^ - },\mathrm{B}{\mathrm{r}^ - },\mathrm{C}{\mathrm{l}^ - },{\mathrm{F}^ - }

Option C (

{\mathrm{I}^ - },\mathrm{B}{\mathrm{r}^ - },\mathrm{C}{\mathrm{l}^ - },{\mathrm{F}^ - }

) is correct because polarizability increases with size. Larger ions are more easily polarized.

Solution approach

Correct Answer + Core Reason: ${\mathrm{I}^ - },\mathrm{B}{\mathrm{r}^ - },\mathrm{C}{\mathrm{l}^ - },{\mathrm{F}^ - }$ . Polarisability increases with size.

Step-by-Step Solution:

Step 1: Recall the definition of polarizability. It is the ability of an ion to be distorted by an electric field.

Step 2: Larger ions are more easily polarized because their valence electrons are further from the nucleus and are less tightly held.

Step 3: The ionic radii increase down a group. Therefore, the order of increasing size is: F- < Cl- < Br- < I-

Step 4: The order of decreasing polarizability is: I- > Br- > Cl- > F-

Why other options are wrong:
- Option A: Incorrect order of polarizability.
- Option B: Incorrect order of polarizability.
- Option D: Incorrect order of polarizability.

Key Concepts:
- Polarisability
- Ionic size

Memory Aids:
- Remember that larger ions are more polarizable.

Quick Tips:
- Consider the size of the ion when comparing polarizabilities.

208 Medium

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Application +4 | -1

In a polar molecule, the ionic charge is

4.8 \times {10^{ - 10}}

e.s.u. If the inter ionic distance is

1{A^0}

unit, then the dipole moment is

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {4.8\mathrm{ }} \end{array}

debye

Option C (4.8) is correct because the dipole moment is calculated directly from the given charge and distance.

Solution approach

Correct Answer + Core Reason: $\begin{array}{llllllllllllllllllll} {4.8\mathrm{ }} \end{array}$ debye. The dipole moment is calculated by multiplying the ionic charge by the interionic distance.

Step-by-Step Solution:

Step 1: Given ionic charge: $q = 4.8 \times 10^{-10} \text{ esu}$

Step 2: Given interionic distance: $d = 1 \text{ A}^0 = 10^{-8} \text{ cm}$

Step 3: Calculate the dipole moment in esu cm: $\mu = q \times d = (4.8 \times 10^{-10} \text{ esu}) \times (10^{-8} \text{ cm}) = 4.8 \times 10^{-18} \text{ esu cm}$

Step 4: Convert dipole moment from esu cm to Debye: $\mu = 4.8 \times 10^{-18} \text{ esu cm} = 4.8 \text{ D}$

Why other options are wrong:
- Option A: $\begin{array}{llllllllllllllllllll} {41.8\mathrm{ }} \end{array}$ debye: This is incorrect due to a calculation error.
- Option B: $\begin{array}{llllllllllllllllllll} {4018\mathrm{ }} \end{array}$ debye: This is incorrect due to a calculation error.
- Option D: $\begin{array}{llllllllllllllllllll} {0.48\mathrm{ }} \end{array}$ debye: This is incorrect due to a calculation error.

Key Concepts:
- Dipole moment
- Conversion of units (esu cm to Debye)

Memory Aids:
- Remember the formula: $\mu = q \times d$
- 1 Debye = $10^{-18}$ esu cm

Quick Tips:
- Ensure consistent units before calculations.
- Pay attention to the units given in the problem.

209 Medium

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Remembering +4 | -1

The element which exists in both hard and soft form is:

View answer and explanation

Correct option: C

\mathrm{C}

Option C (

\mathrm{C}

) is correct because carbon exists in both hard (diamond) and soft (graphite) allotropic forms.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}$ . Carbon exists as both diamond (hard) and graphite (soft).

Step-by-Step Solution:

Step 1: Recall the allotropes of the given elements.

Step 2: Carbon exists as diamond (hard) and graphite (soft).

Step 3: Iron, silicon, and aluminum do not have common allotropes with such drastically different hardness.

Step 4: Final Answer: $\mathrm{C}$

Why other options are wrong:
- Option A: Iron primarily exists in metallic forms.
- Option B: Silicon is a semiconductor and doesn't have drastically different forms.
- Option D: Aluminum is a metal and doesn't have drastically different forms.

Key Concepts:
- Allotropes
- Hardness
- Carbon allotropes

Memory Aids:
- Carbon: Diamond (hard), Graphite (soft)

Quick Tips:
- Remember the common allotropes of elements.

210 Medium

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Understanding +4 | -1

Which of the following is not tetrahedral?

View answer and explanation

Correct option: C

\mathrm{CO}_3^{2 - }

Option C (

\mathrm{CO}_3^{2 - }

) is correct because it has a trigonal planar geometry, not tetrahedral.

Solution approach

Correct Answer + Core Reason: $\mathrm{CO}_3^{2 - }$ . Carbonate ion has a trigonal planar geometry, not tetrahedral.

Step-by-Step Solution:

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

Step 2: $\mathrm{BF}_4^ -$ : Boron has 4 bond pairs and no lone pairs, so it is $sp^3$ hybridized and tetrahedral.

Step 3: $\mathrm{NH}_4^ +$ : Nitrogen has 4 bond pairs and no lone pairs, so it is $sp^3$ hybridized and tetrahedral.

Step 4: $\mathrm{CO}_3^{2 - }$ : Carbon has 3 bond pairs and no lone pairs, so it is $sp^2$ hybridized and trigonal planar.

Step 5: $\mathrm{SO}_4^{2 - }$ : Sulfur has 4 bond pairs and no lone pairs, so it is $sp^3$ hybridized and tetrahedral.

Step 6: Identify the molecule/ion that is not tetrahedral.

Step 7: $\mathrm{CO}_3^{2 - }$ is trigonal planar, not tetrahedral.

Why other options are wrong:
- Option A: $\mathrm{BF}_4^ -$ is tetrahedral.
- Option B: $\mathrm{NH}_4^ +$ is tetrahedral.
- Option D: $\mathrm{SO}_4^{2 - }$ is tetrahedral.

Key Concepts:
- Hybridization
- VSEPR theory
- Tetrahedral geometry
- Trigonal planar geometry

Memory Aids:
- $sp^3$ hybridization leads to tetrahedral geometry.
- $sp^2$ hybridization leads to trigonal planar geometry.

Quick Tips:
- Calculate the number of bond pairs and lone pairs to determine the hybridization.

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