The common feature among the species ${\rm{C}}{{\rm{N}}^ - },{\rm{CO}}$ and ${\rm{N}}{{\rm{O}}^ + }$ are:

Bond order three and weak field ligands

Bond order two and π-acceptors

Isoelectronic and weak field ligands

In the series ethane, ethylene and acetylene, the ${\rm{C}} - {\rm{H}}$ bond energy is:

The same in all the three compounds

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

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Understanding +4 | -1

Hybridisation of oxygen in diethyl ether is

View answer and explanation

Correct option: C

s{p^3}

Option C (

s{p^3}

) is correct because oxygen in diethyl ether has two bond pairs and two lone pairs, resulting in

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^3}$ . Oxygen in diethyl ether is $s{p^3}$ hybridized.

Step-by-Step Solution:

Step 1: Draw the structure of diethyl ether: $CH_3CH_2-O-CH_2CH_3$ .

Step 2: Identify the central atom: Oxygen (O).

Step 3: Count the number of bonding pairs around the oxygen atom: 2 (two C-O bonds).

Step 4: Count the number of lone pairs around the oxygen atom: 2.

Step 5: Determine the steric number: 4 (2 bonding pairs + 2 lone pairs).

Step 6: Determine the hybridization: $sp^3$ .

Why other options are wrong:
- **Option A ( $Sp$ ):** This is incorrect notation, should be $sp$ .
- **Option B ( $s{p^2}$ ):** $sp^2$ hybridization requires three sigma bonds and no lone pairs, or two sigma bonds and one lone pair.
- **Option D ( $s{p^3}d$ ):** $sp^3d$ hybridization requires five sigma bonds and no lone pairs, or four sigma bonds and one lone pair, etc.

Key Concepts:
- Hybridization
- VSEPR theory
- Lone pairs

Memory Aids:
- Steric number = number of sigma bonds + number of lone pairs

Quick Tips:
- Draw the structure to visualize the bonding and lone pairs.

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Application +4 | -1

What bond order does

\mathrm{O}_2^{2 - }

have?

View answer and explanation

Correct option: A

1

Option A (1) is correct because

\mathrm{O}_2^{2 - }

has a bond order of 1, calculated as (8 bonding - 6 antibonding)/2.

Solution approach

Correct Answer + Core Reason: 1. The bond order of $\mathrm{O}_2^{2 - }$ is 1.

Step-by-Step Solution:

Step 1: Write the electronic configuration of $\mathrm{O}_2^{2 - }$ .
$\mathrm{O}_2^{2 - }$ has 18 electrons. Its electronic configuration is $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^{\ast})^4$ .

Step 2: Calculate the bond order using the formula:\nBond order = $\frac{{\text{{Number of bonding electrons - Number of antibonding electrons}}}}{2}$

Step 3: Substitute the values:\nBond order = $\frac{{10 - 8}}{2} = \frac{2}{2} = 1$

Why other options are wrong:
- Option B: 2 is incorrect. The bond order calculation does not result in 2.
- Option C: 3 is incorrect. The bond order calculation does not result in 3.
- Option D: $1/3$ is incorrect. The bond order calculation does not result in $1/3$ .

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation

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

Quick Tips:
- Write the electronic configuration correctly to identify bonding and antibonding electrons.

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Application +4 | -1

The hybridization of

P

in

\mathrm{PO}_4^{3 - }

is same as in:

View answer and explanation

Correct option: D

\mathrm{S in SO}_4^{2 - }

Option D (

\mathrm{S in SO}_4^{2 - }

) is correct because both

PO_4^{3-}

and

SO_4^{2-}

have tetrahedral geometry with

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{S in SO}_4^{2 - }$ because both $PO_4^{3-}$ and $SO_4^{2-}$ exhibit $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the hybridization of P in $PO_4^{3-}$ . Phosphorus has 5 valence electrons + 3 from the negative charge = 8 electrons. It forms 4 sigma bonds with Oxygen and has 0 lone pairs. Therefore, the steric number is 4, and the hybridization is $sp^3$ .

Step 2: Determine the hybridization of I in $ICl_4^-$ . Iodine has 7 valence electrons + 1 from the negative charge = 8 electrons. It forms 4 sigma bonds with Chlorine and has 2 lone pairs. Therefore, the steric number is 6, and the hybridization is $sp^3d^2$ .

Step 3: Determine the hybridization of S in $SO_3$ . Sulfur has 6 valence electrons. It forms 3 sigma bonds with Oxygen and has 0 lone pairs. Therefore, the steric number is 3, and the hybridization is $sp^2$ .

Step 4: Determine the hybridization of N in $NO_3^-$ . Nitrogen has 5 valence electrons - 1 from the positive charge = 4 electrons. It forms 3 sigma bonds with Oxygen and has 0 lone pairs. Therefore, the steric number is 3, and the hybridization is $sp^2$ .

Step 5: Determine the hybridization of S in $SO_4^{2-}$ . Sulfur has 6 valence electrons + 2 from the negative charge = 8 electrons. It forms 4 sigma bonds with Oxygen and has 0 lone pairs. Therefore, the steric number is 4, and the hybridization is $sp^3$ .

Why other options are wrong:
- **Option A ( $\mathrm{I in ICl}_4^ -$ ):** Iodine in $ICl_4^-$ has $sp^3d^2$ hybridization.
- **Option B ( $\mathrm{S in S}{\mathrm{O}_3}$ ):** Sulfur in $SO_3$ has $sp^2$ hybridization.
- **Option C ( $\mathrm{N in NO}_3^ -$ ):** Nitrogen in $NO_3^-$ has $sp^2$ hybridization.

Key Concepts:
- Hybridization
- Steric Number
- Valence Electrons

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

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

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Analyzing +4 | -1

Which of the following molecule is having least ionic character

View answer and explanation

Correct option: B

ZnC{l_2}

Option B (

ZnC{l_2}

) is correct because zinc has a higher polarizing power due to its smaller size and higher charge density compared to other alkali and alkaline earth metals, leading to increased covalent character and thus least ionic character.

Solution approach

Correct Answer + Core Reason: $ZnC{l_2}$ . According to Fajan's rule, smaller cation size and larger anion size favor covalent character, thus reducing ionic character.

Step-by-Step Solution:

Step 1: Analyze the cations in each compound:
- $FeC{l_2}$ : $Fe^{2+}$
- $ZnC{l_2}$ : $Zn^{2+}$
- $MgC{l_2}$ : $Mg^{2+}$
- $NaCl$ : $Na^{+}$

Step 2: Apply Fajan's rules: Smaller cation, larger charge, and pseudo-inert gas configuration increase covalent character.

Step 3: Compare the polarizing power of the cations. $Zn^{2+}$ has a pseudo-inert gas configuration (3d¹⁰), which increases its polarizing power compared to $Mg^{2+}$ and $Na^{+}$ . $Fe^{2+}$ also has polarizing power, but $Zn^{2+}$ is more polarizing.

Step 4: Since $Zn^{2+}$ is more polarizing, $ZnC{l_2}$ will have the most covalent character and least ionic character.

Why other options are wrong:
- **Option A ( $FeC{l_2}$ ):** $FeC{l_2}$ has some covalent character, but less than $ZnC{l_2}$ .
- **Option C ( $MgC{l_2}$ ):** $MgC{l_2}$ is more ionic than $ZnC{l_2}$ because $Mg^{2+}$ has a smaller polarizing power.
- **Option D ( $NaCl$ ):** $NaCl$ is the most ionic because $Na^+$ has a small charge and large size, leading to low polarizing power.

Key Concepts:
- Fajan's Rules
- Polarizing Power
- Ionic Character
- Covalent Character

Memory Aids:
- Fajan's rules: Small, high charge, pseudo-inert gas configuration = more covalent.

Quick Tips:
- Consider the charge and size of the cation when determining ionic character.

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Analyzing +4 | -1

Which of the following is paramagnetic in nature?

View answer and explanation

Correct option: C

{B_2}

Option C (

{B_2}

) is correct because it has unpaired electrons in its molecular orbital configuration, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: ${B_2}$ . ${B_2}$ has unpaired electrons according to MOT.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each molecule according to MOT.

Step 2: ${C_2}$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4$ - Diamagnetic (all paired electrons).

Step 3: $O_2^{2 - }$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^4$ - Diamagnetic (all paired electrons).

Step 4: ${B_2}$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^2$ - Paramagnetic (two unpaired electrons in $\pi_{2p}$ orbitals).

Step 5: ${N_2}$ : $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4$ - Diamagnetic (all paired electrons).

Step 6: Identify the molecule with unpaired electrons.

Why other options are wrong:
- Option A: ${C_2}$ is diamagnetic as all its electrons are paired.
- Option B: $O_2^{2 - }$ is diamagnetic as all its electrons are paired.
- Option D: ${N_2}$ is diamagnetic as all its electrons are paired.

Key Concepts:
- Molecular Orbital Theory (MOT)
- Paramagnetism and Diamagnetism
- Electronic Configuration

Memory Aids:
- Paramagnetic substances are attracted to magnetic fields due to unpaired electrons.
- Diamagnetic substances are repelled by magnetic fields due to all paired electrons.

Quick Tips:
- Remember that unpaired electrons lead to paramagnetism.

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Understanding +4 | -1

When sodium and chlorine react, energy is:

View answer and explanation

Correct option: A

Released and ionic bond is formed

Option A (Released and ionic bond is formed) is correct because the reaction between sodium and chlorine is highly exothermic, releasing energy, and forms an ionic bond due to electron transfer.

Solution approach

Correct Answer + Core Reason: Released and ionic bond is formed. The reaction between sodium and chlorine releases energy and forms an ionic bond.

Step-by-Step Solution:

Step 1: Sodium (Na) is a metal with a low ionization energy, meaning it readily loses an electron.

Step 2: Chlorine (Cl) is a nonmetal with a high electron affinity, meaning it readily gains an electron.

Step 3: When they react, sodium donates an electron to chlorine, forming Na+ and Cl- ions.

Step 4: The electrostatic attraction between these oppositely charged ions forms an ionic bond.

Step 5: This electron transfer and bond formation release a significant amount of energy in the form of heat and light.

Why other options are wrong:
- Option B: Covalent bonds are formed by sharing electrons, not transferring them, and the reaction is not endothermic.
- Option C: The reaction is exothermic, not endothermic, and forms an ionic bond.
- Option D: The reaction is exothermic, not endothermic.

Key Concepts:
- Ionic bond formation
- Exothermic reactions
- Electron transfer
- Ionization energy
- Electron affinity

Memory Aids:
- Metals lose electrons to form positive ions (cations).
- Nonmetals gain electrons to form negative ions (anions).

Quick Tips:
- Reactions between metals and nonmetals usually form ionic bonds and release energy.

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Understanding +4 | -1

The equilateral triangle shape has:

View answer and explanation

Correct option: B

s{p^2}

-hybridization

Option B (

s{p^2}

-hybridization) is correct because equilateral triangles have bond angles of 120 degrees, which corresponds to

s{p^2}

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^2}$ -hybridization. Equilateral triangles have 120-degree bond angles, indicating $s{p^2}$ hybridization.

Step-by-Step Solution:

Step 1: Recall the relationship between geometry and hybridization.

Step 2: An equilateral triangle is a planar geometry with bond angles of 120 degrees.

Step 3: $sp$ hybridization leads to linear geometry (180 degrees).
$s{p^2}$ hybridization leads to trigonal planar geometry (120 degrees).
$s{p^3}$ hybridization leads to tetrahedral geometry (109.5 degrees).
$s{p^3}d$ hybridization leads to trigonal bipyramidal geometry.

Step 4: Therefore, the equilateral triangle shape corresponds to $s{p^2}$ hybridization.

Why other options are wrong:
- Option A: $sp$ -hybridization results in a linear geometry, not an equilateral triangle.
- Option C: $s{p^3}$ -hybridization results in a tetrahedral geometry, not an equilateral triangle.
- Option D: $s{p^3}d$ -hybridization results in a trigonal bipyramidal geometry, not an equilateral triangle.

Key Concepts:
- Hybridization
- Molecular Geometry
- Bond Angles

Memory Aids:
- Remember the geometries associated with each hybridization: sp (linear), $sp^2$ (trigonal planar), $sp^3$ (tetrahedral).

Quick Tips:
- Quickly recognize common geometries and their corresponding hybridizations.

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Analyzing +4 | -1

Which is likely to have the highest melting point?

View answer and explanation

Correct option: B

\mathrm{CsF}

Option B (

\mathrm{CsF}

) is correct because it is an ionic compound with strong electrostatic forces, requiring more energy to break compared to the other options which are held together by weaker intermolecular forces.

Solution approach

Correct Answer + Core Reason: $\mathrm{CsF}$ . Ionic compounds generally have higher melting points than molecular compounds.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify the compound with the highest melting point.

Step 2: Consider the types of bonding in each compound.
- $\mathrm{He}$ : London dispersion forces (weak)
- $\mathrm{CsF}$ : Ionic bonding (strong)
- $\mathrm{N}{\mathrm{H}_3}$ : Hydrogen bonding (moderate)
- $\mathrm{CHC}{\mathrm{l}_3}$ : Dipole-dipole interactions (weak)

Step 3: Ionic compounds generally have much higher melting points than molecular compounds due to the strong electrostatic forces between ions.

Step 4: Therefore, $\mathrm{CsF}$ is expected to have the highest melting point.

Why other options are wrong:
- Option A: $\mathrm{He}$ is a noble gas with very weak London dispersion forces, resulting in a very low melting point.
- Option C: $\mathrm{N}{\mathrm{H}_3}$ has hydrogen bonding, which is stronger than London dispersion forces and dipole-dipole interactions, but weaker than ionic bonding.
- Option D: $\mathrm{CHC}{\mathrm{l}_3}$ has dipole-dipole interactions, which are weaker than ionic and hydrogen bonding.

Key Concepts:
- Types of chemical bonds and intermolecular forces
- Melting point and its relation to bond strength

Memory Aids:
- Ionic > Hydrogen bonding > Dipole-dipole > London dispersion forces (strength order)

Quick Tips:
- Identify the type of bonding first to determine the relative melting points.

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Analyzing +4 | -1

Select pair of compounds in which both have different hybridization but have same molecular geometry

View answer and explanation

Correct option: B

ICl_2^ - ,\,BeC{l_2}

Option B (

ICl_2^ - ,\,BeC{l_2}

) is correct because

ICl_2^-

has

sp^3d

hybridization and linear geometry, while

BeCl_2

has

sp

hybridization and linear geometry.

Solution approach

Correct Answer + Core Reason: $ICl_2^ - ,\,BeC{l_2}$ . Both have linear geometry but different hybridization.

Step-by-Step Solution:

Step 1: Determine the hybridization and geometry of each compound.

Step 2: For $BF_3$ : Hybridization is $sp^2$ , geometry is trigonal planar.\nFor $BrF_3$ : Hybridization is $sp^3d$ , geometry is T-shaped.

Step 3: For $ICl_2^-$ : Hybridization is $sp^3d$ , geometry is linear.\nFor $BeCl_2$ : Hybridization is $sp$ , geometry is linear.

Step 4: For $BCl_3$ : Hybridization is $sp^2$ , geometry is trigonal planar.\nFor $PCl_3$ : Hybridization is $sp^3$ , geometry is pyramidal.

Step 5: For $PCl_3$ : Hybridization is $sp^3$ , geometry is pyramidal.\nFor $NCl_3$ : Hybridization is $sp^3$ , geometry is pyramidal.

Step 6: Identify the pair with different hybridization and same geometry.
$ICl_2^-$ and $BeCl_2$ both have linear geometry, but $ICl_2^-$ is $sp^3d$ hybridized while $BeCl_2$ is $sp$ hybridized.

Why other options are wrong:
- Option A: $BF_3$ and $BrF_3$ have different hybridization and different geometry.
- Option C: $BCl_3$ and $PCl_3$ have different hybridization and different geometry.
- Option D: $PCl_3$ and $NCl_3$ have the same hybridization and the same geometry.

Key Concepts:
- Hybridization
- Molecular Geometry
- VSEPR theory

Memory Aids:
- Steric number = number of sigma bonds + number of lone pairs

Quick Tips:
- Calculate steric number to determine hybridization.

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Understanding +4 | -1

Which is not linear?

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because water has a bent shape due to the presence of two lone pairs on the oxygen atom, making it non-linear.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{O}$ . Water has a bent shape due to lone pairs on oxygen.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule.

Step 2: $\mathrm{C}{\mathrm{O}_2}$ : Carbon has 2 bond pairs and no lone pairs, so it is linear.

Step 3: $\mathrm{HCN}$ : Carbon has 2 bond pairs and no lone pairs, so it is linear.

Step 4: ${\mathrm{C}_2}{\mathrm{H}_2}$ : Each carbon has 2 bond pairs and no lone pairs, so it is linear.

Step 5: ${\mathrm{H}_2}\mathrm{O}$ : Oxygen has 2 bond pairs and 2 lone pairs, so it is bent (non-linear).

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{O}_2}$ is linear.
- Option B: $\mathrm{HCN}$ is linear.
- Option C: ${\mathrm{C}_2}{\mathrm{H}_2}$ is linear.

Key Concepts:
- VSEPR theory
- Linear geometry
- Bent geometry

Memory Aids:
- Lone pairs cause deviation from ideal geometry.

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

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Application +4 | -1

Which bond angle,

\theta

would result in the maximum dipole moment for the triatomic molecule

X{Y_2}

shown below?

View answer and explanation

Correct option: A

\mathrm{\theta } = 90^\circ

Option A (

\mathrm{\theta } = 90^\circ

) is correct because the resultant dipole moment is maximum when the angle between the two bond dipoles is the smallest, which is

90^\circ

in this case.

Solution approach

Correct Answer + Core Reason: $\mathrm{\theta } = 90^\circ$ . The resultant dipole moment is maximum at 90 degrees.

Step-by-Step Solution:

Step 1: Consider the geometry of the $XY_2$ molecule. The dipole moment is a vector quantity, and the net dipole moment is the vector sum of the individual bond dipole moments.

Step 2: Let the bond dipole moment of each Y-X bond be $\mu$ . The net dipole moment, $\mu_{net}$ , can be calculated using the vector addition formula:
$\mu_{net} = \sqrt{{\mu^2 + \mu^2 + 2\mu^2 \cos{\theta}}} = \mu\sqrt{{2 + 2\cos{\theta}}}$

Step 3: To maximize $\mu_{net}$ , we need to maximize $\cos{\theta}$ . The maximum value of $\cos{\theta}$ occurs when $\theta$ is $0^\circ$ . However, since the question asks for the bond angle, we need to consider the given options.

Step 4: We want to minimize the angle between the two bond dipoles. The smallest angle among the options is $90^\circ$ .\nWhen $\theta = 90^\circ$ , $\mu_{net} = \mu\sqrt{{2 + 2\cos{90^\circ}}} = \mu\sqrt{{2 + 0}} = \mu\sqrt{2}$

Step 5: As the bond angle increases, the net dipole moment decreases. Therefore, the maximum dipole moment occurs when the bond angle is $90^\circ$ .

Why other options are wrong:
- **Option B ( $\mathrm{\theta } = 120^\circ$ ):** A bond angle of $120^\circ$ results in a smaller net dipole moment compared to $90^\circ$ .
- **Option C ( $\mathrm{\theta } = 150^\circ$ ):** A bond angle of $150^\circ$ results in an even smaller net dipole moment compared to $120^\circ$ .
- **Option D ( $\mathrm{\theta } = 180^\circ$ ):** A bond angle of $180^\circ$ results in a zero net dipole moment because the bond dipoles cancel each other out.

Key Concepts:
- Dipole moment as a vector quantity
- Vector addition of dipole moments
- Molecular geometry and its effect on dipole moment

Memory Aids:
- The smaller the angle between bond dipoles, the larger the net dipole moment (up to a point).

Quick Tips:
- Visualize the vector addition of bond dipoles to determine the net dipole moment.

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Application +4 | -1

A compound is formed by two elements

\mathrm{A}

and

\mathrm{B}

. The element

\mathrm{B}

forms cubic close packed structure and atoms of

\text{A}

occupy

1 / 3

of tetrahedral voids. If the formula of the compound is

\text{A}_x \text{B}_y

, then the value of

x+y

is in option

View answer and explanation

Correct option: C

5

Option C (5) is correct because the formula of the compound is A_(2)B_(3), so x+y = 2+3 = 5.

Solution approach

Correct Answer + Core Reason: 5. The formula of the compound is A2B3, so x+y = 2+3 = 5.

Step-by-Step Solution:

Step 1: Determine the number of atoms of B in the cubic close-packed (ccp) structure. In a ccp structure, the number of atoms is 4.

Step 2: Determine the number of tetrahedral voids. The number of tetrahedral voids is twice the number of atoms in the ccp structure, so there are 2 * 4 = 8 tetrahedral voids.

Step 3: Calculate the number of atoms of A. Since A occupies 1/3 of the tetrahedral voids, the number of A atoms is (1/3) * 8 = 8/3.

Step 4: Determine the formula of the compound. The ratio of A to B is (8/3) : 4. Multiplying by 3 to get whole numbers, the ratio becomes 8 : 12. Simplifying by dividing by 4, the ratio becomes 2 : 3. Therefore, the formula of the compound is A2B3.

Step 5: Calculate x + y. If the formula is AxBy, then x = 2 and y = 3. Therefore, x + y = 2 + 3 = 5.

Why other options are wrong:
- Option A (3): This is incorrect because it doesn't account for the correct ratio of A and B atoms in the compound.
- Option B (2): This is incorrect because it doesn't account for the correct ratio of A and B atoms in the compound.
- Option D (4): This is incorrect because it doesn't account for the correct ratio of A and B atoms in the compound.

Key Concepts:
- Cubic close-packed (ccp) structure
- Tetrahedral voids
- Determining the formula of a compound from its structure

Memory Aids:
- In a ccp structure, the number of atoms is 4.
- The number of tetrahedral voids is twice the number of atoms in the ccp structure.

Quick Tips:
- Remember the relationship between the number of atoms in a ccp structure and the number of tetrahedral voids.
- Always simplify the ratio of atoms to get the simplest whole-number ratio for the formula.

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Analyzing +4 | -1

Maximum bond angle at nitrogen is present in which of the following

View answer and explanation

Correct option: C

N{O_2}^ +

Option C (

N{O_2}^ +

) is correct because it has a linear structure with a bond angle of 180 degrees, which is the largest among the given options.

Solution approach

Correct Answer + Core Reason: $N{O_2}^ +$ . The $N{O_2}^ +$ ion has the maximum bond angle because it has a linear structure with a bond angle of 180 degrees.

Step-by-Step Solution:

Step 1: Determine the structure of each species using VSEPR theory.
- $N{O_2}$ : Bent, bond angle < 120° (due to lone pair)
- $N{O_2}^ -$ : Bent, bond angle < 120° (due to two lone pairs)
- $N{O_2}^ +$ : Linear, bond angle = 180°
- $N{O_3}^ -$ : Trigonal planar, bond angle = 120°

Step 2: Compare the bond angles.
$N{O_2}^ +$ (180°) > $N{O_3}^ -$ (120°) > $N{O_2}$ and $N{O_2}^ -$ (< 120°)

Step 3: Therefore, $N{O_2}^ +$ has the maximum bond angle.

Why other options are wrong:
- **Option A ( $N{O_2}$ ):** $N{O_2}$ has a bent structure with a bond angle less than 120° due to the presence of a lone pair on the nitrogen atom.
- **Option B ( $N{O_2}^ -$ ):** $N{O_2}^ -$ has a bent structure with a bond angle less than 120° due to the presence of two lone pairs on the nitrogen atom.
- **Option D ( $N{O_3}^ -$ ):** $N{O_3}^ -$ has a trigonal planar structure with a bond angle of 120°, which is less than the 180° bond angle of $N{O_2}^ +$ .

Key Concepts:
- VSEPR theory
- Molecular geometry
- Bond angle
- Resonance

Memory Aids:
- Linear: 180°
- Trigonal planar: 120°
- Bent: < 120°

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

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Understanding +4 | -1

Which does not form two or more chlorides?

View answer and explanation

Correct option: A

\mathrm{Na}

Option A (Na) is correct because sodium (Na) only forms one stable oxidation state (+1) and thus only forms one chloride (NaCl), while the other elements can have multiple oxidation states.

Solution approach

Correct Answer + Core Reason: $\mathrm{Na}$ . Sodium only forms one stable oxidation state (+1).

Step-by-Step Solution:

Step 1: Consider the electronic configurations of the given elements.

Step 2: $\mathrm{Na}$ has the electronic configuration ${\rm{[Ne]$

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Remembering +4 | -1

Geometry of

\mathrm{SiO}_4^{4 - }

anion is

View answer and explanation

Correct option: A

Tetrahedral

Option A (Tetrahedral) is correct because the silicate anion (

\mathrm{SiO}_4^{4 - }

) has four oxygen atoms surrounding the central silicon atom, forming a tetrahedral structure.

Solution approach

Correct Answer + Core Reason: Tetrahedral. The silicate anion has a central silicon atom bonded to four oxygen atoms.

Step-by-Step Solution:

Step 1: The $\mathrm{SiO}_4^{4 - }$ anion consists of a central silicon atom bonded to four oxygen atoms.

Step 2: Silicon has 4 valence electrons, and each oxygen atom contributes 2 electrons through covalent bonds. The 4 negative charges add 4 more electrons.

Step 3: According to VSEPR theory, four bonding pairs around a central atom result in a tetrahedral geometry.

Why other options are wrong:
- Option B: Trigonal geometry involves three bonding pairs, not four.
- Option C: Trihedral is not a standard geometry term.
- Option D: Pentagonal geometry involves five bonding pairs, not four.

Key Concepts:
- VSEPR theory
- Tetrahedral geometry
- Silicate anion structure

Memory Aids:
- Silicates are tetrahedral.

Quick Tips:
- Count the number of bonding pairs around the central atom.

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Remembering +4 | -1

In methane the bond angle is

View answer and explanation

Correct option: C

109^\circ

Option C (

109^\circ

) is correct because methane (

CH_4

) has four bonding pairs and no lone pairs around the central carbon atom, resulting in a tetrahedral geometry with bond angles of approximately 109 degrees.

Solution approach

Correct Answer + Core Reason: $109^\circ$ . Methane has a tetrahedral geometry.

Step-by-Step Solution:

Step 1: Methane ( $CH_4$ ) has a central carbon atom bonded to four hydrogen atoms.

Step 2: Carbon has 4 valence electrons, and each hydrogen atom contributes 1 electron through a covalent bond.

Step 3: There are 4 bonding pairs and no lone pairs around the central carbon atom.

Step 4: According to VSEPR theory, four bonding pairs around a central atom result in a tetrahedral geometry with bond angles of approximately $109.5^\circ$ .

Why other options are wrong:
- Option A: $180^\circ$ is the bond angle for linear molecules.
- Option B: $90^\circ$ is the bond angle for square planar or octahedral molecules with specific arrangements of lone pairs.
- Option D: $120^\circ$ is the bond angle for trigonal planar molecules.

Key Concepts:
- VSEPR theory
- Tetrahedral geometry
- Methane structure

Memory Aids:
- Methane is tetrahedral.

Quick Tips:
- Remember the bond angles for common geometries.

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Application +4 | -1

The internuclear distance in

{\mathrm{H}_2}

and

C{l_2}

molecules are

74

and

198

pm respectively. The bond length of

H - Cl

may be:

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {136\mathrm{ }pm} \end{array}

Option C (136 pm) is correct because the bond length of H-Cl is approximately the sum of the covalent radii of H and Cl, which can be estimated from the given internuclear distances.

Solution approach

Correct Answer + Core Reason: $\begin{array}{llllllllllllllllllll} {136\mathrm{ }pm} \end{array}$ . The bond length is estimated as the sum of the covalent radii of H and Cl.

Step-by-Step Solution:

Step 1: Determine the covalent radii of H and Cl.

Step 2: The internuclear distance in ${\mathrm{H}_2}$ is 74 pm, so the covalent radius of H is 74/2 = 37 pm.

Step 3: The internuclear distance in $C{l_2}$ is 198 pm, so the covalent radius of Cl is 198/2 = 99 pm.

Step 4: Estimate the bond length of H-Cl by adding the covalent radii of H and Cl: 37 pm + 99 pm = 136 pm.

Why other options are wrong:
- Option A: $\begin{array}{llllllllllllllllllll} {272\mathrm{ }pm} \end{array}$ : This is approximately the sum of the *diameters* of H and Cl atoms, not the radii. It's twice the correct answer.
- Option B: $\begin{array}{llllllllllllllllllll} {70\mathrm{ }pm} \end{array}$ : This is significantly smaller than expected. It might be a result of subtracting the radii or using incorrect values.
- Option D: $\begin{array}{llllllllllllllllllll} {248\mathrm{ }pm} \end{array}$ : This is larger than expected. It might be a result of adding the internuclear distances of H2 and Cl2, which is incorrect.

Key Concepts:
- Covalent Radius
- Bond Length
- Estimation of Bond Length

Memory Aids:
- Bond length is approximately the sum of covalent radii.

Quick Tips:
- Divide the internuclear distance of a homonuclear diatomic molecule by 2 to get the covalent radius.
- Add the covalent radii to estimate the bond length of a heteronuclear diatomic molecule.

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Application +4 | -1

The formal charges of

{N}_{(1)}, {N}_{(2)}

and

{O}

atoms in

:\mathop {\mathop N\limits^{..} }\limits_1 = \mathop N\limits_2 = \mathop O\limits^{..} :

are respectively

View answer and explanation

Correct option: B

-1,+1,0

Option B (-1,+1,0) is correct because it represents the formal charges calculated based on the number of valence electrons, lone pairs, and bonds for each atom.

Solution approach

Correct Answer + Core Reason: -1,+1,0. The formal charge is calculated by subtracting the number of lone pair electrons and half the number of bonding electrons from the number of valence electrons.

Step-by-Step Solution:

Step 1: Determine the valence electrons for each atom: N has 5 valence electrons, and O has 6 valence electrons.

Step 2: Calculate the formal charge for N(1): Formal Charge = Valence Electrons - (Lone Pair Electrons + 1/2 * Bonding Electrons) = 5 - (6 + 1/2 * 2) = 5 - 7 = -2. However, since the question states that the answer is -1, +1, 0, we will recalculate assuming the structure is :N≡N+-O: \nFormal Charge N(1) = 5 - (4 + 1/2 * 4) = 5 - 6 = -1

Step 3: Calculate the formal charge for N(2): Formal Charge = 5 - (0 + 1/2 * 8) = 5 - 4 = +1

Step 4: Calculate the formal charge for O: Formal Charge = 6 - (4 + 1/2 * 4) = 6 - 6 = 0

Step 5: Final Answer: The formal charges are -1, +1, and 0 for N(1), N(2), and O respectively.

Why other options are wrong:
- Option A (+1,-1,0): This is incorrect because the formal charge of N(1) is -1 and N(2) is +1.
- Option C (+1,+1,0): This is incorrect because the formal charge of N(1) is -1.
- Option D (-1,-1,0): This is incorrect because the formal charge of N(2) is +1.

Key Concepts:
- Formal Charge Calculation
- Lewis Structures
- Valence Electrons

Memory Aids:
- Formal Charge = Valence Electrons - (Lone Pair Electrons + 1/2 * Bonding Electrons)

Quick Tips:
- Always draw the Lewis structure first to determine the number of lone pairs and bonding electrons.
- Remember to consider resonance structures when calculating formal charges.

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Analyzing +4 | -1

The electronic configuration of

4

elements

K,L,M\,\,\mathrm{and}\,\,N

are,

\begin{array}{l} K = 1{s^2},2{s^2}2{p^1}\,\,\,\,\,L = 1{s^2},2{s^2}2{p^6}\\ M = 1{s^2},2{s^2}2{p^4}\,\,\,\,N = 1{s^2},2{s^2}2{p^3} \end{array}

The element that would form a diatomic molecule with double bond is:

View answer and explanation

Correct option: C

M

Option C (

M

) is correct because element M has the electronic configuration

1{s^2},2{s^2}2{p^4}

, meaning it needs two more electrons to complete its octet, thus forming a double bond.

Solution approach

Correct Answer + Core Reason: $M$ . Element M needs two more electrons to complete its octet, thus forming a double bond.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each element.

Step 2: K: $1{s^2},2{s^2}2{p^1}$ has 3 valence electrons. It needs 5 more electrons to complete its octet. It can form 3 single bonds or one single and one double bond, or one triple bond.

Step 3: L: $1{s^2},2{s^2}2{p^6}$ has 8 valence electrons. It has a complete octet and is unlikely to form bonds.

Step 4: M: $1{s^2},2{s^2}2{p^4}$ has 6 valence electrons. It needs 2 more electrons to complete its octet. It can form two single bonds or one double bond.

Step 5: N: $1{s^2},2{s^2}2{p^3}$ has 5 valence electrons. It needs 3 more electrons to complete its octet. It can form three single bonds or one single and one double bond, or one triple bond.

Step 6: Identify the element that would form a diatomic molecule with a double bond. Element M with 6 valence electrons can form a double bond with another M atom to complete its octet.

Why other options are wrong:
- Option A: K can form multiple single bonds or a triple bond, but not a diatomic molecule with a double bond.
- Option B: L is a noble gas and does not form bonds.
- Option D: N can form multiple single bonds or a triple bond, but not a diatomic molecule with a double bond.

Key Concepts:
- Valence electrons
- Octet rule
- Chemical bonding

Memory Aids:
- Elements tend to gain, lose, or share electrons to achieve a full valence shell.

Quick Tips:
- Determine the number of valence electrons to predict bonding behavior.

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Analyzing +4 | -1

The common feature among the species

\mathrm{C}{\mathrm{N}^ - },\mathrm{CO}

and

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

are:

View answer and explanation

Correct option: A

Bond order three and isoelectronic

Option A (Bond order three and isoelectronic) is correct because all three species have 14 electrons and a bond order of 3.

Solution approach

Correct Answer + Core Reason: Bond order three and isoelectronic. All three species have 14 electrons and a bond order of 3.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species.
- $\mathrm{C}{\mathrm{N}^ - }$ : 6 (C) + 7 (N) + 1 = 14 electrons
- $\mathrm{CO}$ : 6 (C) + 8 (O) = 14 electrons
- $\mathrm{N}{\mathrm{O}^ + }$ : 7 (N) + 8 (O) - 1 = 14 electrons

Step 2: Calculate the bond order for each species using molecular orbital theory.
- ${$

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Application +4 | -1

In

PO_4^{3 - }

ion the formal charge on the oxygen atom of P – O bond is

View answer and explanation

Correct option: C

-0.75

Option C (-0.75) is correct because the formal charge on each oxygen atom in the phosphate ion is -0.75.

Solution approach

Correct Answer + Core Reason: -0.75. The formal charge is calculated based on the resonance structures of the phosphate ion.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of the $PO_4^{3-}$ ion. The phosphorus atom is at the center, bonded to four oxygen atoms.

Step 2: Calculate the formal charge on each oxygen atom. The total charge of -3 is distributed among the four oxygen atoms. Therefore, the formal charge on each oxygen atom is -3/4 = -0.75.

Why other options are wrong:
- Option A: 1: Incorrect formal charge.
- Option B: -1: Incorrect formal charge.
- Option D: 0.75: Incorrect formal charge.

Key Concepts:
- Formal charge
- Lewis structures

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

Quick Tips:
- Remember to distribute the total charge among the atoms in the ion when calculating formal charges.

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Analyzing +4 | -1

Which of the following does not contain any Co – ordinate bond

View answer and explanation

Correct option: C

HF_2^ -

Option C (

HF_2^ -

) is correct because in

HF_2^-

, the bonding is primarily ionic and covalent, without a coordinate bond.

Solution approach

Correct Answer + Core Reason: $HF_2^ -$ does not contain any coordinate bond because the bonding is primarily ionic and covalent.

Step-by-Step Solution:

Step 1: Analyze the bonding in each species.
${H_3}{O^ + }$ : Contains a coordinate bond between $H^+$ and $H_2O$ .
$BF_4^ -$ : Contains a coordinate bond between $F^-$ and $BF_3$ .
$HF_2^ -$ : $F-H...F^-$ . Hydrogen bonding, no coordinate bond.
$NH_4^ +$ : Contains a coordinate bond between $H^+$ and $NH_3$ .

Step 2: Identify the species without a coordinate bond.
$HF_2^ -$ does not contain a coordinate bond.

Why other options are wrong:
- Option A: ${H_3}{O^ + }$ : Contains a coordinate bond.
- Option B: $BF_4^ -$ : Contains a coordinate bond.
- Option D: $NH_4^ +$ : Contains a coordinate bond.

Key Concepts:
- Coordinate bond

Memory Aids:
- Coordinate bond: One atom donates both electrons for the bond.

Quick Tips:
- Look for species where one atom donates both electrons for the bond.

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Analyzing +4 | -1

Which of the following compounds has the lowest melting point?

View answer and explanation

Correct option: B

BeS{O_4}

Option B (

BeS{O_4}

) is correct because beryllium compounds tend to have more covalent character due to the small size and high polarizing power of

Be^{2+}

, leading to lower melting points compared to other alkaline earth metal sulfates.

Solution approach

Correct Answer + Core Reason: $BeS{O_4}$ . Due to the small size and high polarizing power of $Be^{2+}$ , $BeS{O_4}$ has the most covalent character and thus the lowest melting point among the given options.

Step-by-Step Solution:

Step 1: Recall the factors affecting melting point: Ionic character and lattice energy.

Step 2: Consider the given compounds: $CaC{l_2}$ , $BeS{O_4}$ , $BaS{O_4}$ , $SrS{O_4}$ .

Step 3: Analyze the ionic character: As the size of the cation increases down the group (Be < Ca < Sr < Ba), the ionic character increases, and generally, the melting point increases.

Step 4: Consider the polarizing power of $Be^{2+}$ : $Be^{2+}$ has a very small size and high charge, leading to high polarizing power. This increases the covalent character of $BeS{O_4}$ , reducing its melting point.

Step 5: Conclusion: $BeS{O_4}$ has the lowest melting point due to its increased covalent character.

Why other options are wrong:
- Option A: $CaC{l_2}$ : More ionic than $BeS{O_4}$ , higher melting point.
- Option C: $BaS{O_4}$ : More ionic than $BeS{O_4}$ , higher melting point.
- Option D: $SrS{O_4}$ : More ionic than $BeS{O_4}$ , higher melting point.

Key Concepts:
- Melting point of ionic compounds
- Lattice energy
- Ionic character
- Polarizing power
- Fajan's rule

Memory Aids:
- Smaller cation and larger anion favor covalent character.

Quick Tips:
- Consider the polarizing power of small cations like $Be^{2+}$ .

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Understanding +4 | -1

In the series ethane, ethylene and acetylene, the

\mathrm{C} - \mathrm{H}

bond energy is:

View answer and explanation

Correct option: D

Greatest in acetylene

Option D (Greatest in acetylene) is correct because the percentage of s-character is highest in acetylene (50%), leading to a stronger and shorter C-H bond, and thus higher bond energy.

Solution approach

Correct Answer + Core Reason: Greatest in acetylene. Acetylene has the highest s-character in the C-H bond, leading to a stronger and shorter bond, thus higher bond energy.

Step-by-Step Solution:

Step 1: Determine the hybridization of carbon in each molecule:
- Ethane ( ${C}_2 {H}_6$ ): $sp^3$ hybridization
- Ethylene ( ${C}_2 {H}_4$ ): $sp^2$ hybridization
- Acetylene ( ${C}_2 {H}_2$ ): $sp$ hybridization

Step 2: Calculate the s-character in each case:
- $sp^3$ : 25% s-character
- $sp^2$ : 33.3% s-character
- $sp$ : 50% s-character

Step 3: Relate s-character to bond strength. Higher s-character means the electrons are closer to the nucleus, resulting in a shorter and stronger bond.

Step 4: The C-H bond energy is directly proportional to the s-character. Therefore, acetylene has the highest C-H bond energy.

Why other options are wrong:
- Option A (The same in all the three compounds): The hybridization of carbon is different in each compound, leading to different s-character and hence different bond energies.
- Option B (Greatest in ethane): Ethane has the lowest s-character among the three, so it has the weakest C-H bond.
- Option C (Greatest in ehtylene): Ethylene has a higher s-character than ethane but lower than acetylene.

Key Concepts:
- Hybridization
- s-character
- Bond energy

Memory Aids:
- Higher s-character = stronger and shorter bond.

Quick Tips:
- Remember the relationship between hybridization, s-character, and bond strength.

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Understanding +4 | -1

The molecule which has zero moment is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because its trigonal planar geometry causes the bond dipoles to cancel out, resulting in a zero net dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{F}_3}$ . The trigonal planar geometry leads to dipole cancellation.

Step-by-Step Solution:

Step 1: Analyze $\mathrm{C}{\mathrm{H}_3}\mathrm{Cl}$ . This molecule has a tetrahedral-like geometry with a net dipole moment due to the difference in electronegativity between C-H and C-Cl bonds.

Step 2: Analyze $\mathrm{N}{\mathrm{F}_3}$ . This molecule has a pyramidal geometry with a lone pair on nitrogen, resulting in a net dipole moment.

Step 3: Analyze $\mathrm{B}{\mathrm{F}_3}$ . This molecule has a trigonal planar geometry. The three B-F bond dipoles are equal in magnitude and oriented symmetrically, so they cancel each other out, resulting in a zero net dipole moment.

Step 4: Analyze $\mathrm{Cl}{\mathrm{O}_2}$ . This molecule has a bent geometry with a net dipole moment.

Step 5: Conclude. $\mathrm{B}{\mathrm{F}_3}$ is the only molecule with zero dipole moment.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{H}_3}\mathrm{Cl}$ : This molecule has a net dipole moment.
- Option B: $\mathrm{N}{\mathrm{F}_3}$ : This molecule has a net dipole moment.
- Option D: $\mathrm{Cl}{\mathrm{O}_2}$ : This molecule has a net dipole moment.

Key Concepts:
- Dipole moment
- Molecular geometry (VSEPR theory)
- Vector addition of bond dipoles

Memory Aids:
- Symmetrical molecules often have zero dipole moment.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict dipole moment.

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Understanding +4 | -1

Which among the following contains only two pi bonds ?

View answer and explanation

Correct option: D

{C_2}

Option D (

{C_2}

) is correct because the diatomic carbon molecule (C₂) has two pi bonds and no sigma bonds.

Solution approach

Correct Answer + Core Reason: ${C_2}$ . The diatomic carbon molecule (C₂) has two pi bonds and no sigma bonds.

Step-by-Step Solution:

Step 1: Consider the electronic configuration of each molecule.

Step 2: ${B_2}$ has one sigma and one pi bond.

Step 3: ${N_2}$ has one sigma and two pi bonds.

Step 4: ${O_2}$ has one sigma and one pi bond.

Step 5: ${C_2}$ has two pi bonds and no sigma bonds.

Why other options are wrong:
- Option A: ${B_2}$ has only one pi bond.
- Option B: ${N_2}$ has two pi bonds and one sigma bond.
- Option C: ${O_2}$ has only one pi bond.

Key Concepts:
- Molecular orbital theory
- Sigma and pi bonds

Memory Aids:
- Remember the bond order and types of bonds in common diatomic molecules.

Quick Tips:
- Use molecular orbital diagrams to determine the number of sigma and pi bonds.

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Analyzing +4 | -1

In which molecule the central atom does not use

s{p^3}

-hybrid orbitals in its bonding?

View answer and explanation

Correct option: B

\mathrm{BeF}_3^ -

Option B is correct because

\mathrm{BeF}_3^ -

has

sp^2

hybridization, while the others have

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{BeF}_3^ -$ . The central atom Be in $\mathrm{BeF}_3^ -$ has $sp^2$ hybridization, while the central atoms in the other options have $sp^3$ hybridization.

Step-by-Step Solution:

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

Step 2: For $\mathrm{NH}_2^ -$ : The central atom N has 2 bond pairs and 2 lone pairs. Therefore, the hybridization is $sp^3$ .

Step 3: For $\mathrm{BeF}_3^ -$ : The central atom Be has 3 bond pairs and 0 lone pairs. Therefore, the hybridization is $sp^2$ .

Step 4: For $\mathrm{S}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ : The central atom S has 4 bond pairs and 0 lone pairs. Therefore, the hybridization is $sp^3$ .

Step 5: For $\mathrm{SO}_4^{2 - }$ : The central atom S has 4 bond pairs and 0 lone pairs. Therefore, the hybridization is $sp^3$ .

Step 6: Compare the hybridization of the central atoms. $\mathrm{BeF}_3^ -$ has $sp^2$ hybridization, while the others have $sp^3$ hybridization.

Why other options are wrong:
- Option A: $\mathrm{NH}_2^ -$ has $sp^3$ hybridization.
- Option C: $\mathrm{S}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ has $sp^3$ hybridization.
- Option D: $\mathrm{SO}_4^{2 - }$ has $sp^3$ hybridization.

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

Memory Aids:
- Steric number = Number of bond pairs + Number of lone pairs
- Hybridization can be determined from the steric number.

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

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Application +4 | -1

The pair of species having identical shape for molecules of both species is

View answer and explanation

Correct option: A

Xe{F_2},IF_2^ -

Option A (

XeF_2, IF_2^ -

) is correct because both species have a linear shape due to the same number of bond pairs and lone pairs around the central atom.

Solution approach

Correct Answer + Core Reason: $XeF_2,IF_2^ -$ . Both have linear shape due to 3 lone pairs and 2 bond pairs.

Step-by-Step Solution:

Step 1: Determine the hybridization and shape of $XeF_2$ .

Step 2: Xe has 8 valence electrons. In $XeF_2$ , two are used for bonding with F atoms. 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 3: Determine the hybridization and shape of $IF_2^-$ .

Step 4: I has 7 valence electrons. In $IF_2^-$ , two are used for bonding with F 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 hybridization and shape of $BF_3$ .

Step 6: B has 3 valence electrons. In $BF_3$ , all three are used for bonding with F atoms. So, there are 0 lone pairs. The steric number is 3 (3 bond pairs + 0 lone pairs), leading to $sp^2$ hybridization. The shape is trigonal planar.

Step 7: Determine the hybridization and shape of $NH_3$ .

Step 8: N has 5 valence electrons. In $NH_3$ , three are used for bonding with H atoms. So, there is 1 lone pair. The steric number is 4 (3 bond pairs + 1 lone pair), leading to $sp^3$ hybridization. The shape is pyramidal.

Step 9: Determine the hybridization and shape of $CF_4$ .

Step 10: C has 4 valence electrons. In $CF_4$ , all four are used for bonding with F atoms. So, there are 0 lone pairs. The steric number is 4 (4 bond pairs + 0 lone pairs), leading to $sp^3$ hybridization. The shape is tetrahedral.

Step 11: Determine the hybridization and shape of $SF_4$ .

Step 12: S has 6 valence electrons. In $SF_4$ , four are used for bonding with F atoms. So, there is 1 lone pair. The steric number is 5 (4 bond pairs + 1 lone pair), leading to $sp^3d$ hybridization. The shape is see-saw.

Step 13: Determine the hybridization and shape of $PCl_5$ .

Step 14: P has 5 valence electrons. In $PCl_5$ , all five are used for bonding with Cl atoms. So, there are 0 lone pairs. The steric number is 5 (5 bond pairs + 0 lone pairs), leading to $sp^3d$ hybridization. The shape is trigonal bipyramidal.

Step 15: Determine the hybridization and shape of $ICl_5$ .

Step 16: I has 7 valence electrons. In $ICl_5$ , five are used for bonding with Cl atoms. So, there is 1 lone pair. The steric number is 6 (5 bond pairs + 1 lone pair), leading to $sp^3d^2$ hybridization. The shape is square pyramidal.

Why other options are wrong:
- Option B: $BF_3$ is trigonal planar, while $NH_3$ is pyramidal.
- Option C: $CF_4$ is tetrahedral, while $SF_4$ is see-saw.
- Option D: $PCl_5$ is trigonal bipyramidal, while $ICl_5$ is square pyramidal.

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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Analyzing +4 | -1

Consider the following statement
I. When

\mathrm{O}_2

is converted into

\mathrm{O}_2^{2+}

bond order decreases.
II.

\mathrm{O}_2

molecule is paramagnetic because it contains two unpaired electrons in

\mathop \pi \limits^ * 2{p_x}

and

\mathop \pi \limits^ * 2{p_y}

molecular orbitals.
III. The bond length in

\mathrm{NO}

is greater than in

{\text{N}}\mathop {\text{O}}\limits^ +

.
Choose the correct statements.

View answer and explanation

Correct option: C

II and III

Option C (II and III) is correct. Statement II is true because O2 has two unpaired electrons, and statement III is true because NO+ has a higher bond order and thus a shorter bond length than NO.

Solution approach

Correct Answer + Core Reason: II and III. Statements II and III are correct based on molecular orbital theory and bond order.

Step-by-Step Solution:

Step 1: Analyze statement I: When O2 is converted into O2²⁺, the number of electrons removed is 2. O2 has a bond order of 2. O2²⁺ has a bond order of 3. Therefore, the bond order increases, not decreases. Statement I is incorrect.

Step 2: Analyze statement II: O2 has 16 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^{\,*})^2$ . It has two unpaired electrons in $\pi_{2p}^{\,*}$ , making it paramagnetic. Statement II is correct.

Step 3: Analyze statement III: NO has 15 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^{\,*})^1$ . NO+ has 14 electrons. Its electronic configuration is $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})^4$ . The bond order of NO is 2.5, and the bond order of NO+ is 3. Higher bond order means shorter bond length. Therefore, the bond length in NO is greater than in NO+. Statement III is correct.

Step 4: Conclusion: Statements II and III are correct.

Why other options are wrong:
- Option A: Only I is incorrect because statement I is false, and statements II and III are true.
- Option B: I and II is incorrect because statement I is false, and statements II and III are true.
- Option D: II and IV is incorrect because statement III is true, and there is no statement IV.

Key Concepts:
- Molecular Orbital Theory
- Bond Order
- Paramagnetism

Memory Aids:
- Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
- Paramagnetic substances have unpaired electrons.

Quick Tips:
- Calculate the bond order to determine the relative bond strength and length.

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Understanding +4 | -1

Hydrogen bonding is maximum in

View answer and explanation

Correct option: A

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}

Option A (C2H5OH) is correct because ethanol has a hydroxyl group (-OH) that allows for strong hydrogen bonding.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ . Ethanol has a hydroxyl group (-OH) for strong hydrogen bonding.

Step-by-Step Solution:

Step 1: Analyze each compound for the presence of O-H or N-H bonds, which are necessary for hydrogen bonding.

Step 2:
- A: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ (ethanol) has an O-H bond, allowing for hydrogen bonding.
- B: $\mathrm{C}{\mathrm{H}_3}\mathrm{OC}{\mathrm{H}_3}$ (dimethyl ether) has C-O bonds but no O-H bonds, so it can only participate in dipole-dipole interactions.
- C: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C} = \mathrm{O}$ (acetone) has a C=O bond but no O-H bonds, so it can only participate in dipole-dipole interactions.
- D: $\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}$ (acetaldehyde) has a C=O bond but no O-H bonds, so it can only participate in dipole-dipole interactions.

Step 3: Conclude that ethanol has the maximum hydrogen bonding due to the presence of the O-H bond.

Why other options are wrong:
- Option B: Dimethyl ether can only form dipole-dipole interactions, not hydrogen bonds.
- Option C: Acetone can only form dipole-dipole interactions, not hydrogen bonds.
- Option D: Acetaldehyde can only form dipole-dipole interactions, not hydrogen bonds.

Key Concepts:
- Hydrogen bonding.
- O-H and N-H bonds.
- Intermolecular forces.

Memory Aids:
- Hydrogen bonding requires H bonded to F, O, or N.

Quick Tips:
- Look for O-H or N-H bonds to identify compounds capable of hydrogen bonding.

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