${\rm{PC}}{{\rm{l}}_5}$ exists but ${\rm{NC}}{{\rm{l}}_5}$ does not because:

${\rm{NC}}{{\rm{l}}_5}$ is unstable

Nitrogen atom is much smaller than phosphorus

Bond energies in $NO$, ${\rm{N}}{{\rm{O}}^ + }$ and ${\rm{N}}{{\rm{O}}^ - }$are such as

${\rm{N}}{{\rm{O}}^ + } > {\rm{NO}} > {\rm{N}}{{\rm{O}}^ - }$

${\rm{N}}{{\rm{O}}^ - } > {\rm{NO}} > N{{\rm{O}}^ + }$

${\rm{NO}} > {\rm{N}}{{\rm{O}}^ - } > N{{\rm{O}}^ + }$

${\rm{N}}{{\rm{O}}^ + } > {\rm{N}}{{\rm{O}}^ - } > {\rm{NO}}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 8

211 Medium

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Analyzing +4 | -1

In which molecule Sulphur atom is not

s{p^3}

-hybridized?

View answer and explanation

Correct option: B

\mathrm{S}{\mathrm{F}_4}

Option D (None of these) is incorrect, and the correct answer is B (

\mathrm{S}{\mathrm{F}_4}

) because in

SF_4

, the sulfur atom is

sp^3d

hybridized due to the presence of a lone pair, while in

\mathrm{SO}_4^{2 - }

and

\mathrm{S}{\mathrm{F}_2}

, the sulfur atom is

sp^3

hybridized.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}{\mathrm{F}_4}$ . In $SF_4$ , the sulfur atom is $sp^3d$ hybridized, not $sp^3$ .

Step-by-Step Solution:

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

Step 2: For $\mathrm{SO}_4^{2 - }$ , the central atom S has 6+4(6)+2 = 32 valence electrons. The Lewis structure has four S-O bonds and no lone pair on S. Steric number = 4, so it is $sp^3$ hybridized.

Step 3: For $\mathrm{S}{\mathrm{F}_4}$ , the central atom S has 6+4(7) = 34 valence electrons. The Lewis structure has four S-F bonds and one lone pair on S. Steric number = 5, so it is $sp^3d$ hybridized.

Step 4: For $\mathrm{S}{\mathrm{F}_2}$ , the central atom S has 6+2(7) = 20 valence electrons. The Lewis structure has two S-F bonds and two lone pairs on S. Steric number = 4, so it is $sp^3$ hybridized.

Step 5: Identify the molecule where the sulfur atom is not $sp^3$ hybridized.

Why other options are wrong:
- Option A: $\mathrm{SO}_4^{2 - }$ has $sp^3$ hybridized sulfur atom.
- Option C: $\mathrm{S}{\mathrm{F}_2}$ has $sp^3$ hybridized sulfur atom.
- Option D: None of these is incorrect because option B is the correct answer.

Key Concepts:
- Hybridization
- Steric number

Memory Aids:
- Steric Number = Number of Bond Pairs + Number of Lone Pairs

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

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Application +4 | -1

The bond order in

{CN}^{-}

ion is

View answer and explanation

Correct option: B

3

Option B (3) is correct because the bond order in

{CN}^{-}

ion is 3, calculated from its molecular orbital configuration.

Solution approach

Correct Answer + Core Reason: 3. The bond order is calculated using the molecular orbital configuration.

Step-by-Step Solution:

Step 1: Determine the number of electrons in ${CN}^{-}$ .
- Carbon has 6 electrons, and Nitrogen has 7 electrons. $CN^-$ has 6 + 7 + 1 = 14 electrons.

Step 2: Write the electronic configuration for ${CN}^{-}$ using molecular orbital theory.
- $(\sigma_{2s})(\sigma_{2s}^*)(\pi_{2p})^4(\sigma_{2p})^2$

Step 3: Calculate the bond order using the formula: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
- Bonding electrons: 2 ( $\sigma_{2s}$ ) + 4 ( $\pi_{2p}$ ) + 2 ( $\sigma_{2p}$ ) = 8
- Antibonding electrons: 2 ( $\sigma_{2s}^*$ ) = 2
- Bond order = (10 - 4) / 2 = 3

Why other options are wrong:
- Option A: 2.5: Incorrect calculation of bond order.
- Option C: 2: Incorrect calculation of bond order.
- Option D: 3.5: Incorrect calculation of bond order.

Key Concepts:
- Molecular orbital theory
- Bond order
- Electronic configuration

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

Quick Tips:
- Remember the filling order of molecular orbitals.

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Application +4 | -1

In

M{g_2}{C_3},

the number and type of bonds present between carbon atoms are

View answer and explanation

Correct option: C

2 - \sigma ,\,2 - \pi

Option C (

2 - \sigma ,\,2 - \pi

) is correct because in

Mg_2C_3

, the

C_3^{4-}

ion has the structure

^{-}C \equiv C - C^{-}

, which contains two sigma and two pi bonds.

Solution approach

Correct Answer + Core Reason: $2 - \sigma ,\\,2 - \pi$ . The $C_3^{4-}$ ion has the structure $^{-}C \equiv C - C^{-}$ , which contains two sigma and two pi bonds.

Step-by-Step Solution:

Step 1: Determine the structure of $Mg_2C_3$ .

Step 2: $Mg_2C_3$ can be thought of as an ionic compound containing $Mg^{2+}$ and $C_3^{4-}$ ions.

Step 3: The $C_3^{4-}$ ion has the structure $^{-}C \equiv C - C^{-}$ .

Step 4: Count the number of sigma and pi bonds in the $C_3^{4-}$ ion.

Step 5: The triple bond between the first two carbon atoms consists of one sigma and two pi bonds. The single bond between the second and third carbon atoms consists of one sigma bond.

Step 6: Therefore, there are two sigma bonds and two pi bonds in the $C_3^{4-}$ ion.

Why other options are wrong:
- Option A: Incorrect number of sigma and pi bonds.
- Option B: Incorrect number of sigma and pi bonds.
- Option D: Incorrect number of sigma and pi bonds.

Key Concepts:
- Ionic compounds
- Bonding in carbon compounds
- Sigma and pi bonds

Memory Aids:
- A single bond is a sigma bond.
- A double bond is one sigma and one pi bond.
- A triple bond is one sigma and two pi bonds.

Quick Tips:
- Draw the structure of the ion to determine the number of sigma and pi bonds.

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Understanding +4 | -1

\mathrm{PC}{\mathrm{l}_5}

exists but

\mathrm{NC}{\mathrm{l}_5}

does not because:

View answer and explanation

Correct option: A

Nitrogen has no vacant 2

d

-orbitals

The correct answer is A because nitrogen, being a second-period element, lacks available 2d orbitals to expand its octet and form five bonds, unlike phosphorus which has vacant 3d orbitals.

Solution approach

Correct Answer + Core Reason: Nitrogen has no vacant 2 $d$ -orbitals. Nitrogen cannot expand its octet due to the absence of available d-orbitals in its valence shell.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Nitrogen and Phosphorus.

Step 2: Nitrogen (N) has an electronic configuration of $1s^22s^22p^3$ . It has no 2d orbitals.

Step 3: Phosphorus (P) has an electronic configuration of $1s^22s^22p^63s^23p^3$ . It has vacant 3d orbitals.

Step 4: PCl5 can form because phosphorus can expand its octet by utilizing its 3d orbitals to accommodate more than 8 electrons.

Step 5: NCl5 cannot form because nitrogen cannot expand its octet due to the absence of 2d orbitals.

Why other options are wrong:
- Option B: While NCl5 might be unstable if it existed, the primary reason for its non-existence is the lack of available d-orbitals.
- Option C: The size difference between nitrogen and phosphorus is a contributing factor but not the primary reason. The absence of d-orbitals is more crucial.
- Option D: Nitrogen's inertness is not the main reason. It readily forms compounds with other elements.

Key Concepts:
- Electronic configuration
- Octet rule
- Expansion of octet
- Availability of d-orbitals

Memory Aids:
- Remember that second-row elements (like N) cannot have more than 8 electrons in their valence shell.

Quick Tips:
- Consider the electronic configuration and the availability of d-orbitals when determining the ability of an element to form compounds with more than 4 bonds.

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Application +4 | -1

Which ionic compound has the largest lattice energy?

View answer and explanation

Correct option: C

{A} l {~N}

Option C (

{A} l {~N}

) is correct because lattice energy is directly proportional to the product of the charges of the ions. AlN has the highest charges (+3 and -3) among the given options.

Solution approach

Correct Answer + Core Reason: ${A} l {~N}$ . Highest charge leads to the largest lattice energy.

Step-by-Step Solution:

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

Step 2: Lattice energy is directly proportional to the product of the charges of the ions: $U \propto Q_1Q_2$ .

Step 3: Determine the charges of the ions in each compound:
- NaF: +1 and -1
- ${A} l {~F}_3$ : +3 and -1 (x3)
- ${A} l {~N}$ : +3 and -3
- ${MgF}_2$ : +2 and -1 (x2)

Step 4: Calculate the product of the charges for each compound:
- NaF: (1)(1) = 1
- ${A} l {~F}_3$ : (3)(1) = 3
- ${A} l {~N}$ : (3)(3) = 9
- ${MgF}_2$ : (2)(1) = 2

Step 5: AlN has the highest product of charges (9), so it has the largest lattice energy.

Why other options are wrong:
- Option A: NaF has lower charges compared to AlN, resulting in a smaller lattice energy.
- Option B: ${A} l {~F}_3$ has lower charges compared to AlN, resulting in a smaller lattice energy.
- Option D: ${MgF}_2$ has lower charges compared to AlN, resulting in a smaller lattice energy.

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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Understanding +4 | -1

Compound with maximum value of lattice energy is

View answer and explanation

Correct option: C

LiCl

Option C (

LiCl

) is correct because lattice energy is inversely proportional to the size of the ions, and Li+ is the smallest cation among the given options.

Solution approach

Correct Answer + Core Reason: $LiCl$ . Lattice energy is inversely proportional to the size of the ions; $Li^+$ is the smallest cation.

Step-by-Step Solution:

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

Step 2: Identify the factors affecting lattice energy:
- Charge of the ions: Higher charge leads to higher lattice energy.
- Size of the ions: Smaller size leads to higher lattice energy.

Step 3: Compare the given compounds:
- $CsCl$ : $Cs^+$ and $Cl^-$ ions
- $NaCl$ : $Na^+$ and $Cl^-$ ions
- $LiCl$ : $Li^+$ and $Cl^-$ ions
- $KCl$ : $K^+$ and $Cl^-$ ions

Step 4: All compounds have the same anion ( $Cl^-$ ), so the lattice energy depends on the size of the cation.

Step 5: The ionic radii are in the order: $Li^+ < Na^+ < K^+ < Cs^+$ .

Step 6: Therefore, $LiCl$ has the highest lattice energy because $Li^+$ is the smallest cation.

Why other options are wrong:
- **Option A ( $CsCl$ ): Cesium is the largest cation among the given options, leading to the lowest lattice energy.
- Option B ( $NaCl$ ):** Sodium is larger than lithium, so $NaCl$ has lower lattice energy than $LiCl$ .
- **Option D ( $KCl$ ):** Potassium is larger than lithium, so $KCl$ has lower lattice energy than $LiCl$ .

Key Concepts:
- Lattice energy
- Ionic radii
- Charge of ions

Memory Aids:
- Smaller ions = higher lattice energy.

Quick Tips:
- Remember the trends in ionic radii in the periodic table.

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Application +4 | -1

Which oxide of nitrogen is isoelectronic with

\mathrm{C}{\mathrm{O}_2}

?

View answer and explanation

Correct option: B

{\mathrm{N}_2}\mathrm{O}

Option B (

{\mathrm{N}_2}\mathrm{O}

) is correct because it has the same number of electrons as

\mathrm{C}{\mathrm{O}_2}

.

Solution approach

Correct Answer + Core Reason: ${\mathrm{N}_2}\mathrm{O}$ is isoelectronic with $\mathrm{C}{\mathrm{O}_2}$ because they have the same number of electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in $\mathrm{C}{\mathrm{O}_2}$ .\nCarbon (C) has 6 electrons, and Oxygen (O) has 8 electrons. So, $\mathrm{C}{\mathrm{O}_2}$ has $6 + 2 \times 8 = 6 + 16 = 22$ electrons.

Step 2: Calculate the number of electrons in each option.
A. $\mathrm{N}{\mathrm{O}_2}$ : $7 + 2 \times 8 = 7 + 16 = 23$ electrons
B. ${\mathrm{N}_2}\mathrm{O}$ : $2 \times 7 + 8 = 14 + 8 = 22$ electrons
C. $\mathrm{NO}$ : $7 + 8 = 15$ electrons
D. ${\mathrm{N}_2}{\mathrm{O}_2}$ : $2 \times 7 + 2 \times 8 = 14 + 16 = 30$ electrons

Step 3: Identify the species with 22 electrons.
${\mathrm{N}_2}\mathrm{O}$ has 22 electrons, which is the same as $\mathrm{C}{\mathrm{O}_2}$ .

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{O}_2}$ has 23 electrons, not 22.
- Option C: $\mathrm{NO}$ has 15 electrons, not 22.
- Option D: ${\mathrm{N}_2}{\mathrm{O}_2}$ has 30 electrons, not 22.

Key Concepts:
- Isoelectronic species have the same number of electrons.
- Atomic number determines the number of electrons in a neutral atom.

Memory Aids:
- Remember to multiply the number of electrons by the number of atoms of that element in the molecule.

Quick Tips:
- Quickly calculate the number of electrons for each species to find the isoelectronic one.

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Analyzing +4 | -1

Pick the odd one out (The one having zero dipole moment):

View answer and explanation

Correct option: C

\mathrm{BC}{\mathrm{l}_3}

Option C (

\mathrm{BC}{\mathrm{l}_3}

) is correct because it is trigonal planar and symmetrical, resulting in the bond dipoles canceling each other out, leading to a zero dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{BC}{\mathrm{l}_3}$ . It is trigonal planar and symmetrical.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule.
- NH3: Pyramidal
- H2O: Bent
- BCl3: Trigonal planar
- SO2: Bent

Step 2: Determine the polarity of each bond and the overall dipole moment of each molecule.
- NH3: Polar bonds, non-zero dipole moment.
- H2O: Polar bonds, non-zero dipole moment.
- BCl3: Polar bonds, but the molecule is symmetrical, so the bond dipoles cancel out, resulting in a zero dipole moment.
- SO2: Polar bonds, non-zero dipole moment.

Step 3: Identify the molecule with a zero dipole moment.

Why other options are wrong:
- **Option A ( $\mathrm{N}{\mathrm{H}_3}$ ): NH3 has a pyramidal shape, and the bond dipoles do not cancel out, resulting in a non-zero dipole moment.
- Option B ( ${\mathrm{H}_2}\mathrm{O}$ ): H2O has a bent shape, and the bond dipoles do not cancel out, resulting in a non-zero dipole moment.
- Option D ( $\mathrm{S}{\mathrm{O}_2}$ ): SO2 has a bent shape, and the bond dipoles do not cancel out, resulting in a non-zero dipole moment.

Key Concepts:
- Dipole moment
- Molecular geometry
- Bond polarity

Memory Aids:
- Symmetrical molecules with polar bonds can have a zero dipole moment if the bond dipoles cancel out.

Quick Tips:**
- Draw the Lewis structure and determine the geometry of the molecule.
- Consider the polarity of the bonds and the symmetry of the molecule.

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Analyzing +4 | -1

The bond order is maximum in

View answer and explanation

Correct option: B

\mathrm{O}_2^ +

Option B (

{\mathrm{O}_2^ + }

) is correct because it has the highest bond order (2.5) compared to the other oxygen species.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ +$ . It has the highest bond order among the given species.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: ${\mathrm{O}_2}$ : Bond order = (10 - 6)/2 = 2

Step 3: $\mathrm{O}_2^ +$ : Bond order = (10 - 5)/2 = 2.5

Step 4: $\mathrm{O}_2^ -$ : Bond order = (10 - 7)/2 = 1.5

Step 5: $\mathrm{O}_2^{2 - }$ : Bond order = (10 - 8)/2 = 1

Step 6: Identify the species with the maximum bond order.

Why other options are wrong:
- Option A: ${\mathrm{O}_2}$ has a bond order of 2, which is less than that of $\mathrm{O}_2^ +$ .
- Option C: $\mathrm{O}_2^ -$ has a bond order of 1.5, which is less than that of $\mathrm{O}_2^ +$ .
- Option D: $\mathrm{O}_2^{2 - }$ has a bond order of 1, which is less than that of $\mathrm{O}_2^ +$ .

Key Concepts:
- Bond Order Calculation

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

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

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Application +4 | -1

According to

\mathrm{MO}

theory the bond orders for

\mathrm{O}_2^{2-}

,

\mathrm{CO}

and

\mathrm{NO}^{+}

respectively, are

View answer and explanation

Correct option: D

1,3 and 3

Option D (1,3 and 3) is correct because the bond orders for

\mathrm{O}_2^{2-}

,

\mathrm{CO}

and

\mathrm{NO}^{+}

are 1, 3, and 3 respectively, based on their molecular orbital configurations.

Solution approach

Correct Answer + Core Reason: 1,3 and 3. The bond order is calculated using molecular orbital theory.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species.
- $\mathrm{O}_2^{2-}$ : 18 electrons
- $\mathrm{CO}$ : 14 electrons
- $\mathrm{NO}^{+}$ : 14 electrons

Step 2: Write the electronic configuration for each species using molecular orbital theory.
- $\mathrm{O}_2^{2-}$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^4$
- $\mathrm{CO}$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4$
- $\mathrm{NO}^{+}$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4$

Step 3: Calculate the bond order for each species using the formula: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
- $\mathrm{O}_2^{2-}$ : (10 - 8) / 2 = 1
- $\mathrm{CO}$ : (10 - 4) / 2 = 3
- $\mathrm{NO}^{+}$ : (10 - 4) / 2 = 3

Step 4: The bond orders are 1, 3, and 3 respectively.

Why other options are wrong:
- Option A: 1,2 and 3: Incorrect bond order for CO.
- Option B: 1,3 and 2: Incorrect bond order for $NO^+$ .
- Option C: 2,3 and 3: Incorrect bond order for $O_2^{2-}$ .

Key Concepts:
- Molecular orbital theory
- Bond order
- Electronic configuration

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

Quick Tips:
- Remember the filling order of molecular orbitals.

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Remembering +4 | -1

Which of the following methods is used for measuring bond length ?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because X-ray diffraction, electron diffraction, and spectroscopic techniques are all used to measure bond length.

Solution approach

Correct Answer + Core Reason: All of these. Bond length can be measured using X-ray diffraction, electron diffraction and spectroscopic techniques.

Step-by-Step Solution:

Step 1: Identify the question type: This is a conceptual question about methods for measuring bond length.

Step 2: Recall the methods used for bond length measurement. X-ray diffraction, electron diffraction, and spectroscopic techniques are all established methods.

Step 3: Conclude that all the given options are correct.

Why other options are wrong:
- Option A: X-ray diffraction is a valid method, but not the only one.
- Option B: Electron diffraction is a valid method, but not the only one.
- Option C: Spectroscopic techniques are valid methods, but not the only ones.
- Option D: This is the correct answer as it encompasses all valid methods.

Key Concepts:
- Bond length measurement
- X-ray diffraction
- Electron diffraction
- Spectroscopic techniques

Memory Aids:
- Remember that multiple techniques are used to determine bond lengths.

Quick Tips:
- When a question offers "All of these" as an option, consider whether multiple options could be correct.

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Understanding +4 | -1

What is the nature of the bond between

B

and

O

in

{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_2}\mathrm{OB}{\mathrm{H}_3}

?

View answer and explanation

Correct option: B

Coordinate covalent

Option B (Coordinate covalent) is correct because in the given compound, the oxygen atom donates a lone pair of electrons to the boron atom, forming a coordinate covalent bond.

Solution approach

Correct Answer + Core Reason: Coordinate covalent. The oxygen atom donates a lone pair of electrons to the boron atom.

Step-by-Step Solution:

Step 1: Analyze the structure of the compound ${\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_2}\mathrm{OB}{\mathrm{H}_3}$ .

Step 2: Identify the atoms involved in bonding: Oxygen (O) and Boron (B).

Step 3: Consider the electronic configuration and bonding capabilities of oxygen and boron.
- Oxygen has 6 valence electrons and can form two covalent bonds. In this case, it is bonded to two ethyl groups ( ${\mathrm{C}_2}{\mathrm{H}_5}$ ). It still has two lone pairs.
- Boron has 3 valence electrons and can form three covalent bonds. In $BH_3$ , it has only 6 electrons in its valence shell and is electron-deficient.

Step 4: The oxygen atom donates a lone pair of electrons to the boron atom to form a coordinate covalent bond.

Why other options are wrong:
- Option A (Covalent): While there are covalent bonds within the ethyl groups and between B and H, the B-O bond is formed by donation from O to B.
- Option C (Ionic bond): The electronegativity difference between B and O is not large enough to form an ionic bond.
- Option D (Banana shaped bond): Banana bonds are typically found in molecules with strained rings, like cyclopropane, and are not relevant here.

Key Concepts:
- Coordinate covalent bond
- Lewis acid and base
- Electron deficiency

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

Quick Tips:
- Look for electron-deficient atoms like B, Al, etc., that can accept electron pairs.

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Analyzing +4 | -1

Bond energies in

NO

,

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

and

\mathrm{N}{\mathrm{O}^ - }

are such as

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{O}^ + } > \mathrm{NO} > \mathrm{N}{\mathrm{O}^ - }

Option C is correct because the bond order increases in

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

and decreases in

\mathrm{N}{\mathrm{O}^ - }

, affecting bond energy.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{O}^ + } > \mathrm{NO} > \mathrm{N}{\mathrm{O}^ - }$ . Bond energy is directly proportional to bond order.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of NO: The total number of electrons in NO is 7 (N) + 8 (O) = 15.

Step 2: Write the molecular orbital configuration of NO: $(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1$

Step 3: Calculate the bond order of NO: Bond order = $\frac{{Number\,of\,bonding\,electrons - Number\,of\,antibonding\,electrons}}{{2}} = \frac{{10 - 5}}{{2}} = 2.5$

Step 4: Determine the electronic configuration of $\mathrm{N}{\mathrm{O}^ + }$ : The total number of electrons in $\mathrm{N}{\mathrm{O}^ + }$ is 15 - 1 = 14.

Step 5: Write the molecular orbital configuration of $\mathrm{N}{\mathrm{O}^ + }$ : $(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2$

Step 6: Calculate the bond order of $\mathrm{N}{\mathrm{O}^ + }$ : Bond order = $\frac{{10 - 4}}{{2}} = 3$

Step 7: Determine the electronic configuration of $\mathrm{N}{\mathrm{O}^ - }$ : The total number of electrons in $\mathrm{N}{\mathrm{O}^ - }$ is 15 + 1 = 16.

Step 8: Write the molecular orbital configuration of $\mathrm{N}{\mathrm{O}^ - }$ : $(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1$

Step 9: Calculate the bond order of $\mathrm{N}{\mathrm{O}^ - }$ : Bond order = $\frac{{10 - 6}}{{2}} = 2$

Step 10: Compare the bond orders: $\mathrm{N}{\mathrm{O}^ + }$ (3) > NO (2.5) > $\mathrm{N}{\mathrm{O}^ - }$ (2).

Step 11: Since bond energy is directly proportional to bond order, the correct order of bond energies is: $\mathrm{N}{\mathrm{O}^ + } > \mathrm{NO} > \mathrm{N}{\mathrm{O}^ - }$ .

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

Key Concepts:
- Molecular orbital theory
- Bond order
- Relationship between bond order and bond energy

Memory Aids:
- Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
- Higher bond order means stronger bond and higher bond energy.

Quick Tips:
- Calculate bond order to compare bond strengths.
- Remember the molecular orbital filling order.

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Application +4 | -1

The interatomic distances in

{H_2}

and

C{l_2}

molecules are 74 and 198 pm respectively. The bond length of HCl is:

View answer and explanation

Correct option: B

136 pm

Option B (136 pm) is correct because the bond length of HCl can be approximated by summing the covalent radii of H and Cl, derived from the given bond lengths of H2 and Cl2.

Solution approach

Correct Answer + Core Reason: 136 pm. 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 radius of Hydrogen ( ${H_2}$ ).

Step 2: Given the interatomic distance in ${H_2}$ is 74 pm, the covalent radius of H is 74/2 = 37 pm.

Step 3: Determine the covalent radius of Chlorine ( $C{l_2}$ ).

Step 4: Given the interatomic distance in $C{l_2}$ is 198 pm, the covalent radius of Cl is 198/2 = 99 pm.

Step 5: Calculate the bond length of HCl by adding the covalent radii of H and Cl.\nBond length of HCl = Covalent radius of H + Covalent radius of Cl = 37 pm + 99 pm = 136 pm.

Why other options are wrong:
- Option A: 272 pm: This is roughly twice the correct bond length, suggesting the sum of diameters instead of radii was used.
- Option C: 124 pm: This value is incorrect and doesn't follow from a simple addition or subtraction of the given values.
- Option D: 248 pm: This is close to the sum of the bond lengths of H2 and Cl2, which is not the correct approach to estimate the HCl bond length.

Key Concepts:
- Covalent Radius
- Bond Length
- Estimation of Bond Length in Diatomic Molecules

Memory Aids:
- Remember that covalent radius is half the bond length in a homonuclear diatomic molecule.
- Bond length of a heteronuclear diatomic molecule is approximately the sum of the covalent radii of the constituent atoms.

Quick Tips:
- Always divide the bond length of homonuclear diatomic molecules by 2 to find the covalent radius before estimating the bond length of a heteronuclear diatomic molecule.
- Ensure units are consistent throughout the calculation.

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Remembering +4 | -1

The energy of antibonding molecular orbitals is:

View answer and explanation

Correct option: A

Greater than the bonding

M.O.

Option A (Greater than the bonding

M.O.

) is correct because antibonding molecular orbitals are higher in energy than their corresponding bonding molecular orbitals.

Solution approach

Correct Answer + Core Reason: Greater than the bonding $M.O.$ . Antibonding orbitals are always higher in energy than bonding orbitals.

Step-by-Step Solution:

Step 1: Recall the energy levels of bonding and antibonding molecular orbitals.

Step 2: When atomic orbitals combine to form molecular orbitals, they form both bonding and antibonding orbitals.

Step 3: Bonding orbitals are lower in energy and contribute to the stability of the molecule.

Step 4: Antibonding orbitals are higher in energy and decrease the stability of the molecule.

Why other options are wrong:
- **Option B (Smaller than the bonding $M.O.$ ): This is incorrect because bonding orbitals are lower in energy.
- Option C (Equal to that of bonding $M.O.$ ): This is incorrect because bonding and antibonding orbitals have different energy levels.
- Option D (None of the above): This is incorrect because antibonding orbitals have higher energy than bonding orbitals.

Key Concepts:
- Molecular Orbital Theory
- Bonding and Antibonding Orbitals
- Energy Levels

Memory Aids:
- Antibonding = Anti-stability = Higher Energy

Quick Tips:**
- Remember that antibonding orbitals are always higher in energy than bonding orbitals.

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Application +4 | -1

The order of bond order of

C_2^{2 - },\,N_2^{2 - },O_2^{2 - }

is

View answer and explanation

Correct option: A

C_2^{2 - } > \,N_2^{2 - } > O_2^{2 - }

Option A (

C_2^{2 - } > \,N_2^{2 - } > O_2^{2 - }

) is correct because bond order is inversely proportional to the number of antibonding electrons.

Solution approach

Correct Answer + Core Reason: $C_2^{2 - } > \,N_2^{2 - } > O_2^{2 - }$ . Bond order is inversely proportional to the number of antibonding electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each species:
- $C_2^{2 - }$ : 12 + 2 = 14 electrons
- $N_2^{2 - }$ : 14 + 2 = 16 electrons
- $O_2^{2 - }$ : 16 + 2 = 18 electrons

Step 2: Write the electronic configuration for each species:
- $C_2^{2 - }$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})$ - Bond order = (10-4)/2 = 3
- $N_2^{2 - }$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})(\pi_{2p}^{\,*})$ - Bond order = (10-6)/2 = 2
- $O_2^{2 - }$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})(\pi_{2p}^{\,*})(\sigma_{2p}^{\,*})$ - Bond order = (10-8)/2 = 1

Step 3: Compare the bond orders: 3 > 2 > 1

Step 4: Arrange the species in decreasing order of bond order: $C_2^{2 - } > N_2^{2 - } > O_2^{2 - }$

Why other options are wrong:
- Option B: $C_2^{2 - } < N_2^{2 - } < O_2^{2 - }$ : Incorrect order, bond order decreases.
- Option C: $C_2^{2 - } < O_2^{2 - } < N_2^{2 - }$ : Incorrect order.
- Option D: $N_2^{2 - } > O_2^{2 - } > C_2^{2 - }$ : Incorrect order.

Key Concepts:
- Molecular orbital theory
- Bond order calculation
- Electronic configuration of molecules

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

Quick Tips:
- Calculate the number of electrons and write the electronic configuration to determine the bond order.

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Analyzing +4 | -1

The pair of species having identical shape of both species:

View answer and explanation

Correct option: D

\mathrm{Xe}{\mathrm{F}_2},\mathrm{C}{\mathrm{O}_2}

Option D (

\mathrm{Xe}{\mathrm{F}_2},\mathrm{C}{\mathrm{O}_2}

) is correct because both

XeF_2

and

CO_2

have a linear shape.

Solution approach

Correct Answer + Core Reason: $\mathrm{Xe}{\mathrm{F}_2},\mathrm{C}{\mathrm{O}_2}$ because both have a linear shape.

Step-by-Step Solution:

Step 1: Determine the shape of each species.
$\mathrm{B}{\mathrm{F}_3}$ : Trigonal planar
$\mathrm{PC}{\mathrm{l}_3}$ : Pyramidal
$\mathrm{P}{\mathrm{F}_5}$ : Trigonal bipyramidal
$\mathrm{I}{\mathrm{F}_5}$ : Square pyramidal
$\mathrm{C}{\mathrm{F}_4}$ : Tetrahedral
$\mathrm{S}{\mathrm{F}_4}$ : See-saw
$\mathrm{Xe}{\mathrm{F}_2}$ : Linear
$\mathrm{C}{\mathrm{O}_2}$ : Linear

Step 2: Identify the pair with identical shapes.
$\mathrm{Xe}{\mathrm{F}_2}$ and $\mathrm{C}{\mathrm{O}_2}$ are both linear.

Why other options are wrong:
- Option A: $\mathrm{B}{\mathrm{F}_3},\mathrm{PC}{\mathrm{l}_3}$ : Trigonal planar and pyramidal, respectively.
- Option B: $\mathrm{P}{\mathrm{F}_5},\mathrm{I}{\mathrm{F}_5}$ : Trigonal bipyramidal and square pyramidal, respectively.
- Option C: $\mathrm{C}{\mathrm{F}_4},\mathrm{S}{\mathrm{F}_4}$ : Tetrahedral and see-saw, respectively.

Key Concepts:
- VSEPR theory
- Molecular shapes

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

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

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Remembering +4 | -1

The C–C bond dissociation energy in kcal/mol
is–

View answer and explanation

Correct option: C

81

Option C (81) is correct because the typical C-C single bond dissociation energy is approximately 81 kcal/mol.

Solution approach

Correct Answer + Core Reason: 81. The typical C-C single bond dissociation energy is approximately 81 kcal/mol.

Step-by-Step Solution:

Step 1: Recall the definition of bond dissociation energy. Bond dissociation energy is the energy required to break a specific bond in one mole of gaseous molecules.

Step 2: Consider the typical bond dissociation energy of a C-C single bond. The bond dissociation energy of a C-C single bond is approximately 81 kcal/mol.

Why other options are wrong:
- Option A (8.1): This value is too low for a C-C single bond dissociation energy.
- Option B (0.8): This value is extremely low and not representative of any typical chemical bond.
- Option D (810): This value is too high for a C-C single bond dissociation energy. It's more in the range of multiple bonds or very strong single bonds.

Key Concepts:
- Bond Dissociation Energy
- C-C Single Bond

Memory Aids:
- Remember the approximate bond dissociation energy of common bonds.

Quick Tips:
- The C-C single bond dissociation energy is a common value to remember.

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Remembering +4 | -1

Electrons in a paramagnetic compound are:

View answer and explanation

Correct option: B

Unpaired

Option B (Unpaired) is correct because paramagnetic compounds have unpaired electrons, which cause them to be attracted to a magnetic field.

Solution approach

Correct Answer + Core Reason: Unpaired. Paramagnetic compounds contain unpaired electrons.

Step-by-Step Solution:

Step 1: Define paramagnetism.

Step 2: Paramagnetism is a property of materials that are attracted to a magnetic field.

Step 3: This attraction is due to the presence of unpaired electrons in the material.

Step 4: Unpaired electrons have a net magnetic moment, which interacts with the external magnetic field.

Why other options are wrong:
- Option A (Shared): Sharing of electrons occurs in covalent bonding, but it doesn't directly relate to paramagnetism.
- Option C (Donated): Donation of electrons occurs in ionic bonding, but it doesn't directly relate to paramagnetism.
- Option D (Paired): Paired electrons have opposite spins, which cancel out their magnetic moments, leading to diamagnetism (repulsion from a magnetic field).

Key Concepts:
- Paramagnetism
- Unpaired Electrons
- Magnetic Moment
- Diamagnetism

Memory Aids:
- Para = Unpaired
- Dia = Paired

Quick Tips:
- Remember that unpaired electrons are responsible for paramagnetism.

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Application +4 | -1

\mathrm{O}_2^{2 + }

has a bond order of

View answer and explanation

Correct option: C

3

Option C (3) is correct because the bond order of

\mathrm{O}_2^{2 + }

is calculated using Molecular Orbital Theory, resulting in a bond order of 3.

Solution approach

Correct Answer + Core Reason: 3. The bond order of $\mathrm{O}_2^{2 + }$ is calculated using Molecular Orbital Theory.

Step-by-Step Solution:

Step 1: Determine the number of electrons in the $\mathrm{O}_2^{2 + }$ ion.

Step 2: Oxygen (O) has 8 electrons. Therefore, $\mathrm{O}_2$ has 16 electrons. $\mathrm{O}_2^{2 + }$ has 16 - 2 = 14 electrons.

Step 3: Write the electronic configuration of $\mathrm{O}_2^{2 + }$ using Molecular Orbital Theory (MOT).

Step 4: The electronic configuration of $\mathrm{O}_2^{2 + }$ is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4$ .

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

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

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

Step 8: Bond order = (8 - 2) / 2 = 6 / 2 = 3.

Why other options are wrong:
- Option A: A bond order of 1 is too low for $\mathrm{O}_2^{2 + }$ , which has a significant amount of multiple bonding character.
- Option B: A bond order of 2 is also too low for $\mathrm{O}_2^{2 + }$ , which has a significant amount of multiple bonding character.
- Option D: A bond order of 4 is too high for $\mathrm{O}_2^{2 + }$ , as it does not have a quadruple bond.

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

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

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

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Understanding +4 | -1

Which is not an exception to octet rule?

View answer and explanation

Correct option: B

\mathrm{SnC}{\mathrm{l}_4}

Option B (

\mathrm{SnC}{\mathrm{l}_4}

) is correct because Sn in SnCl4 has a complete octet and does not violate the octet rule.

Solution approach

Correct Answer + Core Reason: $\mathrm{SnC}{\mathrm{l}_4}$ . Tin has a complete octet in $SnCl_4$ .

Step-by-Step Solution:

Step 1: Define the octet rule: Atoms tend to gain, lose, or share electrons to achieve a full outer shell of eight electrons.

Step 2: Analyze each compound:
- $\mathrm{B}{\mathrm{F}_3}$ : Boron has only 6 electrons around it (octet incomplete).
- $\mathrm{SnC}{\mathrm{l}_4}$ : Tin has 8 electrons around it (octet complete).
- $\mathrm{Be}{\mathrm{I}_2}$ : Beryllium has only 4 electrons around it (octet incomplete).
- $\mathrm{Cl}{\mathrm{O}_2}$ : Chlorine has 7+2 = 9 electrons around it (odd electron molecule).

Step 3: Identify the compound that follows the octet rule: $\mathrm{SnC}{\mathrm{l}_4}$ .

Why other options are wrong:
- Option A: $\mathrm{B}{\mathrm{F}_3}$ : Boron is electron deficient and does not have a complete octet.
- Option C: $\mathrm{Be}{\mathrm{I}_2}$ : Beryllium is electron deficient and does not have a complete octet.
- Option D: $\mathrm{Cl}{\mathrm{O}_2}$ : Chlorine has an odd number of electrons and cannot satisfy the octet rule.

Key Concepts:
- Octet rule
- Electron deficiency
- Odd electron molecules

Memory Aids:
- Remember exceptions to the octet rule: electron deficient species (Be, B), expanded octets (P, S, Cl), and odd electron species.

Quick Tips:
- Draw Lewis structures to determine the number of electrons around the central atom.

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Understanding +4 | -1

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

is:

View answer and explanation

Correct option: B

Pentagonal pyramidal

Option B (Pentagonal pyramidal) is correct because

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

has a distorted octahedral geometry due to the presence of a lone pair, resulting in a pentagonal pyramidal shape.

Solution approach

Correct Answer + Core Reason: Pentagonal pyramidal. XeF6 has a lone pair which distorts the octahedral shape into a pentagonal pyramidal shape.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons of the central atom (Xe).

Step 2: Xenon has 8 valence electrons.

Step 3: Determine the number of atoms bonded to the central atom (F).

Step 4: Xenon is bonded to 6 fluorine atoms.

Step 5: Calculate the steric number: Steric number = number of bonded atoms + number of lone pairs = 6 + 1 = 7.

Step 6: Determine the geometry based on the steric number and the presence of a lone pair.

Step 7: A steric number of 7 with one lone pair results in a distorted octahedral geometry, which is described as pentagonal pyramidal.

Why other options are wrong:
- Option A (Octahedral): An ideal octahedral geometry would have a steric number of 6 with no lone pairs.
- Option C (Planar): Planar geometries typically have steric numbers of 3 or 4 with specific arrangements of bonded atoms and lone pairs.
- Option D (tetrahedral): Tetrahedral geometry corresponds to a steric number of 4 with no lone pairs.

Key Concepts:
- VSEPR Theory
- Steric Number
- Molecular Geometry
- Lone Pairs

Memory Aids:
- Steric Number 7 with 1 lone pair = Pentagonal Pyramidal

Quick Tips:
- Consider the effect of lone pairs on the geometry of the molecule.

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Remembering +4 | -1

Which is the weakest among the following types of bond

View answer and explanation

Correct option: D

Hydrogen bond

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

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the types of bonds listed: Ionic, Metallic, Covalent, and Hydrogen.

Step 2: Recall the relative strengths of these bonds: Covalent and Ionic bonds are generally the strongest, followed by Metallic bonds, and then Hydrogen bonds.

Step 3: Compare the bond strengths. Hydrogen bonds are intermolecular forces, while the others are intramolecular forces or metallic bonds.

Step 4: Conclude that hydrogen bonds are the weakest.

Why other options are wrong:
- Option A: Ionic bonds are formed through electrostatic attraction between ions and are relatively strong.
- Option B: Metallic bonds are formed through the sharing of electrons in a metal lattice and are also relatively strong.
- Option C: Covalent bonds are formed through the sharing of electrons between atoms and are relatively strong.
- Option D: Hydrogen bonds are weak intermolecular forces.

Key Concepts:
- Types of chemical bonds: Ionic, Covalent, Metallic, Hydrogen.
- Relative bond strengths.
- Intermolecular vs. Intramolecular forces.

Memory Aids:
- Remember the order of bond strengths: Covalent ≈ Ionic > Metallic > Hydrogen.

Quick Tips:
- When comparing bond strengths, always consider whether the bond is intramolecular (within a molecule) or intermolecular (between molecules).

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Application +4 | -1

The pair of species with the same bond order is:

View answer and explanation

Correct option: C

\mathrm{O}_2^{2 - },{\mathrm{B}_2}

Option C (

\mathrm{O}_2^{2 - },{\mathrm{B}_2}

) is correct because both species have a bond order of 1.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^{2 - },{\mathrm{B}_2}$ have the same bond order of 1.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species in each pair.\nFor Option A: $\mathrm{NO}$ has 15 electrons, bond order = 2.5. $\mathrm{CO}$ has 14 electrons, bond order = 3.\nFor Option B: ${\mathrm{N}_2}$ has 14 electrons, bond order = 3. ${\mathrm{O}_2}$ has 16 electrons, bond order = 2.\nFor Option C: $\mathrm{O}_2^{2 - }$ has 18 electrons, bond order = 1. ${\mathrm{B}_2}$ has 6 electrons, bond order = 1.\nFor Option D: $\mathrm{O}_2^ +$ has 15 electrons, bond order = 2.5. $\mathrm{N}{\mathrm{O}^ + }$ has 14 electrons, bond order = 3.

Step 2: Identify the pair with the same bond order.\nOnly Option C has the same bond order for both species.

Why other options are wrong:
- Option A: $\mathrm{NO}$ and $\mathrm{CO}$ have different bond orders (2.5 and 3 respectively).
- Option B: ${\mathrm{N}_2}$ and ${\mathrm{O}_2}$ have different bond orders (3 and 2 respectively).
- Option D: $\mathrm{O}_2^ +$ and $\mathrm{N}{\mathrm{O}^ + }$ have different bond orders (2.5 and 3 respectively).

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation

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

Quick Tips:
- Practice calculating bond orders for common diatomic species.

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Remembering +4 | -1

In

{H}_2 {O}_2

molecule, the angle between the two

{O}-{H}

planes is

View answer and explanation

Correct option: A

90^{\circ}

Option A (

90^{\circ}

) is correct because the angle between the two O-H planes in the H2O2 molecule is approximately 90 degrees due to its open book structure.

Solution approach

Correct Answer + Core Reason: $90^{\circ}$ . The angle between the two O-H planes in H2O2 is approximately 90 degrees.

Step-by-Step Solution:

Step 1: Recall the structure of H2O2. It has an open book structure.

Step 2: The dihedral angle between the two O-H planes is approximately 90 degrees.

Why other options are wrong:
- Option B: 101 degrees is not the correct dihedral angle for H2O2.
- Option C: 103 degrees is not the correct dihedral angle for H2O2.
- Option D: 105 degrees is not the correct dihedral angle for H2O2.

Key Concepts:
- Structure of Hydrogen Peroxide (H2O2)
- Dihedral angle

Memory Aids:
- Visualize the open book structure of H2O2 to remember the 90-degree angle.

Quick Tips:
- Remember the specific structure and properties of common molecules like H2O2.

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Application +4 | -1

The electronegativity values of C, H, O, N and S are 2.5, 2.1, 3.5, 3.0 and 2.5 respectively. Which of the following bonds is most polar?

View answer and explanation

Correct option: D

\mathrm{O}

-

\mathrm{H}

Option D (

\mathrm{O}

-

\mathrm{H}

) is correct because the electronegativity difference between oxygen (3.5) and hydrogen (2.1) is the largest among the given options, resulting in the most polar bond.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}$ - $\mathrm{H}$ . The largest electronegativity difference results in the most polar bond.

Step-by-Step Solution:

Step 1: Bond polarity is determined by the difference in electronegativity between the two bonded atoms.

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

Step 3: Calculate the electronegativity differences for each bond:
- $\mathrm{C}$ - $\mathrm{H}$ : |2.5 - 2.1| = 0.4
- $\mathrm{N}$ - $\mathrm{H}$ : |3.0 - 2.1| = 0.9
- $\mathrm{S}$ - $\mathrm{H}$ : |2.5 - 2.1| = 0.4
- $\mathrm{O}$ - $\mathrm{H}$ : |3.5 - 2.1| = 1.4

Step 4: The $\mathrm{O}$ - $\mathrm{H}$ bond has the largest electronegativity difference (1.4), making it the most polar.

Why other options are wrong:
- Option A: The electronegativity difference between C and H is only 0.4, which is the smallest among the options.
- Option B: The electronegativity difference between N and H is 0.9, which is smaller than that between O and H.
- Option C: The electronegativity difference between S and H is 0.4, which is the smallest among the options.

Key Concepts:
- Electronegativity
- Bond polarity

Memory Aids:
- Higher electronegativity difference = more polar bond.

Quick Tips:
- Calculate the electronegativity difference for each bond to determine polarity.

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Remembering +4 | -1

The bond length of

{Br}-{Br}

in

{Br}_2

molecule is equal to

View answer and explanation

Correct option: D

2.28 \,\mathop {\text{A}}\limits^ \circ

Option D (

2.28 \,\mathop {\text{A}}\limits^ \circ

) is the correct bond length for

{Br}-{Br}

in the

{Br}_2

molecule, as it is the experimentally determined value.

Solution approach

Correct Answer + Core Reason: $2.28 \,\mathop {\text{A}}\limits^ \circ$ . The bond length is an experimentally determined value for the ${Br}-{Br}$ bond in ${Br}_2$ .

Step-by-Step Solution:

Step 1: Recall the definition of bond length. Bond length is the average distance between the nuclei of two bonded atoms in a molecule.

Step 2: Recognize that bond lengths are experimentally determined values and are specific to each bond and molecule.

Step 3: The bond length of ${Br}-{Br}$ in ${Br}_2$ is a known value, which is approximately $2.28 \,\mathop {\text{A}}\limits^ \circ$ .

Why other options are wrong:
- Option A: $0.74\,\mathop {\text{A}}\limits^ \circ$ is too small for a ${Br}-{Br}$ bond. This value is closer to the bond length of ${H}-{H}$ .
- Option B: $1.54\,\mathop {\text{A}}\limits^ \circ$ is too small for a ${Br}-{Br}$ bond. This value is closer to the bond length of ${C}-{C}$ single bond.
- Option C: $1.98\,\mathop {\text{A}}\limits^ \circ$ is smaller than the actual bond length of ${Br}-{Br}$ .

Key Concepts:
- Bond length
- Experimental determination of molecular properties

Memory Aids:
- None applicable for specific bond lengths, these are facts to memorize or look up.

Quick Tips:
- Remember that bond lengths are specific to the atoms involved and the type of bond (single, double, triple).

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Remembering +4 | -1

In a crystal cations and anions are held together by

View answer and explanation

Correct option: B

Electrostatic forces

Option B (Electrostatic forces) is correct because cations and anions in a crystal are held together by electrostatic forces (ionic bonds).

Solution approach

Correct Answer + Core Reason: Electrostatic forces. Cations and anions are held together by electrostatic forces.

Step-by-Step Solution:

Step 1: Identify the type of bonding in ionic crystals. Ionic compounds are formed by electrostatic attraction between oppositely charged ions.

Step 2: Cations are positively charged, and anions are negatively charged. The electrostatic force between them is the primary force holding the crystal together.

Why other options are wrong:
- Option A: Electrons are involved in the formation of ions, but they are not the primary force holding the crystal together.
- Option C: Nuclear forces are strong forces within the nucleus of an atom and are not relevant to bonding between ions.
- Option D: Covalent bonds involve sharing of electrons and are not the primary type of bonding in ionic crystals.

Key Concepts:
- Ionic bonding
- Electrostatic forces
- Cations and anions

Memory Aids:
- Remember that ionic bonds are formed due to electrostatic attraction between ions.

Quick Tips:
- Distinguish between ionic, covalent, and metallic bonding.

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Understanding +4 | -1

Which has highest melting point?

View answer and explanation

Correct option: A

\mathrm{LiCl}

Option A (

\mathrm{LiCl}

) is correct because ionic compounds generally have higher melting points than covalent compounds, and LiCl has the most ionic character among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{LiCl}$ . It has the highest ionic character among the given options, leading to a higher melting point.

Step-by-Step Solution:

Step 1: Analyze the Compounds: Consider the bond types in each compound: ionic vs. covalent.

Step 2: Relate Bond Type to Melting Point: Ionic compounds generally have higher melting points due to strong electrostatic forces between ions.

Step 3: Evaluate Each Compound:
- $\mathrm{LiCl}$ : Lithium chloride is primarily ionic.
- $\mathrm{BeCl}_2$ : Beryllium chloride has some covalent character due to the polarizing power of the $Be^{2+}$ ion.
- $\mathrm{BCl}_3$ : Boron trichloride is primarily covalent.
- $\mathrm{CCl}_4$ : Carbon tetrachloride is purely covalent.

Step 4: Conclusion: $\mathrm{LiCl}$ has the highest melting point because it is the most ionic compound among the given options.

Why other options are wrong:
- Option B: $\mathrm{BeCl}_2$ : Has some covalent character, so lower melting point than LiCl.
- Option C: $\mathrm{BCl}_3$ : Primarily covalent, so lower melting point than LiCl.
- Option D: $\mathrm{CCl}_4$ : Purely covalent, so the lowest melting point.

Key Concepts:
- Ionic Bonds
- Covalent Bonds
- Melting Point
- Fajan's Rule

Memory Aids:
- Ionic compounds = high melting point

Quick Tips:
- Remember that ionic compounds generally have higher melting points than covalent compounds.

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Application +4 | -1

Which of the following has fractional bond order?

View answer and explanation

Correct option: D

\mathrm{H}_2^ -

Option D (

\mathrm{H}_2^ -

) is correct because it has a bond order of 0.5, which is a fractional bond order.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_2^ -$ . It has a bond order of 0.5.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species using molecular orbital theory.

Step 2: The electronic configurations are:
$\mathrm{O}_2^{2 + }$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2$
$\mathrm{O}_2^{2 - }$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2$
$\mathrm{F}_2^{2 - }$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2(\sigma_{2p_z}^*)^2$
$\mathrm{H}_2^ -$ : $(\sigma_{1s})^2(\sigma_{1s}^*)^1$

Step 3: Calculate the bond order for each species.\nBond order of $O_2^{2+}$ = (10 - 4) / 2 = 3\nBond order of $O_2^{2-}$ = (10 - 8) / 2 = 1\nBond order of $F_2^{2-}$ = (10 - 8)/2 = 1\nBond order of $H_2^-$ = (2 - 1) / 2 = 0.5

Step 4: Identify the species with a fractional bond order.

Why other options are wrong:
- Option A: $\mathrm{O}_2^{2 + }$ has a bond order of 3.
- Option B: $\mathrm{O}_2^{2 - }$ has a bond order of 1.
- Option C: $\mathrm{F}_2^{2 - }$ has a bond order of 1.

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Bond order

Memory Aids:
- Remember the formula for calculating bond order.

Quick Tips:
- Calculate the bond order to determine if it is fractional.

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