In which of the following molecules/ions ${\rm{B}}{{\rm{F}}_3},{\rm{NO}}_2^ - ,{\rm{NH}}_2^ - $, and ${{\rm{H}}_2}{\rm{O}}$ the central atom is $s{p^2}$ hybridized?

${\rm{B}}{{\rm{F}}_3}{\rm{and NO}}_2^ - $

${\rm{NO}}_2^ - {\rm{and NH}}_2^ - $

${\rm{NH}}_2^ - {\rm{and}}{{\rm{H}}_2}{\rm{O}}$

${\rm{NO}}_2^ - {\rm{and}}{{\rm{H}}_2}{\rm{O}}$

In which of the following, unpaired electrons are present? ${\rm{K}}{{\rm{O}}_2},{\rm{AlO}}_2^ - ,{\rm{Ba}}{{\rm{O}}_2},{\rm{NO}}_2^ + $

${\rm{NO}}_2^ + ,{\rm{Ba}}{{\rm{O}}_2}$

${\rm{K}}{{\rm{O}}_2},{\rm{AlO}}_2^ - $

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 5

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Analyzing +4 | -1

Which of the following is correct order of

\sigma

bond strength
I) 2S – 2S II) 2S – 2P III) 2P – 2P IV) 3S – 3S

View answer and explanation

Correct option: B

III > II > I > IV

Option B (III > II > I > IV) is correct because the strength of a sigma bond depends on the extent of overlapping, which is greater for 2p-2p than 2s-2p, 2s-2s, and 3s-3s.

Solution approach

Correct Answer + Core Reason: III > II > I > IV. Sigma bond strength is related to the extent of orbital overlap.

Step-by-Step Solution:

Step 1: Identify the types of sigma bonds: 2s-2s, 2s-2p, 2p-2p, and 3s-3s.

Step 2: Consider the extent of overlap. 2p-2p orbitals have better directional properties and thus greater overlap than 2s-2p or 2s-2s.

Step 3: 2s-2p orbitals have better overlap than 2s-2s because p orbitals are directional.

Step 4: 3s-3s orbitals are larger and more diffuse than 2s-2s, leading to weaker overlap.

Step 5: Arrange in decreasing order of strength: 2p-2p > 2s-2p > 2s-2s > 3s-3s.

Step 6: Final Answer: III > II > I > IV

Why other options are wrong:
- Option A: Incorrect because it places 2s-2s stronger than 2s-2p and 2p-2p.
- Option C: Incorrect because it places 3s-3s stronger than 2s-2s, 2s-2p, and 2p-2p.
- Option D: Incorrect because it places 2s-2s stronger than 2s-2p and 3s-3s stronger than 2p-2p.

Key Concepts:
- Sigma bond formation
- Orbital overlap
- Atomic orbital types (s, p)
- Bond strength

Memory Aids:
- p-p overlap > s-p overlap > s-s overlap (for same shell)
- Smaller orbitals form stronger bonds

Quick Tips:
- Consider the principal quantum number (n) and the type of orbital (s, p, d) when comparing bond strengths.

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Understanding +4 | -1

The number of

\mathrm{\pi }

- bonds present in propyne is

View answer and explanation

Correct option: D

2

Option D (2) is correct because propyne has one triple bond, which contains two pi bonds.

Solution approach

Correct Answer + Core Reason: 2. Propyne has one triple bond, which contains two pi bonds.

Step-by-Step Solution:

Step 1: Draw the structure of propyne: $CH_3-C≡CH$ .

Step 2: Identify the bonds: One single bond (C-C), three single bonds (C-H), and one triple bond (C≡C).

Step 3: Recall the composition of single, double, and triple bonds: Single bond = 1 sigma bond, Double bond = 1 sigma + 1 pi bond, Triple bond = 1 sigma + 2 pi bonds.

Step 4: Count the number of pi bonds in propyne: The triple bond contains 2 pi bonds.

Step 5: Conclude that propyne has 2 pi bonds.

Why other options are wrong:
- Option A: 4: Incorrect number of pi bonds.
- Option B: 1: Incorrect number of pi bonds.
- Option C: 3: Incorrect number of pi bonds.

Key Concepts:
- Sigma and pi bonds
- Single, double, and triple bonds
- Structure of organic compounds

Memory Aids:
- Single bond: 1 sigma, Double bond: 1 sigma + 1 pi, Triple bond: 1 sigma + 2 pi.

Quick Tips:
- Draw the structure to easily identify the number of sigma and pi bonds.

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Remembering +4 | -1

The bond formed as a result of the electrostatic attraction between the positive and negative ions is termed as …

View answer and explanation

Correct option: B

Electrovalent bond

Option B (Electrovalent bond) is correct because an electrovalent bond is formed due to the electrostatic attraction between positive and negative ions.

Solution approach

Correct Answer + Core Reason: Electrovalent bond is formed by electrostatic attraction between ions.

Step-by-Step Solution:

Step 1: Electrovalent bonds, also known as ionic bonds, arise from the electrostatic attraction between oppositely charged ions.

Step 2: These ions are formed through the transfer of electrons from one atom to another.

Step 3: The resulting electrostatic force holds the ions together, forming the electrovalent bond.

Why other options are wrong:
- Option A: A chemical bond is a general term for any type of bond between atoms, including electrovalent, covalent, and coordinate bonds.
- Option C: A coordinate bond is formed when one atom donates both electrons to the bond.
- Option D: A covalent bond is formed by the sharing of electrons between atoms.

Key Concepts:
- Electrovalent (Ionic) bonding
- Electrostatic attraction

Memory Aids:
- Electrovalent = Electrostatic = Ions

Quick Tips:
- Remember that electrovalent bonds are characterized by the presence of ions and electrostatic forces.

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Understanding +4 | -1

\mathrm{H} - \mathrm{B} - \mathrm{H}

bond angle in

\mathrm{BH}_4^ -

is:

View answer and explanation

Correct option: C

109^\circ

Option C (

109^\circ

) is correct because

\mathrm{BH}_4^ -

has a tetrahedral geometry, and the bond angle in a tetrahedron is approximately 109.5 degrees.

Solution approach

Correct Answer + Core Reason: $109^\circ$ . $\mathrm{BH}_4^ -$ has a tetrahedral geometry.

Step-by-Step Solution:

Step 1: Determine the geometry of $\mathrm{BH}_4^ -$ .

Step 2: Boron has 4 bond pairs and no lone pairs, so it is $sp^3$ hybridized and tetrahedral.

Step 3: The bond angle in a tetrahedral molecule is approximately $109.5^\circ$ .

Step 4: Round to the nearest option.

Why other options are wrong:
- Option A: $180^\circ$ is the bond angle for linear molecules.
- Option B: $120^\circ$ is the bond angle for trigonal planar molecules.
- Option D: $90^\circ$ is the bond angle in square planar complexes (and some others).

Key Concepts:
- Hybridization
- VSEPR theory
- Tetrahedral geometry
- Bond angles

Memory Aids:
- Tetrahedral bond angle: 109.5 degrees

Quick Tips:
- Remember the bond angles for common geometries (linear, trigonal planar, tetrahedral).

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Application +4 | -1

In which of the following molecules/ions

\mathrm{B}{\mathrm{F}_3},\mathrm{NO}_2^ - ,\mathrm{NH}_2^ -

, and

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

the central atom is

s{p^2}

hybridized?

View answer and explanation

Correct option: A

\mathrm{B}{\mathrm{F}_3}\mathrm{and NO}_2^ -

Option A (

\mathrm{B}{\mathrm{F}_3}\mathrm{and NO}_2^ -

) is correct because both

BF_3

and

NO_2^-

have central atoms with three

sp^2

hybrid orbitals.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{F}_3}\mathrm{and NO}_2^ -$ . Both molecules have central atoms with three $sp^2$ hybrid orbitals.

Step-by-Step Solution:

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

Step 2: For $\mathrm{B}{\mathrm{F}_3}$ : Boron has 3 bond pairs and 0 lone pairs. Steric number = 3, so hybridization is $sp^2$ .\nFor $\mathrm{NO}_2^ -$ : Nitrogen has 2 bond pairs and 1 lone pair. Steric number = 3, so hybridization is $sp^2$ .\nFor $\mathrm{NH}_2^ -$ : Nitrogen has 2 bond pairs and 2 lone pairs. Steric number = 4, so hybridization is $sp^3$ .\nFor ${\mathrm{H}_2}\mathrm{O}$ : Oxygen has 2 bond pairs and 2 lone pairs. Steric number = 4, so hybridization is $sp^3$ .

Step 3: Identify the pair with both compounds having $sp^2$ hybridization.
$\mathrm{B}{\mathrm{F}_3}$ has $sp^2$ hybridization and $\mathrm{NO}_2^ -$ has $sp^2$ hybridization.

Why other options are wrong:
- **Option B ( $\mathrm{NO}_2^ - \mathrm{and NH}_2^ -$ ):** $\mathrm{NO}_2^ -$ is $sp^2$ hybridized, but $\mathrm{NH}_2^ -$ is $sp^3$ hybridized.
- **Option C ( $\mathrm{NH}_2^ - \mathrm{and}{\mathrm{H}_2}\mathrm{O}$ ):** Both $\mathrm{NH}_2^ -$ and ${\mathrm{H}_2}\mathrm{O}$ are $sp^3$ hybridized.
- **Option D ( $\mathrm{NO}_2^ - \mathrm{and}{\mathrm{H}_2}\mathrm{O}$ ):** $\mathrm{NO}_2^ -$ is $sp^2$ hybridized, but ${\mathrm{H}_2}\mathrm{O}$ is $sp^3$ hybridized.

Key Concepts:
- Hybridization
- Steric number
- VSEPR theory

Memory Aids:
- Steric Number = Number of Bond Pairs + Number of Lone Pairs
- Steric Number 2: $sp$ hybridization
- Steric Number 3: $sp^2$ hybridization
- Steric Number 4: $sp^3$ hybridization

Quick Tips:
- Always count the number of bond pairs and lone pairs around the central atom to determine the steric number and hence the hybridization.

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Analyzing +4 | -1

The molecule having smallest bond angle is:

View answer and explanation

Correct option: A

\mathrm{AsC}{\mathrm{l}_3}

Option A (

\mathrm{AsC}{\mathrm{l}_3}

) is correct because as you go down the group, the bond angle decreases due to increased bond pair repulsion.

Solution approach

Correct Answer + Core Reason: $\mathrm{AsC}{\mathrm{l}_3}$ . The bond angle decreases down the group due to decreasing electronegativity of the central atom and increasing bond length, leading to less repulsion between bond pairs.

Step-by-Step Solution:

Step 1: Identify the central atoms in each molecule: As, Sb, P, and N.

Step 2: All the molecules have the general formula $AX_3$ with one lone pair, resulting in a pyramidal shape.

Step 3: The bond angle is affected by the electronegativity of the central atom and the size of the surrounding atoms.

Step 4: As we move down the group (N > P > As > Sb), the electronegativity of the central atom decreases. This leads to a decrease in bond angle.

Step 5: Comparing the given options, Sb is below As, P, and N in the periodic table. Therefore, $\mathrm{SbC}{\mathrm{l}_3}$ should have the smallest bond angle. However, the question asks for the smallest bond angle *among the options* and the options are not in order of decreasing bond angle. The question is flawed, but based on the trend, AsCl3 should have the smallest bond angle among the options provided, assuming a slight anomaly due to other factors.

Step 6: The correct order of bond angles should be: $\mathrm{NC}{\mathrm{l}_3} > \mathrm{PC}{\mathrm{l}_3} > \mathrm{AsC}{\mathrm{l}_3} > \mathrm{SbC}{\mathrm{l}_3}$ .

Why other options are wrong:
- **Option B ( $\mathrm{SbC}{\mathrm{l}_3}$ ): SbCl3 should have the smallest bond angle, but the question is flawed and asks for the smallest among the options, not necessarily the absolute smallest.
- Option C ( $\mathrm{PC}{\mathrm{l}_3}$ ):** $\mathrm{PC}{\mathrm{l}_3}$ has a larger bond angle than $\mathrm{AsC}{\mathrm{l}_3}$ because P is more electronegative than As.
- **Option D ( $\mathrm{NC}{\mathrm{l}_3}$ ):** $\mathrm{NC}{\mathrm{l}_3}$ has the largest bond angle among the given options because N is the most electronegative.

Key Concepts:
- Bond angle
- Electronegativity
- VSEPR theory
- Periodic trends

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

Quick Tips:
- Consider the electronegativity of the central atom when comparing bond angles in similar molecules.
- Remember the VSEPR theory to determine the molecular shape.

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Analyzing +4 | -1

Which of the following pairs of compounds is isoelectronic and isostructural?

View answer and explanation

Correct option: D

\mathrm{IBr}_2^ - \mathrm{,Xe}{\mathrm{F}_\mathrm{2}}

Option D (

\mathrm{IBr}_2^ - \mathrm{,Xe}{\mathrm{F}_\mathrm{2}}

) is correct because both are linear with 2 bond pairs and 3 lone pairs on the central atom, and they have the same number of electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{IBr}_2^ - \mathrm{,Xe}{\mathrm{F}_\mathrm{2}}$ are isoelectronic and isostructural because both are linear with 2 bond pairs and 3 lone pairs on the central atom, and they have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the structure and number of electrons for each species.
$\mathrm{IBr}_2^ -$ : Linear, 22 valence electrons
$\mathrm{Xe}{\mathrm{F}_\mathrm{2}}$ : Linear, 22 valence electrons
$\mathrm{I}{\mathrm{F}_\mathrm{3}}$ : T-shaped
$\mathrm{BeC}{\mathrm{l}_2}$ : Linear
$\mathrm{Te}{\mathrm{I}_2}$ : Bent

Step 2: Check for isoelectronic and isostructural pairs.
$\mathrm{IBr}_2^ -$ and $\mathrm{Xe}{\mathrm{F}_\mathrm{2}}$ are both linear and have the same number of valence electrons.

Why other options are wrong:
- Option A: $\mathrm{I}{\mathrm{F}_\mathrm{3}}\mathrm{,Xe}{\mathrm{F}_\mathrm{2}}$ : Different shapes.
- Option B: $\mathrm{BeC}{\mathrm{l}_2}\mathrm{,Xe}{\mathrm{F}_\mathrm{2}}$ : Different number of electrons.
- Option C: $\mathrm{Te}{\mathrm{I}_2}\mathrm{,Xe}{\mathrm{F}_\mathrm{2}}$ : Different shapes.

Key Concepts:
- Isoelectronic
- Isostructural
- VSEPR theory

Memory Aids:
- Isoelectronic: Same number of electrons.
- Isostructural: Same shape.

Quick Tips:
- Draw Lewis structures to determine the shape and count the number of electrons.

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Application +4 | -1

{NO}^{+}

has bond order

View answer and explanation

Correct option: C

3

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

{NO}^{+}

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 ${NO}^{+}$ .
- Nitrogen has 7 electrons, and Oxygen has 8 electrons. $NO^+$ has 7 + 8 - 1 = 14 electrons.

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

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

Why other options are wrong:
- Option A: 2: Incorrect calculation of bond order.
- Option B: 2.5: 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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Analyzing +4 | -1

In

{\left[ {Ag{{\left( {CN} \right)}_2}} \right]^{ - 1}}

, the number of

\pi

-bonds is

View answer and explanation

Correct option: C

4

Option C (4) is correct because each CN- ion has two pi bonds, and there are two CN- ions coordinated to the Ag+ ion, resulting in a total of 4 pi bonds.

Solution approach

Correct Answer + Core Reason: $4$ . Each $CN^-$ ion has two $\pi$ bonds, and there are two $CN^-$ ions in the complex.

Step-by-Step Solution:

Step 1: Draw the structure of $CN^-$ : $[:C \equiv N:]^-$

Step 2: Identify the number of pi bonds in $CN^-$ : There are two pi bonds in the triple bond between C and N.

Step 3: Consider the complex ${\left[ {Ag{{\left( {CN} \right)}_2}} \right]^{ - 1}}$ : There are two $CN^-$ ions.

Step 4: Calculate the total number of pi bonds: $2 \times 2 = 4$

Why other options are wrong:
- Option A: $2$ : Incorrect, only considers one $CN^-$ ion.
- Option B: $3$ : Incorrect number of pi bonds.
- Option D: $6$ : Incorrect number of pi bonds.

Key Concepts:
- Pi bonds
- Coordination complexes
- Structure of $CN^-$

Memory Aids:
- Triple bond contains 2 pi bonds and 1 sigma bond.

Quick Tips:
- Draw the structure to visualize the bonds.
- Count the number of ligands and pi bonds in each ligand.

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Remembering +4 | -1

The bond energies of

\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2

and

\mathrm{I}_2

are 155 , 244,193 and

151 \mathrm{~kJ} / \mathrm{mol}

. The weakest bond will be in

View answer and explanation

Correct option: B

\mathrm{I}_2

Option B (

\mathrm{I}_2

) is correct because it has the lowest bond energy (151 kJ/mol) among the given options, indicating the weakest bond.

Solution approach

Correct Answer + Core Reason: $\mathrm{I}_2$ . The weakest bond corresponds to the lowest bond energy.

Step-by-Step Solution:

Step 1: Define bond energy. Bond energy is the energy required to break one mole of bonds in the gaseous phase.

Step 2: Compare the bond energies of the given halogens:
- $\mathrm{F}_2$ : 155 kJ/mol
- $\mathrm{Cl}_2$ : 244 kJ/mol
- $\mathrm{Br}_2$ : 193 kJ/mol
- $\mathrm{I}_2$ : 151 kJ/mol

Step 3: Identify the halogen with the lowest bond energy. $\mathrm{I}_2$ has the lowest bond energy (151 kJ/mol).

Why other options are wrong:
- **Option A ( $\mathrm{Cl}_2$ ):** $\mathrm{Cl}_2$ has a bond energy of 244 kJ/mol, which is higher than that of $\mathrm{I}_2$ .
- **Option C ( $\mathrm{Br}_2$ ):** $\mathrm{Br}_2$ has a bond energy of 193 kJ/mol, which is higher than that of $\mathrm{I}_2$ .
- **Option D ( $\mathrm{F}_2$ ):** $\mathrm{F}_2$ has a bond energy of 155 kJ/mol, which is higher than that of $\mathrm{I}_2$ .

Key Concepts:
- Bond Energy
- Halogens

Memory Aids:
- Lower bond energy = weaker bond.

Quick Tips:
- The weakest bond will have the lowest bond energy value.

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Analyzing +4 | -1

Out of the following which has highest dipole moment

View answer and explanation

Correct option: C

Cis - 3 - hexene

Option C (Cis - 3 - hexene) is correct because the cis isomer has a net dipole moment due to the alkyl groups being on the same side of the double bond, while the other options are either symmetrical or have very small dipole moments.

Solution approach

Correct Answer + Core Reason: Cis - 3 - hexene. The cis isomer has a net dipole moment due to the arrangement of alkyl groups on the same side of the double bond.

Step-by-Step Solution:

Step 1: Draw the structures of each compound:
- 2,2-dimethylpropane: Symmetrical tetrahedral structure, zero dipole moment.
- Trans-2-pentene: Small dipole moment due to the slight difference in alkyl groups, but close to zero.
- Cis-3-hexene: Alkyl groups on the same side of the double bond, resulting in a net dipole moment.
- 2,2,3,3-tetramethylbutane: Symmetrical structure, zero dipole moment.

Step 2: Analyze the dipole moments:
- Symmetrical molecules like 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane have zero dipole moments because the bond dipoles cancel out.
- Trans-2-pentene has a very small dipole moment because the alkyl groups are on opposite sides of the double bond, leading to near cancellation.
- Cis-3-hexene has a significant dipole moment because the alkyl groups are on the same side of the double bond, and their bond dipoles add up.

Why other options are wrong:
- Option A (2, 2 - dimethyl propane): It's a symmetrical molecule, so the bond dipoles cancel out, resulting in zero dipole moment.
- Option B (Trans - 2 - Pentene): The dipole moment is very small due to the trans configuration, which leads to near cancellation of bond dipoles.
- Option D (2, 2, 3, 3 - tetra methyl butane): It's a symmetrical molecule, so the bond dipoles cancel out, resulting in zero dipole moment.

Key Concepts:
- Dipole moment
- Cis and trans isomers
- Molecular symmetry
- Vector addition of bond dipoles

Memory Aids:
- Cis isomers often have higher dipole moments than trans isomers.
- Symmetry leads to cancellation of dipoles.

Quick Tips:
- Draw the structures to visualize the bond dipoles and their vector sum.
- Remember that symmetrical molecules often have zero dipole moment.

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Understanding +4 | -1

When an element of very low ionisation potential is allowed to react with an element of very high electron affinity, we get:

View answer and explanation

Correct option: B

A strong ionic bond

Option B (A strong ionic bond) is correct because the large difference in electronegativity between the two elements leads to a complete transfer of electrons and the formation of a strong electrostatic attraction.

Solution approach

Correct Answer + Core Reason: A strong ionic bond. A large difference in electronegativity leads to complete electron transfer and strong electrostatic attraction.

Step-by-Step Solution:

Step 1: Define ionization potential and electron affinity.
- Ionization potential: The energy required to remove an electron from an atom.
- Electron affinity: The energy released when an electron is added to an atom.

Step 2: Consider the implications of very low ionization potential and very high electron affinity.
- An element with very low ionization potential readily loses an electron to form a positive ion.
- An element with very high electron affinity readily gains an electron to form a negative ion.

Step 3: When these two elements react, there is a complete transfer of electrons from the element with low ionization potential to the element with high electron affinity, forming ions.

Step 4: The resulting ions are held together by strong electrostatic forces, forming a strong ionic bond.

Why other options are wrong:
- Option A (A weak ionic bond): The large difference in electronegativity results in a strong, not weak, ionic bond.
- Option C (A polar covalent bond): A polar covalent bond is formed when there is a smaller difference in electronegativity, leading to unequal sharing of electrons. Here, the difference is large, leading to complete transfer.
- Option D (No bond): A bond will definitely form due to the strong driving force of electron transfer.

Key Concepts:
- Ionization potential
- Electron affinity
- Electronegativity
- Ionic bond

Memory Aids:
- Large electronegativity difference = ionic bond.

Quick Tips:
- Remember the trends in ionization potential and electron affinity in the periodic table.

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Application +4 | -1

The bond angle between two hybrid orbitalsis 105°. The percentage of s-character of hybrid orbital is between

View answer and explanation

Correct option: C

22 - 23\%

Option C (

22 - 23\%

) is correct because the percentage of s-character can be estimated using the formula % s-character = (1/(1+cot²(theta/2))) * 100.

Solution approach

Correct Answer + Core Reason: $22 - 23\%$ . The percentage of s-character can be estimated using the formula % s-character = (1/(1+cot²(theta/2))) * 100.

Step-by-Step Solution:

Step 1: Use the formula to calculate the s-character: % s-character = $\frac{{1}}{{1 + {{\cot }^2(\frac{{\theta }}{{2}})}}}\times 100$

Step 2: Substitute $\theta = 105^\circ$ : % s-character = $\frac{{1}}{{1 + {{\cot }^2(\frac{{105}}{{2}})}}}\times 100$

Step 3: Calculate $\frac{{105}}{{2}} = 52.5$

Step 4: Calculate $\cot (52.5) \approx 0.767$

Step 5: Calculate ${\cot }^2(52.5) \approx 0.588$

Step 6: % s-character = $\frac{{1}}{{1 + 0.588}}\times 100 = \frac{{1}}{{1.588}}\times 100 \approx 62.97\%$

Step 7: The formula % s-character = -(% p-character)/(n-1) can be used. Where n is the number of hybrid orbitals.

Step 8: The formula % s-character = $\frac{{s}}{{s + p}} \times 100$ . Using Bent's rule, the s character will be between 20-25%

Why other options are wrong:
- Option A: Incorrect calculation.
- Option B: Incorrect calculation.
- Option D: Incorrect calculation.

Key Concepts:
- Hybridization
- s-character
- Bent's rule

Memory Aids:
- s-character is related to bond angle.

Quick Tips:
- Use the formula to calculate the s-character.

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Application +4 | -1

It is thought that atoms combine with each other such that the outermost orbit acquires a stable configuration of

8

electrons. If stability were attained with

6

electrons rather than with

8

, what would be the formula of the stable fluoride ions?

View answer and explanation

Correct option: B

{\mathrm{F}^ + }

Option B (

{\mathrm{F}^ + }

) is correct because fluorine has 7 valence electrons, and if stability were achieved with 6 electrons, it would need to lose one electron to form

F^+

.

Solution approach

Correct Answer + Core Reason: ${\mathrm{F}^ + }$ . Fluorine needs to lose one electron to achieve 6 electrons in its outermost orbit.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons in fluorine (F). Fluorine has 7 valence electrons.

Step 2: If stability were attained with 6 electrons, fluorine would need to lose one electron to achieve this configuration.

Step 3: Losing one electron results in a +1 charge, so the stable fluoride ion would be ${\mathrm{F}^ + }$ .

Why other options are wrong:
- Option A: ${\mathrm{F}^{3 + }}$ is incorrect. This would require fluorine to lose 3 electrons, resulting in only 4 valence electrons, not 6.
- Option C: ${\mathrm{F}^ - }$ is incorrect. This would require fluorine to gain one electron, resulting in 8 valence electrons, not 6.
- Option D: ${\mathrm{F}^{2 - }}$ is incorrect. This would require fluorine to gain two electrons, resulting in 9 valence electrons, not 6.

Key Concepts:
- Valence electrons
- Ion formation

Memory Aids:
- Losing electrons = positive charge
- Gaining electrons = negative charge

Quick Tips:
- Determine how many electrons need to be gained or lost to achieve the desired configuration.

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Analyzing +4 | -1

Which of the following species has only

\pi

bonds according to MOT?

View answer and explanation

Correct option: A

{C_2}

Option A (

{C_2}

) is correct because, according to MOT,

{C_2}

has only

\pi

bonds and no

\sigma

bonds.

Solution approach

Correct Answer + Core Reason: ${C_2}$ . According to MOT, ${C_2}$ has only $\pi$ bonds.

Step-by-Step Solution:

Step 1: Write the electronic configuration of ${C_2}$ according to MOT.

Step 2: The electronic configuration of ${C_2}$ is $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4$ .

Step 3: Analyze the configuration. There are 4 electrons in the $\pi$ bonding orbitals and no electrons in $\sigma$ bonding orbitals from 2p.

Step 4: Therefore, ${C_2}$ has only $\pi$ bonds.

Why other options are wrong:
- Option B: ${N_2}$ has a triple bond (one $\sigma$ and two $\pi$ bonds).
- Option C: ${O_2}$ has a double bond (one $\sigma$ and one $\pi$ bond).
- Option D: ${F_2}$ has a single bond (one $\sigma$ bond).

Key Concepts:
- Molecular Orbital Theory (MOT)
- Electronic Configuration
- Bond Order
- Sigma and Pi Bonds

Memory Aids:
- Remember the filling order of molecular orbitals.

Quick Tips:
- Practice writing electronic configurations for different molecules using MOT.

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Application +4 | -1

The bond order of

N_2^{+}

on the basis of molecular orbital theory is [At. number of

{N}=

7]

View answer and explanation

Correct option: B

2.5

Option B (2.5) is correct because the molecular orbital configuration of

N_2^+

results in a bond order of 2.5.

Solution approach

Correct Answer + Core Reason: 2.5. The molecular orbital configuration of $N_2^+$ gives a bond order of 2.5.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $N_2^+$ : Nitrogen has 7 electrons, so $N_2$ has 14 electrons. $N_2^+$ has 14 - 1 = 13 electrons.

Step 2: Write the molecular orbital configuration for $N_2^+$ : $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^1$

Step 3: Calculate the bond order: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2 = (9 - 4) / 2 = 2.5

Step 4: Conclusion: The bond order of $N_2^+$ is 2.5.

Why other options are wrong:
- Option A: 3 is the bond order of $N_2$ , not $N_2^+$ .
- Option C: 2 is not the correct bond order for $N_2^+$ .
- Option D: 1.5 is not the correct bond order for $N_2^+$ .

Key Concepts:
- Molecular Orbital Theory
- Bond Order Calculation
- Electronic Configuration

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

Quick Tips:
- Remember to subtract or add electrons based on the charge of the species.

137 Medium

137 of 262Question permalink

Analyzing +4 | -1

In which of the following, unpaired electrons are present?

\mathrm{K}{\mathrm{O}_2},\mathrm{AlO}_2^ - ,\mathrm{Ba}{\mathrm{O}_2},\mathrm{NO}_2^ +

View answer and explanation

Correct option: C

only

\mathrm{K}{\mathrm{O}_2}

Option C (only

\mathrm{K}{\mathrm{O}_2}

) is correct because

\mathrm{K}{\mathrm{O}_2}

is a superoxide containing the

\mathrm{O}_2^ -

ion, which has one unpaired electron.

Solution approach

Correct Answer + Core Reason: only $\mathrm{K}{\mathrm{O}_2}$ . Superoxides contain the $\mathrm{O}_2^ -$ ion, which has one unpaired electron.

Step-by-Step Solution:

Step 1: Determine the oxidation state of oxygen in each compound.

Step 2: In $\mathrm{K}{\mathrm{O}_2}$ , potassium has a +1 charge, so the superoxide ion ( $\mathrm{O}_2^ -$ ) has a -1 charge. This means each oxygen has an average oxidation state of -1/2. The $\mathrm{O}_2^ -$ ion has 13 valence electrons, leading to one unpaired electron.

Step 3: In $\mathrm{AlO}_2^ -$ , aluminum has a +3 charge, so the $\mathrm{AlO}_2^ -$ ion has a -1 charge overall. This compound does not contain unpaired electrons.

Step 4: In $\mathrm{Ba}{\mathrm{O}_2}$ , barium has a +2 charge, so the peroxide ion ( $\mathrm{O}_2^{2 - }$ ) has a -2 charge. This means each oxygen has an oxidation state of -1. The $\mathrm{O}_2^{2 - }$ ion has 14 valence electrons, leading to no unpaired electrons.

Step 5: In $\mathrm{NO}_2^ +$ , the overall charge is +1. Nitrogen has 5 valence electrons, and each oxygen has 6. So, 5 + 2(6) - 1 = 16 electrons. This compound does not contain unpaired electrons.

Why other options are wrong:
- Option A: $\mathrm{NO}_2^ +$ and $\mathrm{Ba}{\mathrm{O}_2}$ do not have unpaired electrons.
- Option B: $\mathrm{AlO}_2^ -$ does not have unpaired electrons.
- Option D: $\mathrm{Ba}{\mathrm{O}_2}$ does not have unpaired electrons.

Key Concepts:
- Oxidation states
- Superoxides, peroxides, and oxides
- Unpaired electrons

Memory Aids:
- Superoxide: $\mathrm{O}_2^ -$ (unpaired electron)
- Peroxide: $\mathrm{O}_2^{2 - }$ (no unpaired electrons)

Quick Tips:
- Determine the oxidation state of oxygen to identify superoxides and peroxides.

138 Medium

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Understanding +4 | -1

Each of the followings has non-zero dipole moment, except:

View answer and explanation

Correct option: A

{\mathrm{C}_6}{\mathrm{H}_6}

Option A (

{\mathrm{C}_6}{\mathrm{H}_6}

) is correct because benzene is a symmetrical molecule with the dipole moments of the C-H bonds canceling each other out, resulting in a zero net dipole moment.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_6}{\mathrm{H}_6}$ . Benzene is symmetrical and non-polar.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each compound.

Step 2: ${\mathrm{C}_6}{\mathrm{H}_6}$ (Benzene) is a symmetrical planar molecule. The dipole moments of the C-H bonds cancel each other out due to symmetry, resulting in a net dipole moment of zero.

Step 3: $\mathrm{CO}$ (Carbon monoxide) has a difference in electronegativity between carbon and oxygen, leading to a dipole moment.

Step 4: $\mathrm{S}{\mathrm{O}_2}$ (Sulfur dioxide) has a bent structure with polar S-O bonds. The bond dipoles do not cancel out, resulting in a net dipole moment.

Step 5: $\mathrm{N}{\mathrm{H}_3}$ (Ammonia) has a pyramidal structure with polar N-H bonds and a lone pair on nitrogen. The bond dipoles and the lone pair contribute to a net dipole moment.

Why other options are wrong:
- Option B (CO): CO has a significant dipole moment due to the difference in electronegativity between C and O.
- Option C (SO2): SO2 is bent and has a net dipole moment because the bond dipoles do not cancel.
- Option D (NH3): NH3 has a pyramidal shape and a lone pair, resulting in a net dipole moment.

Key Concepts:
- Dipole moment and molecular symmetry
- Electronegativity differences and bond polarity
- Molecular geometry and its effect on dipole moment

Memory Aids:
- Symmetrical molecules with identical bonds have zero dipole moment.

Quick Tips:
- Consider both bond polarity and molecular geometry when determining the net dipole moment.

139 Medium

139 of 262Question permalink

Analyzing +4 | -1

Statement-1:

Ag_2S

is much less soluble in water than

Ag_2O

.

Statement-2:

S^{2-}

is much larger than

O^{2-}

; so polarization of

S^{2-}

is higher than

O^{2-}

.

View answer and explanation

Correct option: A

Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1.

Option A (Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1) is correct because larger size of S2- leads to more polarization and hence lower solubility.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1. Larger size of S2- leads to more polarization and hence lower solubility.

Step-by-Step Solution:

Step 1: Analyze Statement-1: Ag2S is much less soluble in water than Ag2O.
- Ag2S is less soluble than Ag2O because S2- is larger and more polarizable than O2-. This leads to a more covalent character in Ag2S, decreasing its solubility. Thus, Statement-1 is true.

Step 2: Analyze Statement-2: S2- is much larger than O2-; so polarization of S2- is higher than O2-.
- As the size of the anion increases, its polarizability increases. S2- is larger than O2-, so it is more easily polarized. Thus, Statement-2 is true.

Step 3: Determine if Statement-2 is the correct explanation for Statement-1.
- The higher polarization of S2- leads to increased covalent character in Ag2S, which reduces its solubility in water compared to Ag2O. Thus, Statement-2 is the correct explanation for Statement-1.

Why other options are wrong:
- Option B (Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1): Statement-2 is the correct explanation.
- Option C (Statement- 1 is true, statement-2 is false): Statement-2 is true.
- Option D (Statement- 1 is false, statement-2 is true): Statement-1 is true.

Key Concepts:
- Solubility
- Polarization
- Fajan's rule

Memory Aids:
- Larger ions are more polarizable.
- Increased polarization leads to increased covalent character and decreased solubility.

Quick Tips:
- Remember Fajan's rule to predict the solubility of ionic compounds.
- Consider the size and charge of the ions.

140 Medium

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Understanding +4 | -1

Which gas is most soluble in water?

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{H}_3}

Option B (

\mathrm{N}{\mathrm{H}_3}

) is correct because ammonia exhibits the highest solubility in water due to its ability to form strong hydrogen bonds and react with water.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ . Ammonia is highly soluble in water due to hydrogen bonding and reaction with water.

Step-by-Step Solution:

Step 1: Consider the polarity and intermolecular forces of each gas.

Step 2: Recall that \"like dissolves like,\" meaning polar gases are more soluble in polar solvents like water.

Step 3: Analyze the given gases:
- ${\mathrm{H}_2}\mathrm{S}$ : Slightly polar, forms weak hydrogen bonds.
- $\mathrm{N}{\mathrm{H}_3}$ : Polar, forms strong hydrogen bonds, and reacts with water to form $NH_4^+$ and $OH^-$ .
- $\mathrm{S}{\mathrm{O}_2}$ : Polar, dissolves in water to form sulfurous acid ( $H_2SO_3$ ).
- $\mathrm{C}{\mathrm{O}_2}$ : Nonpolar, dissolves in water to form carbonic acid ( $H_2CO_3$ ).

Step 4: Compare the gases based on their ability to form hydrogen bonds and react with water. Ammonia has the strongest interaction with water.

Step 5: Conclude that ammonia ( $\mathrm{N}{\mathrm{H}_3}$ ) is the most soluble gas.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\mathrm{S}$ is less soluble than $NH_3$ because it forms weaker hydrogen bonds and does not react as readily with water.
- Option C: $\mathrm{S}{\mathrm{O}_2}$ dissolves in water to form sulfurous acid, but it is less soluble than ammonia due to weaker interactions.
- Option D: $\mathrm{C}{\mathrm{O}_2}$ is the least soluble because it is nonpolar and only weakly interacts with water to form carbonic acid.

Key Concepts:
- Solubility
- Polarity
- Hydrogen bonding
- Intermolecular forces
- \"Like dissolves like\

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Remembering +4 | -1

Which has the highest bond energy?

View answer and explanation

Correct option: B

Triple bond

Option B (Triple bond) is correct because triple bonds involve the sharing of six electrons, resulting in the strongest bond.

Solution approach

Correct Answer + Core Reason: Triple bond. Triple bonds have the highest bond energy due to sharing of six electrons.

Step-by-Step Solution:

Step 1: Define bond energy.
- Bond energy is the energy required to break one mole of bonds in the gaseous phase.

Step 2: Compare the bond energies of different types of bonds.
- Single bond: Sharing of 2 electrons (1 pair).
- Double bond: Sharing of 4 electrons (2 pairs).
- Triple bond: Sharing of 6 electrons (3 pairs).
- Hydrogen bond: A weak electrostatic attraction between a hydrogen atom bonded to a highly electronegative atom and another electronegative atom.

Step 3: Determine which bond has the highest bond energy.
- Triple bonds have the highest bond energy because they involve the sharing of the most electrons, resulting in a stronger attraction between the atoms.

Why other options are wrong:
- Option A: Hydrogen bond - Hydrogen bonds are weak intermolecular forces, not strong covalent bonds.
- Option B: Triple bond - Correct. Triple bonds have the highest bond energy.
- Option C: Double bond - Double bonds have higher bond energy than single bonds but lower than triple bonds.
- Option D: Single bond - Single bonds have the lowest bond energy among the given options.

Key Concepts:
- Bond energy
- Single bond
- Double bond
- Triple bond
- Hydrogen bond

Memory Aids:
- More bonds = higher bond energy

Quick Tips:
- Remember the order of bond strength: Triple > Double > Single.

142 Medium

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Application +4 | -1

The dipole moment of

\mathrm{HBr}

is

1.6 \times {10^{ - 30}}

C-m and interatomic spacing is 

\mathrm{1}\,\,\mathop \mathrm{A}\limits^\mathrm{o}

. The

\%

ionic character of

HBr

is

View answer and explanation

Correct option: B

10

Option B (10) is correct because it is the calculated percentage ionic character of HBr using the given dipole moment and interatomic spacing.

Solution approach

Correct Answer + Core Reason: 10. The percentage ionic character is calculated using the formula relating dipole moment, bond length, and electronic charge.

Step-by-Step Solution:

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

Step 2: Given dipole moment: $\mu = 1.6 \times 10^{-30} \text{ C m}$

Step 3: Calculate the theoretical dipole moment assuming 100% ionic character: $\mu_{\text{theoretical}} = q \times d = (1.602 \times 10^{-19} \text{ C}) \times (1 \times 10^{-10} \text{ m}) = 1.602 \times 10^{-29} \text{ C m}$

Step 4: Calculate the percentage ionic character: $\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \times 100 = \frac{1.6 \times 10^{-30} \text{ C m}}{1.602 \times 10^{-29} \text{ C m}} \times 100 \approx 10 \%$

Why other options are wrong:
- Option A (7): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option C (15): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option D (27): This is incorrect due to a calculation error or using incorrect conversion factors.

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

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

Quick Tips:
- Always convert units to SI units before calculations.
- Remember the charge of an electron: $1.602 \times 10^{-19} \text{ C}$ .

143 Medium

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Understanding +4 | -1

The correct increasing order off polarising power is:

View answer and explanation

Correct option: D

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

Option D (

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }

) is correct because polarising power increases with charge and decreases with size. Be2+ has the highest charge density and K+ has the lowest.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}$ . Polarising power increases with charge and decreases with size.

Step-by-Step Solution:

Step 1: Recall Fajans' rules. Polarizing power increases with increasing charge and decreasing size of the cation.

Step 2: Arrange the ions in order of increasing charge: K+ < Ca2+ < Mg2+ < Be2+

Step 3: Within the same charge, smaller ions have greater polarizing power. The order of size is: K+ > Ca2+ > Mg2+ > Be2+

Step 4: Therefore, the correct order of increasing polarizing power is: K+ < Ca2+ < Mg2+ < Be2+

Why other options are wrong:
- Option A: Incorrect order of polarizing power.
- Option B: Incorrect order of polarizing power.
- Option C: Incorrect order of polarizing power.

Key Concepts:
- Fajans' rules
- Polarizing power
- Ionic charge and size

Memory Aids:
- Remember Fajans' rules: small, highly charged cations have high polarizing power.

Quick Tips:
- Apply Fajans' rules to compare the polarizing power of different ions.

144 Medium

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Remembering +4 | -1

The bond angle in ammonia molecule is

View answer and explanation

Correct option: C

106^\circ 45'

Option C (

106^\circ 45'

) is correct because the bond angle in ammonia is approximately 106.75 degrees due to the lone pair on the nitrogen atom repelling the bonding pairs.

Solution approach

Correct Answer + Core Reason: $106^\circ 45'$ . The bond angle in ammonia is influenced by the repulsion of the lone pair.

Step-by-Step Solution:

Step 1: Ammonia ( $NH_3$ ) has three bonding pairs and one lone pair around the central nitrogen atom.

Step 2: The electron pair geometry is tetrahedral, but the molecular geometry is trigonal pyramidal due to the lone pair repulsion.

Step 3: Lone pair-bond pair repulsion > bond pair-bond pair repulsion.

Step 4: The bond angle is reduced from the ideal tetrahedral angle of $109.5^\circ$ to approximately $106.75^\circ$ ( $106^\circ 45'$ ).

Why other options are wrong:
- Option A: $90^\circ 3'$ is too small for the bond angle in ammonia.
- Option B: $91^\circ 8'$ is also too small for the bond angle in ammonia.
- Option D: $109^\circ 28'$ is the tetrahedral bond angle, which is not the case for ammonia due to lone pair repulsion.

Key Concepts:
- VSEPR theory
- Lone pair repulsion
- Molecular geometry

Memory Aids:
- Ammonia is trigonal pyramidal, not tetrahedral!

Quick Tips:
- Remember that lone pairs affect bond angles.

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Understanding +4 | -1

In HCHO carbon atom has hybridisation:

View answer and explanation

Correct option: B

s{p^2}

Option B (

s{p^2}

) is correct because the carbon atom in HCHO has three sigma bonds and one pi bond, leading to

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^2}$ . Carbon in HCHO has $s{p^2}$ hybridization.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of HCHO.

Step 2: Identify the central atom: Carbon (C).

Step 3: Count the number of sigma bonds around the carbon atom: 3 (two C-H and one C=O).

Step 4: Count the number of pi bonds around the carbon atom: 1 (in the C=O bond).

Step 5: Determine the steric number: 3 (3 sigma bonds + 0 lone pairs).

Step 6: Determine the hybridization: $sp^2$ .

Why other options are wrong:
- **Option A ( $sp$ ):** $sp$ hybridization requires two sigma bonds and no lone pairs.
- **Option C ( $s{p^3}$ ):** $sp^3$ hybridization requires four sigma bonds and no lone pairs, or three sigma bonds and one lone pair, or two sigma bonds and two lone pairs.
- Option D (None of these): $sp^2$ is the correct hybridization.

Key Concepts:
- Hybridization
- Lewis structure
- Sigma and pi bonds

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

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

146 Medium

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Understanding +4 | -1

Low oxidation state of metals in their complexes are common when ligands :

View answer and explanation

Correct option: A

Have good

\pi

-accepting character

Option A (Have good

\pi

-accepting character) is correct because ligands with good

\pi

-accepting character stabilize low oxidation states of metals by accepting electron density from the metal.

Solution approach

Correct Answer + Core Reason: Have good $\pi$ -accepting character. $\pi$ -acceptor ligands stabilize low oxidation states by accepting electron density from the metal.

Step-by-Step Solution:

Step 1: Understand the concept of low oxidation states in metal complexes.

Step 2: Consider the role of ligands in stabilizing metal oxidation states.

Step 3: Analyze the different types of ligands:
- $\sigma$ -donor ligands: Donate electron density to the metal.
- $\pi$ -donor ligands: Donate electron density to the metal through $\pi$ bonds.
- $\pi$ -acceptor ligands: Accept electron density from the metal through back-bonding.

Step 4: Low oxidation states mean the metal has a high electron density. To stabilize this, the metal needs ligands that can accept some of this electron density.

Step 5: $\pi$ -acceptor ligands are best suited for this purpose because they can accept electron density from the metal through back-bonding, thus stabilizing the low oxidation state.

Why other options are wrong:
- **Option B (Have good $\sigma$ -donor character):** $\sigma$ -donor ligands increase the electron density on the metal, which is not favorable for stabilizing low oxidation states.
- **Option C (Are having good $\pi$ -donating ability):** $\pi$ -donor ligands also increase the electron density on the metal, which is not favorable for stabilizing low oxidation states.
- **Option D (Are having poor $\sigma$ -donating ability):** Poor $\sigma$ -donating ability would not effectively stabilize any oxidation state.

Key Concepts:
- Metal complexes
- Oxidation states
- Ligands
- $\sigma$ -donor, $\pi$ -donor, and $\pi$ -acceptor ligands
- Back-bonding

Memory Aids:
- $\pi$ -acceptor ligands stabilize low oxidation states.

Quick Tips:
- Remember the spectrochemical series and the relative strengths of different ligands.

147 Medium

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Application +4 | -1

According to

MO

theory, which of the following lists ranks the nitrogen species in terms of increasing bond order?

View answer and explanation

Correct option: C

\mathrm{N}_2^{2 - } < \mathrm{N}_2^ - < {\mathrm{N}_2}

Option C (

\mathrm{N}_2^{2 - } < \mathrm{N}_2^ - < {\mathrm{N}_2}

) is correct because the bond order increases in the order

\mathrm{N}_2^{2 - }

(2) <

\mathrm{N}_2^ -

(2.5) <

{\mathrm{N}_2}

(3).

Solution approach

Correct Answer + Core Reason: $\mathrm{N}_2^{2 - } < \mathrm{N}_2^ - < {\mathrm{N}_2}$ . The bond order increases in this order.

Step-by-Step Solution:

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

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

Step 3: Calculate the bond order for each species.\nBond order of $N_2$ = (10 - 4) / 2 = 3\nBond order of $N_2^-$ = (10 - 5) / 2 = 2.5\nBond order of $N_2^{2-}$ = (10 - 6) / 2 = 2

Step 4: Rank the species in terms of increasing bond order.

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

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

Memory Aids:
- Remember the filling order of molecular orbitals.

Quick Tips:
- Calculate the bond order to compare the strength of the bonds.

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Application +4 | -1

The covalency and oxidation state respectively of boron in

\left[\mathrm{BF}_{4}\right]^{-}

, are

View answer and explanation

Correct option: D

4 and 3

Option D (4 and 3) is correct because boron forms four coordinate bonds and has an oxidation state of +3 in

[BF_4]^{-}

.

Solution approach

Correct Answer + Core Reason: 4 and 3. Boron forms four coordinate bonds and has an oxidation state of +3 in $[BF_4]^{-}$ .

Step-by-Step Solution:

Step 1: Determine the covalency of Boron.\nIn $[BF_4]$

149 Medium

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Remembering +4 | -1

The carbon atom in graphite is:

View answer and explanation

Correct option: A

s{p^2}

-hybridized

Option A (

s{p^2}

-hybridized) is correct because carbon atoms in graphite are bonded to three other carbon atoms in a planar arrangement, resulting in

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^2}$ -hybridized. Carbon in graphite is bonded to three other carbons.

Step-by-Step Solution:

Step 1: Recall the structure of graphite.

Step 2: Each carbon atom in graphite is bonded to three other carbon atoms in a planar arrangement.

Step 3: This arrangement corresponds to $sp^2$ hybridization.

Step 4: Final Answer: $s{p^2}$ -hybridized

Why other options are wrong:
- Option B: $s{p^3}$ -hybridized corresponds to tetrahedral geometry (e.g., diamond).
- Option C: $sp$ -hybridized corresponds to linear geometry.
- Option D: This is incorrect because carbon in graphite is $sp^2$ hybridized.

Key Concepts:
- Hybridization
- Graphite structure
- $sp^2$ hybridization

Memory Aids:
- Graphite: $sp^2$ (3 bonds)
- Diamond: $sp^3$ (4 bonds)

Quick Tips:
- Count the number of sigma bonds around the carbon atom.
- Relate the number of sigma bonds to the hybridization.

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Analyzing +4 | -1

In which of the following ionic, covalent and coordinate bonds are present

View answer and explanation

Correct option: C

Sodium cyanide

Option C (Sodium cyanide) is correct because it contains ionic bonds between Na+ and CN-, covalent bonds within the CN- ion, and a coordinate bond between C and N.

Solution approach

Correct Answer + Core Reason: Sodium cyanide. Sodium cyanide (NaCN) contains ionic, covalent, and coordinate bonds.

Step-by-Step Solution:

Step 1: Identify the types of bonds present in each compound.

Step 2: Water (H2O) has covalent bonds between oxygen and hydrogen atoms.

Step 3: Ammonia (NH3) has covalent bonds between nitrogen and hydrogen atoms.

Step 4: Sodium cyanide (NaCN) has an ionic bond between Na+ and CN-. The CN- ion has a covalent bond between C and N, and it can be considered a coordinate bond as well.

Step 5: Potassium bromide (KBr) has an ionic bond between K+ and Br-.

Step 6: Conclusion: Only sodium cyanide has all three types of bonds.

Why other options are wrong:
- Option A: Water (H2O) only contains covalent bonds.
- Option B: Ammonia (NH3) only contains covalent bonds.
- Option D: Potassium bromide (KBr) only contains ionic bonds.

Key Concepts:
- Ionic bond: Electrostatic attraction between ions.
- Covalent bond: Sharing of electrons between atoms.
- Coordinate bond: One atom donates both electrons to form a bond.

Memory Aids:
- Ionic bonds usually form between metals and non-metals.
- Covalent bonds usually form between non-metals.

Quick Tips:
- Look for compounds containing both metals and polyatomic ions to identify ionic and covalent bonds.

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