CBSE Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure NCERT Solutions

To determine if species are isostructural, we need to examine their shapes and hybridization. This is typically done by determining the number of bond pairs and lone pairs around the central atom.

(a) NF₃ has a central nitrogen atom with 3 bond pairs and 1 lone pair, resulting in a trigonal pyramidal shape and sp³ hybridization. BF₃ has a central boron atom with 3 bond pairs and 0 lone pairs, resulting in a trigonal planar shape and sp² hybridization. Thus, NF₃ and BF₃ are not isostructural.

(b) BF₄^- has a central boron atom with 4 bond pairs and 0 lone pairs, resulting in a tetrahedral shape and sp³ hybridization. NH₄⁺ has a central nitrogen atom with 4 bond pairs and 0 lone pairs, also resulting in a tetrahedral shape and sp³ hybridization. Therefore, BF₄^- and NH₄⁺ are isostructural.

(c) BCl₃ has a central boron atom with 3 bond pairs and 0 lone pairs, giving it a trigonal planar shape and sp² hybridization. BrCl₃ has a central bromine atom with 3 bond pairs and 2 lone pairs, resulting in a T-shaped geometry and sp³d hybridization. Thus, BCl₃ and BrCl₃ are not isostructural.

(d) NH₃ has a central nitrogen atom with 3 bond pairs and 1 lone pair, giving it a trigonal pyramidal shape and sp³ hybridization. NO₃^- has a central nitrogen atom with 3 bond pairs and 0 lone pairs (considering resonance), resulting in a trigonal planar shape and sp² hybridization. Thus, NH₃ and NO₃^- are not isostructural.

The correct isostructural pair is (b) [BF₄^- and NH₄⁺].

The dipole moment ($\mu$) is a measure of the polarity of a molecule. It depends on the bond polarity and the molecular geometry. If the bond dipoles cancel each other out due to symmetry, the net dipole moment is zero.

(a) CO₂ is a linear molecule. Although the C=O bonds are polar, the bond dipoles are equal and opposite, cancelling each other out. Therefore, CO₂ has a zero dipole moment ($\mu = 0$).

(b) HI is a diatomic molecule. The bond is polar due to the electronegativity difference between H and I, resulting in a small dipole moment ($\mu = 0.38 D$).

(c) H₂O is a bent molecule. The O-H bonds are polar, and due to the bent shape and the presence of two lone pairs on the oxygen atom, the bond dipoles do not cancel out. This results in a significant net dipole moment ($\mu = 1.84 D$).

(d) SO₂ is a bent molecule. The S-O bonds are polar. Similar to water, the bent shape prevents the cancellation of bond dipoles, resulting in a net dipole moment ($\mu = 1.62 D$).

Comparing the dipole moments: H₂O (1.84 D) > SO₂ (1.62 D) > HI (0.38 D) > CO₂ (0 D).

Therefore, H₂O has the highest dipole moment among the given options.

We can determine the hybridization of the central nitrogen atom in each species by considering the number of sigma bonds and lone pairs (steric number = sigma bonds + lone pairs).

1. NO₂⁺: The Lewis structure shows a central nitrogen atom double-bonded to two oxygen atoms, with a positive charge. The nitrogen atom forms two double bonds. Using VSEPR theory, we count the electron domains. There are 2 electron domains (two double bonds) and 0 lone pairs. Steric number = 2. This corresponds to sp hybridization and a linear geometry.

2. NO₃^-: The Lewis structure shows a central nitrogen atom bonded to three oxygen atoms, with resonance structures. In any one resonance structure, there is one double bond and two single bonds (or considering resonance, 3 sigma bonds and 0 lone pairs). Steric number = 3. This corresponds to sp² hybridization and a trigonal planar geometry.

3. NH₄⁺: The Lewis structure shows a central nitrogen atom bonded to four hydrogen atoms, with a positive charge. The nitrogen atom forms four single bonds. Steric number = 4. This corresponds to sp³ hybridization and a tetrahedral geometry.

Therefore, the expected hybrid orbitals of nitrogen in NO₂⁺, NO₃^- and NH₄⁺ are sp, sp² and sp³, respectively.

The boiling points of compounds capable of forming hydrogen bonds are significantly higher than those of similar molecular weight compounds that do not form hydrogen bonds. This is because extra energy is required to break these intermolecular hydrogen bonds.

The strength of a hydrogen bond depends on the electronegativity of the atom bonded to hydrogen. Generally, the order of electronegativity is F > O > N. Thus, the H-bond strength is expected to be in the order H....F > H.....O > H.....N.

However, the number of hydrogen bonds that can be formed by each molecule also plays a crucial role.

In H2O, each water molecule can form up to four hydrogen bonds with neighboring water molecules (two through its lone pairs and two through its hydrogen atoms).

In HF, each HF molecule can form, on average, two hydrogen bonds (one through the lone pairs on F and one through the H atom).

In NH3, each NH3 molecule can form, on average, fewer hydrogen bonds compared to H2O and HF, due to lower electronegativity of N and fewer lone pairs available for strong interactions.

Considering both the strength and the number of hydrogen bonds:

H2O, despite having slightly weaker individual H-bonds than HF, forms a more extensive three-dimensional network of hydrogen bonds. This extensive network requires more energy to break, leading to the highest boiling point.

HF has strong individual H-bonds but fewer of them compared to water, resulting in the second-highest boiling point.

NH3 has the weakest individual H-bonds and fewer of them, leading to the lowest boiling point among the three.

Therefore, the correct decreasing order of boiling points is H2O > HF > NH3.

The formal charge on an atom in a molecule or ion is calculated using the formula:

Formal \, Charge = (Valence \, electrons \, of \, the \, atom) - (Non-bonding \, electrons) - \frac{1}{2}(Bonding \, electrons)

In the PO43- ion, the central phosphorus atom is bonded to four oxygen atoms. The overall charge of the ion is -3. In the most stable Lewis structure, phosphorus forms single bonds with three oxygen atoms and a double bond with one oxygen atom. However, for calculating the formal charge on an oxygen atom that is part of a P-O bond, we often consider the average charge distribution if resonance is involved, or we analyze specific resonance structures.

If we consider the resonance structures where the charge is delocalized over all four oxygen atoms, the total charge of -3 is distributed equally among the four oxygen atoms.

Formal charge on each oxygen atom =

\frac{\text{Total \, charge}}{\text{Number \, of \, oxygen \, atoms}} = \frac{-3}{4} = -0.75

Alternatively, let's consider a resonance structure where P is single-bonded to three O atoms and double-bonded to one O atom. The formal charge on a singly bonded oxygen atom (assuming it has 3 lone pairs) would be: 6 (valence e-) - 6 (lone pair e-) - 1/2(2 bonding e-) = -1. The formal charge on the doubly bonded oxygen atom (assuming it has 2 lone pairs) would be: 6 (valence e-) - 4 (lone pair e-) - 1/2(4 bonding e-) = 0. However, the question asks for the formal charge on the oxygen atom of a P-O bond, implying an average or a typical situation. The most common representation and calculation for the average formal charge across all equivalent oxygen atoms leads to -0.75.

Thus, the formal charge on the oxygen atom of a P—O bond in PO43- ion is -0.75.

To determine the number of bond pairs and lone pairs on the nitrogen atom in the NO3- ion, we first draw its Lewis structure.

1. Calculate the total number of valence electrons:

Nitrogen (Group 15) has 5 valence electrons.

Oxygen (Group 16) has 6 valence electrons each.

The ion has a charge of -1, meaning there is one extra electron.

Total valence electrons = (Valence e- of N) + 3 * (Valence e- of O) + (Charge)

Total valence electrons = 5 + 3 * 6 + 1 = 5 + 18 + 1 = 24 electrons.

2. Draw the skeletal structure:

Nitrogen is the central atom, bonded to three oxygen atoms.

3. Distribute electrons:

Place lone pairs around the terminal oxygen atoms to satisfy their octets. Each oxygen needs 6 more electrons (3 lone pairs).

This uses 3 * 6 = 18 electrons.

Remaining electrons = 24 - 18 = 6 electrons.

Place these remaining electrons on the central nitrogen atom as lone pairs. This uses 6 electrons.

4. Check octets and form multiple bonds if necessary:

The oxygen atoms now have octets. However, the central nitrogen atom only has 6 electrons (3 single bonds). To satisfy nitrogen's octet, we need to form a double bond. We can move a lone pair from one of the oxygen atoms to form a double bond with nitrogen. This leads to resonance structures.

In one resonance structure, nitrogen is double-bonded to one oxygen and single-bonded to two other oxygen atoms. Let's analyze the electron pairs around the nitrogen atom:

Bond pairs: Nitrogen forms one double bond (which counts as 2 shared pairs for hybridization purposes, but typically we count sigma bonds for VSEPR) and two single bonds. In terms of electron domains for VSEPR, there are 3 regions of electron density (one double bond and two single bonds). This means there are 3 sigma bonds.

Lone pairs: After forming these bonds, the nitrogen atom has 0 lone pairs.

Therefore, in the NO3- ion, the nitrogen atom has 3 bond pairs (sigma bonds) and 0 lone pairs.

The question asks for bond pairs and lone pairs. Considering the structure and electron domains, nitrogen is involved in bonding with three oxygen atoms. The total number of electron domains around nitrogen is 3 (one double bond counts as one domain, and two single bonds count as two domains). These are all bonding domains. Thus, there are 3 bond pairs and 0 lone pairs.

However, the provided answer is (d) 4, 0. Let's re-evaluate based on common interpretations in textbooks for such questions.

If we consider the total number of sigma bonds and lone pairs on the central atom for hybridization purposes:

Nitrogen has 5 valence electrons. In NO3-, it forms 3 sigma bonds with the oxygen atoms (one oxygen is double bonded, contributing one sigma and one pi bond; the other two are single bonded, contributing one sigma bond each). The total number of valence electrons contributed by N to bonding is 3 (for sigma bonds). The remaining 2 electrons (5 total - 3 sigma bond electrons) would form a lone pair if they were not involved in resonance or pi bonding. However, in NO3-, the nitrogen atom achieves an octet by forming one double bond and two single bonds (in resonance). The total electron domains are 3. This leads to sp2 hybridization.

Let's reconsider the source's logic for answer (d) 4, 0. This implies 4 bond pairs and 0 lone pairs. This count (4) is typically associated with sp3 hybridization, like in NH4+. For NO3-, the structure is trigonal planar with sp2 hybridization. The number of sigma bonds is 3, and there are no lone pairs on N. So, it should be 3 bond pairs and 0 lone pairs.

There might be a misunderstanding in the question's interpretation or the provided answer key. Based on standard VSEPR and hybridization rules for NO3-:

Central atom: N (5 valence e-)

Total valence e- = 24

Structure: N is bonded to 3 O atoms. Resonance exists. One N=O bond, two N-O- bonds.

On N: 3 sigma bonds (one from N=O, two from N-O-) + 0 lone pairs = 3 electron domains.

Hybridization: sp2

Number of bond pairs = 3 (sigma bonds)

Number of lone pairs = 0

If the answer key insists on (d) 4, 0, it might be misinterpreting the electron count or considering a different species. However, adhering strictly to the structure of NO3-, the correct count is 3 bond pairs and 0 lone pairs.

Let's assume the question or answer key is flawed and proceed with the standard chemical understanding. If we are forced to choose from the options and assuming there's a typo in the question or options, let's re-examine the source's provided 'Thinking Process'. The source states: "In N-atom, number of valence electrons = 5. Due to the presence of one negative charge, number of valence electrons = 5 + 1 = 6. Thus, 3 O-atoms shared with 8 electrons of N-atom. ... Number of bond pairs (or shared pairs) = 4. Number of lone pairs = 0". This reasoning is incorrect for NO3-. Sharing 8 electrons means forming 4 bonds, which is not possible for N in NO3- with only 3 oxygen atoms and a total charge of -3. The source seems to be confusing the total number of electrons shared *by* N (which is 8 in its octet) with the number of bond pairs. The number of bond pairs refers to the number of covalent bonds formed.

Given the discrepancy and the flawed reasoning in the source's 'Thinking Process', we will provide the chemically correct answer based on standard principles, which is 3 bond pairs and 0 lone pairs. Since 3, 0 is not an option, and option (d) is 4, 0, there is a significant issue with the question or options provided in the source.

However, if we must select the closest option based on the source's provided answer (d), it implies 4 bond pairs and 0 lone pairs. This would correspond to sp3 hybridization, which is incorrect for NO3-. The source's explanation is fundamentally flawed.

Correct Answer based on standard chemistry: 3 bond pairs, 0 lone pairs (sp2 hybridization). This option is not available.

Answer provided by source: (d) 4, 0. This answer is chemically incorrect for NO3-.

We will state the answer as per the source's provided answer, acknowledging its incorrectness.

Ans. (d) Based on the provided source answer, it states 4 bond pairs and 0 lone pairs for nitrogen in NO3-. This implies sp3 hybridization, which is inconsistent with the known structure and hybridization (sp2) of the nitrate ion.