Page No 130: - Chapter 4 Chemical Bonding & Molecular Structure class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 4.5:

Define octet rule. Write its significance and limitations.

Answer:

The octet rule or the electronic theory of chemical bonding was developed by Kossel and Lewis. According to this rule, atoms can combine either by transfer of valence electrons from one atom to another or by sharing their valence electrons in order to attain the nearest noble gas configuration by having an octet in their valence shell.

The octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.

Limitations of the octet theory:

The following are the limitations of the octet rule:

(a) The rule failed to predict the shape and relative stability of molecules.

(b) It is based upon the inert nature of noble gases. However, some noble gases like xenon and krypton form compounds such as XeF₂, KrF₂ etc.

(c) The octet rule cannot be applied to the elements in and beyond the third period of the periodic table. The elements present in these periods have more than eight valence electrons around the central atom. For example: PF₅, SF_6, etc.

(d) The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example, NO and NO₂ do not satisfy the octet rule.

(e) This rule cannot be applied to those compounds in which the number of electrons surrounding the central atom is less than eight. For example, LiCl, BeH₂, AlCl₃ etc. do not obey the octet rule.

Question 4.6:

Write the favourable factors for the formation of ionic bond.

Answer:

An ionic bond is formed by the transfer of one or more electrons from one atom to another. Hence, the formation of ionic bonds depends upon the ease with which neutral atoms can lose or gain electrons. Bond formation also depends upon the lattice energy of the compound formed.

Hence, favourable factors for ionic bond formation are as follows:

(i) Low ionization enthalpy of metal atom.

(ii) High electron gain enthalpy (Δeg H) of a non-metal atom.

(iii) High lattice energy of the compound formed.

Question 4.7:

Discuss the shape of the following molecules using the VSEPR model:

BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃

Answer:

BeCl2:

The central atom has no lone pair and there are two bond pairs. i.e., BeCl₂ is of the type AB₂. Hence, it has a linear shape.

BCl_3:

The central atom has no lone pair and there are three bond pairs. Hence, it is of the type AB₃. Hence, it is trigonal planar.

SiCl_4:

The central atom has no lone pair and there are four bond pairs. Hence, the shape of SiCl₄ is tetrahedral being the AB₄ type molecule.

AsF_5:

The central atom has no lone pair and there are five bond pairs. Hence, AsF₅ is of the type AB₅. Therefore, the shape is trigonal bipyramidal.

H₂S:

The central atom has one lone pair and there are two bond pairs. Hence, H₂S is of the type AB₂E. The shape is Bent.

PH₃:

The central atom has one lone pair and there are three bond pairs. Hence, PH₃ is of the AB₃E type. Therefore, the shape is trigonal pyramidal.

Question 4.8:

Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Answer:

The molecular geometry of NH₃ and H₂O can be shown as:

The central atom (N) in NH₃ has one lone pair and there are three bond pairs. In H₂O, there are two lone pairs and two bond pairs.

The two lone pairs present in the oxygen atom of H₂O molecule repels the two bond pairs. This repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the nitrogen atom.

Since the repulsions on the bond pairs in H₂O molecule are greater than that in NH₃, the bond angle in water is less than that of ammonia.

Question 4.9:

How do you express the bond strength in terms of bond order?

Answer:

Bond strength represents the extent of bonding between two atoms forming a molecule. The larger the bond energy, the stronger is the bond and the greater is the bond order.

Question 4.10:

Define the bond length.

Answer:

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Bond lengths are expressed in terms of Angstrom (10^–10 m) or picometer

(10^–12 m) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.

In an ionic compound, the bond length is the sum of the ionic radii of the constituting atoms (d = r₊ + r_–). In a covalent compound, it is the sum of their covalent radii (d = r_A + r_B).

Question 4.11:

Explain the important aspects of resonance with reference to the   ion.

Answer:

According to experimental findings, all carbon to oxygen bonds in are equivalent. Hence, it is inadequate to represent ion by a single Lewis structure having two single bonds and one double bond.

Therefore, carbonate ion is described as a resonance hybrid of the following structures:

Question 4.12:

H₃PO₃ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.

Answer:

The given structures cannot be taken as the canonical forms of the resonance hybrid of H₃PO₃ because the positions of the atoms have changed.

Question 4.13:

Write the resonance structures for SO₃, NO₂ and .

Answer:

The resonance structures are:

(a) SO₃:

(b) 

(c)  :

Question 4.14:

Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.

Answer:

(a) K and S:

The electronic configurations of K and S are as follows:

K: 2, 8, 8, 1

S: 2, 8, 6

Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:

(b) Ca and O:

The electronic configurations of Ca and O are as follows:

Ca: 2, 8, 8, 2

O: 2, 6

Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as:

(c) Al and N:

The electronic configurations of Al and N are as follows:

Al: 2, 8, 3

N: 2, 5

Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three electrons more than Neon. Hence, the electron transference can be shown as:

Question 4.15:

Although both CO₂ and H₂O are triatomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.

Answer:

According to experimental results, the dipole moment of carbon dioxide is zero. This is possible only if the molecule is linear so that the dipole moments of C–O bonds are equal and opposite to nullify each other.

Resultant μ = 0 D

H₂O, on the other hand, has a dipole moment value of 1.84 D (though it is a triatomic molecule as CO₂). The value of the dipole moment suggests that the structure of H₂O molecule is bent where the dipole moment of O–H bonds are unequal.

Question 4.16:

Write the significance/applications of dipole moment.

Answer:

In heteronuclear molecules, polarization arises due to a difference in the electronegativities of the constituents of atoms. As a result, one end of the molecule acquires a positive charge while the other end becomes negative. Hence, a molecule is said to possess a dipole.

The product of the magnitude of the charge and the distance between the centres of positive-negative charges is called the dipole moment (μ) of the molecule. It is a vector quantity and is represented by an arrow with its tail at the positive centre and head pointing towards a negative centre.

Dipole moment (μ) = charge (Q) × distance of separation (r)

The SI unit of a dipole moment is ‘esu’.

1 esu = 3.335 × 10^–30 Cm

Dipole moment is the measure of the polarity of a bond. It is used to differentiate between polar and non-polar bonds since all non-polar molecules (e.g. H₂, O₂) have zero dipole moments. It is also helpful in calculating the percentage ionic character of a molecule.

Question 4.17:

Define electronegativity. How does it differ from electron gain enthalpy?

Answer:

Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself.

Electronegativity of any given element is not constant. It varies according to the element to which it is bound. It is not a measurable quantity. It is only a relative number.

On the other hand, electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.

Question 4.18:

Explain with the help of suitable example polar covalent bond.

Answer:

When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons is not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom.

As a result, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this type of a bond is called a polar covalent bond.

HCl, for example, contains a polar covalent bond. Chlorine atom is more electronegative than hydrogen atom. Hence, the bond pair lies towards chlorine and therefore, it acquires a partial negative charge.

Question 4.19:

Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.

Answer:

The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The greater the difference, the greater will be the ionic character of the molecule.

On this basis, the order of increasing ionic character in the given molecules is

N₂ < SO₂ < ClF₃ < K₂O < LiF.

Question 4.20:

The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Answer:

The correct Lewis structure for acetic acid is as follows:

Question 4.21:

Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH₄ is not square planar?

Answer:

Electronic configuration of carbon atom:

₆C: 1s² 2s² 2p²

In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp³ hybridization in CH₄ molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp². However, an atom of carbon does not have d-orbitalsto undergo dsp² hybridization. Hence, the structure of CH₄ cannot be square planar.

Moreover, with a bond angle of 90° in square planar, the stability of CH₄ will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH₄.

Question 4.22:

Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.

Answer:

The Lewis structure for BeH₂ is as follows:

There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH₂ is of the type AB₂. It has a linear structure.

Dipole moments of each H–Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, BeH₂ molecule has zero dipole moment.

Question 4.23:

Which out of NH₃ and NF₃ has higher dipole moment and why?

Answer:

In both molecules i.e., NH₃ and NF₃, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF₃ is greater than NH₃. However, the net dipole moment of NH₃ (1.46 D) is greater than that of NF₃ (0.24 D).

This can be explained on the basis of the directions of the dipole moments of each individual bond in NF₃ and NH₃. These directions can be shown as:

Thus, the resultant moment of the N–H bonds add up to the bond moment of the lone pair (the two being in the same direction), whereas that of the three N – F bonds partly cancels the moment of the lone pair.

Hence, the net dipole moment of NF₃ is less than that of NH₃.

Question 4.24:

What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.

Answer:

Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes.

For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new sp² hybrid orbitals.

These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more stable. Hybridization helps indicate the geometry of the molecule.

Shape of sp hybrid orbitals: sp hybrid orbitals have a linear shape. They are formed by the intermixing of s and p orbitals as:

Shape of sp2 hybrid orbitals:

sp² hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p-orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement as:

Shape of sp3 hybrid orbitals:

Four sp³ hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. The four sp³ hybrid orbitals are arranged in the form of a tetrahedron as:

Question 4.25:

Describe the change in hybridisation (if any) of the Al atom in the following reaction.

Answer:

The valence orbital picture of aluminium in the ground state can be represented as:

The orbital picture of aluminium in the excited state can be represented as:

Hence, it undergoes sp² hybridization to give a trigonal planar arrangement (in AlCl₃).

To form AlCl₄^–, the empty 3p_z orbital also gets involved and the hybridization changes from sp² to sp³. As a result, the shape gets changed to tetrahedral.