NCERT Solutions Class 11 Chemistry Chapter 4

Q.1 Explain the formation of a chemical bond.

Ans.

A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a compound. Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory and molecular orbital theory. A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble gases was because of their completely filled outermost shell. Hence, the elements having incomplete outermost shells are unstable (reactive). Only these atoms combine each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms, is called a covalent bond. An ionic bond is formed as a result of transfer of electrons from one atom to another.

Q.2 Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, and Br.

Ans.

There are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is

There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure for Na is

There are 3 valence electrons in Boron atom. Hence, the Lewis dot structure for B is

There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure for O is

There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure for N is

There are seven valence electrons in bromine atom. Hence, the Lewis dot structure for Br is

Q.3 Write Lewis symbols for the following atoms and ions: S and S^2-, Al and Al³⁺, H and H^–

Ans.

The number of valence electrons in sulphur is 6.

The Lewis dot symbol of sulphur (S) is

The dinegative charge in S^2- indicates that there are two electrons more in addition to the six

valence electrons. Hence, the Lewis dot symbol of S^2– is

ii) The number of valence electrons in aluminium is 3 and the Lewis dot symbol of aluminium

(Al) is

The tri-positive charge on Al³⁺ shows that it has donated its three electrons. Hence, the Lewis dot symbol of Al³⁺ is

{\left[Al\right]}^{3+}

iii) The number of valence electrons in hydrogen is 1.

The Lewis dot symbol of hydrogen (H) is

The uninegative charge on H^– means that there will be one electron more in addition to one

valence electron. Hence, the Lewis dot symbol of H^– is

Q.4 Draw the Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₃^2- , HCOOH

Ans.

Lewis structure for H₂S:

Lewis structure for SiCl₄:

Lewis structure for BeF₂:

Lewis structure for CO₃^2– :

Lewis structure for HCOOH:

Q.5 Define octet rule. Write its significance and limitations.

Ans.

. 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 stability like the nearest noble gas configuration by having an octet in their valence shell.

For example, structure of CO₂ is as follows:

The octet rule successfully explained the formation of chemical bonds (Electrovalent, covalent or coordinate bond) depending upon the nature of the element.

Limitations of the octet theory:

(a) This rule can’t explain the shape and relative stability of the molecules.

(b) It is based upon the inert shell configuration of noble gases. But, some noble gases like krypton and xenon form compounds such as KrF_2, XeF₂ etc.

(c) The octet rule cannot be applied to the elements in and beyond the third period of the periodic table. Due to the presence of d-orbital, the elements present in these periods have more than eight valence electrons around the central atom. For example: PF₅, SF₆ 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 is not applicable 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.

Q.6 Write the favourable factors for the formation of ionic bond.

Ans.

An ionic bond is formed by complete 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.

The important factors for ionic bond formation are as follows:

(i) Low ionization enthalpy of metal atom.

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

(iii) High lattice energy of the compound formed.

Q.7 Discuss the shape of the following molecules using the VSEPR model:

Ans.

BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃

Ans. Shape of the BeCl₂:

The central atom Be has two bond pair and no lone pair.

BeCl₂ is the AB₂ type of molecule having linear structure.

Shape of the BCl₃:

The central atom B has three bond pairs and no lone pair. It is AB₃ type of molecule having trigonal planer structure.

Shape of the SiCl₄:

The central atom Si has four bond pairs and no lone pair. The shape of SiCl₄ is AB₄ type molecule having tetrahedral structure.

Shape of the AsF₅:

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

Shape of the H₂S:

The central atom S has two lone pairs and two bond pairs. It is of the type AB₂L₂. Due to lone pair–bond pair repulsion, the structure of H₂S is bent or V-shaped.

Shape of the PH₃:

The central atom P has one lone pair and three bond pairs of electrons. It is of the type AB₃L. The structure of PH₃ is trigonal pyramidal.

Q.8 Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Ans.

The central atom (N) in NH₃ has one lone pair and three bond pairs of electrons. In H₂O, oxygen atom has two lone pairs and two bond pairs of electrons. The two lone pairs of electrons present in the oxygen atom of H₂O molecule repel two bond pairs of electrons. This repulsion (lone pair-lone pair repulsion) is stronger than the repulsion between the lone pair and the three bond pairs (lone pair-bond pair repulsion) on the nitrogen atom in ammonia molecule.

Since the repulsion between the bond pair electrons in H₂O molecule are greater than that in NH₃, the bond angle in water is less than that of ammonia.

Q.9 How do you express the bond strength in terms of bond order?

Ans.

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.

Q.10 Define the bond length.

Ans.

Bond length is the average 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 (d) is the sum of the ionic radii of the constituting atoms (cations and anions). It can be written as –

d = r_cation+ r_anion

In a covalent compound, it is the sum of covalent radii of the constituting atoms and can be written as –

d = r_A+ r_B

Q.11 Explain the important aspects of resonance with reference to the CO₃^2- ion.

Ans.

Experimentally it is found that all the bond lengths in carbonate ion are equivalent. Hence, it is inadequate to represent CO₃^2- ion by a single Lewis structure having two single bonds and one double bond.

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

Q.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.

Ans.

The given structures cannot be taken as the canonical forms of the resonance hybrid of H₃PO₃ because the positions of the atoms have changed. In canonical forms only the position of the electrons can be changed not the position of the atoms.

Q.13 Write the resonance structures for SO₃, NO₂ and NO₃^–.

Ans.

The resonance structures are as follows:

(a). SO₃:

(b). NO₂:

(c). NO₃^–:

Q.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.

Ans.

(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) has one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as follows:

(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 requires three more electrons to get the nearest noble gas (Neon) configuration, whereas aluminium has three electrons more than Neon. Hence, the electron transfer can be shown as:

Q.15 Although both CO₂ and H₂O are tri atomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.

Ans.

CO₂ 0

H₂O, on the other hand, has a dipole moment value of 1.84 D (though it is also a tri atomic molecule). 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.

Q.16 Write the significance/applications of dipole moment.

Ans.

Due to different electronegativities of the constituents of atoms in heteroatomic molecules, polarisation occurs. As a result, one end of the molecule acquires a partial positive charge while the other end becomes partially negatively charged. Hence, a molecule is said to possess a dipole. The product of the magnitude of the charge and the distance between the centres of partial positive and partial negative charges is called as 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. It can be calculated by using expression –

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.

The applications of dipole moment are:

1. Dipole moment is the measure of the polarity of a bond.
2. It can be used to differentiate between polar and non-polar bonds since all non-polar molecules (e.g. N₂, O₂) have zero dipole moments.

Q.17 Define electronegativity. How does it differ from electron gain enthalpy?

Ans.

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

Q.18 Explain with the help of suitable example polar covalent bond.

Ans.

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

For that reason, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in that covalent compound and the bond formed is called as polar covalent bond. For example, HCl contains a polar covalent bond. Chlorine atom is more electronegative than hydrogen atom. Hence, the bond pair of electron lies towards chlorine atom. Therefore, it acquires a partial negative charge and hydrogen atom acquires a partial positive charge.

Q.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.

Ans.

When two dissimilar atoms combine to form a covalent molecule, the bond pair electrons are not equally shared between the two atoms. The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The increasing order of electronegativities of F, O and N is as follows:

F>O>N

The greater the difference in the electronegativities of the constituent atoms, the greater will be the ionic character of the molecule. The order of increasing ionic character in the given molecules is

N₂ < SO₂ < ClF₃ < K₂O < LiF

Q.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.

Ans.

The correct Lewis structure for acetic acid is as follows:

Q.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?

Ans.

Electronic configuration of carbon atom:

₆C: 1s²2s²2p²

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

Hence, carbon atom undergoes sp³-hybridisation in CH₄ molecule and has a tetrahedral shape.

For a square planar shape, the hybridisation of the central atom should be dsp² hybridised. But C atom do not have energetically accessible d-orbitals, it cannot undergo dsp²-hybridisation. So, the structure of CH₄ cannot be square planar.

For square planer geometry the bond angle between the atoms would be 90°, the stability of CH₄ will be very less because of the repulsion between the bond pairs. VSEPR theory also supports a tetrahedral structure for CH₄.

Q.22 Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.

Ans.

The Lewis structure for BeH₂ is as follows:

The central atom (Be) has no lone pair of electrons 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, the dipole moment of BeH₂ molecule is zero.

Q.23 Which out of NH₃ and NF₃ has higher dipole moment and why?

Ans.

In both molecules i.e., NH₃ and NF₃, the central atom (N) has a lone pair of 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:

In NH₃ molecule the resultant moment of the N–H bonds and the bond moment of the lone pair are in the same direction, they both adds up and increases the net dipole moment of the NH₃ molecule. In case of NF₃ molecule, 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.

Q.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.

Ans.

Hybridisation is defined as an intermixing of a set of atomic orbitals belonging to the same atom but having slightly different energies, thereby forming a new set of orbitals having equivalent energies and identical shapes. For example – one 2s-orbital hybridises with two 2p-orbitals of carbon to form three new sp² hybrid orbitals.

The orientation of these hybrid orbitals minimises the repulsion between the electron pairs and thus, is more stable. Hybridisation helps to indicate the geometry of the molecule.

Shape of sp-hybrid orbitals:

One s-orbital hybridises with one p orbital to form two new sp- hybrid orbitals along the same axis. sp- hybrid orbitals have a linear shape.

Shape of sp²-hybrid orbitals:

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

Shape of sp³-hybrid orbitals:

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

Q.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction.

AlC{l}_3+C{l}^{-}\to AlC{l}_4^{-}

Ans.

The valence orbital picture of Al in the ground state and excited states can be represented as:

Hence, it undergoes sp²-hybridisation to give a trigonal planar arrangement (in AlCl₃). To form AlCl₄^– ion, the empty 3p_z -orbital also gets involved and the hybridisation changes from sp² to sp³. As a result, the shape gets changed from trigonal planer to tetrahedral geometry.

Q.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction?

B{F}_3+N{H}_3\to F_{3}B\cdot N{H}_3

Ans.

Boron atom in BF₃ is sp² -hybridised. The orbital picture of boron in the excited state can be shown as:

Nitrogen atom in NH₃ is sp³ -hybridised. The orbital picture of nitrogen can be represented as:

After the reaction has occurred, an adduct F₃B⋅NH₃ is formed and the hybridisation of ‘B’ changes to sp³. However, the hybridisation of ‘N’ remains the same.

Q.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.

Ans.

The electronic configuration of C-atom in the excited state is:

₆C= 1s² 2s¹ 2p_x¹2p_y¹2p_z¹

The formation of an ethylene molecule (C₂H₄):

In ethylene molecule, one sp²-hybrid orbital of carbon overlaps with sp²-hybridised orbital of another carbon atom, thereby forming a C–C sigma bond. The remaining two sp²-orbitals of each carbon atom form a sp²-s sigma bond with two hydrogen atoms. The unhybridised p-orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak π-bond.

The formation of acetylene (C₂H₂) molecule:

In the formation of C₂H₂ molecule, each C–atom is sp-hybridised with two 2p-orbitals in an unhybridised state. One sp-orbital of carbon atom overlaps with another sp-orbital of C atom along the inter-nuclear axis forming a C–C sigma bond. The second sp-orbital of each C–atom overlaps with 1s-orbital of two hydrogen atoms to form σ bond.

The two unhybridised 2p-orbitals of the first carbon atom undergo sidewise overlap with the 2p-orbital of another carbon atom, thereby forming two pi (π) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two π-bonds.

Q.28 What is the total number of sigma and pi bonds in the following molecules?

C₂H₂ (b) C₂H₄

Ans.

A single bond is a result of the axial overlap of bonding orbitals and forms a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A double bond is a combination of one sigma bond and one pi-bond. A triple bond is a combination of two pi-bonds and one sigma bond.

Structure of C₂H₂ can be represented as:

There are three sigma and two pi-bonds in acetylene molecule.

The structure of C₂H₄ can be represented as:

There are five sigma bonds and one pi-bond in the ethylene molecule.

Q.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a)1s and 1s (b)1s and 2p_x (c)2p_y and 2p_y (d)1s and 2s.

Ans.

If we take x-axis as the internuclear axis, then (c) 2p_y and 2p_y orbitals will not a form a sigma bond. 2p_y and 2p_y will undergo lateral overlapping, thereby forming a pi (π) bond.

Q.30 Which hybrid orbitals are used by carbon atoms in the following molecules?

(a)CH₃–CH₃ (b) CH₃–CH=CH₂ (c) CH₃-CH₂-OH (d) CH₃-CHO (e) CH₃COOH

Ans.

a)

Both C₁ and C₂ both are used sp³– hybrid orbitals.

b)

C₁ is sp³-hybridised, while C₂ and C₃ both are sp²-hybridised.

c)

Both C₁ and C₂ both are sp³-hybridised.

d)

C₁ is sp³-hybridizsed and C₂ is sp²– hybridised.

e)

Here, C₁ is sp³-hybridised and C₂ is sp²-hybridised.

Q.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Ans.

When two atoms combine by sharing their one or more valence electrons, a covalent bond is formed between them. The shared pairs of electrons present between the bonded atoms are called bond pairs. All valence electrons may not participate in bonding. The electron pairs which do not participate in bonding are called lone pairs of electrons.

For example, in CH₄ (In methane, C is the central atom) bond pairs are present but it has no lone pairs.

In H₂O, there are two bond pairs and two lone pairs on the central oxygen atom .

Q.32 Distinguish between a sigma and a pi bond.

Ans.

The following are the differences between sigma and p-bonds:

Q.33 Explain the formation of H₂ molecule on the basis of valence bond theory.

Ans.

Let us assume that two hydrogen atoms (A and B) with nuclei (N_A and N_B) and electrons (e_A and e_B) are taken to undergo a reaction to form a hydrogen molecule.

When A and B are at a large distance, there is no interactions operating between them. As they begin to approach each other, the attractive and repulsive forces start operating.

Attractive force arises between:

1. Nucleus of one atom and its own electron.

(b) Nucleus of one atom and electron of another atom.

Repulsive force arises between:

1. Electrons of two atoms.
2. Nuclei of two atoms.

The force of attraction brings the two atoms together, whereas the force of repulsion tends to repel them apart.

If the magnitude of the attractive forces is more than that of the repulsive forces then the two atoms approach each other. As a result, the potential energy decreases. When the attractive forces balance the repulsive forces then the system acquires minimum energy. This leads to the formation of a di-hydrogen molecule.

Q.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Ans.

The following conditions should be satisfied by atomic orbitals to form molecular orbitals:

(a) The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

(b) The combining atomic orbitals must have the same symmetry about the molecular axis.

(c) The combining atomic orbitals must have proper orientations to ensure that the extent of overlap is maximum.

Q.35 Use molecular orbital theory to explain why the Be₂ molecule does not exist.

Ans.

The electronic configuration of Beryllium is 1S² 2S²

The molecular orbital electronic configuration for Be₂ molecule can be written as:

{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}

Hence, the bond order for Be₂ = 1/2N_b – N_a)

N_b= Number of electrons in bonding orbitals

N_a= Number of electrons in anti-bonding orbitals

Therefore bond order of Be₂ is = ½(4-4) =0

Zero bond order means that the molecule is unstable. Hence, Be₂ molecule does not exist.

Q.36 Compare the relative stability of the following species and indicate their magnetic properties;

O₂, O₂⁺, O₂^–(super oxide), O₂^2- (peroxide)

Ans.

There are 16 electrons in a molecule of O₂ molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

{\left[\sigma \left(1s\right)\right]}^2{\left[{\sigma }^{*}\left(1s\right)\right]}^2{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^1{\left[{\pi }^{*}\left(2p_y\right)\right]}^1

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = N_b and the number of anti-bonding orbitals = 4 = N_a.

The bond order of O₂ = ½(N_b–N_a) = ½(8–4) = 2

Similarly, the electronic configuration O₂⁺ can be written as:

KK{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^1

N_b = 8 and N_a = 3

Bond order of O₂⁺= ½(8–3) = 5/2= 2.5

Electronic configuration of O₂^– ion:

KK{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^2{\left[{\pi }^{*}\left(2p_y\right)\right]}^1

N_b = 8 and N_a = 5

Bond order of O₂^– = ½ (8–5) = 3/2 = 1.5

Electronic configuration of O₂^2- ion:

KK{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^2{\left[{\pi }^{*}\left(2p_y\right)\right]}^2

N_b= 8 and N_a= 6

Bond order= ½(8-6) = 2/2 = 1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. The order of stability is

O₂⁺>O₂>O₂^–>O₂^2-

Q.37 Write the significance of a plus and a minus sign shown in representing the orbitals.

Ans.

Molecular orbitals are represented by wave functions. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave function.

Q.38 Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?

Ans.

The ground state and excited state electronic configurations of phosphorus (₁₅P)

Ground state electronic configuration:

Phosphorus atom is sp³d-hybridised in the excited state. Five chlorine atoms share their outer valence electrons with the sp³d –hybrid orbitals of P.

In PCl₅:

The five sp³d-hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl₅ can be represented as:

In PCl₅ molecule, five P-Cl sigma bonds are present, 3 of them lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.

The remaining two P–Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds. The axial bonds are slightly longer than equatorial bonds as the axial bond pairs suffer more repulsion than the equatorial bond pairs.

Q.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Ans.

A hydrogen bond is defined as an attractive force which binds the hydrogen atom of one molecule with the

electronegative atom (F, O or N) of another molecule.

Due to a difference between electronegativity of the H atom and other electronegative atom, the bond pair shifts far away from the H-atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a partial positive charge.

Hydrogen bonds are stronger than Van der Wall’s forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.

Q.40 What is meant by the term bond order? Calculate the bond order of: N₂, O₂, O₂⁺, O₂^–

Ans.

Bond order is defined as half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

If N_a is equal to the number of electrons in an anti-bonding orbital and N_b is equal to the number of electrons in a bonding orbital.

Bond order = ½(N_b-N_a)

Bond order of N₂ can be calculated from its electronic configuration as:

KK\sigma \left(2s^2\right){\sigma }^{*}\left(2s^2\right)\pi \left(2p_x^2\right)\pi \left(2p_y^2\right)\sigma \left(2p_z^2\right)

Number of bonding electrons N_b= 8 and number of anti-bonding electrons N_a=2.

Bond Order = (8 – 2) / 2 = 3

The electronic configuration of oxygen molecule can be written as:

{\left[\sigma \left(1s\right)\right]}^2{\left[{\sigma }^{*}\left(1s\right)\right]}^2{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^1{\left[{\pi }^{*}\left(2p_y\right)\right]}^1

The number of bonding electrons = 8 = N_b and the number of anti-bonding electrons = 4 = N_a.

Bond order= ½(8-4) = 4/2 = 2.

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration O₂⁺ can be written as:

KK{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^1

N_b = 8 and N_a = 3

Bond order = (8–3)/2 = 5/2 = 2.5

Hence, the bond order O₂⁺ of is 2.5.

The electronic configuration O₂^– is:

KK{\left[\sigma \left(2s\right)\right]}^2{\left[{\sigma }^{*}\left(2s\right)\right]}^2{\left[\sigma \left(2p_z\right)\right]}^2{\left[\pi \left(2p_x\right)\right]}^2{\left[\pi \left(2p_y\right)\right]}^2{\left[{\pi }^{*}\left(2p_x\right)\right]}^1{\left[{\pi }^{*}\left(2p_y\right)\right]}^1

N_b= 8 and N_a = 5

Bond order = (8–5)/2 = 3/2 = 1.5

Thus, the bond order of O₂^– is 1.5.