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Page No 131: - Chapter 4 Chemical Bonding & Molecular Structure class 11 ncert solutions Chemistry - SaraNextGen [2024]


Question 4.26:

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

BF3 + NH3 → F3B.NH3

Answer:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3645/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_64a2055a.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3645/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_3aa78f80.jpg

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

Question 4.27:

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

Answer:

C2H:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3647/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m930dd8e.gif

In the formation of an ethane molecule (C2H4), one sp2 hybrid orbital of carbon overlaps a sp2 hybridized orbital of another carbon atom, thereby forming a C-C sigma bond.

The remaining two sp2 orbitals of each carbon atom form a sp2s sigma bond with two hydrogen atoms. The unhybridized 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3647/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_350d83dd.jpg

C2H2 :

In the formation of C2H2 molecule, each C–atom is sp hybridized with two 2p-orbitals in an unhybridized state.

One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a C–C sigma bond. The second sp orbital of each C–atom overlaps a half-filled 1s-orbital to form a σ bond.

The two unhybridized 2p-orbitals of the first carbon 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3647/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_49aad63c.jpg

Question 4.28:

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

(a) C2H2 (b) C2H4

Answer:

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

Structure of C2H2 can be represented as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8179/Grade%2011_Chapter%204_html_m4b4628fb.jpg

Hence, there are three sigma and two pi-bonds in C2H2.

The structure of C2H4 can be represented as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8179/Grade%2011_Chapter%204_html_m73e163ea.jpg

Hence, there are five sigma bonds and one pi-bond in C2H4.

Question 4.29:

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

Answer:

2py and 2py orbitals will not a form a sigma bond. Taking x-axis as the internuclear axis, 2py and 2py orbitals will undergo lateral overlapping, thereby forming a pi (π) bond.

Question 4.30:

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

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

Answer:

(a)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3652/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m6eba1fdd.jpg

Both C1 and C2 are sp3 hybridized.

(b)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3652/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m1ea4b904.jpg

C1 is sp3 hybridized, while C2 and C3 are sp2 hybridized.

(c)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3652/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_f3933a9.jpg

Both C1 and C2 are sp3 hybridized.

(d)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3652/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m63b3856e.jpg

C1 is sp3 hybridized and C2 is sp2 hybridized.

(e)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3652/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_278d2fb5.jpg

C1 is sp3 hybridized and C2 is sp2 hybridized.

Question 4.31:

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

Answer:

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 that do not participate in bonding are called lone pairs of electrons.

For example, in C2H6 (ethane), there are seven bond pairs but no lone pair present.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8180/Grade%2011_Chapter%204_html_5a319ab2.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8180/Grade%2011_Chapter%204_html_m6fe55b72.jpg

Question 4.32:

Distinguish between a sigma and a pi bond.

Answer:

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

Sigma (σ) Bond

Pi (π) Bond

(a) It is formed by the end to end overlap of orbitals.

It is formed by the lateral overlap of orbitals.

(b) The orbitals involved in the overlapping are s–ss–p, or p–p.

These bonds are formed by the overlap of p–p orbitals only.

(c) It is a strong bond.

It is weak bond.

(d) The electron cloud is symmetrical about the line joining the two nuclei.

The electron cloud is not symmetrical.

(e) It consists of one electron cloud, which is symmetrical about the internuclear axis.

There are two electron clouds lying above and below the plane of the atomic nuclei.

(f) Free rotation about σ bonds is possible.

Rotation is restricted in case of pi-bonds.

Explain the formation of H2 molecule on the basis of valence bond theory.Question 4.33:

Answer:

Let us assume that two hydrogen atoms (A and B) with nuclei (NA and NB) and electrons (eand eB) are taken to undergo a reaction to form a hydrogen molecule.

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

Attractive force arises between:

(a) Nucleus of one atom and its own electron i.e., NA – eA and NB – eB.

(b) Nucleus of one atom and electron of another atom i.e., NA – eB and NB – eA.

Repulsive force arises between:

(a) Electrons of two atoms i.e., eA – eB.

(b) Nuclei of two atoms i.e., NA – NB.

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3656/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m1365400b.jpg

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

Question 4.34:

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

Answer:

The given 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 proper orientations to ensure that the overlap is maximum.

(c) The extent of overlapping should be large.

Question 4.35:

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

Answer:

The electronic configuration of Beryllium ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3659/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m77b9d882.gif .

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3659/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_201a81e1.jpg

Hence, the bond order for Beis https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3659/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_15d2533e.gif

Where,

Nb = Number of electrons in bonding orbitals

Na = Number of electrons in anti-bonding orbitals

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3659/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_4dd19828.gif Bond order of Be2 https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3659/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_7f2467e5.gif = 0

A negative or zero bond order means that the molecule is unstable. Hence, Be2 molecule does not exist.

Question 4.36:

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

O2,https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_2f39e620.gif ,https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_5d3147f9.gif  (superoxide), https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_md1d97a4.gif (peroxide)

Answer:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_6fce3cab.gif

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

Bond order https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_m3d1acebf.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_m5806fd34.gif

= 2

Similarly, the electronic configuration of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_2f39e620.gif  can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_mc3b724c.gif

Nb = 8

Na = 3

Bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_m691b4e4e.gif

= 2.5

Electronic configuration of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_5d3147f9.gif  ion will be:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_m7d29cd54.gif

Nb = 8

Na = 5

Bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_5d3147f9.gif  = https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_m12456b0b.gif

= 1.5

Electronic configuration of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_md1d97a4.gif ion will be:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_48792bc2.gif

Nb = 8

Na = 6

Bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_md1d97a4.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_6c18e554.gif

= 1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8181/Grade%2011_Chapter%204_html_m9a59bef.gif .

Question 4.37:

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

Answer:

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.

Question 4.38:

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

Answer:

The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are:

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/2(1280).png

Phosphorus atom is sp3d hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:

PCl5

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3662/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m635e0813.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3662/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_b4c45cb.jpg

There are five P–Cl sigma bonds in PCl5. Three P–Cl bonds 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.

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

Question 4.39:

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

Answer:

A hydrogen bond is defined as an attractive force acting between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule (may be of the same kind).

Due to a difference between electronegativities, the bond pair between hydrogen and the electronegative atom gets drifted far away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3663/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_90f4da.gif

The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state.

There are two types of H-bonds:

(i) Intermolecular H-bond e.g., HF, H2O etc.

(ii) Intramolecular H-bond e.g., o-nitrophenol

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/3663/NCERT%20Solution_21-10-08_Vidushi_11_Chemistry_4_40_SJT_html_m4e1ae483.jpg

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

Question 4.40:

What is meant by the term bond order? Calculate the bond order of: N2, O2https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_2f39e620.gif andhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_5d3147f9.gif .

Answer:

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

If Na is equal to the number of electrons in an anti-bonding orbital, then Nb is equal to the number of electrons in a bonding orbital.

Bond order = https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m48bcf8ac.gif

If Nb > Na, then the molecule is said be stable. However, if Nb ≤ Na, then the molecule is considered to be unstable.

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_3f346f65.gif

Number of bonding electrons, Nb = 10

Number of anti-bonding electrons, Na = 4

Bond order of nitrogen molecule https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m77a44e58.gif

= 3

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_6fce3cab.gif

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

Bond order https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m3d1acebf.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m5806fd34.gif

= 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_2f39e620.gif  can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_mc3b724c.gif

Nb = 8

Na = 3

Bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m691b4e4e.gif

= 2.5

Thus, the bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_2f39e620.gif  is 2.5.

The electronic configuration of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_5d3147f9.gif  ion will be:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m7d29cd54.gif

Nb = 8

Na = 5

Bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_5d3147f9.gif  = https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_m12456b0b.gif

= 1.5

Thus, the bond order of https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/197/8182/Grade%2011_Chapter%204_html_5d3147f9.gif  ion is 1.5.

Also Read : INTRODUCTION-Chapter-5-States-Of-Matter-class-11-ncert-solutions-Chemistry

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