Chapter 4: Chemistry Chemical Bonding and Molecular Structure

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 chemical species. 

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 fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with 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 the transfer of electrons from one atom to another.

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

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

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

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

⋅B˙⋅⋅𝐵˙⋅

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

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

Br: There are seven valence electrons in bromine. Hence, the Lewis dot structure is::Br⋅¨⋅:𝐵𝑟⋅¨⋅

3. Write Lewis symbols for the following atoms and ions 

1.   S and S2−𝑆2−

Ans: The number of valence electrons in sulphur is 6. 

The Lewis dot symbol of sulphur (S) is :S¨::𝑆¨:

The dinegative charge infers that there will be two electrons more in addition to the six valence electrons. Hence, that Lewis dot symbol of S2−𝑆2− is [:S¨:..]−2[:𝑆¨:..]−2

2.  Al and Al+3𝐴𝑙+3

 Ans: The number of valence electrons in aluminium is 3. 

The Lewis dot symbol of aluminum (Al) is ⋅AI˙⋅⋅𝐴𝐼˙⋅

The tripositive charge on a species infers that it has donated its three electrons. Hence, the Lewis do symbol is [Al]+3[𝐴𝑙]+3 

3.  H and H−𝐻−

Ans: The number of valence electrons in hydrogen is 1. 

The Lewis do the symbol of hydrogen (H) is  H⋅𝐻⋅

This negative charge infers that there will be one electron more in addition to the one valence electron. Hence, the Lewis dot symbol is [H¨]−[𝐻¨]−

4. Draw the Lewis structure for the following molecules and ions:

1.  H2S𝐻2𝑆 

Ans: 

2.  SiCl4𝑆𝑖𝐶𝑙4 

Ans: 

3.  BeF2𝐵𝑒𝐹2

Ans: 

4.  CO32−𝐶𝑂32− 

Ans: 

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

Ans: 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.

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:

1. The rule failed to predict the shape and relative stability of molecules. 
2. It is based upon the inert nature of noble gases. However, some noble gases like xenon and krypton form compounds such as XeF2𝑋𝑒𝐹2 , KrF2𝐾𝑟𝐹2  etc.
3. 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: PF5𝑃𝐹5, SF6𝑆𝐹6 etc.

4. The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example, NO and NO2𝑁𝑂2 do not satisfy the octet rule. 

5. 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,AlCl3𝐿𝑖𝐶𝑙,𝐵𝑒𝐻2,𝐴𝑙𝐶𝑙3  etc. do not obey that octet rule. 

6. Write the favourable factors for the formation of ionic bonds.

Ans: 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. The bond formation also depends upon the lattice energy of the compound formed. 

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

1. Low ionization enthalpy of metal atoms. 
2. High electrons gain enthalpy of a nonmetal atom. 
3. The high lattice energy of the compound formed

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

1. BeCl2𝐵𝑒𝐶𝑙2

Ans: Cl:BeCl𝐶𝑙:𝐵𝑒𝐶𝑙

The central atom has no lone pair and there are two bond pairs. i.e., BeCl2𝐵𝑒𝐶𝑙2 is of the typeAB2𝐴𝐵2 . Hence, it has a linear shape.

2. BCl3𝐵𝐶𝑙3 

Ans: C⋅⋅lCl:Be:Cl𝐶⋅⋅𝑙𝐶𝑙:𝐵𝑒:𝐶𝑙

The central atom has no lone pair and there are three bond pairs. Hence, it is of the type AB3𝐴𝐵3 . Hence, it is trigonal planar.

3. SiCl4𝑆𝑖𝐶𝑙4 

Ans: 

The central atom has no lone pair and there are four bond pairs. Hence, the shape of SiCl4𝑆𝑖𝐶𝑙4 is tetrahedral being the AB4𝐴𝐵4 type molecule.

4. AsF5𝐴𝑠𝐹5 

Ans: 

The central atom has no lone pair and there are five bond pairs. Hence, AsF₅ is of the type AB5𝐴𝐵5 . Therefore, the shape is trigonal bipyramidal.

5. H2S𝐻2𝑆 

Ans: 

The central atom has one lone pair and there are two bond pairs. Hence, H2S𝐻2𝑆 is of the type AB2E𝐴𝐵2𝐸 . The shape is Bent

6. PH3𝑃𝐻3 

Ans: 

The central atom has one lone pair and there are three bond pairs. Hence, PH3𝑃𝐻3 is of the AB3E𝐴𝐵3𝐸 type. Therefore, the shape is trigonal bipyramidal.

8. Although geometries of NH3𝑁𝐻3  and H2O𝐻2𝑂  molecules are distorted tetrahedral, the bond angle in water is less than that of ammonia. Discuss.

Ans: The molecular geometry of NH3𝑁𝐻3 and H2O𝐻2𝑂 can be shown as:

The central atom (N) in NH3𝑁𝐻3 has one lone pair and three bond pairs. In H2O𝐻2𝑂, there are two lone pairs and two bond pairs. 

The two lone pairs present on the oxygen atom of H2O𝐻2𝑂 molecule repel 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 H2O𝐻2𝑂 molecule are greater than that in NH3𝑁𝐻3, the bond angle in water is less than that of ammonia.

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.

10. Define Bond Length.

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

Bond length are expressed in terms of Angstrom (10−10m10−10𝑚) or picometer (10−12pm10−12𝑝𝑚) 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−𝑑=𝑟++𝑟−).

In a covalent compound, it is the sum of their covalent radii (d=rA+rB)(𝑑=𝑟𝐴+𝑟𝐵) 

11. Explain the important aspects of resonance with reference to the CO32−𝐶𝑂32− ion.

Ans: According to experimental findings, all carbon to oxygen bonds in CO32−𝐶𝑂32− are equivalent.

Hence, it is inadequate to represent CO32−𝐶𝑂32− 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

12. H3PO3𝐻3𝑃𝑂3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representingH3PO3𝐻3𝑃𝑂3? If not, give a reason for the same.

Ans: The given structures cannot be taken as the canonical forms of the resonance hybrid of H3PO3𝐻3𝑃𝑂3 because the positions of the atoms have changed.

13. Write the resonance structures for the following molecules. 

1. SO3𝑆𝑂3

Ans: 

2. NO2𝑁𝑂2 

Ans:

3. NO3−𝑁𝑂3− 

Ans: 

14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:

1. K and S

Ans: the electronic configuration of K and S are as follows:

K: 2, 8, 8, 1

S: 2, 8, 6

   K⋅:S¨:𝐾⋅:𝑆¨:  

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:

2. Ca and O

Ans: 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:

3. Al and N

Ans: 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 aluminum has three electrons more than Neon. Hence, the electron transference can be shown as:

15. Although both CO2𝐶𝑂2  and H2O𝐻2𝑂 w are triatomic molecules, the shape of  H2O𝐻2𝑂molecule is bent while that of CO2𝐶𝑂2 is linear. Explain this on the basis of the dipole moment.

Ans: 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=0D𝜇=0𝐷 

H2O𝐻2𝑂, on the other hand, has a dipole moment value of 1.84 D (though it is a triatomic molecule as CO2𝐶𝑂2 ). The value of the dipole moment suggests that the structure of H2O𝐻2𝑂 molecule is bent where the dipole moment of O-H bonds are unequal.

16. Write the significance/applications of dipole moment.

Ans: 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 centers of positive-negative charges is called the dipole moment (μ𝜇) of the molecule. It is a vector quantity and is represented by an arrow with its tail at the positive center and head pointing towards the negative center.

Dipole moment (μ𝜇) = charge (Q) X distance separation (r)

The S I unit of dipole moment is ‘esu’

1esu=3.335×10−30cm1𝑒𝑠𝑢=3.335×10−30𝑐𝑚

Dipole moment is the measure of the polarity of a bond. It is used to differentiate between polar and nonpolar bonds since all non-polar molecules (e.g, H2,O2𝐻2,𝑂2) have zero dipole moments. It is also helpful in calculating the percentage ionic character of a molecule.

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 a bond pair of electrons towards itself. The electronegativity of any given element is not constant. It varies according to the element to which it is bonded. 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.

18. Explain with the help of suitable example covalent bonds.

Ans: When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons are 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 bond is called a polar covalent bond. 

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

19. Arrange the bonds in order of increasing ionic character in the molecules:

LiF,K2O,N2,SO2&ClF3𝐿𝑖𝐹,𝐾2𝑂,𝑁2,𝑆𝑂2&𝐶𝑙𝐹3

Ans: 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<LiF𝑁2<𝑆𝑂2<𝐶𝑙𝐹3<𝐾2𝑂<𝐿𝑖𝐹

20. The skeletal structure of CH3COOH𝐶𝐻3𝐶𝑂𝑂𝐻 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:

21. Apart from tetrahedral geometry, another possible geometry for CH4𝐶𝐻4  is square planar with the four H atoms at the corners of the square and the C atom at its center. Explain why CH4𝐶𝐻4 is not square planar?

Ans: The electronic configuration of carbon atom is:

6C=1s22s22p26𝐶=1𝑠22𝑠22𝑝2

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

Hence, the carbon atom undergoes Sp3𝑆𝑝3 hybridization in CH4𝐶𝐻4 molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp2𝑑𝑠𝑝2. 

However, an atom of carbon does not have d-orbitals to undergo dsp2𝑑𝑠𝑝2 hybridization. 

Hence, the structure of CH4𝐶𝐻4 cannot be square planar. 

Moreover, with a bond angle of 900900  in square planar, the stability of CH4𝐶𝐻4 will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure forCH4𝐶𝐻4.

22. Explain why the BeH2𝐵𝑒𝐻2  molecule has a zero dipole moment although the Be-H bonds are polar.

Ans: the Lewis structure for BeH2𝐵𝑒𝐻2is as follows:

 There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH2𝐵𝑒𝐻2 is of the type AB2𝐴𝐵2 . 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, BeH2𝐵𝑒𝐻2 molecule has zero dipole moment.

23. Which out of NH3𝑁𝐻3  and NF3𝑁𝐹3  has a higher dipole moment and why?

Ans: In both molecules, i.e NH3𝑁𝐻3 and NF3𝑁𝐹3, 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 NF3𝑁𝐹3 is greater than NH3𝑁𝐻3. However, the net dipole moment of NH3𝑁𝐻3 (1.46 D) is greater than that of NF3𝑁𝐹3 (0.24 D). 

This can be explained on the basis of directions of the dipole moments of each individual bond is NH3𝑁𝐻3 and NF3𝑁𝐹3. These directions can be shown as:

Thus, the resultant moment of the N-H bonds adds 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 NF3𝑁𝐹3 is less than that of NH3𝑁𝐻3.

24. What is meant by hybridization of atomic orbitals? Describe the shapes of sp,sp2,sp3𝑠𝑝,𝑠𝑝2,𝑠𝑝3  hybrid orbitals.

Ans: 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 sp2𝑠𝑝2 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. 

The 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 sp2𝑠𝑝2 hybrid orbitals: 

sp2𝑠𝑝2 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 sp3𝑠𝑝3  hybrid orbitals: 

Four sp3𝑠𝑝3 hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. 

The four sp3𝑠𝑝3 hybrid orbitals are arranged in the form of a tetrahedron as:

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

AlCl3+Cl−→AlCl4−𝐴𝑙𝐶𝑙3+𝐶𝑙−→𝐴𝑙𝐶𝑙4−

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

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

Hence, it undergoes sp2𝑠𝑝2  hybridization to give a trigonal planar arrangement (in AlCl3𝐴𝑙𝐶𝑙3 ). To form, the empty 3pz3𝑝𝑧  orbital also gets involved and the hybridization changes from sp2𝑠𝑝2 to sp3𝑠𝑝3. As a result, the shape gets changed to tetrahedral.

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

BF3+NH3→F3B.NH3𝐵𝐹3+𝑁𝐻3→𝐹3𝐵.𝑁𝐻3

Ans: Boron atoms in BF3𝐵𝐹3 is sp2𝑠𝑝2 hybridized. The orbital picture of boron in the excited state can be shown as:

Nitrogen atom in NH3𝑁𝐻3  is sp3𝑠𝑝3  hybridized. The orbital picture of nitrogen can be represented as: 

After the reaction has occurred, an adduct F3B.NH3𝐹3𝐵.𝑁𝐻3  is formed as hybridization of ‘B’ changes to sp3𝑠𝑝3 . However, the hybridization of ‘N’ remains intact.

27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H2𝐶2𝐻2 and C2H4𝐶2𝐻4 molecules.

Ans: 

C2H4𝐶2𝐻4 :

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

6C=1s22s12p1x2p1y2p1z6𝐶=1𝑠22𝑠12𝑝𝑥12𝑝𝑦12𝑝𝑧1

In the formation of an ethane molecule (C2H4𝐶2𝐻4), one sp2 hybrid orbital of carbon overlaps a sp2𝑠𝑝2  hybridized orbital of another carbon atom, thereby forming a C-C sigma bond. The remaining two sp2𝑠𝑝2 orbitals of each carbon atom from a sp2−s𝑠𝑝2−𝑠 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 n-bond.

C2H2𝐶2𝐻2:

In the formation of C2H2𝐶2𝐻2 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 if each C-atom overlaps a half-filled 1s-orbital to form a σ𝜎 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 (n) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two n-bonds.

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

1. C2H2𝐶2𝐻2

Ans: A single bond is a result of that 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 C2H2𝐶2𝐻2 can be represented as:

Hence, there are three sigma bonds and two pi-bonds in C2H2𝐶2𝐻2.

2. C2H4𝐶2𝐻4

The structure of C2H4𝐶2𝐻4 can be represented as:

Hence, there are five sigma bonds and one pi-bond in C2H4𝐶2𝐻4 .

29. Considering x-axis as the inter-nuclear which out of the following will not form a sigma bond and why? 

1. 1s and 1s 
2. 1s and 2px2𝑝𝑥 
3. 2py2𝑝𝑦 and 2pz2𝑝𝑧 
4. 1s and 2s.

Ans: 2py2𝑝𝑦 and 2pz2𝑝𝑧 orbitals will not form a sigma bond. Taking x-axis as the internuclear axis, 2py2𝑝𝑦 and 2pz2𝑝𝑧 orbitals will undergo lateral overlapping, thereby is forming a pi (n) bond. Option (c)

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

1. CH3−CH3𝐶𝐻3−𝐶𝐻3 

Ans: 

Both C1𝐶1 and C2𝐶2 are sp3𝑠𝑝3 hybridized

2. CH3−CH=CH2𝐶𝐻3−𝐶𝐻=𝐶𝐻2 

Ans: 

C1𝐶1 is sp3𝑠𝑝3 hybridized, C2𝐶2and C3𝐶3 are sp2𝑠𝑝2 hybridized

3. CH3−CH2−OH𝐶𝐻3−𝐶𝐻2−𝑂𝐻 

Ans: 

Both C1𝐶1 and C2𝐶2 are sp3𝑠𝑝3 hybridized

4. CH3CHO𝐶𝐻3𝐶𝐻𝑂 

Ans: 

C1𝐶1 is sp3𝑠𝑝3 hybridized, C2𝐶2 is sp2𝑠𝑝2 hybridized. As the second carbon is present as an aldehyde group which contains double bond with oxygen atom and single bond with hydrogen atom.

5. CH3COOH𝐶𝐻3𝐶𝑂𝑂𝐻 

Ans: 

C1𝐶1 is sp3𝑠𝑝3 hybridized, C2𝐶2 is sp2𝑠𝑝2 hybridized

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

For example, in C2H6𝐶2𝐻6  (ethane), there are seven bond pairs but no lone pair present.

In H2O𝐻2𝑂 , there are two bond pairs and two lone pairs on the central atom (oxygen).

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

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

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

Ans: Let us assume the two hydrogen atoms (A and B) with nuclei (NA𝑁𝐴  and NB𝑁𝐵 ) and electrons (eA𝑒𝐴  and eB𝑒𝐵 ) 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:

1. Nucleus of one atom and its own electron i.e., NA𝑁𝐴 – eA𝑒𝐴 and NB𝑁𝐵- eB𝑒𝐵. 
2. The nucleus of one atom and electron of another atom i.e., NA𝑁𝐴 – eB𝑒𝐵 and NB𝑁𝐵 –eA𝑒𝐴 . 

Repulsive Force Arises Between: 

1. Electrons of two atoms i.e., eA – eB.
2.  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.

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.

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

Ans: The given conditions should be satisfied by atomic orbitals of form molecular orbitals:

1. The combining atomic orbitals must have an equivalent or nearly an equivalent energy. This means that during 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.
2. The combining atomic orbitals must have proper orientations to make sure that the overlap is maximum.
3. The extent of overlapping should be large

35. Use molecular orbital theory to explain why the Be2𝐵𝑒2  molecule does not exist.

Ans: The electronic configuration of Beryllium is 1s22s21𝑠22𝑠2 

The molecular orbital electronic configuration for Be2𝐵𝑒2 molecule can be written as:

σ21sσ∗21sσ22sσ∗22s𝜎1𝑠2𝜎1𝑠∗2𝜎2𝑠2𝜎2𝑠∗2

Hence, the bond order for Be2𝐵𝑒2is 12(Nb−Na)12(𝑁𝑏−𝑁𝑎) 

Where, Nb𝑁𝑏 = number of electrons in bonding orbitals

Na𝑁𝑎 = number of electrons in non-bonding orbitals

∴∴ Bond order of Be2=12(4−4)=0𝐵𝑒2=12(4−4)=0 

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

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

O2,O+2,O2−2(superoxide),O2−2(peroxide)𝑂2,𝑂2+,𝑂22−(superoxide),𝑂22−(𝑝𝑒𝑟𝑜𝑥𝑖𝑑𝑒)

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

[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1[π∗(2py)]1[𝜎(1𝑠)]2[𝜎∗(1𝑠)]2[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(1𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]1[𝜋∗(2𝑝𝑦)]1

Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = Nb𝑁𝑏  and the number of antibonding orbitals = 4 = Na𝑁𝑎 .

Bond order = 12(Nb−Na)=12(8−4)=212(𝑁𝑏−𝑁𝑎)=12(8−4)=2

Similarly the electronic configuration of O+2𝑂2+ can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1𝐾𝐾[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(2𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]1

Nb𝑁𝑏=8

Na𝑁𝑎 = 3

Bond order of O+2𝑂2+= 12(8−3)=2.512(8−3)=2.5 

Similarly the electronic configuration of O−2𝑂2− can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]1𝐾𝐾[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(2𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]2[𝜋∗(2𝑝𝑦)]1

Nb𝑁𝑏=8

Na𝑁𝑎 = 5

Bond order of O−2𝑂2−= 12(8−5)=1.512(8−5)=1.5

Similarly the electronic configuration of O2−2𝑂22−  can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]2𝐾𝐾[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(2𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]2[𝜋∗(2𝑝𝑦)]2

Nb𝑁𝑏=8

Na𝑁𝑎 = 6

Bond order of O2−2𝑂22−= 12(8−6)=112(8−6)=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

O+2>O2>O−2>O2−2𝑂2+>𝑂2>𝑂2−>𝑂22−

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.

38. Describe the hybridization in case of PCl5𝑃𝐶𝑙5. Why are the axial bonds longer as compared to equatorial bonds?

Ans: the ground state and excited state outer electronic configuration of phosphorus (Z=15) are:

Phosphorus atom is sp3d𝑠𝑝3𝑑  hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:

The five sp3d𝑠𝑝3𝑑 hybrid orbitals are directed towards the five corners of the trigonal bipyramidal. Hence, the geometry of PCl5𝑃𝐶𝑙5  can be represented as

There are five P-Cl sigma bonds in PCl5𝑃𝐶𝑙5. Three P-Cl bonds lie in one plane and make an angle of 12001200  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 900900  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.

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

Ans: 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 electro-negativities, the bond pair between hydrogen and the electronegative atom gets drifted away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

Hδ+−Xδ−𝐻𝛿+−𝑋𝛿− ……..Hδ+−Xδ−𝐻𝛿+−𝑋𝛿−……. Hδ+−Xδ−𝐻𝛿+−𝑋𝛿−

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

There are two types of H-Bonds: 

1. Intermolecular H-bond e.g., HF,H2O𝐻𝐹,𝐻2𝑂  etc.
2. Intramolecular H-bond e.g., o-nitrophenol 

Hydrogen bonds are stronger than Vander Waals forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.

40. What is meant by the term bond order? Calculate the bond order of:

N2,O2,O+2,O−2𝑁2,𝑂2,𝑂2+,𝑂2− 

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

If  Na𝑁𝑎 is equal to the number of electrons in an antibonding orbital, then  Nb𝑁𝑏 is equal to the number of electrons in a bonding orbital.

Bond order = 12(Nb−Na)12(𝑁𝑏−𝑁𝑎)

If Nb>Na𝑁𝑏>𝑁𝑎 , then the molecule is said to be stable.

However, Nb≤Na𝑁𝑏≤𝑁𝑎 , then the molecule is considered to be unstable. 

Bond order of N2𝑁2  can be calculated from its electronic configuration as:

[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[π(2px)]2[π(2py)]2[π(2pz)]2[𝜎(1𝑠)]2[𝜎∗(1𝑠)]2[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋(2𝑝𝑧)]2

Nb𝑁𝑏=10

Na𝑁𝑎 = 4

Bond order of nitrogen molecule = 12(10−4)=312(10−4)=3

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

[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1[π∗(2py)]1[𝜎(1𝑠)]2[𝜎∗(1𝑠)]2[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(1𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]1[𝜋∗(2𝑝𝑦)]1

Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = Nb𝑁𝑏  and the number of antibonding orbitals = 4 = Na𝑁𝑎 .

Bond order = 12(Nb−Na)=12(8−4)=212(𝑁𝑏−𝑁𝑎)=12(8−4)=2

Hence, the bond order of oxygen molecule is 2

Similarly the electronic configuration of O+2𝑂2+ can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1𝐾𝐾[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(2𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]1

Nb𝑁𝑏=8

Na𝑁𝑎 = 3

Bond order of O+2𝑂2+= 12(8−3)=2.512(8−3)=2.5 

Similarly the electronic configuration of O−2𝑂2− can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]1𝐾𝐾[𝜎(2𝑠)]2[𝜎∗(2𝑠)]2[𝜎(2𝑝𝑧)]2[𝜋(2𝑝𝑥)]2[𝜋(2𝑝𝑦)]2[𝜋∗(2𝑝𝑥)]2[𝜋∗(2𝑝𝑦)]1

Nb𝑁𝑏=8

Na𝑁𝑎 = 5

Bond order of O−2𝑂2−= 12(8−5)=1.5