5.Chemical bonding

Qno.1. Select and write the most appropriate alternatives from the given choices.

A. Which molecule is linear?

a. SO3      b. CO2

c. H2S      d. Cl2O

B. When the following bond types are

listed in decreasing order of strength

(strongest first). Which is the correct

order ?

a. covalent > hydrogen > vander waals’

b. covalent > vander waal’s > hydrogen

c. hydrogen > covalent > vander waal’s

d. vander waal’s > hydrogen > covalent.

C. Valence Shell Electron Pair repulsion

(VSEPR) theory is used to predict

which of the following :

a. Energy levels in an atom

b. the shapes of molecules and ions.

c. the electrone getivities of elements.

d. the type of bonding in compounds.

D. Which of the following is true for CO2?

Ans:- (a) polar- nonpolar

E. Which O2 molecule is pargmagnetic. It

is explained on the basis of :

a. Hybridisation.       b. VBT

c. MOT                       d. VSEPR

F. The angle between two covalent bonds

is minimum in:

a.CH4               b. C2H2

c. NH3              d. H2O

Qno.2. Draw

A. Lewis dot diagrams for the folowing

a. Hydrogen (H2)

b. Water (H2O)

c. Carbon dioxide (CO2)

d. Methane (CH4)

e. Lifthium Fluoride (LiF)(1 mark each)

Ans:-

B. Diagram for bonding in ethene with sp2

Hybridisation.

Ans:-

C. Lewis electron dot structures of(1 mark each)

a. HF 

b. C2H6

c. C2H4

d. CF3Cl

e. SO2

Ans:-

D. Draw orbital diagrams of

a. Fluorine molecule

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b. Hydrogen fluoxide molecule

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Qno.3. Answer the following questions

A. Distinguish between sigma and pi bond.

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B. Display electron distribution around the

oxygen atom in water molecule and

state shape of the molecule, also write

H-O-H bond angle.

Ans:-

C. State octel rule. Explain its inadequecies with respect to

a. Incomplete octel 

Ans:-

Molecules with incomplete octet:

e.g. BF3, BeCl2, LiCl

In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.

Li in LiCl has only two electrons, Be in BeCl2 has four electrons while B in BF3 has six electrons in the valence shell.

b. Expanded octel

Ans:-

Molecules with expanded octet: Some molecules like SF6,PCl5,H2SO4 have more than eight electrons around the central atom.

D. Explain in brief with one example:

a. Ionic bond

Ans:-

The bond formed by complete transfer of one or more electrons from an electropositive atom to an electronegative atom, leading to formation of ions which are held together by electrostatic attraction is called ionic bond or electrovalent bond.

Formation of sodium chloride (NaCl):

i. The electronic configurations of sodium and chlorine are:

Na (Z = 11) 1s²2s²2p⁶3s¹ or (2, 8, 1) Cl (Z = 17) : 1s²2s²2p⁶3s²3p⁵ or (2, 8, 7)

ii. Sodium has one electron in its valence shell. It has tendency to lose one electron to acquire the electronic configuration of the nearest inert gas, neon (2, 8).

iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).

iv. During the combination of sshum and chilerine atrens, the sadisus ann amfors in valence slectron to the chlorine atom.

v. Sodium atom changes into Na+ ion while the chlorine stom changes imo Cl- ion. The two ions are held together by strong electrostatic force of attraction.

vi. The formation of ionic bond between Na and Cl can be shown as follows:

Formation of calcium chloride (CaCl2):

i. The electronic contigurations of calcium and chlorine are:

Na (Z = 11); 1s² 2s²2p⁴3s²3p⁶4s² or  (2, 8, 8. 2) CI (Z=17); 1s²2s²2p⁶3s² 3p⁵ or (2,8, 7)

ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8)

iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).

iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms

v. Calcium atom changes into Ca2 ion while the two chlorine atoms change into two CT ions. These ions are held together by strong electrostatic force of attraction.

vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. covalend bond

Ans:-

The attractive force which exists due to the mutual sharing of electrons between the two atoms of similar electronegativity or having small difference in electronegativities is called a covalent bond.

Formation of H2 molecule:

i. The electronic configuration of H atom is 1s¹.

ii. It needs one more electron to complete its valence shell.

iii. When two hydrogen atoms approach each other at a certain internuclear distance, they share their valence electrons.

iv. The shared pair of electrons belongs equally to both the hydrogen atoms. The two atoms are said to be linked by a single covalent bond and a H2 molecule is formed.

Formation of Cl2 molecule:

i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.

ii. It needs one more electron to complete its valence shell.

iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.

iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl2 molecule is formed.

c. co-ordinate bond

Ans:- 

i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom.

ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.

iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is

two electrons short of an octet.

The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

E. Give reasons for need of Hybridisation.(3marks)

Ans:- The concept of hybridization was introduced because the valence bond theory failed to explain the following points:

1.Valencies of certain elements: 

The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.

a. Beryllium: The electronic configuration of beryllium is 1s²2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl2.

b. Boron: The electronic configuration of boron is 1s²2s²2p¹x. The valency is expected to be 1 but it is 3 as in B*F_{3}

c. Carbon: The electronic configuration of carbon is 1s²2s²2p¹x2p¹y. The valency is expected to be 2, but observed valency is 4 as in CH4

ii. The shapes and geometry of certain molecules:

The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules.

e.g.

b. Tetrahedral shape of methane molecule.

like b. Bond angles in molecules NH3 (107° 18') and H₂O (104° 35'). However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

F. Explain geometry of methane molecule

on the basis of Hybridisation.

Ans:-

Formation of methane (CH4) molecule on the basis of sp³ hybridization:

i. Methane molecule (CH4) has one carbon atom and four hydrogen atoms.

ii. The ground electronic configuration of C (Z = 6) is 1s²2s²2p¹x 2p¹y2p⁰z.

Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.

iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, 2px, 2py and 2pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.

v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH4 molecule, there are four C-H bonds formed by the sp³ - s overlap.

Diagram:

G. In Ammonia molecule the bond angle

is 107° and in water molecule it is

104°35', although in both the central

atoms are sp³ hybridized Explain

Ans:

i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109° 28'. But the actual bond angle is 107° 28'. It is due to the following reasons.

a. One lone pair and three bond pairs are present in ammonia molecule.

b. The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.

c. Due to these repulsions, there is a small decrease in bond angle (~2°) from 109° 28' to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109° 28'. But the actual bond angle is 104° 35'. It is due to the following reasons.

a. Two lone pairs and two bond pairs are present in water molecule.

b. The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.

c. Due to these repulsions, there is a small decrease in bond angle (~5°) from 109° 28' to 104°35'.

H. Give reasons for:

a. Sigma (σ)bond is stronger than Pi

(π)bond.

Ans:-

i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.

ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.

iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less. Hence, sigma bond is stronger than pi bond

b. HF is a polar molecule(2marks)

Ans:-

i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.

ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. Carbon is a tetravalent in nature.

Ans:- Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature

I. Which type of hybridization is present

in ammonia molecule? Write the

geometry and bond angle present in

ammonia.(2marks)

Ans:-The type of hybridization present in ammonia (NH3) molecule is sp³ .

Geometry of ammonia molecule is pyramidal or distorted tetrahedral.

Bond angle in ammonia molecule is 107° 18'.

J. Identify the type of orbital overlap

present in

a. H2

b. F2

c. H-F molecule.

Explain diagramatically.

Ans:-

K. F-Be-F is a liner molecule but H-O-H

is angular. Explain.(2marks)

Ans:-

L. BF3 molecule is planar but NH3

pyramidal. Explain.(2marks)

Ans:-

i. 2 In the BF3 molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three 2 corners of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH3 molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four corners of a tetrahedron. The expected geometry of NH3 molecule is regular tetrahedral with bond angle 109° 28'. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH3 is distorted and it has pyramidal geometry.

M. In case of bond formation in Acetylene

molecule :(1marks)

a. How many covalend bonds are

formed ?

b. State number of sigma and pi bonds

formed.

c. Name the type of Hybridisation.

Ans:-

i. In acetylene molecule, there are five covalent bonds.

ii. In acetylene molecule, there are three sigma bonds and two pi bonds.

iii. In acetylene molecule, each carbon atom undergoes sp hybridization.

N. Define :

a. Bond Enthalpy(1marks)

Ans:- Bond enthalpy is defined as the amount of energy required to break one mole of bond of one type, present between two atoms in a gaseous state.

b. Bond Length (1marks)

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

O. Predict the shape and bond angles in

the following molecules:

a. CF4 b. NF3

c. HCN d. H2

Ans:-

i.CF4: There are four bond pairs on the central atom. Hence, shape of CF4 is tetrahedral and F-C-F bond angle is 109° 28'.

ii.NF3: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF3 is trigonal pyramidal and F-N-F bond angle is less than 109° 28'.

iii.HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.

iv.H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H2S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

Qno.4. Using data from the Table, answer the following.

a. What happens to the bond length when

unsaturation increases?

b. Which is the most stable compound?

c. Indicate the relation between bond strength and Bond enthalpy.

d. Comment on overall relation between

Bond length, Bond Enthalpy and Bond

strength and stability.

Ans:-

Qno.5. Complete the flow chart:

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Qno.6.Complete the following Table:

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Qno.7. Answer in one sentence:

A. Indicate the factor on which stalility of

ionic compound is measured?

Ans:- The stability of an ionic compound is measured by the amount of energy released during lattice formation.

B. Arrange the following compounds on the basis of lattice energies in decreasing

(descending) order: BeF2, AlCl3, LiCl,

CaCl2, NaCl

Ans:- 

C. Give the total number of electrons

around sulphur (S) in SF6

compound.

Ans:-

D. Covalant bond is directional in nature.

Justify.

Ans:-  Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

E. What are the interacting forces present

during formation of a molecule of a

compound ?

Ans:-

i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.

b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.

iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.

iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases and a covalent bond is not formed.

Hence, lowering of energy takes during bond formation.

F. Give the type of overlap by which pi

(π) bond is formed.

Ans:- The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

G. Mention the steps involved in

Hybridization

Ans:-The steps involved in hybridization are:

i. formation of the excited state and 

ii.mixing and recasting of orbitals.

H. Write the formula to calculate bond

order of molecule.

Ans:-

Bond order of a molecule = Nb-Na

                                                    2

where, No is the number of electrons present in bonding MOs and Na is the number of electrons present in antibonding MOs.

I. Why is O2 molecule paramagnetic?

Ans:-

J. What do you mean by formal charge ?

Explain its significance with the help of suitable example.(4marks)

Ans:-