I. Very Short Answer Type Questions
Question 1. How is bond order related to the stability of a molecule?
Answer: Higher the bond order, greater is the stability.
Question 2. Write the type of hybridisation involved in CH4,C2H4 and C2H2.
Answer: CH4= sp3
C2H4 = sp2
C2H2 = sp
Question 3. Out of sigma and Π bonds, which one is stronger and why?
Answer: sigma-bond is stronger. This is because sigma-bond is formed by head-on overlapping of atomic orbitals and Π bond is formed by side wise overlapping.
Question 4. Write the Lewis dot symbols of the following elements and predict their valencies. (i) Cl (ii) P
Question 5. Predict the shapes of the following molecules using VSEPR theory?
- (i) BeCl2
- (ii) SiCl4
- (i) Linear
- (ii) Tetrahedral
Question 6. Write the state of hybridisation of boron in BF3.
Question 7. Arrange O2,O2–,O22-, O2+in increasing order of bond energy.
O22-< O2– < O22-< O2+
Question 8. What is meant by bond pairs of electrons?
Answer: The electron pairs involved in the bond formation are known as bond pairs or shared pairs.
Question 9. Which of the following has larger bond angle in each pair?
- (i) CO2, BF3
- (ii) NH3, CH4
- (i) CO2
- (ii) CH4
Question 10. Arrange the following, according to increasing covalent nature.
NaCl, MgCl2, AlCl3
NaCl < MgCl2 < AlCl3
Question 11. Define covalent bond according to orbital concept?
Answer: Covalent bond can be formed by the overlap of the orbitals belonging to the two atoms having opposite spins of electrons.
Question 12. Why B2 is paramagnetic in nature while C2 is not?
Answer: The molecular orbital electronic configuration of both B2 and C2 are.
Since, B2has two impaired electrons, B2 is paramagnetic.
C2 has no unpaired electron. Thus, C2 is diamagnetic.
Question 13. Why ethyl alcohol is completely miscible with water?
Answer: This is because ethyl alcohol forms H-bonds witfi water.
Question 14. Which is more polar CO2 or N2O? Give reason.
Answer: N2O is more polar than CO2.
This is “because CO2 is linear and symmetrical. Its net dipole moment is zero.
N2O is linear but not symmetrical. It has a net dipole moment of sigma II6D.
Question 15. State the types of hybrid orbitals associated with (i) P in PCl5 and (ii) S in SF6
Answer: (i) sp3d of P in PCl5 (ii) sp3d2 of S in SF6
Question 16. Why N2 is more stable than O2? Explain on the basis of molecular orbital theory.
Answer: Bond order of N2 (= 3) is greater than that of O2 (= 2).
Question 17. How is bond order related to bond length of a molecule?
Answer: Bond length is inversely proportional to bond order.
Question 18. Out of bonding and antibonding molecular orbitals, which one has lower energy and which one has higher stability?
Answer: Bonding molecular orbital has lower energy and higher stability.
Question 19. Define antibonding molecular orbital.
Answer: The molecular orbital formed by the subtractive effect of the electron waves of the combining atomic orbitals, is called antibonding molecular orbital.
Question 20. Name the two conditions which must be satisfied for hydrogen bonding to take place in a molecule.
- (i) The molecule should contain highly electronegative atom like hydrogen atom.
- (ii) The size of electronegative atom should be small.
II. Short Answer Type Questions
Question 1.What is an electrovalent (or ionic) bond? Explain its formation with two examples.
When a chemical bond is formed by the complete transfer of electrons from one atom to another, so as to complete their outermost shell and therefore, aquire the stable noble gas configuration, the bond formed is called ionic bond or electrovalent bond. ‘
Question 2. What are Lewis structures? Write the Lewis structure of H2, BeF2 and H2O.
Answer: The outer shell electrons are shown as dots surrounding the symbol of the atom. These symbols are known as Lewis symbols or Lewis structures.
Question 3. Define Lattice energy. How is Lattice energy influenced by (i) Charge on the ions (ii) Size of the ions?
Answer: Lattice energy is defined as the energy released when one mole of crystalline solid is formed by the combination of oppositely charged ions.
(i)As the magnitude of charge on an ion increases there will be greater force of interionic attraction and hence greater will be the value of Lattice energy,
(ii)Smaller the. size of the ions> lower will be the intemuclear distance and thus greater will be the Lattice energy,
Question 4. Give the shapes of the following molecules:
- (i) AB3
- (ii) AB4
Question 5. Define Hybridisation. Explain sp hybridisation with suitable example.
Answer: Hybridisation: It is the phenomenon of intermixing of atomic orbitals of slightly different energies to form new hybrid orbitals of equivalent energy,
Formation of water. In water (H20)> the atomic number of oxygen is 8 and its orbitals electronic configuration is 1s2 2s2 2px2 2py1 2pz1.The oxygen atom is also SP3 hybridised. However, in this case, the two orbitals with one electron each (half filled) are involved in overlap With the hydrogen orbitals.
Question 6. Account for the following:
- (i) Water is a liquid while H2S is a gas
- (ii) NH3 has higher boiling point than PH3.
(i) In case of water hydrogen bonding causes association of the H2O molecules. There is no such hydrogen bonding in H2S, that’s why it is a gas.
(ii) There is hydrogen bonding in NH3 but not in PH3.
Question 7. What do you mean by Dipole moment? Draw the dipole diagram of H2O.
Answer: The product of magnitude of charges (+ve, or -ve) and distance between them is called dipole moment. It is usually denoted by µ.
Question 8. What are the main postulates of Valence Shell Electron Pair Repulsion (VSEPR) theory?
- The shape of a molecule depends upon the no. of electron pairs around the central atom.
- There is a repulsive force between the electron pairs, which tend to repel one another.
- The electron pairs in space tend to occupy such positions that they are at maximum distance so, that the repulsive force will be minimum.
- A multiple bond is treated as if it is single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.
Question 9. Define bond order. How is it related to the stability of a molecule?
Answer: Bond order is defined as half of the difference between the number of electrons present in bonding and antibonding molecular orbitals.
Bond order (B.O.) = 1/2[Nb – Na ] z
If the bond order is positive (Nb > Na), the molecule or ion will be stable. If it is negative (Nb < Na)the molecule or ion will be unstable.
Question 10. Explain the diamagnetic behaviour of P2 molecule on the basis of molecular orbital theory.
III. Long Answer Type Questions
- (a) Explain the formation of ionic bond with two examples.
- (b) Discuss the conditions which favour the formation of ionic bond.
(a) An ionic or electrovalent bond is formed by the complete transference of one or more electrons from one atom to another.
(b) Conditions favourable for the formation of ionic bond:
(i)Lesser the ionization enthalpy, easier will be the removal of an electron i.e., formation of a positive ion and hence greater the chances of formation of ionic bond.
(ii)Higher is the electron affinity, more is the energy released and stabler will be the negative ion produced. Consequently, the probability of formation of ionic bond will be enhanced.
- (a) Define dipole moment. What are the units of dipole moment?
- (b) Dipole moment values help in predicting the shapes of covalent molecules. Explain.
(a)Dipole moment: In a polar molecule, one end bears a positive charge and the other has a negative charge. Thus, the molecule has two poles with equal magnitude of the charges. The molecule is known as dipolar molecule and possesses dipole moment.
It is defined as the product of the magnitude of the positive or negative charge and the distance between the charges. µ (dipole moment) = q x d
SI unit of dipole moment is coulomb metre (m) or Debye.
(b)The dipole moment values are quite helpful in determining the general shapes of molecules.
For molecules with zero dipole moment, shapes will be either linear or symmetrical. For Example. BeF2 CO2etc. Molecules that possess dipole moments, their shape will not be symmetrical.
Question 3. Discuss the orbital structures of the following molecules on the basis of hybridisation,
- (i) BH3
- (ii) C2H2
B atom gets hybridised to form three equivalent hybrid orbitals directed towards three comers of equilateral triangle with B atoms in the centre. Bond angle = 120°.
Both the carbon atoms are sp hybridised. Both the carbon atoms have also two unhybridised orbitals which overlap sidewise with the similar orbitals of the other carbon atom to form two Jt bonds.
(a) How many a and n bonds are present in
(b) Why Hf is more stable than H2?
(c) Why is B2 molecule paramagnetic?
(a) No. of c bonds = 7
IV. HOTS Questions
Question 1. 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:
Phosphorus atom is sp3 d hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:
The five sp3 d hybrid orbitals are directed towards the five comers of the trigonal bipyramidal. Hence, the geometry of PCl5 can be represented as:
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 front the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.
Question 2. Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the comers of the square and the C atoms at its centre. Explain why CH4 is not square planar?
Answer: Electronic configuration of carbon atom: C: sigma 1s2 2s2 2p2.
In the excited state, the orbital picture of carbon can be represented as:
Hence, carboh atom undergoes sp3 hybridization in CH4 molecule and takes a tetrahedral shape.
For a square planar shape, the hybridization of the central atom has to be dsp3. However, an atom of carbon does not have d-orbitals to undergo dsp3 hybridization. Hence, the structure of the CH4 is tetrahedral.
Question 3. Explain why the BeH2 molecule has a zero dipole moment although the Be-H bonds are polar.
Answer: The Lewis structure for BeH2 is as follows:
There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH2 is of the type AB2. It has a linear structure,
Dipole moments of each H —Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, BeH2 has a zero dipole moment.