CHEMICAL BONDING AND MOLECULAR STRUCTURE

Question 1

State octet rule. Who proposed this rule?

Answer 1

Octet rule states that atoms containing 8 electrons in their valence shell are stable.
OR, atoms undergo chemical reaction in order to attain 8 electrons in the valence shell. It was proposed by Lewis and Kossel.

Question 2

Give any 2 limitations of Octet rule.

Answer 2

(i) It could not explain the stability of molecules containing odd number of electrons.
(ii) It could not account for the shape of molecules.

Question 3

Define dipole moment. Give its unit.

Answer 3

Dipole moment is the product of charge at one end (e) and distance between the
charges (r). i.e. μ = e x r.Its unit is Debye (D)

Question 4

The dipole moment of BeF2 is zero. Why?

Answer 4

: BeF2, the net dipole moment is zero, since the two equal bond dipoles are in opposite directions and cancel each other.

Question 5

The dipole moment of BF3 is zero. Why?

Answer 5

: In BF3, the net dipole moment is zero. Here the resultant of any 2 bond d ipoles is equal and opposite to the third.

Question 6

Ammonia (NH3) has higher dipole moment than NF3, eventhough F is more
electronegative than H. why?

Answer 6

This is because in the case of NH3, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the three N – H bonds. But in NF3, the orbital dipole is in the opposite direction to the resultant dipole moment of the three N–F bonds. So the dipole
moments get partially cancelled.

Question 7

State Fajan’s rule regarding the partial covalent character of an ionic bond.

Answer 7

(i) The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
(ii) The greater the charge on the cation, the greater the covalent character of the ionic bond.
(iii) For cations of the same size and charge, the ion with electronic configuration (n –1)dn
ns0 is more polarising than the ion with a noble gas configuration (ns2 np6).

Question 8

What any 3 postulates of valence shell electron pair repulsion (VSEPR) theory?

Answer 8

(i) The shape of the molecule depends on the number of valence shell electron pairs
(VSEPRs) around the central atom.
(ii) The valence shell electron pairs repel each other.
(iii) As the angle between the electron pairs increases, the repulsion decreases.

Question 9

Explain the shape ammonia and water molecules on the basis of VSEPR theory.

Answer 9

In ammonia, there are 4 VSEPs (3 bond pairs and one lone pair). Hence the expected shape of the molecule is tetrahedral. But due to the presence of lone pairs, the shape is distorted to triangular pyramid and the bond angle changes to 1070.
In water, there are 4 VSEPs (2 bond pairs and 2 lone pairs). Hence the expected shape of the molecule is tetrahedral. But due to the presence of 2 lone pairs, the shape is distorted bent or angular or inverted ‘v’ shape and the bond angle changes to 104.50 .
NH3 Molecule H2O Molecule

Question 10

Write any 3 important characteristics of Hybridisation?

Answer 10

(i) The number of hybrid orbitals formed is equal to the number of atomic orbitals
undergo hybridization.
(ii) The hybrid orbitals are always equivalent in energy and identical in shape.
(iii) The hybrid orbitals are directed to some fixed positions in space. So the type of
hybridization gives the shape of the molecule.

Question 11

Explain the geometry of PCl5 molecule and account for its high
reactivity.

Answer 11

In PCl5, phosphorus atom is in sp3d hybridisation. So the
molecule has trigonal bipyramidal shape with bond angles 120° and
90°.
In PCl5, there are two types of P–Cl bonds – two axial bonds and three equatorial bonds.
Due to the greater repulsion, axial bond length is greater than the equatorial bond length. So
PCl5 is highly unstable and is very reactive.

Question 12

Draw the potential energy curve for the formation of a hydrogen molecule on the basis of
inter nuclear distance of the hydrogen atoms.

Question 13

Define bond order.

Answer 13

It is defined as the half of the difference between the number of bonding electrons (Nb) and
the number of anti-bonding electrons (Na). i.e. Bond order (B.O) = ½ [Nb – Na]

Question 14

He2 does not exist. Why?

Answer 14

For He2 molecule, bond order = 0, the molecule cannot exist.

Question 15

Ne2 does not exist. Why?

Answer 15

For Ne2 molecule bond order = 0, the molecule cannot exist.

Question 16

Based on bond order compare the relative stability of O2 and O2²-

Answer 16

The M.O configuration of O2 is σ1s2 σ* 1s2 σ2s2 σ2s2 σ2pz2 π2px2
= π2py2 π2px1
= π2py1
B.O = ½ [Nb – Na] = ½ [10 – 6] = ½ x 4 = 2
Since bond order is positive, it is stable.
The M.O configuration of O2²–is σ1s2 σ 1s2 σ2s2 σ2s2 σ2pz2 π2px2=π2py2 π2px2= π*2py2
B.O = ½ [Nb – Na] = ½ [10 – 8] = ½ x 2 = 1
Since bond order of O2 is greater than O2²–, it is more stable

Question 17

Draw the M.O diagram for oxygen molecule (O2). Give its magnetic character.

Answer 17

In O2 there are 16 electrons.
The M.O configuration for O2 is
σ1s2 σ* 1s2 σ2s2 σ2s2 σ2pz2 π2px2= π2py2 π2px1= π*2py1
Here there are 2 unpaired electrons. So it is paramagnetic in nature

Question 18

Write the molecular electronic configuration of the N2 molecule. Predict the stability and
magnetic property of N2 with reasons.

Answer 18

The M.O configuration of N2 is σ1s2 σ* 1s2 σ2s2 σ*2s2 π2px2
= π2py2 σ2pz2
Bond Order = ½ [Nb – Na] = ½ [10 – 4] = ½ x 6 = 3
Since bond order is positive N2 is stable. Also, due to the absence of unpaired electrons, N2 is
diamagnetic.

Question 19

Define hydrogen bonding with example. Which are the different types of H–bonding?

Answer 19

The weak attractive force between Hydrogen atom of one molecule and electronegative
atom (like F, O or N) of the same or different molecule is termed as Hydrogen bond.
e.g. Hydrogen bonding in H–F ….H-F……..H-F……H-F……H-F……
There are two types of H bonds–inter molecular H–bonding and intra molecular H–bonding.

Question 20

Which has higher boiling point: o–nitrophenol or p–nitrophenol? Give reason.

Answer 20

p-Nitrophenol. Because of the presence of intermolecular H–bonding in it.