NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure
Topics and Subtopics in NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure :
| Section Name | Topic Name |
| 4 | Chemical Bonding and Molecular Structure |
| 4.1 | Kössel-Lewis Approach to Chemical Bonding |
| 4.2 | Ionic or Electrovalent Bond |
| 4.3 | Bond Parameters |
| 4.4 | The Valence Shell Election Pair Repulsion (VSEPR) Theory |
| 4.5 | Valence Bond Theory |
| 4.6 | Hybridisation |
| 4.7 | Molecular Orbital Theory |
| 4.8 | Bonding in Some Homonuclear Diatomic Molecules |
| 4.9 | Hydrogen Bonding |
NCERT Solutions Class 11 Chemistry Chemistry Lab Manual Chemistry Sample Papers
NCERT TEXTBOOK QUESTIONS SOLVED
Question 1. Explain the formation of a chemical bond.
Answer:
According to Kossel and Lewis, atoms combine together in order to complete their respective octets so as to acquire the stable inert gas configuration. This can occur in two ways; by transfer of one or more electrons from one atom to other or by sharing of electrons between two or more atoms.
Question 2.Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Answer:
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Question 3. Write Lewis symbols for the following atoms and ions: S and S
2
– ; Al and Al
3+
; H and H
–
Answer:
Question 4. Draw the Lewis structures for the following molecules and ions:
H
2
S, SiCl
4 ,
BeF
2
, C0
3
2-
, HCOOH
Answer:
Question 5. Define Octet rule. Write its significance and limitations.
Answer:
Octet rule:
Atoms of elements combine with each other in order to complete their respective octets so as to acquire the stable gas configuration.
Significance:
It helps to explain why different atoms combine with each other to form ionic compounds or covalent compounds.
Limitations of Octet rule:
-
According to Octet rule, atoms take part in chemical combination to achieve the configuration of nearest noble gas elements. However, some of noble gas elements like Xenon have formed compounds with fluorine and oxygen. For example: XeF
2
, XeF
4
etc.
Therefore, validity of the octet rule has been challenged. - This theory does not account for shape of molecules.
Question 6. Write the favourable factors for the formation of ionic bond.
Answer:
- Low ionization enthalpy of metal atoms
- High electron gain enthalpy of non-metal atoms
- High lattice enthalpy of compound formed.
Question 7. Discuss the shape of the following molecules using the VSEPR model:
BeCl
2
, BCl
3
, SiCl
4
, AsF
5
, H
2
S, PH
3
Answer:
Question 8. Although geometries of NH
3
and H
2
0 molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:
Because of two lone pairs of electrons on O-atom, repulsion on bond pairs is greater in H
2
0 in comparison to NH
3
. Thus, the bond angle is less in H
2
0 molecules.
Question 9. How do you express the bond strength in terms of bond order?
Answer:
Bond strength is directly proportional to the bond order. Greater the bond order, more is the bond strength.
Question 10. Define the bond-length.
Answer:
Bond-length: It is the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond-lengths are measured by spectroscopic methods.
Question 11. Explain the important aspects of resonance with reference to the C0
3
2-
ion.
Answer:
Resonance in C0
3
2-
, I, II and III represent the three canonical forms.
- In these structures, the position of nuclei are same.
- All three forms have almost equal energy.
- Same number of paired and impaired electrons, they differ only in their position.
Question 12. H
3
PO
3
can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H
3
PO
3
? If not, give reasons for the same.
Answer:
No, these cannot be taken as canonical forms because the positions of atoms have been changed.
Question 13. Write the resonance structures for SO
3
,NO
2
and NO
3
Answer:
Question 14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions (a) K and S (b) Ca and O (c) Al and N.
Answer:
Question 15. Although both CO
2
and H
2
O are triatomic molecules, the shape of H
2
O molecule is bent while that of CO
2
is linear. Explain this on the basis of dipole moment.
Answer:
In CO
2
, there are two C=O bonds. Each C=O bond is a polar bond.
The net dipole moment of CO
2
molecule is zero. This is possible only if CO
2
is a linear molecule. (O=C=O). The bond dipoles of two C=O bonds cancel the moment of each other.
Whereas, H
2
O molecule has a net dipole moment (1.84 D). H
2
O molecule has a bent structure because here the O—H bonds are oriented at an angle of 104.5° and do not cancel the bond moments of each other.
Question 16. Write the significance/applications of dipole moment.
Answer:
- In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those of zero dipole moments are non-polar in nature.
- In the determination of shapes of molecules.
- In calculating the percentage ionic character.
Question 17. Define electronegativity. How does it differ from electron gain enthalpy?
Answer:
Electronegativity:
Electronegativity is the tendency of an atom to attract shared pair of electrons. It is the property of bonded atom.
Whereas, electron gain enthalpy is the tendency of an atom to attract outside electron. It is the property of an isolated atom.
Question 18. Explain with the help of suitable example polar covalent bond.
Answer:
When two atoms with different electronegativity are linked to each other by covalent bond, the shared electron pair will not in the centre because of the difference in electronegativity. For example, in hydrogen flouride molecule, flouride has greater electronegativity than hydrogen. Thus, the shared electron pair is displaced more towards’flourine atom, the later will acquire a partial negative charge (∂
–
). At the same time hydrogen atom will have a partial positive charge (∂
+
). Such a covalent bond is known as polar covalent bond or simply polar bond.
It is represented as
Question 19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K
2
O, N
2
, SO
2
and ClF
3
.
Answer:
N
2
< SO
2
< ClF
3
< K
2
O < LiF
Question 20. The skeletal structure of CH
3
COOH
as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
Answer:
Question 21. Apart from tetrahedral geometry, another possible geometry for CH
4
is square planar with the four H atoms at the comers of the square and the C atom at its centre. Explain why CH
4
is not square planar?
Answer:
According to VSEPR theory, if CH
4
were square planar, the bond angle would be 90°. For tetrahedral structure, the bond angle is 109°28′. Therefore, in square planar structure, repulsion between bond pairs would be more and thus the stability will be less.
Question 22. Explain why BeH
2
molecule has a zero dipole moment although the Be—H bonds are polar.
Answer:
BeH
2
is a linear molecular (H—Be—H), the bond angle = 180°.
Be—H bonds are polar due to difference in their electronegativity but the bond polarities cancel each other. Thus, molecule has resultant dipole moment of zero.
Question 23. Which out of NH
3
and NF
3
has higher dipole moment and why?
Answer:
In NH
3
and NF
3
, the difference in electronegativity is nearly same but the dipole moment of NH
3
= (1.46D) For Example, NH
3
= (0.24D)
In NH3, the dipole moments of the three N—H bonds are in the same direction as the lone pair of electron. But in NF
3
, the dipole moments of the three N—F bonds are in the direction opposite to that of the lone pair. Therefore, the resultant dipole moment in NH
3
is more than in NF
3
.
Question 24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp
2
, sp
3
hybrid orbitals.
Answer:
Hybridisation:
It is defined as the process of intermixing of atomic oribitals of slightly different energies to give rise to new hybridized orbitals having equivalent energy and identical shapes.
Shapes of Orbitals:
sp hybridisation:
When one s-and one p-orbital, intermix then it is called sp-hybridisation. For example, in BeF2, Be atom undergoes sp-hybridisation. It has linear shape. Bond angle is 180°.
sp
2
hybridisation:
One s-and two p-orbitals get hybridised to form three equivalent hybrid orbitals. The three hybrid orbitals directed towards three corners of an equilateral triangle. It is, therefore, known as trigonal hybridisation.
sp
3
hybridisation:
One s-and three p-orbitals get hybridised to form four equivalent hybrid orbitals. These orbitals are directed towards the four corners of a regular tetrahedron.
Question 25. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl
3
+ Cl
–
——> AlCl
4-
.
Answer:
Electronic configuration of
13
Al = 1s
2
2s
2
2p
6
3s
1
3p
x
1
3p
y
1
(excited state)
Hence, hybridisation will be SP
2
In AlCl
–
4
, the empty 3p
z
orbital is also involved. So, the hybridisation is sp
3
and the shape is tetrahedral.
Question 26. Is there any change in the hybridisation ofB and N atoms as a result of the following reaction ?
BF
3
+ NH
3
——-> F
3
B.NH
3
Answer:
In BF
3
, B atom is sp
2
hybridised. In NH
3
, N is sp
3
hybridised.
After the reaction, hybridisation of B changes from sp
2
to sp
3
.
Question 27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in
C
2
H
4
and C
2
H
2
molecules.
Answer:
Question 28. What is the total number of sigma and pi bonds in the following molecules?
(a) C
2
H
2
(b) C
2
H
4
Answer:
(a) H—C = C—H
Sigma bond = 3 Π bonds = 2
Question 29. Considering X-axis as the intemuclear axis which out of the following will not form a sigma bond and why? (a) Is and Is (b) Is and 2p
x
(c)
2p
y
and 2p
y
(d) Is and 2s
Answer:
(c) It will not form a s-bond because taking x-axis as the intemuclear axis, there will be lateral overlap between the two 2p
y
orbitals forming a Π -bond.
Question 30. Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH
3
-CH
3
(b) CH
3
-CH = CH
2
(c) CH
3
-CH
2
-OH (d) CH
3
-CHO (e) CH
3
COOH.
Answer:
Question 31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two atoms during covalent bonding is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.
Question 32. Distinguish between a sigma bond and a pi bond.
Answer:
Question 33. Explain the formation of H
2
molecule on the basis of valence bond theory.
Answer:
Let us consider the combination between atoms of hydrogen H
A
and H
B
and eA and eB be their respective electrons.
As they tend to come closer, two different forces operate between the nucleus and the electron of the other and vice versa. The nuclei of the atoms as well as their electrons repel each other. Energy is needed to overcome the force of repulsion. Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more. Thus, when two hydrogen atoms approach each other, the overall potential energy of the system decreases. Thus, a stable molecule of hydrogen is formed.
Question 34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
-
The combining atomic orbitals should have comparable energies.
For example, Is orbital of one atom can combine with Is atomic orbital of another atom, 2s can combine with 2s. - The combining atomic orbitals must have proper orientations. So that they are able to overlap to a considerable extent.
- The extent of overlapping should be large.
Question 35. Use molecular orbital theory to explain why the Be
2
molecule does not exist.
Answer:
Question 36. Compare the relative stability of the following species and indicate their magnetic properties: O
2
, O
2
, O
2
–
(Superoxide),O
2
2-
(peroxide)
Answer:
O
2
— Bond order = 2, paramagnetic
O
2
+
— Bond order = 2.5, paramagnetic
O
2
–
— Bond order = 1.5, paramagnetic
O
2
2-
— Bond order = 1, diamagnetic
Order of relative stability is
O
2
+
> O
2
> O
2
–
> O
2
2-
(2.5) (2.0) (1.5) (1.0)
Question 37. Write the significance of plus and minus sign in representing the orbitals,
Answer:
Plus and minus sign is used to indentify the nature of electrons wave. Plus (+ve) sign denotes crest, while (-ve) sign denotes trough.
Question 38. Describe the hybridisation in case of PCl
5
. Why are the axial bonds longer as compared to equatorial bonds?
Answer:
The ground state E.C. and the excited state E.C. of phosphorus are represented as:
The one s, three-p and one d-orbitals hybridise to yield five sets of SP
3
d hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal as in Fig.
Because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than equatorial bonds.
Question 39. Define hydrogen bonds. Is it weaker or stronger than the van der Waals forces?
Answer:
When hydrogen is attached with highly electronegative element in a covalent bonding the electrons of the covalent bond are shifted towards the more electronegative atom. Thus, a partially positively charged hydrogen atom forms a bond with the other more electronegative atom. This bond is known as a hydrogen bond. Hydrogen bond is stronger than the van der Waals forces.
Question 40. What is meant by the term bond order? Calculate the bond order of N
2
, O
2
, O
2
+
,O
2
–
Answer:
Bond order is defined as the half of the difference between the number of electrons present in bonding and antibonding molecular orbitals.
MORE QUESTIONS SOLVED
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 CH
4
,C
2
H
4
and C
2
H
2
.
Answer:
CH
4
= sp
3
C
2
H
4
= sp
2
C
2
H
2
= 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
Answer:
Question 5. Predict the shapes of the following molecules using VSEPR theory?
(i) BeCl
2
(ii) SiCl
4
Answer:
(i) Linear
(ii) Tetrahedral
Question 6. Write the state of hybridisation of boron in BF
3
.
Answer:
SP
2
Question 7. Arrange O
2
,O
2
–
,O
2
2-
, O
2
+
in increasing order of bond energy.
Answer:
O
2
2-
< O
2
–
< O
2
2-
< O
2
+
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) CO
2
, BF
3
(ii) NH
3,
CH
4
Answer:
(i) CO
2
(ii) CH
4
Question 10. Arrange the following, according to increasing covalent nature.
NaCl, MgCl
2
, AlCl
3
Answer:
NaCl < MgCl
2
< AlCl
3
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 B
2
is paramagnetic in nature while C
2
is not?
Answer:
The molecular orbital electronic configuration of both B
2
and C
2
are.
Since, B
2
has two impaired electrons, B
2
is paramagnetic.
C
2
has no unpaired electron. Thus, C
2
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 CO
2
or N
2
O? Give reason.
Answer:
N
2
O is more polar than CO
2
.
This is “because CO
2
is linear and symmetrical. Its net dipole moment is zero.
N
2
O 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 PCl
5
and (ii) S in SF
6
Answer:
(i) sp
3
d
of P in PCl
5
(ii)
sp
3
d
2
of S in
SF
6
Question 16. Why N
2
is more stable than O
2
? Explain on the basis of molecular orbital theory.
Answer:
Bond order of N
2
(= 3) is greater than that of O
2
(= 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.
Answer:
(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.
Answer:
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. ‘
For Example,
Question 2. What are Lewis structures? Write the Lewis structure of H
2
, BeF
2
and H
2
O.
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) AB
3
(ii) AB
4
Answer:
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 (H
2
0)> the atomic number of oxygen is 8 and its orbitals electronic configuration is 1s
2
2s
2
2p
x
2
2p
y
1
2p
z
1
.The oxygen atom is also SP
3
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 H
2
S is a gas
(ii) NH
3
has higher boiling point than PH
3
.
Answer:
(i) In case of water hydrogen bonding causes association of the H
2
O molecules. There is no such hydrogen bonding in H
2
S, that’s why it is a gas.
(ii) There is hydrogen bonding in NH
3
but not in PH
3
.
Question 7. What do you mean by Dipole moment? Draw the dipole diagram of H
2
O.
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?
Answer:
- 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[N
b
– N
a
] z
If the bond order is positive (N
b
> N
a
), the molecule or ion will be stable. If it is negative (N
b
< N
a)
the molecule or ion will be unstable.
Question 10. Explain the diamagnetic behaviour of P
2
molecule on the basis of molecular orbital theory.
Answer:
III. Long Answer Type Questions
Question 1. (a) Explain the formation of ionic bond with two examples.
(b) Discuss the conditions which favour the formation of ionic bond.
Answer:
(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.
Question 2. (a) Define dipole moment. What are the units of dipole moment?
(b) Dipole moment values help in predicting the shapes of covalent molecules. Explain.
Answer:
(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. BeF
2
CO
2
etc. 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) BH
3
(ii) C
2
H
2
Answer:
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.
Question 4. (a) How many a and n bonds are present in
(b) Why Hf is more stable than H
2
?
(c) Why is B
2
molecule paramagnetic?
Answer:
(a) No. of c bonds = 7
IV. Multiple Choice Questions
Question 1. A co-ordinate bond is formed by:
(a)sharing of electrons contributed by both the atoms
(b)complete transfer of electrons
(c)sharing of electrons contributed by one atom only (d) none of these
Question 2. The species CO, CN
–
and N
2
are:
(a) isoelectronic (b) having coordinated bond
(c)having polar bond (d) having low bond energies
Question 3. The axial overlap between the two orbitals leads to the formation of a:
(a) sigma bond (b) pi bond (c) multiple bond (d) none of these
Question 4. In S0
2
molecule, S atom is:
(a)sp
3
hybridized (b) sp hybridized (c) sp
2
hybridized (d) d
sp
2
hybridized
Question 5. A molecule or ion is stable if:
(a) Nb = Na (b) Nb < Na (c) Na < Nb (d) Na – Nb = + ve
Question 6. The molecule Ne2 does not exist because
(a) Nb > Na (b) Nb = Na + (c) Nb < Na (d) None of these
Question 7. Which one is diamagnetic among NO
+
, NO and NO ?
(a)NO
+
(b) NO (c)
NO
–
(d) None of these
Question 8. In sp
3
, sp
2
and sp hybridized carbon atom, the p character is maximum in:
(a) sp
3
(b) sp
2
(c) sp (d) all of the above have same p-character
Question 9. Out of the following, intramolecular hydrogen bonding exists in:
(a) water (b)H
2
S
(c) 4-nitrophenol (d) 2-nitrophenol
Answer:
1. (c) 2. (a) 3. (a) 4. (c) 5. (c)
6. (b) 7. (a) 8. (a) 9. (d)
V. HOTS Questions
Question 1. Describe the hybridisation in case of PCl
5.
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 sp
3
d hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:
The five sp
3
d hybrid orbitals are directed towards the five comers of the trigonal bipyramidal. Hence, the geometry of PCl
5
can be represented as:
There are five P-Cl sigma bonds in PCl
5
. 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 CH
4
is square planar with the four H atoms at the comers of the square and the C atoms at its centre. Explain why CH
4
is not square planar?
Answer:
Electronic configuration of carbon atom: C: sigma 1s
2
2s
2
2p
2
.
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 dsp
3
. However, an atom of carbon does not have d-orbitals to undergo dsp
3
hybridization. Hence, the structure of the CH
4
is tetrahedral.
Question 3. Explain why the BeH
2
molecule has a zero dipole moment although the Be-H bonds are polar.
Answer:
The Lewis structure for BeH
2
is as follows:
There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH
2
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, BeH
2
has a zero dipole moment.