NCERT Solutions Class 11 Chemistry Chapter 4
NCERT Solutions for Class 11 Chemistry Chapter 4 – Chemical Bonding and Molecular Structure
Chemistry is the study of substances and is a field that involves everything around us. According to the American Chemical Society, all that can be seen, touched, tasted, smelled, and heard has chemistry and chemicals involved in it. To help students in preparing for their exams, Extramarks offers NCERT Solutions for Class 11 Chemistry Chapter 4. They can be accessed easily from the link below
Access NCERT Solutions for Class 11 Chemistry Chapter – 4 Chemical Bonding and Molecular Structure
Limitations of the Octet Theory:
Octet theory has the following limitations:
- It applies only to atoms in their ground state.
- It does not take into account the nature of chemical bonding between atoms.
- It does not take into account the number of electrons in an atom.
Chapter 4 explains all these limitations of Octet theory in detail.
NCERT Solutions for Class 11 Chemistry Chapter – 4 Chemical Bonding and Molecular Structure
Chapter 4 of NCERT Class 11 Chemistry covers the formation of chemical bonds, Lewis dot symbols, and other concepts. The chapter has exercise questions at the end to help students gauge their understanding of the concepts. NCERT Solutions for Chapter 4 enables students to get accurate answers to the practise questions listed at the end of the chapter.
NCERT Class 11 Chemistry Chapter 4 – Question 1
Que. Explain the formation of a chemical bond.
Ans. A chemical bond is an attractive force holding the atoms, ions and other constituents together in a chemical species. The formation of a chemical bond can be related to the tendency of a system for attaining stability. The inertness of noble gases is due to their fully filled outermost orbitals. Thus, it was assumed that elements with incomplete outermost shells are unstable or reactive. This is why atoms combine with each other to complete their octets or duplets for attaining stable configuration of the nearest noble gas. This can occur by sharing electrons or transfer of one or more electrons from one atom to another. This results in the formation of a chemical bond.
NCERT Class 11 Chemistry Chapter 4 – Question 2
Que. Write Lewis dot symbols for atoms of following elements: NA, B, Mg, O, Br, and N.
Ans. Since there are two valence electrons in Mg atom, the Lewis dot symbol is –
There is only one valence electron in an atom of sodium, the Lewis dot symbol is –
There are three valence electrons in Boron atom, the Lewis dot symbol is –
There are six valence electrons in an atom of oxygen, the Lewis dot symbol is –
There are five valence electrons in an atom of nitrogen, the Lewis dot symbol is –
There are seven valence electrons in bromine, the Lewis dot symbol is –
NCERT Class 11 Chemistry Chapter 4 – Question 3
Que: Write Lewis symbols for the following atoms and ions –
- S and S2
- Al and Al+3
- H and H-
Ans: S and S2
The number of valence electrons in Sulphur is 6, the Lewis dot symbol is:
The dinegative charge refers that there will be two electrons more along with the six valence electrons, so the Lewis dot symbol of S2 is:
Al and Al+3
The number of valence electrons in aluminium is 3, Lewis dot symbol is:
Al3+ means it has donated its three electrons, Lewis dot symbol is:
H and H-
The number of valence electrons in hydrogen is 1. Lewis dot symbol is:
The uninegative charge means that one more electron in addition to one valence electron, the Lewis dot symbol is:
NCERT Class 11 Chemistry Chapter 4 – Question 4
Que. What are the favourable factors for the formation of ionic bonds?
Ans. The transfer of one or more electrons from one atom to another leads to the formation of ionic bonds. The formation is dependent on the ease with which neutral atoms can gain or lose electrons. Also, the formation of a bond is dependent on the lattice energy of the compound formed.
Here are some of the favourable factors for the formation of ionic bond:
- Low ionisation enthalpy of metal atom
- High lattice energy of the compound that is formed
- High electron gain enthalpy of a non-metal atom
NCERT Class 11 Chemistry Chapter 4 – Other Questions
The other questions focus on the octet rule, ionic bonds, bond order, molecular geometry, resonance, etc. To refer to these concepts and get accurate answers to the questions, students can access NCERT Solutions for Class 11 Chemistry Chapter 4.
NCERT Chemistry Class 11 – Chapter 4 (Chemical Bonding and Molecular Structure)
The subject of Chemistry includes a topic about Chemical Bonding and Molecular Structure in Chapter 4. The NCERT Solutions for Chapter 4 Chemistry by Extramarks are prepared by subject-matter experts to ensure that students understand the concepts thoroughly, and get accurate answers to the questions given at the end of the chapter.
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Q.1 Explain the formation of a chemical bond.
A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a compound. 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 completely filled outermost shell. Hence, the elements having incomplete outermost shells are unstable (reactive). Only these atoms combine 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 transfer of electrons from one atom to another.
Q.2 Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, and Br.
There are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is
There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure for Na is
There are 3 valence electrons in Boron atom. Hence, the Lewis dot structure for B is
There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure for O is
There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure for N is
There are seven valence electrons in bromine atom. Hence, the Lewis dot structure for Br is
Q.3 Write Lewis symbols for the following atoms and ions: S and S2-, Al and Al3+, H and H–
The number of valence electrons in sulphur is 6.
The Lewis dot symbol of sulphur (S) is
The dinegative charge in S2- indicates that there are two electrons more in addition to the six
valence electrons. Hence, the Lewis dot symbol of S2– is
ii) The number of valence electrons in aluminium is 3 and the Lewis dot symbol of aluminium
The tri-positive charge on Al3+ shows that it has donated its three electrons. Hence, the Lewis dot symbol of Al3+ is
iii) The number of valence electrons in hydrogen is 1.
The Lewis dot symbol of hydrogen (H) is
The uninegative charge on H– means that there will be one electron more in addition to one
valence electron. Hence, the Lewis dot symbol of H– is
Q.4 Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, CO32- , HCOOH
Lewis structure for H2S:
Lewis structure for SiCl4:
Lewis structure for BeF2:
Lewis structure for CO32– :
Lewis structure for HCOOH:
Q.5 Define octet rule. Write its significance and limitations.
. 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 stability like the nearest noble gas configuration by having an octet in their valence shell.
For example, structure of CO2 is as follows:
The octet rule successfully explained the formation of chemical bonds (Electrovalent, covalent or coordinate bond) depending upon the nature of the element.
Limitations of the octet theory:
(a) This rule can’t explain the shape and relative stability of the molecules.
(b) It is based upon the inert shell configuration of noble gases. But, some noble gases like krypton and xenon form compounds such as KrF2, XeF2 etc.
(c) The octet rule cannot be applied to the elements in and beyond the third period of the periodic table. Due to the presence of d-orbital, the elements present in these periods have more than eight valence electrons around the central atom. For example: PF5, SF6 etc.
(d) The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example, NO and NO2 do not satisfy the octet rule.
(e) This rule is not applicable to those compounds in which the number of electrons surrounding the central atom is less than eight. For example – LiCl, BeH2, AlCl3 etc. do not obey the octet rule.
Q.6 Write the favourable factors for the formation of ionic bond.
An ionic bond is formed by complete 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. Bond formation also depends upon the lattice energy of the compound formed.
The important factors for ionic bond formation are as follows:
(i) Low ionization enthalpy of metal atom.
(ii) High electron gain enthalpy (DegH) of a non-metal atom.
(iii) High lattice energy of the compound formed.
Q.7 Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Ans. Shape of the BeCl2:
The central atom Be has two bond pair and no lone pair.
BeCl2 is the AB2 type of molecule having linear structure.
Shape of the BCl3:
The central atom B has three bond pairs and no lone pair. It is AB3 type of molecule having trigonal planer structure.
Shape of the SiCl4:
The central atom Si has four bond pairs and no lone pair. The shape of SiCl4 is AB4 type molecule having tetrahedral structure.
Shape of the AsF5:
The central atom As has five bond pairs and no lone pair. Hence, AsF5 is the AB5 type of molecule. Therefore, the shape is trigonal bipyramidal.
Shape of the H2S:
The central atom S has two lone pairs and two bond pairs. It is of the type AB2L2. Due to lone pair–bond pair repulsion, the structure of H2S is bent or V-shaped.
Shape of the PH3:
The central atom P has one lone pair and three bond pairs of electrons. It is of the type AB3L. The structure of PH3 is trigonal pyramidal.
Q.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
The central atom (N) in NH3 has one lone pair and three bond pairs of electrons. In H2O, oxygen atom has two lone pairs and two bond pairs of electrons. The two lone pairs of electrons present in the oxygen atom of H2O molecule repel two bond pairs of electrons. This repulsion (lone pair-lone pair repulsion) is stronger than the repulsion between the lone pair and the three bond pairs (lone pair-bond pair repulsion) on the nitrogen atom in ammonia molecule.
Since the repulsion between the bond pair electrons in H2O molecule are greater than that in NH3, the bond angle in water is less than that of ammonia.
Q.9 How do you express the bond strength in terms of bond order?
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.
Q.10 Define the bond length.
Bond length is the average distance between the nuclei of two bonded atoms in a molecule. Bond lengths are expressed in terms of angstrom (10-10 m) or picometer (10–12 m) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.
In an ionic compound, the bond length (d) is the sum of the ionic radii of the constituting atoms (cations and anions). It can be written as –
d = rcation+ ranion
In a covalent compound, it is the sum of covalent radii of the constituting atoms and can be written as –
d = rA+ rB
Q.11 Explain the important aspects of resonance with reference to the CO32- ion.
Experimentally it is found that all the bond lengths in carbonate ion are equivalent. Hence, it is inadequate to represent CO32- ion by a single Lewis structure having two single bonds and one double bond.
The carbonate ion is described as a resonance hybrid of the following structures:
Q.12 H3PO3 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 H3PO3? If not, give reasons for the same.
The given structures cannot be taken as the canonical forms of the resonance hybrid of H3PO3 because the positions of the atoms have changed. In canonical forms only the position of the electrons can be changed not the position of the atoms.
Q.13 Write the resonance structures for SO3, NO2 and NO3–.
The resonance structures are as follows:
Q.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.
(a) K and S:
The electronic configurations of K and S are as follows:
K: 2, 8, 8, 1
S: 2, 8, 6
Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) has one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as follows:
(b) Ca and O:
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:
(c) Al and N:
The electronic configurations of Al and N are as follows:
Al: 2, 8, 3
N: 2, 5
Nitrogen requires three more electrons to get the nearest noble gas (Neon) configuration, whereas aluminium has three electrons more than Neon. Hence, the electron transfer can be shown as:
Q.15 Although both CO2 and H2O are tri atomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
H2O, on the other hand, has a dipole moment value of 1.84 D (though it is also a tri atomic molecule). The value of the dipole moment suggests that the structure of H2O molecule is bent
where the dipole moment of O–H bonds are unequal.
Q.16 Write the significance/applications of dipole moment.
Due to different electronegativities of the constituents of atoms in heteroatomic molecules, polarisation occurs. As a result, one end of the molecule acquires a partial positive charge while the other end becomes partially negatively charged. Hence, a molecule is said to possess a dipole. The product of the magnitude of the charge and the distance between the centres of partial positive and partial negative charges is called as the dipole moment (µ) of the molecule. It is a vector quantity and is represented by an arrow with its tail at the positive centre and head pointing towards a negative centre. It can be calculated by using expression –
Dipole moment (µ) = charge (Q) × distance of separation (r).
The SI unit of a dipole moment is ‘esu’. 1 esu = 3.335 × 10–30 cm.
The applications of dipole moment are:
- Dipole moment is the measure of the polarity of a bond.
- It can be used to differentiate between polar and non-polar bonds since all non-polar molecules (e.g. N2, O2) have zero dipole moments.
Q.17 Define electronegativity. How does it differ from electron gain enthalpy?
Electronegativity is the ability of an atom in a chemical compound to attract electrons of a bond pair towards itself. Electronegativity varies elements to elements. 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.
Q.18 Explain with the help of suitable example polar covalent bond.
When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of the electrons is not shared equally. The bond pair shifts towards the nucleus of the more electronegative atom. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom.
For that reason, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in that covalent compound and the bond formed is called as polar covalent bond. For example, HCl contains a polar covalent bond. Chlorine atom is more electronegative than hydrogen atom. Hence, the bond pair of electron lies towards chlorine atom. Therefore, it acquires a partial negative charge and hydrogen atom acquires a partial positive charge.
Q.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
When two dissimilar atoms combine to form a covalent molecule, the bond pair electrons are not equally shared between the two atoms. The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The increasing order of electronegativities of F, O and N is as follows:
The greater the difference in the electronegativities of the constituent atoms, the greater will be the ionic character of the molecule. The order of increasing ionic character in the given molecules is
N2 < SO2 < ClF3 < K2O < LiF
Q.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
The correct Lewis structure for acetic acid is as follows:
Q.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?
Electronic configuration of carbon atom:
In the excited state, the orbital diagram of carbon can be represented as:
Hence, carbon atom undergoes sp3-hybridisation in CH4 molecule and has a tetrahedral shape.
For a square planar shape, the hybridisation of the central atom should be dsp2 hybridised. But C atom do not have energetically accessible d-orbitals, it cannot undergo dsp2-hybridisation. So, the structure of CH4 cannot be square planar.
For square planer geometry the bond angle between the atoms would be 90°, the stability of CH4 will be very less because of the repulsion between the bond pairs. VSEPR theory also supports a tetrahedral structure for CH4.
Q.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
The Lewis structure for BeH2 is as follows:
The central atom (Be) has no lone pair of electrons 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, the dipole moment of BeH2 molecule is zero.
Q.23 Which out of NH3 and NF3 has higher dipole moment and why?
In both molecules i.e., NH3 and NF3, the central atom (N) has a lone pair of 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 NF3 is greater than NH3. However, the net dipole moment of NH3 (1.46 D) is greater than that of NF3 (0.24 D).
This can be explained on the basis of the directions of the dipole moments of each individual bond in NF3 and NH3. These directions can be shown as:
In NH3 molecule the resultant moment of the N–H bonds and the bond moment of the lone pair are in the same direction, they both adds up and increases the net dipole moment of the NH3 molecule. In case of NF3 molecule, three N – F bonds partly cancels the moment of the lone pair. Hence, the net dipole moment of NF3 is less than that of NH3.
Q.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Hybridisation is defined as an intermixing of a set of atomic orbitals belonging to the same atom but having slightly different energies, thereby forming a new set of orbitals having equivalent energies and identical shapes. For example – one 2s-orbital hybridises with two 2p-orbitals of carbon to form three new sp2 hybrid orbitals.
The orientation of these hybrid orbitals minimises the repulsion between the electron pairs and thus, is more stable. Hybridisation helps to indicate the geometry of the molecule.
Shape of sp-hybrid orbitals:
One s-orbital hybridises with one p orbital to form two new sp- hybrid orbitals along the same axis. sp- hybrid orbitals have a linear shape.
Shape of sp2-hybrid orbitals:
Three sp2– hybrid orbitals are formed as a result of the intermixing of one s-orbital and two p-orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement as:
Shape of sp3-hybrid orbitals:
Four sp3 hybrid orbitals are formed by the intermixing of one s-orbital with three p-orbitals. The four sp3-hybrid orbitals are arranged in the form of a tetrahedron as:
Q.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction.
The valence orbital picture of Al in the ground state and excited states can be represented as:
Hence, it undergoes sp2-hybridisation to give a trigonal planar arrangement (in AlCl3). To form AlCl4– ion, the empty 3pz -orbital also gets involved and the hybridisation changes from sp2 to sp3. As a result, the shape gets changed from trigonal planer to tetrahedral geometry.
Q.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
Boron atom in BF3 is sp2 -hybridised. The orbital picture of boron in the excited state can be shown as:
Nitrogen atom in NH3 is sp3 -hybridised. The orbital picture of nitrogen can be represented as:
After the reaction has occurred, an adduct F3B⋅NH3 is formed and the hybridisation of ‘B’ changes to sp3. However, the hybridisation of ‘N’ remains the same.
Q.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
The electronic configuration of C-atom in the excited state is:
6C= 1s2 2s1 2px12py12pz1
The formation of an ethylene molecule (C2H4):
In ethylene molecule, one sp2-hybrid orbital of carbon overlaps with sp2-hybridised orbital of another carbon atom, thereby forming a C–C sigma bond. The remaining two sp2-orbitals of each carbon atom form a sp2-s sigma bond with two hydrogen atoms. The unhybridised p-orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak π-bond.
The formation of acetylene (C2H2) molecule:
In the formation of C2H2 molecule, each C–atom is sp-hybridised with two 2p-orbitals in an unhybridised state. One sp-orbital of carbon atom overlaps with another sp-orbital of C atom along the inter-nuclear axis forming a C–C sigma bond. The second sp-orbital of each C–atom overlaps with 1s-orbital of two hydrogen atoms to form σ bond.
The two unhybridised 2p-orbitals of the first carbon atom undergo sidewise overlap with the 2p-orbital of another carbon atom, thereby forming two pi (π) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two π-bonds.
Q.28 What is the total number of sigma and pi bonds in the following molecules?
C2H2 (b) C2H4
A single bond is a result of the axial overlap of bonding orbitals and forms a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A double bond is a combination of one sigma bond and one pi-bond. A triple bond is a combination of two pi-bonds and one sigma bond.
Structure of C2H2 can be represented as:
There are three sigma and two pi-bonds in acetylene molecule.
The structure of C2H4 can be represented as:
There are five sigma bonds and one pi-bond in the ethylene molecule.
Q.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a)1s and 1s (b)1s and 2px (c)2py and 2py (d)1s and 2s.
If we take x-axis as the internuclear axis, then (c) 2py and 2py orbitals will not a form a sigma bond. 2py and 2py will undergo lateral overlapping, thereby forming a pi (π) bond.
Q.30 Which hybrid orbitals are used by carbon atoms in the following molecules?
(a)CH3–CH3 (b) CH3–CH=CH2 (c) CH3-CH2-OH (d) CH3-CHO (e) CH3COOH
Both C1 and C2 both are used sp3– hybrid orbitals.
C1 is sp3-hybridised, while C2 and C3 both are sp2-hybridised.
Both C1 and C2 both are sp3-hybridised.
C1 is sp3-hybridizsed and C2 is sp2– hybridised.
Here, C1 is sp3-hybridised and C2 is sp2-hybridised.
Q.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
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 which do not participate in bonding are called lone pairs of electrons.
For example, in CH4 (In methane, C is the central atom) bond pairs are present but it has no lone pairs.
In H2O, there are two bond pairs and two lone pairs on the central oxygen atom .
Q.32 Distinguish between a sigma and a pi bond.
The following are the differences between sigma and p-bonds:
|Sigma (σ) bond||Pi(π) bond|
Q.33 Explain the formation of H2 molecule on the basis of valence bond theory.
Let us assume that two hydrogen atoms (A and B) with nuclei (NA and NB) and electrons (eA and eB) are taken to undergo a reaction to form a hydrogen molecule.
When A and B are at a large distance, there is no interactions operating between them. As they begin to approach each other, the attractive and repulsive forces start operating.
Attractive force arises between:
- Nucleus of one atom and its own electron.
(b) Nucleus of one atom and electron of another atom.
Repulsive force arises between:
- Electrons of two atoms.
- Nuclei of two atoms.
The force of attraction brings the two atoms together, whereas the force of repulsion tends to repel them apart.
If the magnitude of the attractive forces is more than that of the repulsive forces then the two atoms approach each other. As a result, the potential energy decreases. When the attractive forces balance the repulsive forces then the system acquires minimum energy. This leads to the formation of a di-hydrogen molecule.
Q.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
The following conditions should be satisfied by atomic orbitals to form molecular orbitals:
(a) The combining atomic orbitals must have the same or nearly the same energy. This means that in 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.
(b) The combining atomic orbitals must have the same symmetry about the molecular axis.
(c) The combining atomic orbitals must have proper orientations to ensure that the extent of overlap is maximum.
Q.35 Use molecular orbital theory to explain why the Be2 molecule does not exist.
The electronic configuration of Beryllium is 1S2 2S2
The molecular orbital electronic configuration for Be2 molecule can be written as:
Hence, the bond order for Be2 = 1/2Nb – Na)
Nb= Number of electrons in bonding orbitals
Na= Number of electrons in anti-bonding orbitals
Therefore bond order of Be2 is = ½(4-4) =0
Zero bond order means that the molecule is unstable. Hence, Be2 molecule does not exist.
Q.36 Compare the relative stability of the following species and indicate their magnetic properties;
O2, O2+, O2–(super oxide), O22- (peroxide)
There are 16 electrons in a molecule of O2 molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:
Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = Nb and the number of anti-bonding orbitals = 4 = Na.
The bond order of O2 = ½(Nb–Na) = ½(8–4) = 2
Similarly, the electronic configuration O2+ can be written as:
Nb = 8 and Na = 3
Bond order of O2+= ½(8–3) = 5/2= 2.5
Electronic configuration of O2– ion:
Nb = 8 and Na = 5
Bond order of O2– = ½ (8–5) = 3/2 = 1.5
Electronic configuration of O22- ion:
Nb= 8 and Na= 6
Bond order= ½(8-6) = 2/2 = 1
Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. The order of stability is
Q.37 Write the significance of a plus and a minus sign shown in representing the orbitals.
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.
Q.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
The ground state and excited state electronic configurations of phosphorus (15P)
Ground state electronic configuration:
Phosphorus atom is sp3d-hybridised in the excited state. Five chlorine atoms share their outer valence electrons with the sp3d –hybrid orbitals of P.
The five sp3d-hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl5 can be represented as:
In PCl5 molecule, five P-Cl sigma bonds are present, 3 of them 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. The axial bonds are slightly longer than equatorial bonds as the axial bond pairs suffer more repulsion than the equatorial bond pairs.
Q.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
A hydrogen bond is defined as an attractive force which binds the hydrogen atom of one molecule with the
electronegative atom (F, O or N) of another molecule.
Due to a difference between electronegativity of the H atom and other electronegative atom, the bond pair shifts far away from the H-atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a partial positive charge.
Hydrogen bonds are stronger than Van der Wall’s forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.
Q.40 What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+, O2–
Bond order is defined as half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.
If Na is equal to the number of electrons in an anti-bonding orbital and Nb is equal to the number of electrons in a bonding orbital.
Bond order = ½(Nb-Na)
Bond order of N2 can be calculated from its electronic configuration as:
Number of bonding electrons Nb= 8 and number of anti-bonding electrons Na=2.
Bond Order = (8 – 2) / 2 = 3
The electronic configuration of oxygen molecule can be written as:
The number of bonding electrons = 8 = Nb and the number of anti-bonding electrons = 4 = Na.
Bond order= ½(8-4) = 4/2 = 2.
Hence, the bond order of oxygen molecule is 2.
Similarly, the electronic configuration O2+ can be written as:
Nb = 8 and Na = 3
Bond order = (8–3)/2 = 5/2 = 2.5
Hence, the bond order O2+ of is 2.5.
The electronic configuration O2– is:
Nb= 8 and Na = 5
Bond order = (8–5)/2 = 3/2 = 1.5
Thus, the bond order of O2– is 1.5.
FAQs (Frequently Asked Questions)
1. Give an overview of the chapter Chemical Bonding and Molecular Structure.
The chapter discusses core concepts of molecular structure and chemical bonding. Covalent bonds, ionic bonds, valence electrons, etc. are introduced as a part of this chapter. The students learn about Lewis Structure, the polar character of covalent bonds, the geometry of covalent molecules, valence bond theory, bond parameters, the covalent character of an ionic bond, etc. as well. They will also read about types of chemical equations, balanced chemical equations, combination reactions, displacement reactions, decomposition reactions, rancidity, and corrosion. Furthermore, students will be exposed to hybridisation, s, p, and d orbitals, shapes of molecules, VSEPR theory, etc. as well.
2. What are the topics covered in chapter 4 Chemical Bonding and Molecular Structure?
- Octet Rule
- Covalent Bond
- Lewis Representation of Simple Molecules (The Lewis Structures)
- Formal Charge
- Limitations of the Octet Rule
- Ionic or Electrovalent Bond
- Lattice Enthalpy
- Bond Parameter
- Bond Length
- Bond Angle
- Bond Enthalpy
- Bond Order
- Resonance Structures
- Polarity of Bonds
- The VSEPR (Valence Shell Electron Pair Repulsion) Theory
- Valence Bond Theory
- Orbital Overlap Concept
- Directional Properties Of Bonds
- Overlapping of Atomic Orbitals
- Types of Overlapping and Nature of Covalent Bonds
- The Strength of Sigma and Pi Bonds
- Types of Hybridisation
- Other Examples of Sp3, Sp2, and Sp Hybridisation
- The Hybridisation of Elements Involving d Orbitals
- Molecular Orbital Theory
- Formation of Molecular Orbitals, Linear Combination of Atomic Orbitals (LCAO)
- Conditions For The Combination of Atomic Orbitals
- Types of Molecular Orbitals
- Energy Level Diagram For Molecular Orbitals
- Electronic Configuration and Molecular Behaviour
- Bonding in Some Homonuclear Diatomic Molecules
- Hydrogen Bonding
- Cause of Formation of Hydrogen Bond
- Types of H-bonds.
3. What is Hybridisation?
When two atomic orbitals are mixed, it leads to the rise of new orbitals. The new orbitals are called Hybrid Orbitals. This process of intermixing atomic orbitals is called Hybridization. The intermixing leads to the formation of hybrid orbitals having completely different shapes, energies, etc.
Types of Hybridization:
- sp Hybridisation
- sp2 Hybridisation
- sp3 Hybridisation
- sp3d Hybridisation
- sp3d2 Hybridisation
4. What is the VSEPR Theory?
In order to predict the shapes of covalent molecules, the VSEPR Theory (Valence Shell Electron Pair Repulsion Theory) provides a simple procedure. This theory was proposed by Sidgwick and Powell in 1940. The electron pairs’ repulsive interactions in the valence shell of the atoms were the basis for this theory. Nyholm and Gillespie further developed and redefined this theory and it was stated that a molecule’s shape depends on the valence shell electron pairs around the central atom.
5. How necessary is it to solve all the questions in the NCERT textbook for Class 11 Chemistry Chapter 4?
Solving these questions will assure that students score well in your exams. For help, students must refer to Chemistry Class 11 Chapter 4 NCERT Solutions that have answers to all the questions that are discussed in the NCERT textbook for Class 11 Chemistry.
6. How can I understand the chapter of Chemical Bonding and Molecular structure in a better way?
Students can refer to the study material and solutions by Extramarks to understand the chapter of Chemical Bonding and Molecular Structure. Reading the chapters a couple of times and solving the questions using NCERT Solutions, will further strengthen the concepts of students.
7. Write about the directional properties of bonds.
The overlapping of atomic orbitals leads to the formation of any covalent bond in nature. The overlapping of 1s-orbitals of two hydrogen atoms forms the molecule of hydrogen. The geometry of molecules is also crucial in the case of polyatomic molecules. A part is played by the VSEPR theory in dictating the molecule’s geometry, however, the valence bond theory is more suited to provide an explanation for the directional properties of a bond.