NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

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SOLUTIONS
CLASS - 11TH

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Class : 11th
Subject : Chemistry
Chapter : 4
Chapter Name : Chemical Bond And Molecular Structure

Q4.1 Explain the formation of a chemical bond.

Answer. A chemical bond is de ned as an attractive force that holds the constituents (atoms, ions
etc.) together in a chemical species.
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 fully lled outermost orbitals.
Hence, it was postulated that the elements having incomplete outermost shells are unstable
(reactive). Atoms, therefore, combine with each other and complete their respective octets or
duplets to attain the stable con guration 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 the transference of electrons from one atom to
another.

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Q4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.

Answer.

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Q4.3 Write Lewis symbols for the following atoms and ions: S and S 2 − ; Al and Al 3 + ; H and H − .

Answer. (i) S and S 2 −
The number of valence electrons in sulphur is 6.

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Q4.4 Draw the Lewis structures for the following molecules and ions :
H 2S, SiCl 4, BeF 2, CO 23 − , HCOOH .

Answer.

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Q4.5 De ne octet rule. Write its signi cance and limitations.

Answer. The octet rule or the electronic theory of chemical bonding was developed by Kossel and
Lewis. 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 the nearest noble gas

con guration by having an octet in their valence shell.

The octet rule successfully explained the formation of chemical bonds depending upon the nature
of the element.
Limitations of The Octet theory:
The following are the limitations of the octet rule:
(a) The rule failed to predict the shape and relative stability of molecules.
(b) It is based upon the inert nature of noble gases. However, some noble gases like xenon and
krypton form compounds such as XeF 2, KrF 2 etc.
(c) The octet rule cannot be applied to the elements in and beyond the third period of the periodic
table. The elements present in these periods have more than eight valence electrons around the
central atom. For example: PF 5, SF 6 etc.

(d) The octet rule is not satis ed for all atoms in a molecule having an odd number of electrons.
For example, NO and NO 2 do not satisfy the octet rule.

(e) This rule cannot be applied to those compounds in which the number of electrons surrounding
the central atom is less than eight. For example, LiCl, BeH 2, AlCl 3etc. do not obey the octet rule.

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Q4.6 Write the favourable factors for the formation of ionic bond.

Answer. An ionic bond is formed by the 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.
Hence, favourable factors for ionic bond formation are as follows:
(i) Low ionization enthalpy of metal atom.

( )
(ii) High electron gain enthalpy Δ egH of a nonmetal atom.
(iii) High lattice energy of the compound formed.

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Q4.7 Discuss the shape of the following molecules using the VSEPR model:

BeCl 2, BCl 3, SiCl 4, AsF 5, H 2S, PH 3

Answer. BeCl 2
Cl : Be : Cl
The central atom has no lone pair and there are two bond pairs. i.e., BeCl 2 is of the type AB 2.
Hence, it has a linear shape.
BCl 3

The central atom has no lone pair and there are three bond pairs. Hence, it is of the type AB 3.
Hence, it is trigonal planar.

SiCl 4

The central atom has no lone pair and there are four bond pairs. Hence, the shape of SiCl 4 is
tetrahedral being the AB 4 type molecule.
AsF 5

The central atom has no lone pair and there are ve bond pairs. Hence, AsF 5 is of the type AB 5.
Therefore, the shape is trigonal bipyramidal.
H 2S :
H : s̈ : H

The central atom has one lone pair and there are two bond pairs. Hence, H 2S is of the type AB 2E.
The shape is Bent.
PH 3 :

The central atom has one lone pair and there are three bond pairs. Hence, PH 3 is of the AB 3E type.
Therefore, the shape is trigonal pyramidal.

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Q4.8 Although geometries of NH 3 and H 2O molecules are distorted tetrahedral, bond angle in
water is less than that of ammonia. Discuss.

Answer. The molecular geometry of NH 3 and H 2O can be shown as:

The central atom (N) in NH 3 has one lone pair and there are three bond pairs. In H 2O, there are
two lone pairs and two bond pairs.
The two lone pairs present in the oxygen atom of H 2O molecule repels the two bond pairs. This
repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the
nitrogen atom.
Since the repulsions on the bond pairs in H 2O molecule are greater than that in NH 3 , the bond
angle in water is less than that of ammonia.
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Q4.9 How do you express the bond strength in terms of bond order ?

Answer. 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.

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Q4.10 De ne the bond length.

Answer. Bond length is de ned as the equilibrium distance between the nuclei of two bonded
atoms in a molecule.

( )
Bond lengths are expressed in terms of Angstrom 10 − 10m or picometer

(10 − 10m
) and are measured by spectroscopic X-ray diffractions and electron-diffraction
techniques.
In an ionic compound, the bond length is the sum of the ionic radii of the constituting atoms

(d = r + + r − ).
(
In a covalent compound, it is the sum of their covalent radii d = r A + r B .)

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2−
Q4.11 Explain the important aspects of resonance with reference to the CO 3 .

Answer. According to experimental ndings, all carbon to oxygen bonds in CO 23 − are equivalent.
2−
Hence, it is inadequate to represent CO 3 ion by a single Lewis structure having two single bonds
and one double bond.
Therefore, carbonate ion is described as a resonance hybrid of the following structures:

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Q4.12 H 3PO 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 3PO 3? If not, give reasons for
the same.

Answer. The given structures cannot be taken as the canonical forms of the resonance hybrid of
H 3PO 3 because the positions of the atoms have changed.

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−
Q4.13 Write the resonance structures for SO 3, NO 2 and NO 3 .

Answer. The resonance structures are:

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Q4.14 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. (a) K and S
The electronic con gurations 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) requires one electron
more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:

(b) Ca and O :
The electronic con gurations 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 con gurations of Al and N are as follows:
Al: 2, 8, 3
N: 2, 5
Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three
electrons more than Neon. Hence, the electron transference can be shown as:

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Q4.15 Although both CO 2 and H 2O are triatomic molecules, the shape of H 2O molecule is bent
while that of CO 2is linear. Explain this on the basis of dipole moment.

Answer. According to experimental results, the dipole moment of carbon dioxide is zero. This is
possible only if the molecule is linear so that the dipole moments of C–O bonds are equal and
opposite to nullify each other.

H 2O,on the other hand, has a dipole moment value of 1.84 D (though it is a triatomic molecule as
CO 2). The value of the dipole moment suggests that the structure of H 2O molecule is bent where
the dipole moment of O–H bonds are unequal.

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Q4.16 Write the signi cance/applications of dipole moment.

Answer. In heteronuclear molecules, polarization arises due to a difference in the
electronegativities of the constituents of atoms. As a result, one end of the molecule acquires a
positive charge while the other end becomes negative. Hence, a molecule is said to possess a
dipole.
The product of the magnitude of the charge and the distance between the centres of positive-
negative charges is called 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.
Dipole moment (μ) = charge (Q) × distance of separation ®
The SI unit of a dipole moment is ‘esu’.
1 esu = 3.335 × 10 − 30Cm
Dipole moment is the measure of the polarity of a bond. It is used to differentiate between polar
and non-polar bonds since all non-polar molecules (e.g. H 2, O 2) have zero dipole moments. It is
also helpful in calculating the percentage ionic character of a molecule.

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Q4.17 De ne electronegativity. How does it differ from electron gain enthalpy ?

Answer. Electronegativity is the ability of an atom in a chemical compound to attract a bond pair
of electrons towards itself.
Electronegativity of any given element is not constant. It varies according to the element to which
it is bound. It is not a measurable quantity. 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.

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Q4.18 Explain with the help of suitable example polar covalent bond.

Answer. When two dissimilar atoms having different electronegativities combine to form a
covalent bond, the bond pair of electrons is not shared equally. The bond pair shifts towards the
nucleus of the atom having greater electronegativity. As a result, electron distribution gets
distorted and the electron cloud is displaced towards the electronegative atom.
As a result, the electronegative atom becomes slightly negatively charged while the other atom
becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this
type of a bond is called a polar covalent bond.
HCl, for example, contains a polar covalent bond. Chlorine atom is more electronegative than
hydrogen atom. Hence, the bond pair lies towards chlorine and therefore, it acquires a partial
negative charge.

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Q4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF,
K 2O, N 2, SO 2 and ClF 3

Answer. The ionic character in a molecule is dependent upon the electronegativity difference
between the constituting atoms. The greater the difference, the greater will be the ionic character
of the molecule.
On this basis, the order of increasing ionic character in the given molecules is
N 2 < SO 2 < CIF 3 < K 2O < LiF
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Q4.20 The skeletal structure of CH 3COOH as shown below is correct, but some of the bonds are
shown incorrectly. Write the correct Lewis structure for acetic acid.

Answer. The correct Lewis structure for acetic acid is as follows:

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Q4.21 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 ?

Answer. Electronic con guration of carbon atom:
6C : 1s 22s 22p 2

In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon 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 dsp2. However, an atom
of carbon does not have d-orbitalsto undergo dsp2 hybridization. Hence, the structure of CH4
cannot be square planar.
Moreover, with a bond angle of 900 in square planar, the stability of CH4 will be very less because
of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral
structure for CH4.

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Q4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.

Answer. The Lewis structure for BeH2 is as follows:
H : Be : H
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 molecule has zero dipole moment.

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Q4.23 Which out of NH3 and NF3 has higher dipole moment and why ?

Answer. In both molecules i.e., NH3 and NF3, the central atom (N) has a lone pair electron and
there are three bond pairs. Hence, both molecules have a pyramidal shape. Since uorine 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:

Thus, the resultant moment of the N–H bonds add up to the bond moment of the lone pair (the
two being in the same direction), whereas that of the 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.

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Q4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2 , sp3
hybrid orbitals.

Answer. Hybridization is de ned as an intermixing of a set of atomic orbitals of slightly different
energies, thereby forming a new set of orbitals having equivalent energies and shapes.
For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new sp2 hybrid
orbitals.
These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more
stable. Hybridization helps indicate the geometry of the molecule.
Shape of sp hybrid orbitals: sp hybrid orbitals have a linear shape. They are formed by the
intermixing of s and p orbitals as:

Shape of sp2 hybrid orbitals:
sp2 hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p-orbitals.
The hybrid orbitals are oriented in a trigonal planar arrangement as:

Shape of sp3 hybrid orbitals:
Four sp3 hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. The four sp3
hybrid orbitals are arranged in the form of a tetrahedron as:

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Q4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction.
−
AlCl 3 + Cl − → ACl 4

Answer. The valence orbital picture of aluminium in the ground state can be represented as:

The orbital picture of aluminium in the excited state can be represented as:

Hence, it undergoes sp2 hybridization to give a trigonal planar arrangement (in AlCl3).
To form AlCl3-, the empty 3pz orbital also gets involved and the hybridization changes from sp2 to
sp3. As a result, the shape gets changed to tetrahedral.

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Q4.26 Is there any change in the hybridisation of B and N atoms as a result of the following
reaction?
BF 3 + NH 3 → F 3B. NH 3

Answer. Boron atom in BF3 is sp2 hybridized. The orbital picture of boron in the excited state can
be shown as:

After the reaction has occurred, an adduct F3B⋅NH3 is formed as hybridization of ‘B’ changes to
sp3. However, the hybridization of ‘N’ remains intact.

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Q4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon

atoms in C2H4 and C2H2 molecules.

Answer. C 2H 4 :
The electronic con guration of C-atom in the excited state is:
2 1 1 1 1
6C = 1s 2s 2p x 2p y 2p z
In the formation of an ethane molecule (C2H4), one sp2 hybrid orbital of carbon overlaps a sp2
hybridized 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 unhybridized 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.

C 2H 2 :
In the formation of C 2H 2 : molecule, each C–atom is sp hybridized with two 2p-orbitals in an
unhybridized state.
One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a
C–C sigma bond. The second sp orbital of each C–atom overlaps a half- lled 1s-orbital to form a σ
bond.
The two unhybridized 2p-orbitals of the rst carbon 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.

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Q4.28 What is the total number of sigma and pi bonds in the following molecules?
C 2H 2 (b) C 2H 4

Answer. A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a
sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise
overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-
bonds and one sigma bond.
Structure of C 2H 2 : can be represented as:

Hence, there are three sigma and two pi-bonds in C 2H 2 : .
The structure of C 2H 4 : can be represented as:

Hence, there are ve sigma bonds and one pi-bond inC 2H 4 : .

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Q4.29 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.

Answer. 2py and 2py orbitals will not a form a sigma bond. Taking x-axis as the internuclear axis,
2py and 2py orbitals will undergo lateral overlapping, thereby forming a pi (π) bond.

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Q4.30 Which hybrid orbitals are used by carbon atoms in the following molecules?
CH 3 − CH 3; (b) CH 3 − CH = CH 2; (c) CH 3 − CH 2 − OH; (d) CH 3 − CHO (e) CH 3COOH

Answer.

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Q4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one

example of each type.

Answer. 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 that do not participate in bonding are
called lone pairs of electrons.
For example, in C2H2, (ethane), there are seven bond pairs but no lone pair present.

In H2O, there are two bond pairs and two lone pairs on the central atom (oxygen).

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Q4.32 Distinguish between a sigma and a pi bond.

Answer. The following are the differences between sigma and pi-bonds:

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Q4.33 Explain the formation of H2 molecule on the basis of valence bond theory.

Answer. 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 interaction between them. As they begin to
approach each other, the attractive and repulsive forces start operating.
Attractive force arises between:
(a) Nucleus of one atom and its own electron i.e., NA – eA and NB – eB.
(b) Nucleus of one atom and electron of another atom i.e., NB – eB and NA – eB.
Repulsive force arises between:
(a) Electrons of two atoms i.e., eA – eB.
(b) Nuclei of two atoms i.e., NA – NB.
The force of attraction brings the two atoms together, whereas the force of repulsion tends to push
them apart.

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two
atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached

when the attractive forces balance the repulsive forces and the system acquires minimum energy.
This leads to the formation of a dihydrogen molecule.

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Q4.34 Write the important conditions required for the linear combination of atomic orbitals to
form molecular orbitals.

Answer. The given conditions should be satis ed 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 proper orientations to ensure that the overlap is
maximum.
(c) The extent of overlapping should be large.

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Q4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist.

Answer. The electronic con guration of Beryllium is 1s 22s 2
The molecular orbital electronic con guration for Be2 molecule can be written as:
2 2 2 2
σ 1s σ 1s σ 2s σ 2s
1
(
Hence, the bond order for Be 2 is 2 N b − N a )
Where
Nb = Number of electrons in bonding orbitals
Na = Number of electrons in antibonding orbitals
1
∴ Bond order of Be 2 = 2 (4 − 4) = 0
A negative or zero bond order means that the molecule is unstable. Hence, Be2 molecule does not
exist.

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Q4.36 Compare the relative stability of the following species and indicate their magnetic
properties;
+ − 2−
O 2, O 2 , O 2 (superoxide), O 2 (Peroxide).

Answer. There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The
electronic con guration of oxygen molecule can be written as:

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order,
the greater will be the stability. On this basis, the order of stability is
+ − 2−
O2 > O2 > O2 > O2 .

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Q4.37 Write the signi cance of a plus and a minus sign shown in representing the orbitals.

Answer. 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.

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Q4.38 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 con gurations of phosphorus (Z = 15)
are:

Phosphorus atom is sp3d hybridized in the excited state. These orbitals are lled by the electron

pairs donated by ve Cl atoms as:

The ve sp3d hybrid orbitals are directed towards the ve corners of the trigonal bipyramidal.
Hence, the geometry of PCl5 can be represented as:

There are ve P–Cl sigma bonds in PCl5. Three P–Cl bonds lie in one plane and make an angle of
1200 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 900
with the plane. These bonds are called axial bonds.
As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are
slightly longer than equatorial bonds.

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Q4.39 De ne hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Answer. A hydrogen bond is de ned as an attractive force acting between the hydrogen attached
to an electronegative atom of one molecule and an electronegative atom of a different molecule
(may be of the same kind).
Due to a difference between electronegativities, the bond pair between hydrogen and the
electronegative atom gets drifted far away from the hydrogen atom. As a result, a hydrogen atom

becomes electropositive with respect to the other atom and acquires a positive charge.
4 δ − X δ − ………H δ + − X δ − ………H δ + − X δ −
The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state.
There are two types of H-bonds:
(i) Intermolecular H-bond e.g., HF, HO etc.
(ii) Intramolecular H-bond e.g., o-nitrophenol

Hydrogen bonds are stronger than Van der Waals forces since hydrogen bonds are regarded as an
extreme form of dipole-dipole interaction.

Page : 135 , Block Name : Exercise

+ −
Q4.40 What is meant by the term bond order? Calculate the bond order of : N 2, O 2, O 2 and O 2 .

Answer. Bond order is de ned as one half of the difference between the number of electrons
present in the bonding and antibonding orbitals of a molecule.
If Na is equal to the number of electrons in an antibonding orbital, thenNb is equal to the number
of electrons in a bonding orbital.
1
(
Bond order = 2 N b − N a )
If Nb > Na, then the molecule is said be stable. However, if Nb ≤ Na, then the molecule is
considered to be unstable.
Bond order of N2 can be calculated from its electronic con guration as:

[ ] 2
[
[σ(1s)] 2 σ ∗ (1s) [σ(2s)] 2[σ(2s)] 2 σ ∗ (2s) ] [π (2p )] [π (2p )] [σ (2p )]
2
x
2
y
2
z
2

Number of bonding electrons, Nb = 10
Number of anti-bonding electrons, Na = 4
1
Bond order of nitrogen molecule = 2 (10 − 4)
=3
There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic
con guration of oxygen molecule can be written as:

[ ] 2
[σ − (1s)] 2 σ ∗ (1s) [σ(2s)] 2 σ 1p z [ ( )]2 [π (2px )]2 [π (2py )] 2 [π ∗ (2px ) ] 1 [π ∗ (2py ) ]1

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 electrons = 4 = Na.
1
(
Bond order = 2 N b − N a )
1
= 2 (8 − 4)
=2
Hence, the bond order of oxygen molecule is 2.
Similarly, the electronic con guration of O 2+ can be written as:

[
KK[σ(2s)] 2 σ ∗ (2s) ] [σ (2p )] [π (2p )] [π (2p )] [π (2p ) ]
2
z
2
x
2
y
2 ∗
x
1

Nb = 8
Na = 3
1
Bond order of O 2+ = 2 (8 − 3)
= 2.5
Thus, the bond order of O 2+ is 2.5.
The electronic con guration of O 2− ion will be:

[ ] [ ( )] [π (2p )] [π (2p )] [π (2p ) ] [π (2p ) ]
2
KK[σ(2s)] 2 σ ∗ (2s) ( σ 2p z 2
x
2
y
2 ∗
x
2 ∗
y
1

Na = 8
Na = 5
1
Bond order of O 2− = O 2− = 2 (8 − 5)
= 1.5
−
Thus, the bond order of O 2 ion is 1.5.

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