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1. Explain the formation of a chemical bond.
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Existence of a strong force of binding between two or many atoms is referred to as a chemical bond
and it results in the formation of a stable compound with properties of its own. Molecules having two
identical atoms like H2, O2, Cl2, N2 are called as homo-nuclear diatomic molecules. Molecules
containing two different atoms like CO, HCl, NO, HBr etc., are called as hetero-nuclear diatomic
molecules.
Chemical bonds are basically classified into three types consisting of
i) ionic or electrovalent bond
ii) covalent bond and co-ordinate–covalent bond .
According to the electronic theory of valency, a chemical bond is said to be formed when atoms
interact by losing, gaining or sharing of valence electrons and in doing so, a stable noble gas
electronic configuration is achieved by atoms.
2. Write Lewis dot symbols for atoms of the following elements : Be, Na, B, O, N, Br.
Solution:
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Solution:
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3.
Write Lewis symbols for the following atoms and ions:
S and S2–; Al and Al3+; H and H–
1
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4. Draw the Lewis structures for the following molecules and ions:
H2S, SnCl4, BeF2, 3CO
, HCOOH
Solution:
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5. Define octet rule. Write its significance and limitations.
Solution:
The tendency of atoms to have eight electrons in their outer-shell by interacting with other atoms
through electron sharing or electron transfer is known as the octet rule of chemical bonding.
Significance:
1. Used to explain chemical bonding in various compounds.
2
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2. Able to illustrate various type of bonds like covalent, electrovalent and co-ordinate.
3. Useful for understanding the structures of most of the organic compounds.
Limitations:
The octet rule is not satisfied for all atoms in a compound.
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1. Incomplete octet of the central atom:
LiCl, BeH2, BCl3, AlCl3,BF3,LiBr, AlBr3
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2.Odd number of electrons in the outer configuration of certain atoms
In some compounds like NO, NO2, odd number of electrons are present in the outermost shell of
nitrogen atom.
3.More than eight electrons in the central atom: The expanded octet
In some compounds like PF5, SF6, H2SO4 apart from s and p orbitals, d orbitals are also available for
bonding. This way of bonding will lead to the presence of more than eight valence electrons around
the central atom.
There are 5 bonds surrounding phosphorus atom. Each bond corresponds to 2 electrons. Therefore
10 electrons are present around P atom instead of 8 electrons.
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4. Octet theory was not able to explain various shapes attained by molecules.
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Six bonds around S atom i.e 12 electrons around S atom.
5. Octet theory is based upon the chemical inertness of rare gases instead rare gases do involve in
forming compounds like XeF2, KrF2, XeF4, XeOF2, etc.
6. Write the favourable factors for the formation of ionic bond.
Solution:
Favourable factors for the formation of ionic bond
3
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i) Ease of formation of cation and anion form neutral atoms.
ii) Elements /Atoms with lower ionisation energies.
iii) Atoms with greater negative values of electron gain enthalpy.
iv) Cations and anions together form a stable crystal lattice by releasing large amount of energy.
v) Lattice enthalpy of ionic solids are extremely high.
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7.
Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
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Solution:
VSEPR Theory is based upon the fact that in a polyatomic molecule, the direction of bonds around
the central atom depends upon the total number of electron pairs (lone pairs as well as bond pairs)
in its valence shell. These electron pairs place themselves as far apart as possible in space so as to
have minimum repulsive interactions between them. The presence of lone pairs in addition to bond
pairs in the molecule causes distortion in the geometry of the molecule.
Species
Lewis Dot
Structure
Electron pairs
around central
atom
bp
2
Reason
Shape
lp
0
0
No lone
pairs. Two
bond pairs
take up
opposite
positions
due to bp
repulsion.
Linear
ca
3
Arrangement of
electron pairs
Trigonal
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Tetrahedral
Four bond
pairs
acquire
tetra hedral
geometry.
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0
on
4
No lone
pairs. These
bond pairs
take up a
trigonal
geometry
(maximum
angle of
separation).
4
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AsF5
5
Five bond
pairs
acquire
Trigonal bipyramidal
trigonal bipyramidal
geometry.
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H2S
PH3
2
3
2
1
Bent
Two bond
pairs and
two lone
pairs
present.
The
molecule
acquires
angular
shape
instead of
linear due
to lp-lp, lpbp
repulsion.
Pyramidal
One lp and
three bps
present and
the
molecule
acquires
pyramidal
shape.
ca
8. Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle
in water is less than that of ammonia. Discuss.
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Solution:
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.in
If an atom is surrounded by four bonded pairs of electrons, it must assume a tetrahedral shape due
to electron repulsion. But in the case of ammonia, N is surrounded by one lone pair and three
bonded pairs of electrons. The lp-bp repulsion makes the molecule to assume a pyramidal shape
whose HNH angle gets reduced from 109o 28’ to 107o. In the case of water molecule the molecule
contains 2 lps and 2 bps of electrons. The lp-bp and lp-bp repulsion together exists and makes the
molecule to assume a bent shape and the HOH angle is reduced from 109o28’ to 104.5o . Since the
lp-lp repulsions are much greater than the lp-bp repulsion the HOH bond angle in water is smaller
than HNH bond angle in NH3.
5
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9. How do you express the bond strength in terms of bond order?
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Solution:
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Bond order may be defined as the half difference between number of electrons in bonding molecular
orbitals and the number of electrons in anti-bonding molecular orbitals.
Bond Order =
[Nb – Na]
If the value of bond order is positive, it indicates a stable molecule and if the value of bond order is
negative or zero, it means that the molecules unstable and is not formed.
Bond order cannot be used for quantitative comparison of the strengths of chemical bonds. It can
give us only approximate idea about the strength of the bond. Greater the bond order, greater is the
strength of the bond.
10. Define the bond length.
Solution:
Bond length may be defined as the average equilibrium distance between the centres of the two
bonded atoms.
Solution:
ion.
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11. Explain the important aspects of resonance with reference to the CO
In CO32- ion all C – O bonds are equivalent. This is possible only different Canonical structures of
CO32- undergo resonance to give resonance hybrid in which all C – O bonds are equivalent.
.in
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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.
6
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Solution:
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The given two canonical structures of H3PO3 do not have the same number of single and double
bonds. 1st structure has P = 0 and no lone pair electron on phosphorus atom while the 2nd structure
has no double bond between one of oxygen atom and P-atom and P-atom has one lone pair
electrons.
13. Write the resonance structures for SO3, NO2 and NO3-
Solution:
SO3
NO2
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NO3-
7
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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.
Solution:
(a) K and S
(b) Ca and O
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(c) Al and N
15. Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is
bent while that of CO2 is linear. Explain this on the basis of dipole moment.
Solution:
A polyatomic molecule has more than two atoms bonded by covalent bonds. In such cases the idea
8
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of a dipole can be applied to individual bonds within the molecule. The dipole moment of the
individual bond in a polyatomic molecule is referred to as bond dipole. The dipole moment of the
molecule depends upon the orientations of various bond dipoles.
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Bond dipoles
Cancel
Resultant µ = 0.
Bond dipoles do not
Cancel
Resultant µ = 1.83 D.
Although both CO2 and H2O are both triatomic molecules, the shape of CO2 is linear but H2O is bent.
The shape of the molecules can be predicted by dipole moments. Experiments should that the dipole
moment of CO2 is zero. This is possible only if bond dipoles of two c = 0 bonds cancel each other. In
other words, the two bond dipoles must be oriented in opposite directions. This is possible if the
molecule is linear. Similarly H2O has a dipole moment of 1.83D. Thus its molecule cannot be linear
because the bond dipoles do not cancel each other. Thus, the H2O molecule must have an angular
shape.
16. Write the significance/applications of dipole moment.
Solution:
Applications of dipole moment
1. Distinction between polar and non-polar molecules. The molecules having dipole moment
are called polar molecules whereas molecules having zero dipole moment are said to be non-polar
molecules. For example:
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(i) Molecules such as H2, N2, O2, Cl2 etc., have non-polar bonds and zero value of dipole moment.
(ii) Molecules such as CO2, BF3, CH4, CCl4, PCl5, SF6, etc., have polar bonds but zero value of dipole
moment.
(iii) Molecules such as HF, HCl, NH3, H2O, NF3, have polar bonds and their dipole moment is greater
than zero. In other words, they are polar molecules.
on
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2. Ionic character in a molecule. Dipole moment gives an idea about the ionic character in a bond
or a molecule. For example, let us calculate the percentage of ionic character in HCl molecule.
Experiments have shown that the dipole moment of HCl is 1.03 D and its bond length is 1.257 A° .
Now for 100% ionic character the charge developed on H and Cl atoms would be 4.8 × 10-10 e. s. u.
Therefore, dipole moment in case of 100% ionic character is given as
m ionic = 4.8 × 10-10 ×€1.275 × 10-8 e. s. u. –cm.
= 6.12 × 10-18 e. s. u. –cm = 6.12 D.
The observed dipole moment
m obs = 1.032 D
∴€Percentage Ionic Character =
=
= 16.83%.
9
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In general, larger the value of dipole moment is, more will be the ionic character.
3. Shapes of molecules. A polyatomic molecule has more than two atoms bonded by covalent
bonds. In such cases the idea of a dipole can be applied to individual bonds within the molecule. The
dipole moment of the individual bond in a polyatomic molecule is referred to as bond dipole. The
dipole moment of the molecule depends upon the orientations of various bond dipoles.
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Bond dipoles
Cancel
Resultant µ = 0.
Bond dipoles do not
Cancel
Resultant µ = 1.83 D.
Although both CO2 and H2O are both triatomic molecules, the shape of CO2 is linear but H2O is bent.
The shape of the molecules can be predicted by dipole moments. Experiments should that the dipole
moment of CO2 is zero. This is possible only if bond dipoles of two c = 0 bonds cancel each other. In
other words, the two bond dipoles must be oriented in opposite directions. This is possible if the
molecule is linear. Similarly H2O has a dipole moment of 1.83D. Thus its molecule cannot be linear
because the bond dipoles do not cancel each other. Thus, the H2O molecule must have an angular
shape.
17. Define electro negativity. How does it differ from electron gain enthalpy?
Solution:
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Electronegativity
Electronegativity of an element may be defined as the tendency of its atom to attract the shared pair
of electrons towards it self in a covalent bond.
Electronegativity
1. It is the tendency of an atom to attract
outside electron to receive.
1. It is the tendency of an atom to
attract shared pair of electrons.
2. It is the absolute electron attracting
tendency of the atom
2. It is the relative electron attracting
tendency of an atom.
3. It is the property of an isolated atom.
3. It is the property of bonded atom.
4. The elements with symmetrical
configuration have almost zero electron
affinities.
4. The elements with symmetrical
configuration have specific
electronegativities.
5. It has certain units i.e., KJ mol_1 and
eV/atom.
5. It has no units.
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Electron affinity
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18. Explain with the help of suitable example polar covalent bond.
Solution:
Polar covalent bond
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The bond between two unlike atoms which differ in their electronegativities is said to be a polar
covalent bond. In other words polar bond is due to partial ionic character of covalent bond.
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19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O,
N2, SO2 and ClF3.
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Solution:
N2 < SO2 < ClF3 < LiF < K2O.
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.
Solution:
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?
Solution:
ca
In CH4 the central carbon atom has no lone pair of electron. Had there been two lone pairs on carbon
atom after bonding with four Hydrogen atoms, it would have been square planer in shape according
to VSEPR theory.
ti
Since carbon atom is smaller and has no d-orbital space, it cannot undergo d2SP3 hybridisation which
is square planer in shape. Since carbon atom undergoes SP3 hybridisation the CH4 molecule is
tetrahedral.
on
22.
Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
.in
Solution:
Be and H differ in their electronegativities. So, the Be – H bond should be a polar bond. BeH2 is sp
hybridized molecule and hence it is linear. H – Be-H angle is 1800.
The dipole moment due to one Be – H bond cancels that due to another Be – H bond. So net dipole
moment is zero.
11
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23. Which out of NH3 and NF3 has higher dipole moment and why?
09
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Solution:
In N-H bond direction of dipole moment is towards hydrogen. Resultant dipole moment is in the
same direction of dipole moment due to lone pair of N: But in NF3, resultant dipole moment is in
opposite direction to that of dipole moment due to lone pair of N. Hence dipole moment of NF3 has
less dipole moment than NH3.
24. What is meant by hybridisation of atomic orbitals? Describe the shapes
of sp, sp2, sp3 hybrid orbitals.
Solution:
ca
Hybridisation may be defined as the phenomenon of intermixing of atomic orbitals of slightly
different energies to form new orbitals of equivalent energies and identical shapes.
ti
There are many different types of hybridisation which involve the use of s and p orbitals i.e.,
sp3,sp2 and sp hybridisation.
.in
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(i) sp3 hybridisation: This type of hybridisation involves the mixing of one orbital of s-sub level and
three orbitals of p-sub level of the valence shell to form four sp3 hybrid orbitals of equivalent
energies and shape. Each sp3 hybrid orbital has 25% s character and 75% p-character. These
hybridised orbitals tend to lie as apart in space as possible so that repulsive interactions between
them are minimum. The four sp3 hybrid orbitals are directed towards the four corners of a
tetrahedron. The angle between the sp3 hybrid orbitals is 109.50.
12
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sp3 hybridisation is also known as tetrahedral hybridisation. The molecules in which central atom is
sp3 hybridised and is linked to four other atoms directly, have tetrahedral shape.
ii)sp2 hybridisation: This type of hybridisation involves the mixing of one orbital of ssublevel and two orbitals of p sublevel of the valence shell to form three sp2 hybrid
orbitals. These sp2 hybrid orbitals lie in a plane and are directed towards the corners of equilateral
triangle.
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ca
.in
on
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Each sp2 hybrid orbitals has one-third s- character and two-third p-character. sp2 hybridisation is
also called trigonal hybridisation. The molecule in which central atom is sp2 hybridised and is linked
to other atoms directly have triangular planar shape.
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(i) sp-hybridisation: This type of hybridisation involves the mixing of one orbital of s-sub-level and
one orbital of p-sub level of the valence shell of the atom to form two sp-hybridised orbitals of
equivalent shapes and energies. These sp hybridised orbitals are oriented in space at an angle of
1800. This hybridisation is called diagonal hybridisation.
Each sp hybrid orbital has equal s and p character, i.e., 50% s-character and 50% p-character. The
molecules in which the central atom is sp hybridised and is linked to two other atoms directly have
linear shape.
25.
Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl3 + Cl- →€AlCl4Solution:
ca
The phenomenon of intermixing of two or more Atomic orbitals of an atom, having nearly the same
energies which result in their rearrangement to form equal number of new set orbitals of equal
energy and shape, is called hybridisation.
ti
AlCl3 is supposed to have been formed by sp2 hybridisation.
.in
Al - at. no. 13 - Electronic configuration is 1S22S22S63S23PX13PY03PZ0
on
These hybrid orbitals, formed in hybridisation, orient in three dimensional space, giving specific
shape, depending on the type of hybridisation. For example SP3 hybridisation give tetrahedral shape,
sp2 gives trigonal while sp gives linear or diagonal shapes.
In the excited state 2s electron will go to vacant 2py orbital .
14
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Excited state
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Now 2s, 2px and 2py undergo sp2 hybridisation. The three sp2 hybrid orbitals form σ - bonds with 3
Cl- atoms.
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In the formation of AlCl3, there is no change in hybridisation, because the vacant hybridized orbital
can receive, one lone pair electrons from Chlorine atom.
26.
Is there any change in the hybridisation of B and N atoms as a result of the following
reaction?
BF3 + NH3 → F3B-NH3
Solution:
The B-atom is sp2 hybridised. BF3 molecule has one unhybridised orbital which can accommodate one
lone pair 2S- electrons (by coordinate covalancy) from N-atom in pyramidal structure of
NH3 molecule. So there is no need for change in the hybridisation of either B-atom or N-atom.
ca
27.
Draw diagrams showing the formation of a double bond and a triple bond between carbon
atoms in C2H4 and C2H2 molecules.
ti
Solution:
.in
on
(i) C2H4
(i) C2H4 or H2 – C = C – H2 is formed by sp2 hybridisation of each of two carbon atoms. The two of
the three hybrid orbitals of each C-atom forms two σ -bonds with two hydrogen atoms and the third
hybrid orbital of one carbon atom forms σ -bonds with third hybrid orbitals of second C-atom. Thus
three σ -bonds are in one plane. The unhybridised orbitals (2pz) of one C-atom laterally overlaps with
that of 2nd C-atom forming one π -bond which is at right angles to the three σ -bonds.
Ethene molecule
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The Lewis structure of ethylene is represented as
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(ii) C2H2
CH ≡ CH (Ethyne) molecule is formed by sp hybridisation of each C- atom. One hybrid bond on each
C- atom forms σ -bond with one Hydrogen atom. The other hybrid bond of one C-atom forms σ -bond
with another C-atom. The two unhybridised orbitals on each carbon atom form two π -bonds with
those of second carbon atom.
Ethyne molecule
28.
What is the total number of sigma and pi bonds in the following molecules ?
(a) C2H2 (b) C2H4
Solution:
(a) C2H2 =
ti
ca
(b) C2H4 =
on
Solution:
.in
29.
Considering x-axis as the inter-nuclear 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.
a) s-s overlapping:
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b) s-p overlapping:
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c) p-p overlapping:
2Py and 2Py form π -bond
Since the p-orbitals undergo lateral overlapping it cannot form a σ - bond.
d) 1S and 2S form σ - bond
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ti
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
.in
(a) sp3
(b) sp3, sp2
(c) sp3
(d) sp3,sp2
(e) sp3,sp2
on
Solution:
31.
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one
example of each type.
Solution:
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The pairs of electrons between two atoms in a compound, which are involved in bonding are called
bond pairs.
For example:
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Lone pairs
The pairs of electron surrounding the central atom which belongs to the central atom and is not
involved in bonding are called lone pairs.
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ca
For Eg:
NH3 contains one lone pair of electrons.
H2O contains two lone pair of electrons.
Although CO2 and H2O are both triatomic molecules, the shape of CO2 is linear but H2O is bent. The
shape of the molecules can be predicted by dipole moments. Experiments showed that the dipole
moment of CO2 is zero. This is possible only if bond dipoles of two C=O bonds cancel each other. In
other words, the two bond dipoles must be oriented in opposite directions. This is possible if the
molecule is linear. Similarly H2O has a dipole moment of 1.83 D. Thus its molecule cannot be linear.
Because the bond dipoles do not cancel each other. Thus the H2O molecule must have an angular
shape.
σ€- bond is formed by the head on (or) axial overlap of
s-; s-p and p-p orbitals.
.in
Solution:
The following are the differences between σ€
€and π -bonds.
on
32. Distinguish between a sigma and a pi bond.
1. A π€- bond is formed by the sidewise (or)
lateral overlap of p-orbitals.
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2. It is weaker since the extent of overlapping of
orbitals is minimum.
3. A σ€- molecular orbital consists of a single electron
cloud which is symmetrical along the internuclear axis.
3. There are two electron clouds one above and
another below the plane of the bonded atoms.
4. There is free rotation of atoms about the σ- bond.
4. π- bond is rigid and no free rotation of atoms
is possible about π- bond.
5. There can be only one σ€- bond between two atoms.
5. There can be one or two π€- bonds between
two atoms.
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2. A σ- bond is stronger since there is maximum
overlapping.
33. Explain the formation of H2 molecule on the basis of valence bond theory.
Solution:
The formation of the simplest molecule, hydrogen (H2) molecule can be explained by VB theory.
H2 molecule is a diatomic homo nuclear molecule. Each H atom has a proton as the positive nucleus
and an electron in 1s orbital. According to VB theory, the H – H covalent bond is formed by the
overlap of the two 1s orbitals of the hydrogen atoms, thus sharing the pair of electrons in a common
region of space lying between the two nuclei. Also, the two electrons overlapping orbitals exist with
opposite spins only. Because of spherical and symmetrical nature of 1s orbitals, when the two 1s
orbitals of the two H atoms overlap, the paired electrons are usually located at a region between two
nuclei. Thus both the electrons are attracted equally by both the nuclei.
The formation process of H2 molecule can be explained by considering the potential energy changes
when the two H atoms approach each other from infinity position in following manner:
ca
(i) When the two H atoms are far apart, there is no interaction between them, an the P. E of the
system of two H atoms is taken as zero.
ti
(ii) As the two atoms approach each other, the inter nuclear distance of separation gets reduced and
the attraction of the nuclei for the electrons increases and the repulsion between nuclei as well as
between the electrons increase.
on
.in
(iii) As long as attraction forces are more than the repulsions the potential energy of the system
decreases than zero and becomes negative. This trend continuous till the potential energy of the
system reaches a minimum value. This corresponds to the most stable state of the system.
(iv) Further decrease in the inter nuclear distance causes steep increase in the potential energy due
to the increased inter-nuclear and inter electronic repulsion and the system becomes unstable.
(v) The distance of separation between the two H atoms at which the total energy of the system is
minimum is considered as the inter-nuclear distance or the bond length of the H2 molecule.
(vi) The decrease in the potential energy takes place, by the release of energy during a single bond
formation.
i.e. When H2 molecule is formed heat is given off. Conversely, to break up the H – H bond energy
19
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must be supplied to the molecule.
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34. Write the important conditions required for the linear combination of atomic orbitals
to form molecular orbitals.
Solution:
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For the atomic orbitals to combine resulting in the formation of molecular orbitals the main
conditions are:
1.The combining atomic orbitals should have almost same energies.
For example, in case of homo nuclear diatomic molecules 1s orbital of one atom can combine with 1s
orbital of the other atom but 1s orbital of one atom cannot combine with 2s orbital of the other atom.
2.The extent of overlap between the atomic orbitals of the two atoms should be large.
3.The combining atomic orbitals should have the same symmetry about the molecular axis.
For example, 2px orbital of one atom can combine with 2px orbital of the other atom but not with
2pz orbital.
It must be noted that z-axis is taken as the inter nuclear axis according to modern conventions.
35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
Solution:
Molecular orbital configuration of Be2 is
(σ 1s)2, (σ 1s*)2, (σ 2s)2, (σ 2s*)2
ca
Bond order of Be2 =
(Nb – Na) =
(4-4) = 0.
Since the bond order is 0, Be2 molecule does not exist.
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36. Compare the relative stability of the following species and indicate their magnetic
properties:O2,O2+,O2-,O22-.
on
Solution:
O2: (σ
2
1s)
, (σ
* 2
1s ) ,
(σ
2
2s) ,
(σ€2s*)2, (σ
2
2pz) ,
(π
2
2px) ,
(π
2
2py) ,
(<π
*
1
2px) ,
(π
*
1
2py) .
.in
OXYGEN MOLECULE
The electronic configuration of oxygen is 1s2, 2s2, 2p4. Each oxygen atom has 8 electrons, hence, in
O2 molecule there are 16 electrons. The electronic configuration of O2 molecule, therefore, is
From the electronic configuration of O2 molecule it is clear that ten electrons are present in bonding
molecular orbitals and six electrons are present in anti-bonding molecular orbitals. Its bond order,
therefore, is
Bond order =
[Nb – Na] =
(10-6) =
× 4 = 2.
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So in oxygen molecule, atoms are held by a double covalent bond. The bond dissociation energy of
O2 molecule has been found to be 495 kJ/mole and bond length 121 pm. Moreover, from the
molecular orbital diagram of O2 molecule, it may be noted that it contains two unpaired electrons
in π2px*, π 2py* molecular orbitals. Therefore, O2 molecule has paramagnetic nature.
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Oxygen Molecule Ion (O2+)
This ion is formed by removal of one electron from O2 molecule.
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O2 – e - → O2+
From the M. O. diagram of O2 it is clear that this electron would be removed from one of the antibonding molecular orbitals (π 2px*, π 2py*). Therefore, the electronic configuration of O2+ ion is
O2+: (σ
2
1s)
, (σ
* 2
1s ) ,
2
2s) ,
(σ
And its bond order is
* 2
2s ) ,
(σ
(10-5) =
2
2pz) ,
(σ
2
2px) ,
(π
2
2py) ,
(π
*
(π
1
2px)
× 5 = 2.5.
Since bond order of O2+ is greater than the bond order of O2, therefore, it will have smaller bond
length and larger bond dissociation energy than that of O2 molecule. Its bond length is 112 pm and
bond dissociation energy is 625 kJ/mole. Moreover it exhibits paramagnetic character because of the
presence of unpaired electrons.
Superoxide Ion (O2-)
This ion is formed by addition of one electron to O2 molecule.
O2 + e- →€O2-
This additional electron enters into one of the anti-bonding orbitals (π
electronic configuration of O2- ion is
2
1s)
, (σ
* 2
1s ) ,
(σ
2
2s) ,
* 2
2s ) ,
(σ
2
2pz) ,
(10-7) =
(π
2
2px) ,
(π
2
2py) ,
(π
*
× 3 = 1.5.
2
2px) ,
(π
,π
*
2py
*
). Therefore,
1
2py) .
ti
And its bond order is =
(σ
*
ca
O2-: (σ
2px
on
Since bond order of O2- is less than that of O2 molecule, therefore, it will have longer bond length
and smaller bond dissociation energy than O2molecule. It is also paramagnetic because of the
presence of an unpaired electron.
.in
Peroxide Ion (O2 2-)
This ion is formed by addition of two electrons to O2 molecule.
O2 + 2e- → O2 2-.
These additional electrons enter the two half-filled p anti-bonding orbitals. Therefore,
electronic configuration of O2 2- ion is
(O2 2-): (σ€1s)2 , (σ
* 2
1s ) ,
(σ
2
2s) ,
(σ
* 2
2s ) ,
(σ
2
2pz) ,
(π
2
2px) ,
(π
2
2py) ,
(π
*
2
2px) ,
(π
*
2
2py) .
22
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And its bond order is =
(10-8) =
(2) = 1.
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Since bond order of O22- ion is less than that of O2 molecule, it will have longer bond length and
smaller bond dissociation energy than that of O2molecule. Since there is no unpaired electron in it, it
is diamagnetic in nature.
37. Write the significance of a plus and a minus sign in representing the orbitals.
09
Solution:
The positive and negative signs in an orbital indicate whether the orbital wave function is positive or
negative in a particular region. These signs do not represent the positive or negative charge. They
are just analogous to the signs of amplitude in case of plane waves.
38.
Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to
equatorial bonds?
Solution:
Hybridisation in PCl5
Hybridisation is the phenomenon of intermixing of atomic orbitals of slightly different energies of the
atom (by redistributing their energies) to form new set of orbitals of equivalent energies and
identical shape.
Atomic number of phosphorus P = 15.
Electronic configuration of P
Ground State 1s22s22p63s23p33d0
ti
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Excited State
One electron from 3s orbital has jumped to the higher 3d orbital.
After attained the excited state 1s, 3p and 1d orbital (altogether 5 orbitals) are hybridised to form an
equal set of equivalent five sp3d hybrid orbitals. These 5 sp3d orbitals are directed towards the five
corners of a trigonal bipyramidal geometry.
Cl atom is present at each corner of the trigonal bipyramid. In PCl5 , out of the five hybrid orbitals,
three orbitals form P-Cl bond in one plane making an angle (P-Cl-P) 120° with each other. This plane
is represented as equatorial plane and the bonds formed are equatorial bonds.
Out of the remaining two hybrid orbitals, one lie perpendicularly above and the other lie
perpendicularly below the equatorial plane, making an angle 90° with the plane and forms P-Cl
23
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bonds. These two bonds are called axial bonds. Since the axial bonds suffer more repulsive
interaction from the equatorial bond pairs, they are found to be slightly longer and hence slightly
weaker than the equatorial bonds.
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39. Define Hydrogen bond. Is it weaker or stronger than the Van der Walls forces?
Solution:
It is defined as the electrostatic force of attraction which exists between the covalently bonded
H2 atom of one molecule and the electronegative atom of the other molecule. For example, in case of
HF, the H2 bond exists between H atom of one molecule and F atom of other molecule.
…………. H – F …………. H – F ………….. H - F …………. H – F …………
The attractive forces between the instantaneous dipoles and induced dipoles are called Van der
Waals forces. These forces are quite weak and its strength lies below 12.5 kJ mol-1.
ca
The strength of Hydrogen bond ranges between 12.5 kJ mol-1 to 41.5 kJ mol-1.
Hence it is clear that H2 bonding is stronger than the Van der Waals forces.
ti
40. What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+, O2-.
on
Solution:
.in
Bond order cannot be used for quantitative comparison of the strengths of chemical bonds. It can
give us only approximate idea about the strength of the bond. Greater the bond order, greater is the
strength of the bond.
The molecular orbital configuration and bond order for
N2, O2, O2+, O2- is given as follows:
N2: (σ
2
1s)
, (σ
* 2
1s ) ,
Bond order of N2 =
(σ
2
2s) ,
(σ€2s*)2, (π
(10-4) =
2
2px) ,
(π
2
2py) ,
(σ
2
2pz)
× 6 = 3.
24
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O2: (σ
2
1s)
, (σ
* 2
1s ) ,
Bond order of O2 =
2
1s)
(σ
* 2
2s ) ,
(10-6) =
, (σ€1s*)2, (σ
2
2s) ,
(σ
(σ
2
2pz) ,
(π
2
2px) ,
(π
2
2py) ,
*
(π
1
2px) ,
(π
*
1
2py) .
× 4 = 2.
* 2
2s ) ,
(σ
2
2pz) ,
(π€2px)2, (π
2
2py) ,
(π
*
1
2px) .
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O2+: (σ
2
2s) ,
(σ
09
Bond order of O2+=
O2-: (σ
2
1s)
, (σ
* 2
1s ) ,
Bond order of O2- =
(10-5) =
(σ
2
2s) ,
(σ
* 2
2s ) ,
(10-7) =
× 5 = 2.5.
(σ
2
2pz) ,
(π
2
2px) ,
(π
2
2py) ,
(π
*
2
2px) ,
(π
*
1
2py) .
× 3 = 1.5.
ti
ca
.in
on
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