-
Chemical bonding and molecular structure
1) Explain the formation of a chemical bond.
Solution
A chemical bond is the force of attraction that holds various
constituents (atoms or ions)
together in different chemical species.
Atoms combine in different ways to attain stable configuration
of nearest inert gas. Atoms
combine by either transfer of electrons (ionic bond) or by
sharing of electrons (covalent
bond). Due to the formation of chemical bond, the energy of the
atoms is lowered,
resulting in maximum stability.
Types of bonds: Depending on the nature of formation, bonds are
classified into two
kinds:
1) Ionic bond 2) Covalent bond
Ionic bond: The bond formed by the transfer of electrons from
one atom to another atom
is called an ionic bond.
Example : NaCl
NaCl is formed by the complete transfer of one electron from Na
to Cl.
Then Na+ and Cl
- ions get attracted to each other to form an ionic bond.
Sodium (Na):
Atomic number (Z) =11
Electronic configuration = 1s22s
22p
63s
1
Chlorine (Cl):
Atomic number (Z) =17
Electronic configuration = 1s22s
22p
63s
2 3p
5
-
After electron transfer :
Na+
Cl-
1s22s
22p
6 1s
22s
22p
6 3s
23p
6
(Electronic configuration of Neon) (Electronic configuration of
Argon)
Na+
+ Cl- →NaCl
Covalent bond: The bond formed by sharing of electrons between
the two bonded atoms is
called a covalent bond.
Example : H2
A hydrogen molecule (H2 ) is formed by the sharing of a pair of
electrons between two
hydrogen atoms. The two hydrogen atoms in a hydrogen molecule
are at such a distance
from each other that there is minimum repulsion and maximum
stability.
Hydrogen atom (H):
Atomic number (Z)=1
Electronic configuration = 1s1
Formation of covalent bond in hydrogen molecule:
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2) Write the Lewis dot symbol for the following elements : Mg,
Na, B, O , N, Br.
Solution
Lewis dot symbol for Mg
Lewis dot symbol for Na:
Lewis dot symbol for B:
Lewis dot symbol for O:
Lewis dot symbol for N:
Lewis dot symbol for Br:
3) Write the Lewis symbols for the following atoms and ions:
S and S2-
; Al and Al3+
; H and H- .
Solution
Lewis dot symbols for S and S2-
:
-
Lewis dot symbols for Al and Al3+
:
Lewis dot symbols for H and H- :
4) Draw the Lewis structures for the following molecules and
ions :H2S, SiCl4 ; BeF2, CO32-
,
HCOOH.
Solution
Lewis structure of H2S :
Lewis structure of SiCl4:
Lewis structure of BeF2 :
Lewis structure of CO32-
:
Lewis structure of HCOOH :
-
5) Define the octet rule. Write its significance and
limitations.
Solution
Octet rule: Atoms combine either by the transfer of electrons or
by their sharing to gain
eight electrons in their valence shell (octet). Atoms with eight
electrons in their valence
shell are considered to be stable. This is called the octet
rule.
Significance :
1) The octet rule explains the stability of most covalent
compounds. By counting the
number of valence electrons in the central atom and the other
atoms participating in the
bond, the stability can be explained. If the atoms in a molecule
have an octet in their
valence shell, then the molecule is considered to be stable.
2) The octet rule is very useful in determining the structure of
most organic compounds.
3) The octet rule is mostly applicable to the elements of the
second period.
Limitations : Though the octet rule is applicable to explain the
stability of covalent
molecules, it is not universal.
Compounds without an octet in the valence shell of their atoms
are also stable.
For example :
1) Molecules like BCl3, in which the central atom boron has only
six electrons in its
valence shell. It is 2 electrons short of a valence octet. Even
then, BCl3 is a highly stable
molecule.
2) Molecules like NO and nitrogen dioxide, which ahve an odd
number of electrons, do
not satisfy the octet rule. Even then they are stable.
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3) Molecules like SF6, PF5 expand their octet by involving 3d
orbitals in the bond
formation. They have more electrons than an octet in their
valence shells. In SF6, sulphur
has 12 electrons in its valence shell, and in PF5 , phosphorus
has 10 electrons in its valence
shell.
Though SF6 and PF5, have more electrons than octet, they are
stable.
6) Write the factors favourable for the formation of an ionic
bond.
Solution
Ionic bond: The bond formed by the complete transfer of an
electron from one atom to
another atom, in which there exists an electrostatic force of
attraction between the
positively charged and negatively charged ions, is called an
ionic bond.
An ionic bond is formed between highly electronegative halogens
and highly
electropositive alkali metals.
Conditions favourable for the formation of ionic bond:
1) An atom with low ionisation energy can lose an electron
easily and can form a positive
ion (cation) easily.
2) An atom with high electron affinity can gain electron easily
and form a negative
ion(anion) easily.
3) The negative and positive ions formed attain stable
electronic configuration of the
nearest noble gas.
4) The negative and positive ions are stabilised by
electrostatic attraction, resulting in the
formation of an ionic bond.
-
5) As alkali metals have low ionisation energy and halogen have
high electron affinity,
usually an ionic bond is easily formed between them.
7) Discuss the shape of the following molecules using the VSEPR
model: BeCl2 , BeCl3,
SiCl4.
Solution
The structure of a covalent molecule can be predicted by the
number of electron pairs in
the valence shell of the central atom in the molecule based of
VSEPR theory.
Structure of BeCl2 :
The central atom of BeCl2 is beryllium. In the valence shell of
beryllium in BeCl2 there are
2 bond pairs of electrons. Therefore, the shape of the BeCl2
molecule is linear with a bond
angle of 180°.
Structure of BCl3 :
The central atom is BCl3 is boron. In the valence shell of
boron, there are three bond pairs
of electrons. Therefore, the shape of the molecules is
triangular, with a bond angle of 120°.
Structure of SiCl4 :
The central atom of SiCl4 is silicon. There are four bond pairs
of electrons in the valence
shell of Si. Therefore, the structure of SiCl4 is tetrahedral
with a bond angle of 109°28’.
-
Cl
Structure of AsF5 is arsenic. There are 5 bond pairs of
electrons in the valence shell
of arsenic. Therefore, accoriding to VSEPR, the structure of
AsF5 is trigonal bipyramidal
in the bond angles 90° and 120° .
Structure of H2S:
The central atom of H2S is sulphur. Sulphur has 2 lone pairs and
2 bond pairs of electrons
in its valence shell. According to VSEPR theory, the structure
of the H2S molecule is bent
or angular due to lone pair-lone pair and bond pair-lone pair
repulsions.
-
The lone apirs and the bond pairs of electrons are arranged
tetrahedrally. The bond angle is
92°32’.
Structure of PH3 :
The central atom of the PH3 molecule is phosphorus. The
phosphorus atom has four pairs
of electrons in its valence shell. Three electron pairs are bond
pairs and one is lone pair.
According to VSEPR theory, the structure of PH3 molecule is
trigonal pyramidal. The
bond angle is 93°36’ due to the repulasion between lone pair and
bond pair of electrons,
and bond pair or electrons.
The arrangement of lone pair and bond pair of electrons around
the phosphorus atom is
tetrahedral.
8) Although the geometries of NH3 and H2O molecules are
distorted tetrahedral, the bond
angle in water is less than that of ammonia. Discuss.
Solution
The bond angle in a water molecule is less than that of an
ammonia moleucle due to the
repulsion between lone pair-lone pair and lone pai-bond pair of
electrons. In an ammonia
molecule, repulsion is only present between lone pair-bond pair
of electrons. On the other
hand, in a water molecule, there is repulsion between lone
pair-lone pair and lone pair-
bond pair of electrons. Therefore, in a water moleucle, the bond
angle is less than that of
ammonia. The bond angle in a water moleucle is 104°.5’ , while
in an ammonia molecule
the bond angle is 107°.
9) How do you express the bond length in terms of bond
order?
-
Solution
Bond order is the number of bonds between the two atoms bound in
a molecule. As the
bond order increases, the bond strength increases. This is
because bond enthalpy increases
with an increase in the bond order.
10) Define bond length.
Solution
Bond length is defined as the equilibrium distance between the
nuclei of two bonded
atoms in a molecule.
11) Explain the important aspects of resonance with reference to
the CO32-
ion.
Solution
Lewis structure of some molecules and ions cannot represent its
structural aspects.
Experimentally, it has been proved that, whenever a single Lewis
structure cannot
represent the structure of a molecule accurately, then a number
of structures with similar
energy, position of nuclei, bonding and non-bonding pairs of
electrons are given. These
structures are called canonical structures or resonance
structures. This is called resonance.
For example, the structure of a carbonate ion cannot be
represented by a single Lewis
structure. It is proved experimentally that the structure of
carbonate ion is a resonance
hybrid of the following three resonance structures I, II and
III.
Resonance structures of a carbonate ion:
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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 reason for the same.
Solution
No, the two given structures do not represent the resonance
structures of phosphorous
acid. In phosphorous acid, there is a P-H bond, but in structure
(2), it is missing. There is
double bond between P and O . This is also missing in structures
(1) and (2). Phosphorus
forms five bonds, but in the given two structures , only four
bonds have been shown to
phosphorus atom. Therefore , these two given structures cannot
be considered as
resonance structures of H3PO3.
Structure of H3PO3
13) Write the resonance structures for SO3, NO2 and NO3-.
Solution
Sulphur trioxide (SO3)
Resonance structures of SO3 :
-
Resonance structures of NO2 :
Resonance structures of NO3-:
14) Use Lewis symbols to show the 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
Lewis symbols for K and S atoms:
-
Lewis symbols for K+ and S
2- ions:
b) Ca and O
Lewis symbol for Ca and O atoms:
Lewis symbols for Ca2+
and O2- ions:
c) Al and N:
Lewis symbol for Al and N atoms:
-
Lewis symbols for Al3+
and N3-
ions:
15) Although both CO2 and H2O are triatomic molecules, the shape
of an H2O molecule is
bent, while that of CO2 is linear . Explain this on the basis of
dipole moment.
Solution
A water molecule is a bent molecule in which the two O-H bonds
are oriented at an angle
of 104.5°. The bond moment of O-H bonds give the reultant net
dipole moment of 1.85 D
for a water molecule. On the other hand, a CO2 molecule is a
linear molecule with the two
C-O bonds at 180° in a straight line. The bond moment of one C-O
bond is exactly
cancelled by the ond moment of another C-O bond. Thereofore, a
CO2 molecule has zero
dipole moment and a linear structure.
16) Write the significance/applications of dipole moment.
Solution
Significance of dipole moment:
1. The percentage ionic character of a covalent bond can be
determined from dipole
moment.
2. The symmetry of the moleucle can be decided using dipole
moment value.
3. The bond angle of a covalent molecule can be determined using
dipole moment value.
17) Define electro-negativity. How does it differ from electron
gain enthalpy?
-
Solution
Electronegativity is the tendency of a bonded atom in a molecule
to attract the shared pair
of electrons towards itself.
Electronegativity is the property of a onded atom , but not the
property of an independent
individual atom. On the other hand, electron gain enthlpy is the
energy released when an
electron is added to a gaseous neutral atom.
18) Explain with the help of a suitable example a polar covalent
bond.
Solution
A polar covalent bond is a covalent bond formed by an unequal
sharing of electrons
between two atoms participating in the bond due to the
electronegativity difference
between them.
A polar covalent bond is formed between atoms of different
elements.
For example, a polar covalent bond is formed between H and Cl in
an HCl molecule. In an
HCl molecule, the polar covalent bond is formed by the unequal
sharing of the bond pair
of electrons between H and Cl . The chlorine atom, due to high
electronegativity , drags
the electron pair towards itself. Due to the unequal sharing of
electrons between H and Cl,
H develops a partial and permanent positive charge, while Cl
develops a partial and
permanent negative charge. Thus, a polar covalent bond is formed
between H and Cl in an
HCl molecule.
Formation of polar covalent bond in HCl molecule:
19) Arrange the following bonds in order of increasing ionic
character in the molecules:
LiF, K2O,N2, SO2 and ClF3.
-
Solution
Increasing order of ionic character in the molecules: N2 <
ClF3 < SO2 < K2O < LiF.
20) The skeletal strucutre of CH3COOH as shown below is correct,
but some of the bonds
are shown incorrectly. Write the correct Lewis strcuture for
cetic 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
Square planar geometry arises due to octahedral arrangement with
four ond pairs and two
lone pairs of electrons. But in CH4, there are four bond pairs
of electrons arrnaged in
tetrahedral manner with a bond angle of 109°28’, and there are
no lone pairs of electrons.
Therefore, CH4 is tetrahedral and not square planar.
22)Explain why a BeH2 molecule has a zero dipole moment although
the Be-H bonds are
polar?
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Solution
A BeH2 molecule is linear with 180° bond angle. The two Be-H
bonds are polar and have
bond moment. But the bond moment of both the bonds are equal and
opposite. Due to this,
the ond moment cancels each other and the resultant dipole
moment becomes zero.
23) Which out of NH3 and NF3 has a higher dipole moment and
why?
Solution
NH3 and NF3 both are oyramidal with one lone pair of electrons
on a nitrogen atom.
Though fluorine is more electronegative than nitrogen, the
dipole moment of NH3 is 4.9 x
10-30
cm, which is greater than the dipole moment of NF3(0.8 x
10-30
cm). This is becuase ,
in NH3, the orbital ipole due to the lone pair of electrons is
in the same direction as the
resultant dipole moment of the N-H bonds, whereas in NF3 , the
orbital dipole due to the
lone pair of electrons is in direction opposite to that of the
resultant dipole moment of the
three N-F onds. Due to this , NF3 has a low dipole moment.
24) What is meant y hybridisation of atomic orbitals? Describe
the shapes of sp, sp2
and
sp3 hybrid orbitals.
Solution
-
Hybridisation : The phenomenon of intermixing of the pure atomic
orbitals of nearly same
or slightly different energies, so as to redistribute their
energies, resulting in the formation
of a new set of hybrid orbitals of equivalent energies and
shapes, is called hybridisation.
25) Describe the change in hybridisation (if any) of the Al atom
in the following reaction.
AlCl3 + Cl- →AlCl4
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Solution
When AlCl3 accepts a pair of electrons from Cl- ion to form
AlCl4
-, then the hybridisation
of Al changes from sp2 to sp
3 . The structure of AlCl3 also changes from trigonal planar
to
tetrahedral.
Type of
hybridistion
Shape Structure
sp Linear
Sp
2 Trigonal planar
Sp
3 Tetrahedral
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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
In the formation of ammonia boron trifluroide molecule the
hybridisation of B changes
from sp2
to sp3
. There is no change in the hybridisation of N-atom as a result
of the
formation of dative bond between the BF3 and NH3 molecules. The
electron pair donated
by the nitrogen atom is accepted into the vacant sp3 hybrid
orbital of the boron atom.
27) Draw diagrams showing the formation of a double bond and a
triple bond between the
carbon atoms in C2H4 and C2H2 molecules.
Solution
Formation of double bond in C2H2 (Ethylene):
-
Formation of triple bond in C2H2(Acetylene):
28) What is the total number of sigma and Pi bonds in the
following molecules?
a) C2H2 b) C2H4
Solution
a) C2H2 b) C2H4
σ bonds -3 σ bonds -5
π bonds- 2 π bonds- 1
29) Considering the X-axis as the intermolecular 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
-
Solution
c) 2py and 2py orbitls do not form a sigma bond as the
inter-nuclear axis is the X –axis
and the py orbitals is perpenicular to it. Therefore, 2py and
2py can only for a π-bond by
lateral or sidewise overlap.
30) Which hybrid orbitals are used by the carbon atoms in the
following molecules?
a) CH3 –CH3 b) CH3 –CH = CH2 (Propene)
c) CH3– CH2-OH d) CH3CHO e) CH3-COOH (ethanoic acid)
Solution
a) CH3 –CH3 (ethane)
In ethane, carbon uses sp3 hybrid orbitals.
b) CH3 –CH = CH2 (Propene)
In propene, CH3 –carbon uses sp3
hybrid orbitals, in –CH and –CH2 with double bond,
carbon atom uses sp2 hybrid orbitals.
c) CH3– CH2-OH (ethanol)
In ethanol , carbon uses sp3 hybrid orbitals.
d) CH3CHO (ethanal)
In ethanal, CH3-uses sp3 hybrid orbitals and –CHO carbon atom
uses sp
2 hybrid orbitals.
e) CH3-COOH (ethanoic acid)
In ethanoic acid, CH3- carbon atom uses sp3
hybrid orbitals, and –COOH carbon atom
uses sp2 hybrid orbitals.
31) What do you understand by ond pairs and lone pairs of
electrons? Illustrate by giving
one example of each type.
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Solution
Bond pairs- The electron pairs participating in bond formtion
are called bond pairs.
Lone pairs: The electron pairs present in the valence shell of a
onded atom, which do not
particiapate in bond formation, are called a lone pair of
elctrons or non-bonding electrons.
Example: In a ammonia molecule, in the valence shell of the
nitrogen atom, there are 3
bond pairs of electrons participating in 3 N-H bonds and one
lone pair of electrons.
32) Distinguish between a sigma bond and a Pi bond.
Solution
S.No. Sigma bond (σ bond) S.No. Pi bond(π bond)
1. It is a head-on-head
overlap of atomic
orbitals along the inter-
nuclear axis during the
formation of a covalent
bond.
1. The bond formed by lateral or
side wise overlap of atomic
orbitals is called a Pi bond.
2. It is along the
inter-nuclear axis
2. It is above and below the inter-
nuclear axis.
3. It is a strong bond
as the extent of
overlap is more.
3. It is a weak bond as the extent
of overlap is less.
4. It is formed by
atomic orbitals and
hybrid orbitals.
4. It is not formed by hybrid
orbitals.
5. s-orbitals forms
only σ bonds.
5. s-orbitals cannot form π bonds.
33) Explain the formation of an H2 molecule on the basis of
valence bond theory.
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Solution
According to valence bond theory, a covalent bond is formed by
an overlap of atomic
orbitals with unpaired electrons.
Formation of hydrogen molecule: A hydrogen atom has one unpaired
electron in the 1s
orbital. An H2 molecule is formed by the overlap of the 1s
orbital of one hydrogen atom
with the 1s orbital of another hydrogen atom, resulting in the
formation of a σ bond and a
hydrogen molecule.
34) Write the importanat conditions required for the linear
combination of atomic orbitals
to form molecular orbitals.
Solution
The conditions necessary for the linear combination of atomic
orbitals to form
molecular orbitals are:
1) The atomic orbitals that combine to for molecular orbitals
should have the same or
nearly the same energy.
Example: 1s orbital can combine with another 1s orbital, but not
with 2s orbital has higher
energy.
2) The atomic orbital that combines must have the same symmetry
about the molecular
axis.
Example: The 2pz orbital of one atom can only combine with the
2pz orbital of another
atom, but not with the 2px or 2py orbitals because of their
different symmetry.
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3) The extent of overlap between the combining atoms should be
maximum. As greater the
extent of the overlap, the greater will be the electron density
between the nuclei of a
molecular orbital.
35) Use molecular orbital theory to explain why the Be2 molecule
does not exist.
Solution
The electronic configuration of Be is 1s2
2s2
. Each beryllium atom contains 4 electrons,
and , therefore, a Be2 molecule has 8 electrons. These 8
electrons are arranged in σ1s,
σ∗1s, σ2s, σ∗2s.
Be2 : σ1s, σ∗1s, σ2s2, σ∗2s.
Bond order of Be2 =1/2 (4-4)=0
∴ Be2 molecule is unstable due to zero bond order and does not
exist.
36) Compare the relative stability of the following species and
indicate their magnetic
properties: O2 , O2+
, O2-(superoxide) , O2
2-(peroxide).
Solution
O2 molecule: The electronic configuration of an O2 molecule
is
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ∗2s)
2(σ2pz)
2 (π2px
2≡π2py)
2(π∗2px
1≡π∗2py
1)
The electronic configuration of an O2 molecule indicates that
there are ten electrons in the
bonding molecular orbitals, and six electrons in the
anti-bonding molecular orbitals.
∴Bond order=1/2 (10-6)=2
∴ The oxygen (O2)molecule is stable due to positive bond order.
Due to the presence of
unpaired electrons in π∗O2px and π∗2py, the O2 molecule is
paramagnetic.
O2+ ion:
Electronic configuration of O2+ ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ2s)
2(π22pz)
2 (π2px
2≡π2py
2) (π∗2px
1)
-
The electronic configuration of an O2+ ion indicates that there
are ten electrons in the
bonding molecular orbitals and five electrons in the
anti-bonding molecular orbitals.
∴Bond order=1/2 (10-5)=2.5
O2+ has positive bond order. Therefore , it is stable. Due to
the presence of one unaired
electron in O2+
, it is paramagnetic.
O2- ion:
Electronic configuration of O2- ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ2pz)
2 (π2px
2≡π2py
2) (π2px
2≡π2py
1)
The electronic configuration of O2- ion indicates that there re
10 electrons in the bonding
molecular orbitals and 7 electrons in the anti-bonding molecular
orbitals.
∴Bond order=1/2 (10-7)=1.5
An O2- ion also has positive bond order. Therefore, it is stable
. Due to the presence of one
unpaired electron in π∗2py1 orbital, it is paramagnetic.
O22-
ion:
Electronic configuration of O22-
ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ2s)
2(σ2pz)
2(π2px
2≡π2py
2) (π∗2px
2≡π2py
2)
The electronic configuration of the O22
ion indicates that there are 10 electrons in the
bonding molecular orbitals and 8 electrons in the anti-bonding
molecular orbitals.
∴Bond order=1/2 (10-8)=1
An O22-
ion has positive bond order. Therefore, it is stale. There are
no unpaired electrons
in O22molecular orbitals. Therefore, it is diamagnetic .
Relative order of stability (Decreasing order): O2+
>O2 >O2->O 2
2-.
Higher the bond order more is the stability of the species.
37) Write the significance of a plus and a minus sign shown in
representing orbitals.
-
Solution
The plus and minus sign shown in representing orbitals is the
sign of the wave function
(ψ) of the electron wave or atomic orbital. When atomic orbitals
join to form molecular
orbitals during bond formation, the lobes of the atomic orbitals
with the same sign
combine to form bonding molecular orbitals, while in the
formation of anti-bonding
orbitals, the lobes of the atomic orbitals with the opposite
signs join together.
38) Describe the hybridisation in case of PCl5 . Why are the
bonds longer as compared to
equatorial bonds?
Solution
In PCl5, the central atom phosphorus undergoes sp3d
hybridisation in its excited state. The
five sp3d hybrid orbitals are arranged in a trigonal bipyramidal
structure. These five sp
3d
hybrid orbitals with the pure p-orbital to five chlorine atoms
froming 5P-Cl bonds. Two P-
Cl bonds lie perpendicular to the triangular plane at 90° and
three P-Cl bonds lie in the
plane of the triangle making an angle of 120°.
P-Ground state
P- Excited state
-
PCl5
The axial bonds in PCl5 are longer than the equatorial bonds,
since the axial bond pairs
suffer more repulsive interaction from the equatorial bond
pairs.
39) Define a hydrogen bond. Is it weaker or stronger than vander
waal’s forces?
Solution
A hydrogen bond is defined as the attractive force that binds
the hydrogen atom of one
molecule with the electronegative atom (F, O or N) of another
molecule.
A hydrogen bond is stronger than vander waal’s forces, as in a
hydrogen bond, it is
electrostatic force of attraction between the opposite
charges.
40) What is meant by the term bond order?
i)Calculate the bond order of : N2 ,O2.
Solution
Bond order is defined as one half the difference between the
number of electrons present
in the bonding and the anti-bonding orbitals.
Bond order= ½ (Nb-Na)
Nb = Number of electrons in bonding molecular orbitals.
Na= = Number of electrons in anti- bonding molecular
orbitals.
Bond order N2 :
-
Electronic configuration of N2 ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ2s)
2(σ2pz)
2(π2px
2≡π2py
2)
Nb= 10
Na =4
∴Bond order=1/2 (Nb-Na)
= ½ (10-4)
=6/2=3
Bond order of N2 molecule is 3.
Bond order O2:
Electronic configuration of O2 ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ2s)
2(σ2pz)
2(π2px
2≡π2py
2) (π2px
2≡π2py
2) (π∗2px
2≡π∗2py
1)
Nb =10
Na = 6
Bond order= ½ (Nb-Na)
= ½ (10-6)
=6/2 =2
∴Bond order of O2 molecule is 2.
ii) Calculate the bond order of : O2+
and O2-
Solution
Bond order of O2+
:
Electronic configuration of O2+
ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ2s)
2(σ2pz)
2(π2px
2≡π2py
2) (π2px
2≡π2py
2) (π∗2px
1)
Na =5
-
Nb =10
Bond order = ½ (10-5)
= 5/2 =2.5
∴Bond order of O2+
ion is 2.5.
Bond order of O2- ion:
Electronic configuration of O2- ion:
(σ1s)2 (σ∗1s)
2 (σ2s)
2 (σ∗2s)
2(σ2pz)
2(π2px
2≡π2py
2) (π∗2px
2≡π∗2py
1)
Na =10
Nb =7
Bond order = ½ (10-7)
=3/2 =1.5
∴Bond order of O2- ion is 1.5.