CCB – 1
CHEMICAL BONDING
C1
INTRODUCTION
A molecule will only be formed if it is more stable, and has a lower enrgy than the individual atoms.
Normally only electrons in the outermost shell of an atom are involved in forming bonds. We divide
elements into three classes :
(A)
Electropositive elements, whose atoms give up one or more electrons easily. They have low
ionisation potentials.
(B)
Electronegative elements, which can gain electrons. They have higher value of electronegativity.
(C)
Elements which have little tendency to lose or gain electrons.
C2A Ionic Bonding
An ionic bond is formed when a metal atom transfers one or more electrons to a nonmetal atom.
Na 
 Na   e 
F  e 
 F 
As a result of this transfer the metal atom becomes cation and the nonmetal the anion.
Cations and anions attain noble gas configuration with outer shell octet of electrons. In some cases cations
may not have octet but different outershell.
Properties of Ionic solids
They are good conductor of electricity in fused state and in aq. solution. They are soluble in polar solvents
and insoluble in non polar solvents. They have high m.p. and b.p. They have strong force of attraction
2
between cation and anion (Coulombic force) F  z 1 z 2 e
Dr 2
Practice Problems :
1.
Which of following elements has the strongest tendency to form electrovalent compound ?
(a)
2.
Li
(b)
Na
(c)
Be
(d)
Mg
KCl
(c)
MgCl2
(d)
CaCl2
By the Coulombic force of attraction F 
z 1 z 2e 2
Which of the following is more ionic ?
(a)
NaCl
(b)
[Answers : (1) b (2) b]
C2B
Solubility of Ionic Solids :
. We can study the solubility of ionic solids in different
Dr 2
solvents. Greater the force of attraction between ions, smaller the tendency of the ions to go into solution
and thus smaller the solubility.
For a given solute, greater the value of D then smaller the force of attraction between positive and negative
ions, hence greater the solubility.
D(H2O) > D(CH3CH2OH) > D(CH3OCH3)
hence solubility of ionic solid (say NaCl) in H2O > CH3CH2OH > CH3OCH3
Practice Problems :
1.
MgSO4 is soluble while BaSO4 in insoluble in H2O. This is because
(a)
lattice energy of BaSO4 is greater than MgSO4
(b)
BaSO4 is more covalent than MgSO4
(c)
hydration energy of Mg2+ is greater than Ba2+
(d)
lattice energy of MgSO4 is greater than BaSO4
[Answers : (1) c]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB– 2
C3A Covalent Bonding : If duplet (2) or octet (8) is completed by sharing of electrons between two
electronegative elements, the bond formed is called covalent bond.
C3B
The Octet Rule : For many light atoms a stable arrangement is attained when the atom is surrounded by
eight electrons – the octet rule. (In case of H2, duplet is completed)
Exceptions to the Octet Rule : In so many cases, the octet rule is violated :
In BeF2, octet of Be is not complete, in BF3, octet of B is not complete. Other examples are PCl5, SF6, IF7
where centre atom is having more than eight electrons.
C3C Coordinate Bonding
A covalent bond results from the sharing of pair of electrons between two atoms where each atom
contributes one electron to the bond. It is also possible to have an electron pair bond where both electrons
originate from one atom and none from the other. Such bonds are called coordinate bonds or dative
bonds. Since in coordinate bonds, two electrons are shared by two atoms, they differ from normal covalent
bonds. It is represented as  Atom/ion/molecule donating electron pair is called DONOR or LEWIS
BASE. Atom/ion/molecule accepting electron pair is called ACCEPTOR or LEWIS ACID () points
donor to acceptor.
NH4+ : NH3 has three (N — H) bonds and one lone pair. In NH4+ formation this lone pair is donated to H+
(having no electron).
C4A Valence Shell Electron Pair Repulsion (VSEPR) Theory
(Gillespie Theory)
The shape of the molecule is determined by repulsions between all of the electron pairs represent in the
valence shell.
Repulsion is in order : (lp – lp) > (lp – bp) > (bp – bp)
The magnitude of repulsion between pairs of electrons depends on the electronegativity difference between
the central atom and the other atom. Double bond cause more repulsion than single bonds, and triple bonds
cause more repulsion than a double bond.
C4B
The Effect of Bonding and Lone Pairs of Geometry and Bond Angles
Example
Orbitals on
Central Atom
Theoretical
Shape
Bond Angle
Distorted Geometry
due to Repulsion
BeCl2
2
Linear
1800
Linear
BF3
3
Plane triangle
1200
Plane triangle
Plane triangle
0
Angular
Tetrahedral
0
109 28’
Tetrahedral
Tetrahedral
0
107 48’
Trigonal pyramidal
Tetrahedral
0
102 30’
Pyramidal
Tetrahedral
0
104 27’
Angular
Tetrahedral
0
Angular
SO2
CH4
NH3
NF3
H2 O
F2 O
H2 S
3
4
4
4
4
4
4
Tetrahedral
119
102
90
0
Angular
0
0
PCl5
5
Trigonal bipyramidal
120 and 90
SF4
5
Trigonal bipyramidal
101 36’ and 86 33’ Irregular tetrahedral
CIF3
5
Trigonal bipyramidal
87040’
I3
–
SF6
BrF5
XeF4
Einstein Classes,
5
6
6
6
Trigonal bipyramidal
0
180
0
Trigonal bipyramidal
0
T-shaped
Linear
0
Octahedral
Octahedral
0
84 30’
Square pyramidal
Octahedral
0
Square planar
Octahedral
90
90
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB – 3
Practice Problems :
1.
Molecular shapes of SF4, CF4 and XeF4 are
(a)
the same, with 2, 0 and 1 lone pair of electrons respectively.
(b)
the same, with 1, 1 and 1 lone pair of electrons respectively.
(c)
different, with 0, 1 and 2 lone pairs of electrons respectively.
(d)
different, with 1, 0 and 2 lone pairs of electrons respectively.
[Answers : (1) d]
C5
Isoelectronic Principle : Isoelectronic species usually have the same structure. This may be extended to
species with the same number of valence electrons.
Species
Structures
+
4
CH4, NH , BF
2–
3
–
4
–
3
CO , NO , SO3
–
3
+
2
CO2, N , NO
tetrahedral
planar triangle
linear
C6A Valence Bond Theory : The covalent bond is a region of high electron charge density (high electron
probability) that results from the overlap of atomic orbitals between two atoms. In general, the greater the
amount of overlap between two orbitals, the stronger the bond. For each bond there is a condition of
maximum atomic orbital overlap leading to maximum bond strength at a particular internuclear distance
(bond length). This is called Valence Bond Approach.
C6B
–
This is a localised electron model of bonding.
–
Most of the electrons retain the same orbital locations as in a separated atoms, and the bonding
electrons are localised (fixed) in the region of atomic orbital overlap.
–
In H2 molecules, H – H  bond is by s – s overlapping
–
In F2 molecule : F – F  bond is by axial overlapping of two p-orbitals.
–
In O2 molecule : One  bond is by axial overlap of p-orbitals and  bond is by lateral overlap of
p-orbitals.
–
In N2 molecule : One  bond is by axial overlap of p-orbitals and two  bonds are by lateral
overlap of p-orbitals.
Hybridisation : Hybridisation is defined as the concept of intermixing of orbitals of same energy or of
slightly different energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane. New orbitals formed are called hybrid orbitals.
–
Only the orbitals of an isolated single atom can undergo hybridisation.
–
The hybrid orbitals generated are equal in number to that of the pure atomic orbitals which mix
up.
–
A hybdrid orbitals, like the atomic orbitals, cannot have more than two electrons of opposite
signs.
C6C Sigma and PI Bond There can be following types of overlaping along the axes (end to end) :
–
(s - s) overlapping when s-orbital overlaping with another s-orbital
–
(s - p) overlapping, (p can be px or py or pz)
–
(px – px) overlapping
–
any of the hybrid orbitals overlaps with another hybrid orbitals or s or p orbital.
Bond formed in the manner is called sigma () bond in which electron density is concentrated in between
the two atoms, and on a line joining the two atoms.
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB– 4
Double or triple bonds occur by sideways overlap of orbitals (like (py – py) and (pz – pz) orbitals) giving pi
() bonds in which electron density also concentrates between the atoms, but on either side of the line
joining the atoms.
The shape of the molecule is determined by the  bonds (and lone pairs) but not by the  bonds. Pi bonds
merely shorten bond length. Thus
Practice Problems :
1.
In which of the following molecules would you expect the nitrogen to nitrogen bond to be the
shortest ?
(a)
2.
N2 H 4
(b)
N2
(c)
N2 O 4
(d)
N2 O
Allyl cyanide has
(a)
9  bonds and 4  bonds
(b)
9 -bonds, 3  bonds and one lone pair
(c)
8  bonds and 5  bonds
(d)
8 -bonds, 3  bonds
[Answers : (1) b (2) b]
C6D Hybrid Orbitals and their Geometric Orientation
Hybrid Orbitals Orientation
sp (two)
linear
Example
BeCl2, CO2
Predict Bond Angle
180
0
Actual Shape
linear
C2H 2
2
sp (three)
trigonal planar
1200
trigonal planar
CH 4
1090 28’
tetrahedral
NH3
1070 48’
trigonal pyramidal (due
to lp–bp repulsion)
H2 O
1040 27’
angular (V-shaped) due
to lp – lp and lp – bp
repulsion
PCl5
1200 and 900
BF3,C2H4
SO2, SO3
3
sp (four)
sp3d(five)
tetrahedral
trigonalbi
pyramidal
SF4
I
3 2
sp d
Einstein Classes,
octahedral
–
3
SF6
0
trigonal pyramidal
0
101 , 36’ and 86 33’
180
90
0
0
irregular tetrahedral
linear
octahedral
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB – 5
Practice Problems :
1.
The hybridisation of atomic orbitals of N in NO2+, NO3– and NH4+ are, respectively
(a)
2.
4.
5.
(b)
sp, sp3, sp2
(c)
sp2, sp, sp3
(d)
sp2, sp3, sp
If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M are
(a)
3.
sp, sp2, sp3
pure p
(b)
sp hybrids
(c)
sp2 hybrids
(d)
sp3 hybrids
The shape of sulphate ion is
(a)
hexagonal
(b)
square planar
(c)
trigonal bipyramidal
(d)
tetrahedral
The hybridisation of atomic orbitals of carbon in CH4, C2H4, C2H2
(a)
sp, sp3 and sp2 respectively
(b)
sp, sp2 and sp3 respectively
(c)
sp2, sp and sp3 respectively
(d)
sp2, sp3 and sp respectively
Specify the coordination geometry around and hybridization of N and B atoms in a 1 : 1 complex of
BF3 and NH3
(a)
N : tetrahedral, sp3 ; B : tetrahedral, sp3 (b)
N : pyramidal, sp3 ; B : pyramidal, sp3
(c)
N : pyramidal, sp3 ; B : planar, sp2
N : pyramidal, sp3 ; B : tetrahedral, sp3
(d)
[Answers : (1) a (2) c (3) d (4) d (5) a]
C7
Dipole Moment : In H2, there is no displacement of the electric charge due to same electron affinity of both
H-atoms and the bond is non-polar. In HCl, the Cl atom has a more electron affinity than does the H atom.
Electronic charge distribution is shifted towards the Cl atom. The H-Cl bond is said to be polar.
The magnitude of the charge displacement in a polar covalent bond is measured through a quantity called
the dipole moment µ. It is the product of the magnitude of charges () and the distance separating them (d).
(Here the symbol () suggests a small magnitude of charge, less than the charge on an electron)
µ=×d
If
 = 4.8 × 10–10 esu and d = 1Å = 1 × 10–8 cm
then
µ = 4.8 × 10–10 × 1 × 10–8 = 4.8 × 10–18 esu cm
In S.I. unit, 1 D = 3.33 × 10–30 coulomb meter (when charge  = 3.33 × 10–20 C and d = 1 × 10–10 m).
In diatomic molecule,µ =  × d
but in polyatomic molecule with angle , resultant dipole moment is the vector summation of the vector
moments. Also µresultant
1

 cos   , µ  . Symmetrical molecules without lone pair will have µ = 0.
2

 
% ionic character in a molecule =
observed value of µ
 100
theoretical value of µ
Practice Problems :
1.
A molecule possessing dipole moment is
(a)
2.
CH4
(b)
H2 O
(c)
BF3
(d)
CO2
(d)
II < III < I
Correct order of dipole moment for the following molecule is
(I)
(a)
(II)
I = II = III
Einstein Classes,
(b)
(III)
I < II < III
(c)
I > II > III
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB– 6
3.
Which of the following hydrocarbons has the lowest dipole moment ?
(a)
(c)
4.
CH3CH2C = CH
(b)
CH3C = CCH3
(d)
CH2 – CH – C = CH
Dipole moment is exhibited by
(a)
1, 4-dichlorobenzene
(b)
1, 2-dichlorobenzene
(c)
trans-1, 2 dichloroethene
(d)
trans 1, 2-dichloro-2-butene
[Answers : (1) b (2) d (3) b (4) b]
C8A Hydrogen Bonding : Hydrogen bonding is said to be formed when slightly acidic hydrogen attached to a
strongly electronegative atom such as F, N and O, is held with weak electrostatic forces by the non-bonded
pair of electrons of another atom. The coordination number of hydrogen in such cases is two. It acts as a
bridge between two atoms, to one of which it is covalently bonded, and to the other attached through
electrostatic forces, also called Hydrogen Bond.
Of all the electronegative donor atoms, only F, N and O atoms enter into stable hydrogen bond formation.
The weak electrostatic interaction leading to the hydrogen bond formation is shown by dott (....) lines. Thus
X – H....Y represent hydrogen bonding between hydrogen and Y atom.
C8B
Intramolecular H-Bonding :
This type of H-bonding occurs when polar H and electronegative atom are present in the same molecule.
(a)
(b)
C8C Intermolecular H-Bonding :
This type of H-bonding takes place between H and electronegative element present in the different
molecules of the same substance (as in between H2O and H2O) or different substances (as in between H2O
and NH3)
e.g.
In water molecules :
Due to polar nature of H2O, there is association of water molecules giving a liquid state of abnormally high
b.p.
C9A Molecular Orbital Theory (MOT) : According to Valence Bond Theory only the half-filled orbitals of
one atom overlaps with the half filled orbitals of other atom to form the covalent bond. According to MOT,
on the other hand, the all atomic orbitals of one atom overlap with the all the atomic orbitals of the other
atom provided the overlapping orbitals are of the same symmetry and of similar energy. The resulting
polynuclear molecular orbitals contain the all electrons of the molecule.
C9B
The basic principles of the MOT :
1.
When nuclei of two atoms come close to each other, their atomic orbitals interact leading to the
formation of molecular orbital. The atomic orbitals of the atoms in a molecule completely lose
their identity after the formation of molecular orbital.
2.
Each molecular orbital is described by a wave function  , known as molecular orbital wave
function.
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB – 7
3.
The molecular orbital wave function  is such that  2 represents the probability density or
electron charge density.
4.
Each molecular orbital wave function (  ) is associated with a set of quantum numbers which
determine the energy and shape of the molecular orbital.
5.
Each  is associated with definite energy value.
6.
Electrons fill the molecular orbitals in the same way as they fill the atomic orbital following the
aufbau principle, Pauli’s exclusion principle, and the Hund’s rule of maximum multiplicity.
7.
Each electron in a molecular orbital belongs to all the nuclei present in the molecule.
8.
Each electron moving in a molecular orbital has a spin of 
1
1
or  .
2
2
The basic difference between an atomic orbital and molecular orbital is that while an electron in an atomic
orbital belongs to or influenced by one positive nucleus only, an electron in a molecular orbital is
influenced by all nuclei of atoms contained in a molecule.
Linear combination of Atomic orbitals (LCAO) in case of H2+
1.
A linear combination of two atomic orbitals  A and  B leads to the formation of two
molecular orbitals   and  
2.
The energy E+ of molecular orbital   is lowe than either of EA and EB (energies of isolated
atoms). It is therefore designated as bonding molecular orbital (BMO).
3.
The energy E– of molecular orbital   is higher than either of EA and EB. It is therefore
designated as antibonding molecular orbital (ABMO)
4.
The extent of lowering of energy of the bonding molecular orbital is equal to the extent of
increase of energy of antibonding molecular orbital.
Energies of bonding and antibonding molecular orbitals.
The order of energy of molecular orbital for lighter elements like boron, carbon and nitrogen are as
follows : (1s) < *(1s) < (2s) < *(2s) < 2py = 2pz < 2px < *2py = *2pz < *px
The order of energy of molecular orbitals for heavier elements after nitrogen are :
(1s) < *(1s) < (2s) < *(2s) < 2px < 2py = 2pz < *2py = *2pz < *2px
Practice Problems :
1.
Which of the following molecular orbital in N2 has least energy ?
(a)
(b)
 2p y
 2p z
(c)
 2s
(d)
 *2p z
[Answers : (1) c]
C10
Heteronuclear Diatomic Molecule : The same principle apply when combining atomic orbitals from two
different atoms as applied when atoms are identical that is
(a)
only atomic orbitals of about same energy can combine effectively
(b)
They should have maximum overlap.
(c)
They must have the same symmetry
Since the two atoms are different, the energies of their atomic orbitals are slightly different.
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB– 8
C11
Electronic Configuration of Molecules : While discussing the electronic configuration of molecules, we
shall frequently make use of a term called bond order.
B.O. =
1
[Nb – Na]
2
Nb  number of bonding electrons
Na  number of anti-bonding electrons
If B.O. = 0, 1, 2, 3 so on it means that no bond is formed, one bond is formed, two bonds or three bonds are
formed between the atoms respectively.
1
Bond Length
1.
B.O,  Bond dissociation energy and B.O. 
2.
Electronic configuration helps to predict the magnetic character of the molecule.
If all the electrons in a molecule are paired they are diomagnetic and if unpaired electron is present they are
paramagnetic.
Practice Problems :
1.
2.
3.
Among KO2, AlO2–, BaO2 and NO2+, unpaired electron in present in
(a)
KO2 only
(b)
(c)
KO2 and AlO2–
(d)
+
2
BaO2 only
+
2
When N is formed from N2, bond order ...... and when O is formed from O2, bond order ..........
(a)
increases
(b)
decreases
(c)
increases, decreases
(d)
decreases, increases
(c)
O 2–
(d)
O22–
Which of the following has longest bond length ?
(a)
4.
NO2+ and BaO2
O2
(b)
O 2+
Which of the following has identical bond order ?
(I)
CN–
(II)
O 2–
(III)
NO+
(IV)
CN+
(a)
I, III
(b)
II, IV
(c)
I, II, III
(d)
I, IV
[Answers : (1) a (2) d (3) d (4) a]
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
4.
5.
The electronegativity of cesium is 0.7 and that of
fluorine is 4.0. The bond formed between the two is
(a)
covalent
(b)
electrovalent
(c)
coordinate
(d)
metallic
Delocalised molecular orbitals are found in
(a)
H2
(b)
HS–
(c)
CH4
(d)
CO32–
In which of the following molecules the bond angle
is maximum ?
(a)
CH4
(b)
H2O
(c)
NH3
(d)
CO2
Total number of lone pair of electrons in XeF4 is
(a)
0
(b)
1
(c)
2
(d)
3
CO2 is isostructural with
(a)
SO2
(b)
HgCl2
(c)
C2H2
(d)
SnCl2
Einstein Classes,
6.
7.
A molecule possessing dipole moment is
(a)
CH4
(b)
H2O
(c)
BF3
(d)
CO2
Correct order of dipole moment for the following
molecule is
(I)
(II)
(III)
(a)
(c)
I = II = III
I > II > III
(b)
(d)
I < II < III
II < III < I
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB – 9
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
Which of the following is the most polar bond ?
(a)
Cl – Cl
(b)
N–F
(c)
C–F
(d)
O–F
(A)
Tetracyanomethane
(B)
Carbon dioxide
(C)
Benzene
(D)
1, 3 butadiene
Ratio of  and  bonds is in order
(a)
A = B < C < D (b)
A=B<D<C
(c)
A = B = C = D (d)
C <D<A<B
3
In sp d hybridisation, the d orbital that participates
in hyubridisation is
(a)
dx2 – y2
(b)
dz2
(c)
dxy
(d)
dxz
C2 – C3 sigma single bond in vinyl acetylene is due
to the overlapping of
(a)
sp – sp
(b)
sp2 – sp2
2
(c)
sp – sp
(d)
sp2 – sp
In the compound of the type ECl3, where E = B, P,
As or Bi the angles Cl – E – Cl for different E are in
the order
(a)
B > P = As = Bi (b)
B > P > As > Bi
(c)
B < P = As = Bi (d)
B < P < As < Bi
A -bonds is formed by the overlap of
(a)
s-s orbital
(b)
s-p orbital
(c)
p-p orbital in end to end fashion
(d)
p-p orbital in sidewise manner
Mg2 C3 reacts with water and forms propyne C34–
which has
(a)
two sigma and two pi - bond
(b)
three sigma and one pi - bond
(c)
two sigma and one pi - bond
(d)
two sigma and three pi - bonds
The correct order of the increasing C – O bond
length of CO, CO32–, CO2 is
(a)
CO32– < CO2 < CO
(b)
CO2 < CO32– < CO
(c)
CO < CO32– < CO2
(d)
CO < CO2 < CO32–
Element X is strongly electropositive and Y is
strongly electronegative. Both are univalent. The
compound formed would be
(a)
X+ Y–
(b)
X–Y
–
+
(c)
X Y
(d)
XY
The correct order of the O – O bond length in
O2, H2O2 and O3 is
(a)
O3 > H2O2 > O2 (b)
O2 > H2O2 > O3
(c)
O2 > O3 > H2O2 (d)
H2O2 > O3 > O2
Einstein Classes,
18.
19.
20.
MgSO4 is soluble while BaSO4 in insoluble in H2O.
This is because
(a)
lattice energy of BaSO4 is greater than
MgSO4
(b)
BaSO4 is more covalent than MgSO4
(c)
hydration energy of Mg2+ is greater than
Ba2+
(d)
lattice energy of MgSO4 is greater than
BaSO4
The molecule which has highest dipole moment
amongts the following is
(a)
CH3Cl
(b)
CH2Cl2
(c)
CHCl3
(d)
CCl4
What is the decreasing order of bond order for O2,
O2+, O2–, O22–
(a)
O2+ > O2 > O2– > O22–
(b)
O2 > O2 > O22– > O22–
(c)
O22+ > O2– > O2– > O22–
(d)
O22– > O2 > O2– > O2+
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
1.
b
11.
d
2.
d
12.
b
3.
d
13.
d
4.
c
14.
a
5.
c
15.
d
6.
b
16.
a
7.
d
17.
d
8.
c
18.
c
9.
a
19.
b
10.
b
20.
a
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB– 10
EXCERCISE BASED ON NEW PATTERN
5.
MATRIX-MATCH TYPE
Matching-1
Column - A
Column - B
(P)
sp3d
NO2+
(Q)
sp2
(C)
NH2–
(R)
sp
(D)
PCl5
(S)
sp3
(A)
NO2
(B)
–
6.
Matching-2
Column - A
Column - B
(A)
Pyramidal
(P)
XeF4
(B)
bipyramidal
(Q)
IF5
(C)
square pyramid
(R)
PF5
(D)
Square planner
(S)
H3 O +
7.
Matching-3
Column - A
Column - B
(A)
Cycloheplane (planer)
(P)
1800
(B)
cyclohexane (chain form) (Q)
1200
(C)
propene
(R)
1350
(D)
propyne
(S)
109.50
8.
9.
Matching-4
Column - A
Column - B
(A)
Ice
(P)
paramagnetic
(B)
NO
(Q)
refrigeration
(C)
SO2Cl2
(R)
hydrogen
bonding
(D)
fluorocarbons
(S)
tetrahedral
10.
MULTIPLE CORRECT CHOICE TYPE
1.
2.
3.
4.
Which of the following species have identical bond
order
(a)
CN–
(b)
O2–
(c)
NO+
(d)
CN+
The polar molecules among the following are
(a)
CHCl3
(b)
CF4
(c)
C2F2
(d)
F2 O
The type of chemical bonds present in N 2 O 5
molecule are
(a)
Covalent
(b)
Electrovalent
(c)
Co-ordinate
(d)
H-bonds
In which of the following solid (s) the interparticle
forces are not hydrogen bonds
(a)
dry ice
(b)
Ice
(c)
Anhydrous hydrogen floride
(d)
O-nitrophenol
Einstein Classes,
11.
12.
What is true about the expresion BA (16 + 3.5
BA) where BA is a electronegativity difference
between two atoms A, B forming the bond
(a)
It gives % ionic character in the bond
A-B
(b)
It gives dipole moment of the bond
(c)
It is known as pauling’s expression
(d)
It was given by Hanny and Smith
In which of species, the hybrid state of central atom
is other than sp3d
(a)
I3 +
(b)
SF4
(c)
PF5
(d)
IF5
In which there is change of hybridisation when
(a)
NH3 combine with H+
(b)
AlH3 combine with H–
(c)
CH3+ combine with H–
(d)
CH3– combine with H+
The species which contains odd number of valence
electrons
(a)
NO
(b)
NO2
(c)
ClO2–
(d)
N2O 4
Which of the following shows intramolecular
H-bonding ?
(a)
O-nitrophenol
(b)
p-nitrophenol
(c)
phenol
(d)
salicyldehyde
2–
CO 3 anion has which of the following
characteristics ?
(a)
bonds of equal length
(b)
sp2 hybridization of Cation
(c)
resonance stabilization
(d)
same bond angles
Which of the following has zero dipole moment
(a)
BeCl2
(b)
SO2
(c)
H2 O
(d)
CO2
Intermolecular H-bonding in anhydrous HF makes
it
(a)
high b.Pt. liquid
(b)
Non-electrolyte
(c)
strongest acid out of HF, HCl, HBr, HI
(d)
with unique property of attacking glass
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CCB – 11
1.
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
STATEMENT-1 : Electronic structure of O3 is
5.
6.
7.
STATEMENT-1 : Shape of NH3 is very similar to
CH3–
STATEMENT-2 : N and C in NH3 and CH3– are
sp3 hybridised.
STATEMENT-1 : LiCl has higher boiling point
then HCl compound.
STATEMENT-2 : LiCl is ionic while HCl is a
covalent.
STATEMENT-1 : For a change of O2 to O2+, the
bond dissociation energy increases and bond
length decreases.
STATEMENT-2 : The electron is lost from
antibonding molecular orbital.
STATEMENT-2 :
structure is not
allowed because O atom can not expand its octet.
STATEMENT-1 : The bond angle arround O in
H2O is greater than that arround S in H2S
STATEMENT-2 : H-bonding does not occur in
H2S due to larger size of S
STATEMENT-1 : The bond angle arround B in
BF3 is smaller than that in BF4–
STATEMENT-2 : There is no lone pear arround
B is BF4–
STATEMENT-1 : When a magnet is dipped in
liquid oxygen some of its sticks to the magnet.
STATEMENT-2 : Oxygen is paramagnetic in
nature.
2.
3.
4.
(Answers) EXCERCISE BASED ON NEW PATTERN
MATRIX-MATCH TYPE
1.
[A-Q, B-R, C-S, D-P]
4.
[A-R, B-P, C-S, D-Q]
2.
[A-S, B-R, C-Q, D-P]
3.
[A-R, B-S, C-Q, D-P]
5.
a, d
6.
11.
a, d
A
6.
MULTIPLE CORRCT CHOICE TYPE
1.
a, c
2.
a, d
3.
a, c
4.
a, d
7.
b, c
8.
a, b, c
9.
a, d
10.
a, b, c, d
12.
a, b
3.
D
4.
A
a, d
ASSERTION-REASON TYPE
1.
A
7.
A
2.
Einstein Classes,
B
5.
A
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111