Chemistry. CHEMICAL BONDING – SESSION I Session Opener.
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Transcript of Chemistry. CHEMICAL BONDING – SESSION I Session Opener.
![Page 1: Chemistry. CHEMICAL BONDING – SESSION I Session Opener.](https://reader038.fdocuments.in/reader038/viewer/2022102808/56649e4d5503460f94b42b79/html5/thumbnails/1.jpg)
Chemistry
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CHEMICAL BONDING – SESSION I
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Session Opener
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Session Objectives
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1. Introduction
2. Octet rule
3. Different types of bonding
4. Lewis theory
5. VSEPR theory and shape of molecules
Session Objectives
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Force of attraction holdinggroup(s) of atoms
What is Chemical bonding?
Chemical bonds
Na+ Cl-
Better stability against
chemical reagents
But why bonds are formed ??
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Atoms
two electrons in the valenceshell (1s2)
Octet rule
noble gas configuration attain betterstability.
Na
2 8 1
Very reactive
Na
2 8
+
Ne
Cl
2 8 7
Very reactive
Cl
2 8 8
Ar
-
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SFF
F
F
FF*
***
*
* .....
.
In SF6, ‘S’ has twelve electron in itsvalence shell, leads to minimisation of energy.
Other examples are: PCl5, BF3
Limitation of octet rule
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Questions
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The molecule that deviates from octet rule is
(a) NaCl (b) BeCl2
(c) MgO
(d) NH3
Illustrative Problem
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The no. of valence electrons in different central atoms is:
Hence, the answer is (b).
Solution
Na+ 8 Be+2 2 Mg+2 8
N atom in NH3 (covalent compounds) 8
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Bonding
Ionic
Covalent
Co-ordinate or dative
Metallic
Pi bond Sigma bond
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Formation of ionic bond
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Covalent bond
courtesy:www.lbw.cuny.edu
Formed by mutual
sharing of electrons
Covalent bonds
1 1
H2C CH2
Double bond
HC CHTriple bond
1 2
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non-polar covalent bond between two carbon atomspolar covalent bond between carbon and hydrogen atoms.
Formation of covalent bond
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Question
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Covalent bonds are called directional while ionic bonds are called non-directional -explain
Solution:
Illustrative Problem
p and d-orbitals generate directional covalent bond.
electrostatic force of attraction.
Ionic bond
overlap of atomic orbitals
covalent bond
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Strength of these sigma bonds is in the order:
sigma bond forms due to end-to-end or head-on overlap
p-p+
s-s
+
+
Types of covalent bonds
s-p
p-p > s-p >s-s
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This is formed by lateral or sideways overlap which is possible for p or d-orbitals.
Sigma bond is stronger than pi bond due to greater extent of overlap.
+ or
Types of covalent bonds
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Difference between sigma and pi bonds
Stronger as compared to bond
Weaker as compared to bond
H C C H
Formed by head-on overlapping of s-s or s-p or p-p or any hybrid orbital
Formed by side ways overlapping of unhybridised p-orbital
First bond between any two atoms is always sigma
Rest are bonds
In plane of molecule
Perpendicular to plane of molecule
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Ionic compound(NaCl)
Covalent compound (CHCl3)
MP/BP
Very high Volatile liquid H20 solubility
Highly solubleAlmost insoluble
Benzene solubility
InsolubleHighly Insoluble
Directional nature
Non-directional
Except s-soverlap all are directional
Difference between ionic and covalent compound.
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Question
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How many sigma and pi-bondsare present in a benzene molecule?
Solution:
The structure of benzene molecule is H
H H
H
H
H
no. of pi bonds are 3 [C=C]
no. of sigma bonds are 12 [C-C and C-H]
Illustrative Problem
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•Single atom donating lone pair•Shared by two atoms involved
Coordinate covalent bond
O
H
H+H :
H3O+
H
N
H
H+H :
[NH4]+
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Question
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NH3 and BF3 form an adduct readily-explain.
N- atom in NH3 have one lonepair and BF3
is electron deficient. They form an adductthrough coordinate bond, so BF3 can complete its octet.
Solution:
The adduct.N B
F
H
H
H
F
F:
Illustrative Problem
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Metals lose their valence electrons to form cation in the pool of electrons. This is the Electron Sea Model for metallic bond.
Formation of metallic bond
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Excellent electricaland thermal conductivity
Regular close packed structures
Characteristics of Metallic bond
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Question
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Which of the following has othertype of bonding with covalentbonding?
(a) CCl4 (b)AlI3
(c) NH4Cl (d) HCl
Illustrative Problem
Solution:
covalent bonding is between N and three H-atomsCo-ordinate bond is present between N and one H atom ionic bond is there between NH4
+ and Cl– ions.
Hence, the answer is (c).
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Important aspects
i) Central atom.
[Exception: NH3, H2O more electronegative central atoms.]
ii) Formal charge on ‘each atom’= (valence electron in atom) – (no. of bonds) – (no. of unshared electrons)
Lewis theory
lesselectronegativ
e atom
iii) Multiple bonds complete the octet of atoms.
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n1=4+4 =8;
n2=2x4+8x1 =16;
n3=n2-n1 =8;
no. of bonds= n3/2=4
no. of non-bonding electron= n4
=(n1-n3)=0
no. of lone pairs= 0
Structure and bonding in CH4
C
H
H
H
H
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It cannot explain
Limitations of Lewis theory
Odd electron species NO,NO2
Electron-deficient species BF3 ,BeCl2
Electron-rich species PCl5 ,SF6
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1. Order of repulsionlp-lp > lp-bp > bp-bp
VSEPR theory
3. Decreasing order of repulsion, Triple bond > double bond > single bond.
2.lp-lp repulsion Electro-negativity of central atom
Electro-negativity of other atoms
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Linear
Planar
Tetrahedral
Trigonal bipyramidal
Octahedral
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PCl5
central atom is P.therefore, V= 5+5= 10 V/2=5
Shape will be trigonal bipyramidal.
Application of VSEPR theory
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Central atom is S V= 6 + 6 = 12
V/2=6
No. of atoms attached to central atom is six.
Hence, shape is octahedral.
Shape of SF6
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Question
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The shape of CH3+ is likely to be
(a) Pyramidal (b) tetrahedral
(c) linear (d) planar
Illustrative Problem
According to VSEPR,
N = 4 +3 –1=7
N/2=3
the shape should be planar.
Solution:
Hence, the answer is (d).
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Limitations
Cannot determine the shape
Multiple bonded species CO2,SO4
-2
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Question
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The shape of NH3 is very similar to
(a) CH4
(b) CH3
–
(c) BH3
(d) CH3+
Illustrative Problem
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Hence, the answer is (b).
4
N 4 4for CH , 4
2 2
shape is tetrahedral
3CHN 4 3 1
for , 42 2
shape is pyramidal
3
N 5 3for NH , 4
2 2
shape is pyramidal
3
N 3 3for BH , 3
2 2
shape is planar
3CHN 4 3 1
for , 32 2
shape is planar
Solution
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Class Test
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Class Exercise - 1
Pi bond formation involves ______ overlap.
(a) s-p head-on (b) p-p head-on
(c) s-s head-on (d) p-p sideways
Solution:
Pi bond formation involves only sideways overlap of pand d-orbitals.
Hence the answer is (d)
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Class Exercise - 2
Solution:
According to Lewis theory,n1 = 5 + 1 + 6 × 3 = 24n2 = 2 × 0 + 8 × 4 = 32n3 = n2 – n1; number of bonds = 3n
42
O — N — O
O
24 88
2
Number of lone pairs =
Formal charge on ‘N’ atom = 5 – 4 – 0 = +1
What is the formal charge on ‘N’ atom of ?3NO
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Class Exercise - 3
Molecular structures of SF4 and XeF4
are (a) the same, with 2 and 1 lone pairs respectively (b) the same, with 1 lone pair each (c) different, with 0 and 2 lone pairs respectively (d) different with 1 and 2 lone pairs respectively
Solution:
For 4N 6 4
SF 52 2
Lone pair is one and the structure is trigonal bipyramidal.
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Solution
Lone pairs are two and the structure is octahedral.
Hence, answer is (d)
For 4N 8 4
XeF , 62 2
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Class Exercise - 4
Predict the geometry of H3O+ based on VSEPR theory.
Solution:
For H3O+, central atom is ‘O’
N 6 3 14
2 2
Since 3 atoms are attached to the central atom,geometry will be of pyramidal according toVSEPR to minimize lp-bp repulsion.
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Class Exercise - 5
Solution:
Electronic configuration of Ca metal is 2, 8, 8, 2.While the configuration for Ca+2 is 2, 8, 8 which is a stable noble gas configuration.
Among Ca metal and Ca+2 the morereactive will be (Atomic No. of Ca is20)
(a) Calcium metal(b) Calcium ion(c) both are equally reactive (d) Cannot be predicted
Hence the answer is (a)
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Class Exercise - 6
Pi-bonds in N2 and CN– are due to
(a) p-p overlap for both species(b) p-p and p-d overlap(c) d-d overlap for both species(d) p-d and p-p overlap
Solution:
Since Pi bonds are formed due to overlap of either p ord orbitals only. Both N and C-atoms do not have anyelectrons in d-orbitals. Hence, Pi bonds in both casesare obtained because of p-p overlap only.
Hence, the answer is (a).
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Class Exercise - 7
The geometry of XeF2 accordingto VSEPR is
• angular• linear• pyramidal• None of these
Solution:
Number of atoms attached to the central atom is two.According to VSEPR theory geometry should be linear.
For XeF2N 8 2
52 2
Hence, the answer is (b)
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Class Exercise - 8
Which of the following is a tri-atomicmolecule?
(a) Ammonia(b) Sulphur dioxide(c) Sulphur tri-oxide(d) Phosphine
Solution:
NH3 tetratomic moleculeSO2 tri-atomic moleculeSO3 tetratomic moleculePH3 tetratomic molecule
Hence the answer is (b)
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Thank you