chemistry. chemical bonding – session i session opener
TRANSCRIPT
Chemistry
CHEMICAL BONDING – SESSION I
Session Opener
Session Objectives
1. Introduction
2. Octet rule
3. Different types of bonding
4. Lewis theory
5. VSEPR theory and shape of molecules
Session Objectives
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 ??
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
-
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
Questions
The molecule that deviates from octet rule is
(a) NaCl (b) BeCl2
(c) MgO
(d) NH3
Illustrative Problem
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
Bonding
Ionic
Covalent
Co-ordinate or dative
Metallic
Pi bond Sigma bond
Formation of ionic bond
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
non-polar covalent bond between two carbon atomspolar covalent bond between carbon and hydrogen atoms.
Formation of covalent bond
Question
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
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
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
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
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.
Question
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
•Single atom donating lone pair•Shared by two atoms involved
Coordinate covalent bond
O
H
H+H :
H3O+
H
N
H
H+H :
[NH4]+
Question
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
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
Excellent electricaland thermal conductivity
Regular close packed structures
Characteristics of Metallic bond
Question
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).
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.
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
It cannot explain
Limitations of Lewis theory
Odd electron species NO,NO2
Electron-deficient species BF3 ,BeCl2
Electron-rich species PCl5 ,SF6
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
Linear
Planar
Tetrahedral
Trigonal bipyramidal
Octahedral
PCl5
central atom is P.therefore, V= 5+5= 10 V/2=5
Shape will be trigonal bipyramidal.
Application of VSEPR theory
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
Question
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).
Limitations
Cannot determine the shape
Multiple bonded species CO2,SO4
-2
Question
The shape of NH3 is very similar to
(a) CH4
(b) CH3
–
(c) BH3
(d) CH3+
Illustrative Problem
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
Class Test
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)
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
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.
Solution
Lone pairs are two and the structure is octahedral.
Hence, answer is (d)
For 4N 8 4
XeF , 62 2
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.
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)
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).
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)
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)
Thank you