1. Chemistry

2. CHEMICAL BONDING – SESSION I

3. Session Opener

4. Session Objectives

5. Session Objectives Introduction Octet rule Different types of bonding
Lewis theory
VSEPR theory and shape of molecules

6. What is Chemical bonding?
Chemical bonds
Force of attraction holding group(s) of atoms
Na+
Cl-
But why bonds are formed ??
Better stability against chemical reagents

7. Na Na Cl Cl Octet rule Atoms two electrons in the valence shell (1s2)
noble gas configuration attain better stability.
two electrons in the valence shell (1s2) Na 2 8 + Ne Na 2 8 1
Very reactive Cl 2 8 Ar - Cl 2 8 7
Very reactive

8. Limitation of octet rule
In SF6, ‘S’ has twelve electron in its valence shell, leads to minimisation of energy. Other examples are: PCl5, BF3

9. Questions

10. Illustrative Problem The molecule that deviates from octet rule is
(a) NaCl
(b) BeCl2
(c) MgO
(d) NH3

11. Solution The no. of valence electrons in different central atoms is:
Na Be Mg+2 8
N atom in NH3 (covalent compounds) 8
Hence, the answer is (b).

12. Ionic
Covalent
Pi bond
Sigma bond
Bonding
Co-ordinate or dative
Metallic

13. Formation of ionic bond

14. Formed by mutual sharing of electrons
Covalent bond
Formed by mutual sharing of electrons
Covalent bonds
Double bond
Triple bond
Sigma bond is a primary bond, formed first Pi bond is a secondary bond.
courtesy:

15. Formation of covalent bond
non-polar covalent bond between two carbon atoms polar covalent bond between carbon and hydrogen atoms.

16. Question

17. Illustrative Problem
Covalent bonds are called directional while ionic bonds are called non-directional -explain
Solution:
electrostatic force of attraction.
Ionic bond
overlap of atomic orbitals
covalent bond
p and d-orbitals generate directional covalent bond.

18. Types of covalent bonds
s-s +
s-p +
p-p +
Strength of these sigma bonds is in the order:
p-p > s-p >s-s
sigma bond forms due to end-to-end or head-on overlap

19. Types of covalent bondsor +
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.

20. Difference between sigma and pi bonds
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 p bonds
In plane of molecule
Perpendicular to plane of molecule
Stronger as compared to p bond
Weaker as compared to s bond

21. Difference between ionic and covalent compound.
Ionic compound (NaCl)
Covalent compound (CHCl3)
MP/BP
Very high
Volatile liquid
H20 solubility
Highly soluble
Almost insoluble
Benzene solubility
Insoluble
Highly Insoluble
Directional nature
Non-directional
Except s-s overlap all are directional

22. Question

23. Illustrative Problem
How many sigma and pi-bonds are present in a benzene molecule?
Solution:
The structure of benzene molecule is
no. of pi bonds are 3 [C=C]
no. of sigma bonds are 12 [C-C and C-H]

24. Coordinate covalent bondH N H+ :
[NH4]+ O H H+ :
H3O+
Single atom donating lone pair
Shared by two atoms involved

25. Question

26. Illustrative Problem NH3 and BF3 form an adduct readily -explain.
Solution:
N- atom in NH3 have one lonepair and BF3 is electron deficient. They form an adduct through coordinate bond, so BF3 can complete its octet.
The adduct.

27. Formation of metallic bond
Metals lose their valence electrons to form cation in the pool of electrons. This is the Electron Sea Model for metallic bond.

28. Characteristics of Metallic bond
Regular close packed structures
Excellent electrical and thermal conductivity

29. Question

30. Illustrative Problem Which of the following has other
type of bonding with covalent
bonding?
CCl4 (b)AlI3 (c) NH4Cl (d) HCl
Solution:
covalent bonding is between N and three H-atoms Co-ordinate bond is present between N and one H atom ionic bond is there between NH4+ and Cl– ions.
Hence, the answer is (c).

31. less electronegative atom
Lewis theory
Important aspects
Central atom.
less electronegative 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)
iii) Multiple bonds complete the octet of atoms.

32. Structure and bonding in CH4
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

33. Limitations of Lewis theory
It cannot explain
Electron-rich species PCl5 ,SF6
Electron-deficient species BF3 ,BeCl2
Odd electron species NO,NO2

34. VSEPR theory Order of repulsion lp-lp > lp-bp > bp-bp
Electro-negativity of central atom
2.lp-lp repulsion
Electro-negativity of other atoms
Decreasing order of repulsion, Triple bond > double bond > single bond.

35. Linear
Planar
Tetrahedral
Trigonal bipyramidal
Octahedral

36. Application of VSEPR theory
PCl5
central atom is P.
therefore, V= 5+5= 10
V/2=5
Shape will be trigonal bipyramidal.

37. Shape of SF6 Central atom is S V= 6 + 6 = 12 V/2=6
No. of atoms attached to central atom is six. Hence, shape is octahedral.

38. Question

39. Illustrative Problem The shape of CH3+ is likely to be
Pyramidal (b) tetrahedral
(c) linear (d) planar
Solution:
According to VSEPR, N = 4 +3 –1=7 N/2=3 the shape should be planar.
Hence, the answer is (d).

40. Multiple bonded species CO2,SO4-2
Limitations
Cannot determine the shape
Multiple bonded species CO2,SO4-2

41. Question

42. Illustrative Problem The shape of NH3 is very similar to (a) CH4
(b) CH3–
(c) BH3
(d) CH3+

43. Solution shape is pyramidal shape is tetrahedral shape is planar
Hence, the answer is (b).

44. Class Test

45. 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 p and d-orbitals.
Hence the answer is (d)

46. Class Exercise - 2 What is the formal charge on ‘N’ atom of ?
Solution:
According to Lewis theory,
n1 = × 3 = 24
n2 = 2 × × 4 = 32
n3 = n2 – n1; number of bonds =
Number of lone pairs =
Formal charge on ‘N’ atom = 5 – 4 – 0 = +1

47. 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
Lone pair is one and the structure is trigonal bipyramidal.

48. Solution For Lone pairs are two and the structure is octahedral.
Hence, answer is (d)

49. Class Exercise - 4 Predict the geometry of H3O+ based on VSEPR theory.
Solution:
For H3O+, central atom is ‘O’
Since 3 atoms are attached to the central atom, geometry will be of pyramidal according to VSEPR to minimize lp-bp repulsion.

50. Class Exercise - 5 Among Ca metal and Ca+2 the more
reactive will be (Atomic No. of Ca is
20)
Calcium metal
(b) Calcium ion
(c) both are equally reactive
(d) Cannot be predicted
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.
Hence the answer is (a)

51. Class Exercise - 6 Pi-bonds in N2 and CN– are due to
p-p overlap for both species
p-p and p-d overlap
d-d overlap for both species
p-d and p-p overlap
Solution:
Since Pi bonds are formed due to overlap of either p or
d orbitals only. Both N and C-atoms do not have any
electrons in d-orbitals. Hence, Pi bonds in both cases
are obtained because of p-p overlap only.
Hence, the answer is (a).

52. Class Exercise - 7 The geometry of XeF2 according to VSEPR is angular
linear
pyramidal
None of these
Solution:
For XeF2
Number of atoms attached to the central atom is two.
According to VSEPR theory geometry should be linear.
Hence, the answer is (b)

53. Class Exercise - 8 Which of the following is a tri-atomic molecule?
Ammonia
Sulphur dioxide
Sulphur tri-oxide
Phosphine
Solution:
NH3 tetratomic molecule
SO2  tri-atomic molecule
SO3  tetratomic molecule
PH3  tetratomic molecule
Hence the answer is (b)

54. Thank you