Chemistry

CHEMICAL BONDING – SESSION I

Session Opener

Session Objectives

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

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

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

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

Questions

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

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

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

Formation of ionic bond

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:www.lbw.cuny.edu

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

Question

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.

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

Types of covalent bonds or + 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.

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

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

Question

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]

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

Question

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.

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.

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

Question

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).

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.

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

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

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.

Linear Planar Tetrahedral Trigonal bipyramidal Octahedral

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

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.

Question

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).

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

Question

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

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

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

Class Exercise - 2 What is the formal charge on ‘N’ atom of ? Solution: According to Lewis theory, n1 = 5 + 1 + 6 × 3 = 24 n2 = 2 × 0 + 8 × 4 = 32 n3 = n2 – n1; number of bonds = Number of lone pairs = Formal charge on ‘N’ atom = 5 – 4 – 0 = +1

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.

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

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.

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)

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).

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)

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)

Thank you