Chapter 03: Chemical Bonding

Chemical Bond Definition: A chemical bond is the force that holds together two or more atoms, molecules or ions. The property is of a substance depend on the type of the chemical bond between its atoms. Explanation: The term chemical bond includes infra-molecular forces which are: Ionic bond (i) Covalent bond (ii) Fig: NaCl has ionic bond and is solid, but Dative bond (wii) water is a covalent compound and is liquid. (iv) Metallic bond TYPES OF BONDING Lewis concept • Lewis concept of bonding gives a simple explanation of the formation of all types of bonds. • Atoms make bonds to complete their outermost shells to have noble gas-like configuration. • It is mostly attained through the formation of an octet in the valence shell. 1. Ionic Bond (electrovalent bond) Definition: The ionic bond is formed by the complete transfer of electrons from an, atom with low ionization energy to another atom with high electron affinity. Example: Formation of ionic bond in NaCl • The Na atom (11Na = 2,8,1) tends to lose the outermost electron to form Na* (2,8) ion, which has the electron configuration of oNe, a noble gas nearest to it. • Chlorine atom Cl (2,8,7) gains one electron to form the chloride ion, Cl (2,8,8), also gaining the next noble gas electron arrangement. • The oppositely charged Nat and Cl ions are held together by strong ionic bond in the crystal of NaCl. The formation of ionic bond is shown in the figure. (2,8, 7) (2, 8, 1) Fig: Ionic bond formation between Na and Cl A similar type of bond is expected between group 1 and 2 metals and groups 17 and 16 non-metal elements. 2. Covalent Bond Definition: A covalent bond is formed by the mutual sharing of electrons between two atoms. By sharing electrons, each atom completes its valence shell and attains the nearest noble gas configuration. Types: (i) Polar Covalent bond free i'm. (ii) Non-Polar Covenant bond (i) Non Polar bond Definition: The bond formed due to equal sharing of electrons between two atoms. • In formation of Cl2 molecule, the bond formation between two Cl atoms is shown in figure. • The bond between two Cl atoms purely covalent and non-polar. • The electronegativity of the two atoms is exactly the same. • So, the bonded atoms remain electrically neutral and there is no charge on either atom. ** ** (2,8, 7) (2,8, 7) Fig: Formation of a covalent bond between two Cl atoms The other such molecules are H2, F2, Br2, and Iz. (ii) Polar Covalent bond Definition: The bond formed due to unequal sharing of electrons between two atoms. Examples of polar covalent bonds are H-Cl, H-Br, H2O, etc. [2, 8]+ [2,8, 8] NaC1 (Na C!)) ** *x ** 12, 8, 8] [2, 8, 8] 3. Dative Bond (Co-ordinate Covalent Bond) Definition: A dative bond is formed between two atoms when the shared pair of electrons is donated by one of the bonding atom only. Examples of dative bond: Following are the examples of the dative bond. (i) Hydronium ion (H3Ot) • In the co-ordinate covalent bond formation between H2O and a proton (H*). H2O has two covalent bonds and there are two lone pairs of electrons on the oxygen atom. • The proton is deficient in electrons. • The oxygen atom can donate its lone pair of electrons to the acceptor Ht. • This results in the formation of a dative bond as in the figure. (ii) Carbon monoxide • Carbon monoxide also contains a dative bond between oxygen and carbon atoms. • Oxygen shares its two valence shell electrons with the carbon atom to make a normal double covalent (b) bond. • The valency of carbon is not satisfied with these bonds. • It needs one more pair of electrons to complete its octet. • Oxygen atom donates one of its lone pairs for the formation of a covalent bond, which is a dative bond as exhibited in figure. (ill) Ozone • It is formed by a reaction between the oxygen molecule with an oxygen atom (a free radical (0°). • One of the Oxygen atoms in the molecule donates a pair of electrons to make a coordinate covalent bond with this free radical. • The central atom has total three bonds including a double covalent bond on one side and a coordinate covalent bond on the other. • It carries a positive charge, while the oxygen atom that accepts the lone pair carries a negative charge. • The structure of ozone molecule is presented in figure. Quick Check 3.1 (a) Draw the Lewis structures of N2 and CS molecules? Ans. Lewis structures of N2 NIN Lewis structure of CS2 SC S (b) How many electrons are there in the valence shell of B in BF3? Does it have the ability to accept a lone pair of electrons? Ans. There are six electrons in the valence shell of B in BF3 as shown below + (a) :C::O: :CO: Lewis structure 0+ + 0=0 : (c) 0- oxygen atom Fig: Dative bond formation (a) oxonium (hydronium) ion, (b) carbon monoxide, (c.) ozone molecule. *FוB•×F· Above structure indicates that it have the ability to accept a lone pair of electron to complete its octet because it is deficient of two electrons by octet rule. Show the formation of a dative bond between. NH3 and BF Ans. Formation of dative bond between NH3 and BF3. P H H - N: + H F

Polyatomic Ions Definition: Polyatomic ions are the ions composed of more than one type of atoms. There formal charge is the net charge on them which is calculated based on the number of electrons in their valence shells after the formation of bonds. Except ammonium ion (NH4*), these ions mostly carry a negative charge. Example: The examples of polyatomic ions are carbonate (CO3), sulfate (SO4'), and nitrate (NO32) ions. Expanded Octets • In some polyatomic ions, the central atom violates the octet rule by expanding its electrons density to the higher orbitals. • These are set to have expanded octets. Examples of expanded octets: The prominent examples of expanded of octets are (i) Expanded Octets in SO42-: From Lewis structures of SO4- , the number of electrons around the central S atom can be calculated. The number of electrons in the valence shell of S atom can be no. of calculated as: No: of valence electrons = 2 × (double bond electrons) + 2 × (no. of single bond electrons) = 2(4) + 2(2) = 12 electrons • S has 12 electrons in its valence shell and it exceeds the octet by 4 electrons. • The expansion of octet is caused by the involvement of the d orbital in bonding, which can accommodate the extra electrons. • The following electronic configuration of native S (0) atom shows that it has two unpaired electrons in the p orbitals. • The presence of d orbital allows this configuration to extend to 4 unpaired electrons by the transfer of one electron from 3s and one from the 3p electron pair. *F* + H F B - F H-N°- H 2- Ö: sulfate ion • It. explains not only the variable oxidation states of S, but also the possibility of accepting extra electrons in its d orbital. 11|1/1 3s 3p 111 3p (i) Expanded Octets in Is Ion: The number of valence electrons around the central iodine atom in the try-iodide ion can be calculated. The Lewis structure of the ion is given by: • .. The number of valence electrons of the central atom can be calculated as follows. No of valence electrons = 2 × (no: of single bond electrons) + 2 × (no. of lone pair) = 2(2) + 2(3) =10 electrons The central iodine atom in try-iodide ion I,' has 10 electrons in the valence shell and the octet expands by two electrons. Quick Check 3.2 (a) Can the elements of period 2 of the periodic table have expanded octet? Explain why or why not? Ans. Elements of period 2 of periodic table do not have expanded octet, because their valence shell contains 2s and 2p orbitals which can accommodate maximum of eight electrons. The 3d orbitals needed for expansion which is not present in them. (b) Predict and explain the expanded octets in the following ions: PO,, ClO*. Ans. Lewis structure of PO*- .. :O No. or Valence electrons of P = 3 (no. of single bond electron) +1 (no. of double bond Electron = 3 (2) +1(4) = 6 + 4 = 10 electrons So, central P atom has 10 electrons and the octet expands by two electrons Lewis structure of ClO; No. of electrons = 1(2) + 2(4) 3d 21111 3d - 3 :O: :0? •O = 2+. 8 = 10 electrons The above Lewis structure shows that the central chlorine atom has expanded octet by 2 electrons. ELECTRONEGATIVITY AND THE TYPE OF BONDS

Electronegativity and the type of bonds The difference in electronegativity of two bonded atoms provides an approximate measure of the bond polarity and an indication of the type of bond. Explanation • When this difference is very small or zero, the bond is covalent and non-polar. • When it is large, the bond is polar covalent or ionic. The following figure graphically depicts the change in the nature of the bond with electronegativity difference. • The atoms having electronegativity difference less than 0.4 are said to make a pure covalent bond • An electronegativity difference between, 0.4 and 1.8 corresponds to a polar covalent bond, whereas, above this value the bond between two atoms will be ionic in nature. • The electronegativity differences between the atoms in the bonds such as H-H bond is purely non polar. H-Cl a polar covalent bond, whereas NaCl is an ionic compound. Difference of electronegativity between bonded atoms Pure Covalent Zero Polar Covalent < 1.8 > 1.8 Fig: Variation of the nature of bond with the electronegativity difference Quick Check 3.3 (a) Predict with the help of electronegativity values whether the bonds in these compounds would be non-polar covalent, polar covalent, or ionic. (il) KI (i) HF (üi) CaF2 Ans. Compound Electronegativity HP KI CaF2 ICe Br2 Bond type Covalent character increases Covalent character decreases Ionic (v) Br2 (iv) ICl Nature of Bond 1.8 value Highly Polar 1.7 Highly Polar, Ionic 3 Tonic 0.5 Polar 0 Non-Polar

Dipole Moment and polarity of molecules Definition: It is a quantitative measurement of the polarity of a bond or a molecule. A molecule with &t charge on one part and 8 - charge on the other part is called a dipole and such a molecule is said to have a dipole moment. Direction of dipole moment The dipole moments of diatomic molecules like HF, Hcl, HBr, HI, NO, etc. are directed from the positive ends (8 +) to negative ends (8 -). Unit: Dipole moment is measured in Debye unit (D). Higher dipole moment indicates high polarity in a molecule. F H nucleus F nucleus Fig: Electron density in HF is higher near the F atom and the dipole moment is directed from H to F Dipole Moment and structure of molecules Some molecules have net zero dipole moment as the symmetry in the structure causes the cancelation of the individuals bonds moments. For compounds having more than two atoms, the dipole moment depends on the structure of the molecule. Dipole Moment of H2O: The dipole moment of water is 1.85 D which is directed from the end having two hydrogen atoms to the end with the oxygen atom as in the structure, • A linear H2O molecule (H-O-H) would have zero dipole moment. • The non-zero dipole moment value shows that water is a non-linear molecule. • Experiments reveal that the water molecule has a V-shaped structure. Dipole Moment of SO2: In contrast, SO2, has a dipole moment of 1.61 D and in opposite direction to that in water. Dipole moment of HzS: HS is also a non-linear molecule as the individual bonds are polar, but they don't cancel each other's dipole moment and the overall molecule has some dipole moment. 8- HS+ нб+ P= 1.85 D P= 1.61 D Electron density map of the water molecule. Fig: Vector addition of individual bond moments in angular H20 and SO2 molecules Dipole Moment of BeCl • BeCl is a linear molecule having two similar Cl atoms on both sides of the central atom at 180°. • The individual Be-Cl bond moments are cancelled out as they are opposite in direction and equal in amount. Dipole Moment of CClA • CCl4, has four C-Cl bonds which are expected to have high polarity due to a large electronegativity difference between C and Cl atoms. • The bond moments associated to the four C-Cl bonds are directed in such a way that they cancel each other. • The net dipole moment of the CCl4, molecule is zero making it a non-polar molecule. • The CCl4 molecule is perfectly tetrahedral. Dipole Moment of BF3 • BF3 has a trigonal planner symmetrical structure and its dipole moment is also zero. • As a rule, the molecules that have same ligands (atoms or groups of atoms with the central zero) in a regular geometry, the individual dipole moments may not be zero. : The overal moms sold has zero dipole moment. Co • This molecule is said to be non-polar. 180° 0 St Cl Be P=0D P=OD Fig: The individual bond moments in symmetrical molecules; BeCl2, BF3 and CCl are cancelled out to give zero net dipole moment Quick Check 3.4 (a) Can you explain why CO has a dipole moment but COz does not have any? Ans. CO has a dipole moment. It is a polar molecule with a small dipole moment of 0.122D. In CO Carbon is partially positive while oxygen is partially negative due to difference of electronegativity of carbon and oxygen as shown below. 8+ - 8- C=O P= 0.122D CO2 does not have dipole moment. Carbon dioxide has linear structure as shown below. 0- - 0+ - 0- O=C=O P=QD Two arrows on opposite side indicates that dipoles moments are equal and opposite which cancel the effect of each other. AFE CI C. F8 p=OD (b) Do you think that individual bonds in CCla are polar? Explain in terms of the electronegativity difference. What about the polarity of overall Cla molecules? Ans. CCla has four C-Cl bonds which are expected to have high polarity due to large electronegativity difference between C and Cl atoms. The bond moments associated to the four C-Cl bonds are directed in such a way that they cancel each other. The net dipole moment of CCla molecules is zero making it a non-polar molecule. The CCla molecule is perfectly tetrahedral. б- CI p=0D (c) Are the following molecules polar or non-polar? Briefly give reasons. HF, CH2Cl2, O2, H2S. Ans. HF is a polar molecule due to difference of electronegativity of hydrogen and fluorine. Fluorine is more electronegative so, attracts shared pair of electron more strongly creating a dipole moment with partial negative charge on fluorine and partial positive charge on hydrogen. H-฿: P = 1.83D CH2Cl2 dicholomethane is a polar molecule due to the difference in electronegativity between carbon, hydrogen and chlorine which result in polar bounds. The tetrahedral geometry of CH Cl2 causes these bond dipoles do not cancel each other leading to a net dipole moment. •/ P = 1.60 D Oz is non-polar molecule because it is homotopic molecule and there is no difference of electronegativity between its atoms. Ö=Ö Hes is a polar molecule due to its bent shape and difference in electronegativity between sulphur and hydrogen. H2S hydrogen sulphide is slightly polar due to small electronegativity difference of hydrogen and sulfur. The bent molecule geometry results in a net dipole P = 0.95 D CI Ist CI 8- CI C H P=OD INTERMOLECULAR FORCES

Intermolecular forces Definition: The forces of attraction between two molecules in a compounds are called intermolecular forces. Explanation • Intermolecular forces are also considered as binding forces just like chemical bonds. • The attraction between the molecules is much weaker than the chemical bonds. • These forces are believed to exist between all kinds of atoms and molecules when they are sufficiently close to each other. • Such intermolecular forces are called van der Waals forces. • These intermolecular forces bring the molecules close together and give particular physical properties to the substances in gaseous, liquid, and solid states. Three types of such forces are mentioned here: (i) Instantaneous dipole-induced dipole forces or London dispersion forces (ii) Permanent dipole-permanent dipole forces (ill) Hydrogen Bonding BOND ENERGY AND BOND LENGTH

Bond energy Definition: The Bond energy is the average amount of energy required to break all bonds of a particular type in one mole of a substance. • Measurement of bond energy: It is determined experimentally by measuring the heat involved in a chemical reaction and its unit is kJ mol • Bond energy is a measure of the strength of a bond and its reactivity. • The strength of a bond depends upon the following factors: (a) Electronegativity difference of bonded atoms (b) Size of atoms Table 3.1 Average bond energies of some selected bonds in kJ mol-1 Single Bonds C-I 432 H-H 240 565 C-Cl H-F 339 H-Cl 305 C-N 427 C-O H-Br 358 363 H-I 295 391 N-H 201 N-O 413 C-H C-C 347 O-H 467 Multiple Bonds 154 F-F C=C 614 C=C 839 Ce-Cl 239: Br-Br 0=0 193 495 I-I 149 C=0 745 266 S-S NEN 941 340 Si-Si C=N 615 N=O 452 Si-O 607 (a) Electronegativity difference of bonded atoms • The bond energies of H-X type of compounds can be observed, where X= F, Cl, Br, I. • The data shows that the bond energy of a bond rises with the increasing electronegativity difference between the bonded atoms. • As the difference in electronegativity increases, the bond polarity also becomes greater and this gives rise to additional attractive force for binding the atoms. This is why the bond energy for HF is higher (565 kJ mol) than for HI (295 kJ mol). Multiple bonds: It may be noted that energies of multiple bonds are greater than those of single bonds. C=C> C=C> C-C Bond Length Definition: Bond length is the distance between the nuclei of two atoms forming a covalent bond Measurement of bond length: The bond lengths are experimentally determined by physical techniques, such as electron diffraction, X-ray diffraction, or spectral studies. Factors: The bond length of a bond is electronegativity, size, and the nature of the covalent bond (single, double, or triple). Some selected bond lengths are given in table 3.2. Table 3.2 Bond lengths of some selected bonds Bond Bond length (pm) H-H 74 144 H-Br the C-C 154 C=C 133 C=C 120 C=O 122 146 SI-H Electro-negativity cause shortening of bond. With an increase in size of the atoms, the covalent bond length also increases. The C-Cl bond length is about 180 pm, whereas the C-F bond length is 135 pm. This is because Cl atom is much larger. Ionic character of Si-F bond: The bond length of Si-F bond in SiF4, is found to be near 155pm, whereas the calculation of Si-F bond forms the covalent radii of Si and F (S1-117 pm and F-64 pm) is 181 pm. The electronegativity difference causes an ionic character in the covalent bond. The ionic character results in shortening of the bond length due to the additional attraction between the bonded atoms. Quick Check 3.5 (a) HI is a stronger acid and a robust reducing agent, whereas HF is a weaker acid. Explain. Ans. HI is stronger acids due to weakness of hydrogen and iodine bond. It releases hydrogen easily as compared to HF. The stronger bond between hydrogen and fluorine due to difference of electronegativity makes it a weak acid. governed by many factors including Bond Bond length (pm) Si-F 155 C-F 135 O C-Cl 180 C-Br 196 C-I 214 B-F 130 B-Cl 175 (b) Acetylene (HC = CH) is more stable than ethene (H2C=CH2). Can you explain why? Ans. Acetylene has triple bond in which three bond pairs are shared leading to stronger nuclear attraction and shorter bond length. So triple bond is more stable while Ethene has double bond is which two bond pairs are shared leading to less nuclear attraction so double bond is less stable. Acetylene is more stable than ethene. A COMPARISON AMONG IONIC, COVALENT, METALLIC BONDS AND INTERMOLECULAR FORCES

Chemical bond Chemical bond's, i.e. ionic, covalent, and metallic bond usually termed as true chemical bonds, as they affect the chemical properties of a substance. This is because of certain reasons. Chemical bonds result in the formation a new species through transfer and sharing of electrons. Intermolecular forces Intermolecular forces act to bring molecules closer and influence the physical properties. A clear distinction between the chemical bonds and intermolecular forces is not possible. Comparison of chemical bonds and intermolecular forces The distinctive feature of the chemical bonds and intermolecular forces is their bond 'strength. The strength of a force is measured by the bond energy. Table 3.3 provides a comparison among different chemical bonds and intermolecular forces. Bond energy comparison Ionic bond: The bond energy of ionic bond (sodium chloride, 760 kJ mol) is highest, followed by covalent bond (O-H=464 kJ mol '), and average hydrogen bond energy (20- 25 kJ mol-1). It shows that the ionic bond is the strongest form of chemical bonding. It also reveals that chemical bonds are generally much stronger than intermolecular forces. Intermolecular forces • The permanent dipole-dipole forces (5-20 kJ mol-1) and London dispersion forces (1-20 kJ möl) are even weaker as indicated by their low bond energies. • The metallic bond is mostly elaborated in terms of electrostatic forces, although some theories suggest that it may be of covalent nature. • In any case, the metallic bond is weaker than both the ionic and covalent bonds. The average bond energy of the metallic bond is 100-150 kJ mol-1 the low strength of metallic bond is extensive delocalization of electrons within the metallic crystal. Table 3.3 Relative strengths of chemical bonds and intermolecular forces Bond Energy Bond Type (kJ mol)) 760 Ionic Bond in NaCl 464 O-H bond in water 20-50 Hydrogen bonding Permanent dipole- 5-20 Permanent dipole force 1-20 Van der Waals forces • A satisfactory argument VALENCE BOND THEORY 08. Explain the valence bond theory. How the bonding in the following molecules can be explained with respect to valence bond theory? (wii) Cla 1) 11) 11 Ans. Introduction • The VSEPR model predicts and explain the shapes of molecules but does not reasons for the formation of bonds. • VBT is concerned with both the formations of bonds and the shapes of molecules. • This method describing a covalent bond considers a molecule as a combination of atoms. The postulates of VBT are given below: Postulates: • Formation of Covalent Bond: A covalent bond is formed when half-filled orbitals in the valence shell of two atoms with similar energy overlap. : Strength of Bond: A greater overlap of the orbitals results in a stronger bond. Direction of Covalent Bond: Covalent bonds are directional. The direction of the bond is determined by the shape and mode of the overlapping orbitals. Explanation of VBT Formation of a sigma bond: A sigma bond is formed by the overlap of two half-filled atomic orbitals on adjacent atoms. The orbitals approach each other on the nuclear axis. Both s and p orbitals can overlap head on to form sigma bonds. s-s Overlap: The s-orbital of the one atom overlap with the•s orbital of the other to give a sigma (o) bond. (i) Hydrogen (H2): This type of overlap occurs during the formation of H2 molecule, where each hydrogen atom has a half filled 's' orbital. After bond formation electrons are paired up and the electron density in this molecule is symmetrical around the two nuclei as shown in figure. + Hydrogen Hydrogen. Fig: Overlap between s orbitals of two hydrogen atoms to form the H2 molecule s-p Overlap: The s-orbital of the one atom overlap with the p orbital of the other to form sigma bond. (ii) Hydrogen chloride (HCl): In HCl molecule, a half filled 's' orbital of hydrogen overlaps with the half-filled p. orbital of chlorine as shown in figure. The electron density is higher close to Cl atom due to its higher electronegativity value. This is why HCl is a polar molecule. + 3p 1s Overlap starts Hydrogen atom Chlorine atom Fig: Formation of o bond by s-p overlap p-p Overlap: The p-orbital of the one atom overlap with the p orbital of the other atom to form sigma bond. (v) Oz (vi) HF (vii) N2 Symmetrical Electron Density H, molecule → CI H CI Hydrogen chloride Hcl (wii)Chlorine (Cl2): An example of this type of overlap is the formation of the Cl2 molecule where "p' orbitals of two chlorine atoms overlap on the nuclear axis. The electron density is symmetrical around the nuclei of the two Cl atoms because both have same value of electronegativity. 3p 3p Chlorine atom Chlorine atom Atomic orbitals Fig: Overlap of p and p orbitals to form Cl molecule according to VBT Interesting information! S.Q. Why these is difference in bond angles of H2O and H2S? Ans. The bond angle in H2O is 104:5°, whereas in HzS, it is 92°. This is because the orbitals of S are larger and the lone pair exerts a stronger repulsion. (iv) H2S The H2s molecule is a non-linear molecule which is formed by the combination of one sulfur and two hydrogen atoms. The two 3p (say 3py and 3pz) orbitals of sulfur containing one electron each can overlap. with the Is orbitals of two hydrogen atoms. A v-shaped molecule is thus formed having a bond angle of 92° as in figure. Overlap 90° Partially filled 3p orbitals Filled 3p orbital Fig: The formation of H2S by H and S orbitals overlap (V) 02 Formation of a bond • In the formation of a double covalent bond between oxygen atoms (O=O). • There are two unpaired electrons on each atom in perpendicular p orbitals (say Px and Py). • The px orbitals on the two oxygen atoms are oriented in such a way that they overlap end-to-end linearly. • The linear overlap of the px orbitals gives a o bond • The py orbitals on both atoms are aligned parallel to each other. • They overlap in a parallel way so that the two p lobes overlap above the plane of the nuclei and the other two lobes below the plane as shown in figure. • This results in the formation of a bond. • The oxygen atoms are doubly bounded through o and it bounds. Bond orbital ee il 92° H H Overlap T bond Py O bond I bond Fig: Formation of double bond (o and T) between oxygen atoms (vi) Hydrogen fluoride (HF): In the formation of H-F, the partially filled 1s orbital of H- atom overlap with partially filled orbital of fluorine to form sigma bond 3p 1s Overlap starts Hydrogen atom Chlorine atom Fig: Formation of HF molecule (vii) Nitrogen (N2): In case of N? p orbitals are half filled. One half-filled 2px orbitals of one nitrogen overlap linearly with half-filled 2px orbitals of second nitrogen. There are also two half-filled 2py and 2pz orbitals which undergo sideways overlap to from two ro bonds. -- T-bond 2p, 2p; 2p. 1L Fig: Formation of N2 Molecules Quick Check 3.6 (a) Draw the orbital structure of H20 and N2 molecules. In H2O, two half-filled s orbital of hydrogen overlap with two halt-tilled p orbitals (Py, Pz) of oxygen. A v-shaped molecule is formed as shown below Fig: Orbital overlap in H20 Orbital structure of N2 In N2, there are three unpaired electrons on each atom in perpendicular p orbitals (Px, Py and pz). The px orbitals on the two nitrogen atoms are oriented in such a way that they overlap nd-to-end linearly. The linear overlap of px orbitals give a o bond. However, the py and p orbitals on both atoms are aligned parallel to each other. They overlap in a parallel way t form two a hands. The nitrogen atoms are trinlv handed through o and two t hands. 2p. 2p, 2p. 2p 2p. 2p. 2p, Fig: Orbital overlap in N2 F H Hydrogen fluoride Tt-bond 2p, O-bond 2p, 2p H H 104.5° TT-bond T-bond 2p, 2p 2p, O-bond 2p. 1L 2p (b) Draw the orbital overlap to show the formation of F2 and HF molecules. Ans. Orbital overlap of Fz: In F2, p orbitals of two fluorine atoms overlap to form o bonds. The electron density is symmetrical around the nucleic of the two fluorine atoms because both have same electronegativity value, 3p 3p Chlorine atom Chlorine atom Fig: Orbital overlap in F2 Hr: In HE, a half-filled orbital of hydrogen overlaps with a halt-filled P orbital of fluorine as shown in Figure. The electron density is higher closed to fluorine due to its higher electronegativity value. That is way HF is a polar Molecule. + 2p 1s (H) (F) Fig: Orbital overlap in HF ATOMIC ORBITAL HYBRIDIZATION

Hybridization Detinition: A process in which atomic orbitals of slightly different energies and shapes are mixed together to form a new set of equivalent orbitals of same energy and same shape is called hybridization. Keep in mind! S.Q. Give basic rules of hybridization. • Hybridization is a process of mixing of orbitals in a single atom (or ion). • Only orbitals of comparable (relatively close) energies can be mixed to form hybrid orbitals. • The number of mixing orbitals is always equal to the number of the resulting hybrid orbitals. Types of Hybridization: The types of hybridization are sp sp Hybridization Definition: In sp hybridization, one s (low energy and spherical and one p orbital (high energy and dumb-bell shaped), intermix to give a new set of two hybrid orbitals of same energy and same shape called sp hybrid orbitals. These sp hybrid orbitals are arranged in linear geometry and oriented at 180°. Examples: BeCl2, CH = CH, MgCl2 etc. BeClz: In beryllium chloride the central atom is beryllium. The ground, excited and hybridized states of beryllium are given below. Ground 4Be ground = 2s 2Px 2py 2pz Beexcited = Excited Behybridized 2s 2Px 2py H P HF molecules Overlap starts 3, sp' and sp. - 111 hybridized sp sp The two sp hybrid orbitals lie linearly as in the following diagram. Hybridization + s orbital p orbital 180° Be Fig: Mixing of s and p orbitals to give two hybrid sp orbitals and the formation of BeCl The sp hybridization explains the geometry of linear molecules such as beryllium chloride, BeCl2. BeCle is formed when two sp hybrid orbitals of Be atom overlap with the half-filled p-. orbitals of chlorine atoms.

sp' Hybridization Definition: In sp' hybridization, one s and two p-orbitals of an atom intermix to form three orbitals called sp' hybrid orbitals. It is called sp' hybridization. The three half-filled sp' hybrid orbitals are arranged in a trigonal planar geometry with bond angles of 120°. Examples: BF3, CH2 = CH2 BH3, BCl3, propanone etc. + 2s orbital 2p, orbital 2p, orbital Fig: Formation of sp' hybrid orbitals from atomic orbitals BF3: The sp' hybridization explains the geometry of planar molecules such as BF3. Electric configuration of sB. 5B ground = 2s 2Px Bexcited = 2s 2px 2py The three outermost orbitals of B mix together to give three sp' hybrid orbitals. On the other hand, one of the p orbitals of fluorine is half filled (2p) orbitals. BFs is formed by the overlap of three half-filled sp? hybrid orbitals of boron with 2pz, orbitals of three fluorine atoms. sp hybrid orbitals shown together Two sp hybrid orbitals (Large lobes only) 180° CI Be CI Z 120° Hybridization -x + Shape of one sp' orbital Ground Excited The structure is triangular planer with bond angles equal to 120°, each. Fig: Formation of BF3 molecule through sp'-p overlap

Definition: In sp hybridization, one s-orbital (low energy and spherical and three p- orbitals (high energy and dumbbell shaped), intermix to give a new set of four orbitals of same energy and same shape called sp hybrid orbitals. Each of these hybrid orbitals is a sp hybrid orbital: These hybrid orbitals are arranged in the tetrahedral geometry with angle of 109.5° Example: CH4, NH3, H2O, NHt etc. are compounds which show sp mode of hybridization. Methane (CH4) The central atom is carbon. The electronic configuration of valence shell of 6C in its excited and hybridized states are given as follow. ground = 2s 2pg 2Px 2py crame i'm. Cexcited = Excited Chybridized = Each of these hybrid orbitals is a sp' hybrid orbital and they are arranged in the tetrahedral geometry. The energies of hybrid orbitals are lower than unhybrid orbitals. The outermost four atomic orbitals of carbon mix up to give four hybrid orbitals of same energy and same shape. Z x+ 0+ 2s orbital 2py orbital 2p, orbital Fig: Formation of sp' hybrid orbitals of carbon in methane Did you Know! S.Q. What are the lobes of sp orbitals? Ans. Each sp hybrid orbital consists of two lobes, one larger and the other smaller. For the sake of simplicity, the smaller lobe is usually not shown while representing sp? hybrid orbitals together. 120° F hybridized Shape of one sp' orbital having two lobes •X + Four sp' hybrid orbitals 2p, orbital The four new hybrid orbitals of equal energy have a tetrahedral geometry with carbon nucleus at the center. The methane (CH4) molecule is formed by the overlap of sp hybrid orbitals of carbon with Is orbitals of four hydrogen atoms separately to form four sigma bonds. The four C-H bonds, which result sp -s overlaps, are directed towards the corners of a regular tetrahedron

VSEPR Theory Introduction: The VSER model describes the shapes of molecules based on the electron pairs that surround the central atom. This model was presented by Sidgwick and Powell in 1940. Assumptions: It is based on the assumption that a molecule will take a shape such that the electronic repulsions. among the valence electrons of that atom are minimum. In order to have the minimal repulsions, the electron pairs arrange themselves at farthest possible distances. This arrangement of the electron pairs determines the geometry of the resulting molecule. Postulates (i) Both the lone pairs (non-bonded) as well as the bond pairs participate in determining the geometries of molecules. (ii) The electron pairs are arranged around the central atom so as to remain at maximum distance apart to avoid repulsions. (wii) The electrons of lone pairs occupy more space than the bond pairs. As a result, the non- bonding electron pairs exert greater repulsive forces on bonding electron pairs and, thus, tend to compress the bond pairs. The magnitude of repulsions between the electron pairs in a given molecule decreases in the following order: Repulsion Order: Lone pair-lone pair (Ip-lp) > lone pair - bond pair (Ip-bp) > bond pair - bond pair (bp-bp). An electron pair shared by two nuclei occupies less space than a lone pair bound by a single nucleus. (Iv The two electron pairs of a double bond and three electron pairs of a triple bond, contain a higher electronic charge density, but they are regarded as single pairs. (v) The presence of highly electronegative atoms with the central atom results in decreased electronic repulsions between the bond pairs, but stronger repulsion by the lone pair. Predicting the shapes of molecules • In order to illustrate this model, the central atom is named 'A'. The electron pairs around 'A' are designated as 'B' • There may be different number of electron pairs around the central atom depending upon its valency. • It gives rise to various types of molecules, such as AB2, AB3, AB4, etc. • The electron pair geometry of a molecule is determined by the total number of electron pairs around 'A' • The actual shape of the molecule is determined by the atoms excluding the lone pairs. Following table 3.4 gives the possible shapes of different types of molecules having varying numbers of bond and lone pairs. Table 3.4: Geometries and shapes of different systems of covalently bonded compounds No. of electron No. of pairs Electron pair lone around geometry respect to atoms pairs central atom 2 0 Linear 3 0 Trigonal 1 3 Trigonal Angular, v-shaped 4 Tetrahedral 1 4 Tetrahedral 4 2 Tetrahedral Angular, v-shaped Trigonal 5 0 bipyramidal 6 0 Octahedral Steps to Determine the shapes of molecules. The following steps are followed to predict the shape of a molecule. • The least electronegative atom or the element with the least number of atoms is mostly selected as the central atom. In CCla, the carbon atom is the central atom. • The total number of electron pairs around the central counted (including bond pairs and lone pairs). In CCla, the total number of bond pairs around the C atom are 4 and it has no lone pair. • The total number of electron pairs determines the electron pair geometry (as in the above table) of the molecule. • CCla has total four electron pairs and its geometry is tetrahedral. • The actual shape of the molecule is determined excluding the lone pairs (if any). Angular Shape with separation Examples of atoms 180° Linear CO2, BeClz 120° Trigonal BCl3, SO3 Less than SO2, SnCl 120° 109.5° Tetrahedral CH4, CCl4 Less than Pyramidal NH3, H3O+ 109.5° Less than H2O, OF2 109.5° Trigonal PCl5, Is 120°, 90° bipyramidal 90° Octahedral SF6, XeF4 CI CI •* C *• CI atom are Keep in mind! S.Q. How number of lone pairs of NH3 are calculated? Ans. Number of electron pairs around the central atom are calculated as in the following example: In NH3, Valence electrons in N= 5 Electrons shared by 3H = 3 x 1 = 3 Total number of electrons = 5 + 3 = 8 Total number of electron pairs =

AB type molecules (linear geometry) In such molecules, two electrons pairs around the central atom are arranged at an angle of 180° to minimize repulsions between them. Thus, they form a linear geometry as shown in figure. Example: BeCl molecule is of AB2 type with Be as the central atom and two bond pairs around it, but no lone pair. Be- CI- riel Fig: Linear shape of the BeCk molecule ABs type (Trigonal planar geometry) In AB3 type molecules, the central atom is surrounded by three electron pairs, which are arranged at maximum distance apart at an angle of 120° giving a trigonal geometry. Example: BF molecule has a trigonal planar shape with each F-B-F bond angle of 120° figure. The similar geometries are expected in the hydrides of group 3, ie. AlH3, GaH3, BH3, etc. F 120° F F-B Fig: BF3 in an AB3 system having triangular planar geometry SnClz: In this molecule, the Sn atom has 4 electrons in its outermost shell. It makes two bonds with two Cl atoms and the remaining two electrons exist as a lone pair. One of the corners of the triangle is occupied by this lone pair, giving rise to a distorted trigonal electron pair geometry (in vapor phase). The actual shape of SnCl2 is v-shaped and bond angle less than 120° due to the presence of a lone pair on Sn atom. Calculation for lone pairs on N No. of lone pairs on N= Total no. of pairs - no. of bond pairs No. of lone pairs on N= 4 - 3 = 1 Thus, N has 1 lone pair in NH3 Less than 120° Fig: Shape of SnCl2 ABs-Type with multiple bonds SO3: The molecule of SO has all the three regions occupied by S=0 bonds. The structure of SO is perfectly trigonal with each angle equal to 120° SO2: In this molecule, one corner of the triangle is occupied by a lone pair and the other two corners each by a double bond (S = O). Thus, it makes an angular or v-shaped structure just as SnCla does as in figure.

AB4 Type (Tetrahedral geometry) In this type of molecules the electron pair geometry is a regular tetrahedron. Each of the bond angles is of 109.5° (i) CH4 (Methane) In methane, four bonded electron pairs are directed from the center towards the corners of a regular tetrahedron, as following figure. On the same grounds, SiH4, GeH4, CCl4 possess the similar shape. (ii) NH3 (Ammonia) In the ammonia molecule (NH3), there are three H atoms attached to the nitrogen atom having one lone pair. Due to a lone pair, the ideal angle 109.5° is reduced to 107° as in figure. Ammonia has triangular pyramidal shape. The substitution of hydrogen in NH3 with electronegative atoms like For Cl further reduces the bond angles. (wii) NF3 (Nitrogen trifluoride) In NF3, three highly electronegative F atoms are bonded to the Natom. The first reason being the strong polarity of N-F bond, due to which N atom pulls the lone pair closer to its nucleus. The lone pair exerts a stronger repulsion on bonding electrons. Moreover, the bond pairs (N-F bonds) are closer to F atoms than N atom. The increased distances in these bond pairs make their repulsions weaker and allow the bonds to come closer. The bond angles are further compressed to 102.5°. N Fig: Distorted geometry and pyramidal shape of the NH3 molecule 120° S Less than 120° Fig: Shapes of multiply bonded SO2 and SO3 H $ 109.5° H Fig: Regular tetrahedral shape of the CH4 molecule N H- H 107° (iv) H2O (Water) The water molecule (HO) is angular or v-shaped, although it has four electron pairs around the central atom. Two of the corners of the tetrahedron are occupied by two lone pairs and the remaining two by bond pairs. But, due to the greater repulsions of the lone pairs, the bond angle is reduced to 104.5° as in figure. Fig: Water molecule has two lone pairs and is v-shaped with suppressed bond angle (v) H3O* (Hydronium ion) and NH; (amide ion) The [H3O+] ion is formed by the capture of a proton by a water molecule and it has the tetrahedral trigonal pyramidal shape. On the other hand, amid ion, [NH;], has a V-shaped structure as in figure. free im H H tetrahedral electron pair geometry Hydronium ion Fig: Shapes of hydronium (H30*) and amide (NH2) ions Quick Check 3.7 (a) Calculate the no. of lone pairs on the central atoms in the following: H2O, ICl3, I3. Ans. In H2O, central oxygen has two lone pairs as shown below tI In IC| central iodine has two lone pairs as shown below: H H 104.5° I 104.1° V-shaped amide ion Trigonal pyramidal shape H -Čl: •Cl: In I; central iodine has three lone pairs as shown below: 1) Sult: the number of bond pairs in central si aton are our as it ms four single covalent bonds with hydrogen atoms: It has no lone pair HaSe: The number of bond pairs in central Se atom are two as it forms two single covalent bonds with hydrogen atoms. It has two lone pairs. PH3: The number of bond pairs in central P' atom are three, as it forms three single covalent bonds with hydrogen atoms. It has one lone pair as well. (c) Predict the shapes and angles in SiH4, H2Se, and PH3. Ans. The shape of SiH4 is tetrahedral, and the bond angle is 109.55°. H The shape of HaSe is bent or v-shaped. and the bond angle is 91°. The shape of PH3 is trigonal pyramidal, and the bond angle is 93.5° H' (d) Predict how the bond angle in H2S would be different from that in H2O. Ans. The bond angle in H2O (104.5°) is larger than the bond angle in HS (92°) due to greater electronegativity of oxygen as compared to sulphur causing bonding electrons in H2O closer to oxygen.

ABs type molecules (trigonal bipyramidal geometry) ABs type molecules, have trigonal bipyramidal geometry. PCl5 • This is evident from the geometry of the PCs molecule given in figure. • The angles within the plane are 120º and that between the axial atom and a planar atom is 90°. :CI: - CI :C!: Fig: Lewis structure and shape of PCs molecule H S •Se H 90° 120° Tri-iodide ion [13] • It is an example of ABs system. • The central atom in this ion is the iodine atom. • There are five electron pairs around the iodine atom including two bond pairs and three lone pairs. • The lone pairs occupy the corners of the trigonal region and the iodine atoms occur vertically one being above and the second below the plane. • The actual shape of the try-iodide ion is linear and the iodine atoms are farthest apart at an angle of 180° as in figure. ABo type molecules (octahedral geometry) ABo type molecules have 6 electron pairs around the central atom and the octahedral geometry. All the ABA angles in this type of geometry are of 90° SF6 1s an AB type molecule. The Lewis structure and the shape of SF6 is given in figure. F : F Fig: Lewis structure and octahedral shape of SF6 with each angle of 90° XeF4 is an ABs type molecule with four bond pairs and two lone pairs. the two lone pairs remain at farthest distance at angle of 180° XeF4 is square planar with all F-P-F angles equal to 90° as in figure. : F Xe : F .. Fig: Lewis structure and square planar shape of XeF4 Quick Check 3.8 (a) Draw the Lewis structures for IF and IFs and predict their geometry with reference to the VSEPR model. Ans. Lewis Structure of IF3 180° Fig: Shape of try- iodide ion ([Is]) 90° FIl ". S F F .90 Excluding the lone pairs, the actual shape of lone pair is farthest at 180° 90° xe F: Geometry According to VSEPR Theory, IF has trigonal bipyramidal Geometry because it has three bonding pairs while two lone pairs. Lewis Structure of IF5 Geometry IFs has square pyramidal because central Iodine atom has five bond pairs while only one lone pair of electrons.

Applications of VSEPR Model in Drug Designing Definition of drug: Drugs are chemical substances that prevent, diagnose, or treat a disease. • Drugs interact with human body at specific points, or with specific processes called targets or substrates, through their bioactive molecules called ligands. • A receptor is mostly a protein that receives and responds to a ligand through binding. • The examples of targets in the body are receptors, such as enzymes, proteins, nucleic acids, and cellular pathways. • Each drug interacts with a specific target in the body to a different degree, this feature is called specificity of a drug or a ligand Molecular shape is an important feature that determines how a drug interacts with a biological target. Only the ligands with suitable shape can fit in the active sites of a biological target as a specific key fits in a specific lock. VSEPR model is successfully applied in determining the shapes of various biological systems, such as substrate recognition, ligand specificity or selectivity, and antibody recognition. Aspirin • It is an analgesic drug (painkiller) used for relief from pain as a primary medicine. • It with COX an interacts enzyme (cyclooxygenase) by binding to its active site through the acetyl group (-OCCH3). • The shape of COX active site and that of acetyl → Action Drug A binds to receptor Drug B cannot bind to receptor irreversible acetylating Platelet COX-1 dimer COOH COOH •OH aspirin salicylic acid Fig: An interaction of Aspirin with COX (Published after permission from Tulane University Press) group on Aspirin are compatible with each other as in figure. • The binding is successful to black COX, which is a cause of the pain in the body. MOLECULAR ORBITAL THEORY (MOT)

The molecules orbital approach explains the results of quantum mechanical calculations for covalent bonding. The postulates of this theory: (i) Overlapping of orbitals: During the formation of a molecule, the atomic orbitals of the combining atoms overlap to form new orbitals called 'molecular orbitals', which are characteristic of the whole molecule. (ii) Polarization of orbitals: The atomic orbitals overlap with the lobe having suitable sign of the wave function of the orbital. For example, one lobe of a p orbital is given the + sign and the other is marked with '-'sign. (wii) Types of molecular orbitals: Two atomic orbitals overlap to form two molecular orbitals. When same sign orbitals overlap, the bonding molecular orbital (o or t) is formed that has lower energy than the parent atomic orbital, while with opposite signs, high energy anti-bonding molecular orbital (o * or *) is formed, that has higher energy than the parent atomic orbitals. (iv) Bond order: The number of bonds formed between two atoms after the atomic orbitals overlap, is called the bond order and is taken as half of the difference between the number of bonding electrons (say a) and antibonding electrons (say b). Bond Order = No. of Electrons in B. M.O) - (No. of Electrons in A. B. M.O) s-s Overlap: H2 molecule: In the formation of H2 molecule, two s orbitals of H atoms combine to give two molecular orbitals. The bonding molecular orbital is symmetrical about the axis joining the nuclei of the bonded atoms (nuclear axis), while the anti-bonding molecular orbital has the electron density away from the nuclei of the overlapping atoms. H H H2 O* 1s 1s is Energy Ols a Energy Diagram Fig: Formation of bonding and anti-bonding orbitals for the H2 molecule O* 1s MOz antibonding molecular orbital Ols MOZ bonding molecular orbital Electron Probability Distribution (a) Head on approach The p-orbitals of the two atoms approach along the same axis (say x-axis). This combination of atomic orbitals gives rise to o (2px) bonding and o* (2px) antibonding orbitals. (b) Sideways Approach • When the axes of two p-orbitals (i:e py or pz orbitals) are parallel to each other, they interact to form a molecular orbitals. • The bonding molecular orbitals m(2py) or I(2pz) have zero electron density on the nuclear axis (called the nodal plane). • Anti-bonding molecular orbitals * (2py) and * * (2pz) have the least electron density in the inter-nuclear region as in figure. П2p* 88 2p. - 2p. Antibonding п2p Bonding 2p, + 2p, Fig: Formation of a MOs from py and pz orbitals

Molecular orbital diagram of H2 • In the formation of H2 molecule, two s orbitals of H atoms combine to give two molecular orbitals. • The bonding molecular orbital is symmetrical about the axis joining the nuclei of the bonded atoms (nuclear axis), while the anti-bonding molecular orbital has the electron density away from the nuclei of the overlapping atoms. • One of these molecular orbitals is the low-energy bonding molecular orbital (o 1s) as there are only two electrons to be accommodated, these enter the lower energy o ls B.M.O. o* Is remains empty. • The electrons in the bonding molecular orbital have opposite spin as shown in figure. (No. of Electrons in B. M.O) - (No. of Electrons in A. B.M.O) Bond Order = 2H(1s') → H2(o.15) 02p 2px 2px Antibonding + • •2p Bonding 2px 2Px Fig: Head on overlap of p orbitals T2p Bonding 2p, + 2py. П2p*. 2p, - 2py Antibonding 2 - 0 = = 1 2 2 H H H2 1s Energy OIs • -b Energy Diagram Fig: Formation of bonding and anti-bonding orbitals for the H2 molecule (ii) Molecular orbital diagram of He (Helium) The electron configuration of He is 1s2 For a successful formation of Hez molecule, 1s orbitals of two He atoms must combine to form bonding (o 1s) and anti-bonding (o*Is) orbitals as shown in figure. • Out of four electrons, two enter the bonding molecular orbital (o 1s) and the remaining two occupy the antibonding (o* 1s) molecular orbital. • But on calculation we discover that the bond order for Hez is zero. Hence, Her molecule is not formed. = 0 Bond order of He2 = 2-2 (wii) Molecular orbital diagram of Nitrogen (N2) ", 2s? Electronic configuration of N is N= 1s' The molecular orbital diagram of N2 based on this electron configuration is shown in the figure below. The valence shell 2s on both N atoms give o. 2s, o* 2s orbitals; whereas, 2p orbitals give six molecular orbitals give six molecular orbitals which are arranged in the increasing order of energy as: 02sN ›N= 0(2p,) 2p, 2p, 2p, 2p„2p, 2p2 A.O A.O B.M.O * (2p.) T (2p,) Energy A.B.M.O O'(2s) - N 2s 2s A:O A.O 0(2s) B.M.O Fig: Molecular orbital diagram of N2 molecule (wii) Molecular orbital diagram of oxygen Oг The formation of molecular orbitals in oxygen molecule is shown as follows in figure. The bond energy of the molecular orbitals can be arranged as: 025<0"25<02px

(i) Lewis structure of F2 (ii) Structure of F2 by VBI Priors so a around more or wo murine Because both have some value of electronegativity. 00-0000-00 2p 2p Fluorine Fluorine' Fig: Overlap of p and p orbitals to form F2 molecule according to VBT (wii) Formation of Fz according to MOT 9F = 15? ?, 25 2, 2p5 9F = 2p,2P, 2Pz. .O • Energy 9F = 2s A.O Fig: Molecular orbital diagram of F2 molecule Bond order of F2 ≤ 6=4 Bond order of Fy =1 XX F, O(2px) A.B.M.O it (2p,) T (2p,) - 9F П (2p,) T (2p.) 2p, 2p, 2p, B.M.O 0(2px) A.B.M.O O(2s) - 9F 2s 0(2s) в.M.O 2