Chemical Bonding Revision Notes 2024 PDF

Click on this button Click on this button CHEMICAL BONDING  QUIZRR 3 1. INTRODUCTION A chemical bond is the physical process responsible for the attractive interactions b...

Click on this button Click on this button CHEMICAL BONDING  QUIZRR 3 1. INTRODUCTION A chemical bond is the physical process responsible for the attractive interactions between atoms and molecules, and that which confers stability to diatomic and polyatomic chemical compounds. In general, strong chemical bonding is associated with the sharing or transfer of electrons between the participating atoms. Molecules, crystals and diatomic gasesăindeed most of the physical environment around usăare held together by chemical bonds, which dictate the structure of matter. Bonds vary widely in their strength which is associated both with the energy required to break them, and the forces they exert on the atoms they hold together. Generally covalent and ionic bonds are often described as „strong‰, whereas hydrogen bonds and van der Waal’s bonds are generally considered to be „weak,‰ although there exist overlaps in strength within these bond classes. Since opposite charges attract via a basic electromagnetic force, the negatively-charged orbiting the nucleus and the positively-charged protons in the nucleus attract each other. Also, an electron positioned between two nuclei will be attracted to both of them. Thus, the most stable configuration of nuclei and electrons is one in which the electrons spend more time between nuclei, than anywhere else in space. These electrons cause the nuclei to be attracted to each other. However, this assembly cannot collapse to a size dictated by the volumes of these individual particles. Due to the matter wave nature of electrons and their relatively smaller mass, they occupy a very much larger amount of volume compared with the nuclei, and this volume occupied by the electrons keeps the atomic nuclei relatively far apart, as compared with the size of the nuclei themselves. 1. 1 C ause of C hemical C omb inat ion The atoms interact with each other on account of the following reasons : (i) Decrease in energy : It is a fundamental truth that all natural systems tend to lose potential energy and become more stable. Other things being equal, a system that has stored potential energy is less stable than a system that has none. It is an observed fact that a bonded state is more stable than unbonded stable. This is due to the fact that the bonded state has lower potential energy than unbonded state. Hence, when two atoms approach each other, they combine only under the condition that there is a decrease in potential energy. When two atoms approach each other, new forces of attraction and repulsion come into play. The forces of attraction are between the nucleus of one atom and the electrons of the other. The forces of repulsion are between two nuclei as well as between the electrons of the two atoms. If the net result is attraction, the total potential energy of the system decreases and a chemical bond results. No chemical bonding is possible if net result is repulsion. (ii) Lewis octet rule : The noble gases are known for their lack of chemical activity. There are no known compounds of helium, neon and argon. Why are these elements so unreactive CHEMICAL BONDING  4 QUIZRR towards other elements ? All these elements have electronic structures that consist of filled outermost sheels. Except for helium, whose electronic configuration is 1s2, the s-and p- subshells of the highest energy level contain a total of eight electrons. It is, therefore, concluded that s2p6 configuration in the outer energy level constitutes a-structure of maximum stability and therefore, of minimum energy. The atoms of all elements when enter into chemical combination try to attain noble gas configuration, i.e., they try to attain either 2 electrons (when only one energy shell) or 8 electrons in their outermost energy level which is of maximum stability and hence of minimum energy. The tendency of atoms to achieve eight electrons in their outermost shell is known as Lewis octet rule. Octet rule was the basis of electronic theory of valency. 1. 2 L ewis Symb ols of E lement s Chemical bonding mainly depends on the number of electrons present in the outermost energy level. These electrons are termed as valency electrons. The electronic configuration of sodium (Na) is 2, 8, 1 and that of sulphur has (S) 2, 8, 6. Thus, sodium has one valency electron while sulphur has six valency electrons. The valency electrons in atoms are shown in terms of Lewis symbols. To write Lewis symbol for an element, we write down its symbol surrounded by a number of dots of crosses equal to the number of valency electrons. Paired and unpaired valency electrons are also indicated. The Lewis symbols for hydrogen, sodium, nitrogen, oxygen and chlorine may be written as : H Na N O Cl Generalised, Lewis symbols for the representative elements are given in the following table : 1 2 13 14 15 16 17 Group IA IIA IIIA IVA VA VIA VIIA Lewis symbol X X X X X X X 1. 3 E lect r onic t heor y of Valency The theory of valency explains chemical combination in terms of electrons. The theory was developed independently by W. Kossel and G.N. Lewis (1916) and extended by Irving Langmuir (in 1919). The chemical behaviour of an atom is determined to a large extent by the number and arrangement of electrons in the outer orbitals of the atom. Only these electrons are involved in chemical combination and so these are called the valence electrons. CHEMICAL BONDING  QUIZRR 5 C omp let ed E lect r on O ct et or Dup let Group 0 of the periodic table contains the noble gases. With the exception of helium which has a 1s2 electron arrangement others have ns2 np6 configuration in the outer orbitals. He 1s2 Ne 1s2 2s22p6 Ar 1s2 2s2 2p6 3s2 3p6 Kr 1s2 2s2 2p6 3s2 3p6 3d 10 4s2 4p6 Xe 1s2 2s2 2p6 3s2 sp6 3d10 4s2 4p6 4d 10 5s2 5p6 Since the atoms of the noble gases were not known to form chemical bonds, it was argued that the presence of 8 electrons (an electron octet) in the valence shell makes the atom stable. Therefore all other atoms must undergo bonding by gaining or losing or sharing electrons so as to acquire the electronic configuration of the nearest inert gas. The presence of 8 electrons gives the name octet rule to this concept. In the case of the first few elements such as hydrogen, lithium and beryllium the atoms combine in such a way as to attain the stable structure of helium with 2 electrons (duplet) in its only one valence shell. There are, however, many exceptions to the octet rule. Also compounds of noble gases, especially xenon, have been synthesized. Hence, bond formation can take place in 3 ways. (1) By losing or accepting electrons as known as elec trovalent bond. (2) By equal contribution of electrons from two atoms and these electrons are then shared equally to form covalent bond. (3) Contribution is made by one atom and both the electrons are shared equally by 2 atoms to form co-ordinate bond. Thus it can be summarised as ÂÂThe union of two or more atoms involving redistribution of electrons in their outer shells (either by transference or sharing) in such a way so that all the atoms acquire the stable noble gas configuration of the minimum energy is known as electronic theory of valencyÊÊ. 2. I O NI C BO ND An Ionic bond is a type of chemical bond that involves a metal and a non-metal ion through electrostatic attraction. In short, it is a bond formed by the attraction between two oppositely charged ions. The metal donates one or more electrons, forming a positively charged ion or cation with a stable electron configuration. These electrons then enter the non metal, causing it to form a negatively charged ion or anion which also has a stable electron configuration. The electrostatic attraction between the oppositely charged ions causes them to come together and form a bond. CHEMICAL BONDING  6 QUIZRR For example : Sodium Chloride : The free sodium atom has one valence electron (electronic configugation 2, 8, 1) i.e. 3s1. Whereas the chlorine atom has seven valence electrons (electronic configuration 2, 8, 7) i.e. 3s2 3p5. In forming an ionic bond, the sodium atom loses its valence electron which is accepted by chlorine atom. As a result sodium achieves noble gas configuration of Neon (2, 8) and becomes positive ion (Na+) chlorine achieves moble gas configuration of argon (2, 8, 8) and acquires a negative charge (Clă). The attraction between sodium ion and chloride ion is an ionic bond. A short hand way of showing the formation of sodium chloride using dot symbols. ă ï Na Cl = + Na ï Cl or Na+ Clă (2, 8, 1) (2, 5, 7) (2, 8) (2, 8, 8) C ond it ions for For ming E lect r ovalent or I onic Bond The following conditions favour the formation of an electrovalent bond. (i) Number of valency electrons : One atom should possess 1, 2 or 3 valency electrons while the other atom should have 5, 6 or 7 valency electrons. Thus, they can also form electrovalent bonds but do not acquire inert gas configuration always. (ii) Difference in electronegativity : The formation of an electrovalent bond will be easier if the difference in the electronegativities of the two atoms is high. A difference of about 2 is necessary for the formation of an electrovalent bond. The electronegativity of sodium is 0.9 and that of fluorine is 4.0. Since the difference is 3.1 both will readily form an electrovalent bond. (iii) Overall decrease in energy : In the formation of an electrovalent bond, there must be overall decrease in energy, i.e., energy must be released. Energy changes are involved in the following steps : This is called as Born-Haber cycle. Na(s) + 1 + ă 2 Cl2 (g) Na Cl (s) Sub. BE Na(g) F(g) Lattice Energy IE EA Na+ + Cl ă(g) CHEMICAL BONDING  QUIZRR 7 Where sub is the sublimation energy, BE is the bond energy, IE is ionization energy and EA is electron affinity. (1) Sublimation Energy : The energy required to convert one mole of a solid into gas at constant pressure and temperature. (2) Ionization energy : The energy required by the metal atom to release its valence electron. For example, sodium requires small energy to give up its loosely field electron and form Na+ ion. Ionization energy of an element with a single electron in its valence still is less than that with two electrons. In going across a period of the periodic table from left to right, I.E. increases and the formation of the cation is less likely. On going down a group, the outermost electron gets further away from the nucleous, and hence is more easily removed i.e. I.E. decreases, the formation of the cation becomes more likely lower the value of ionisation energy of an atom, greater will be the ease of formation of the cation from it. (3) Bond energy : The energy required required to break the molecule into atoms is called bond energy or dissociation energy. (4) Electron affinity : the amount of energy released when an atom takes up an electron and forms an anion. For example, chlorine takes up an electron from the Na atom and forms Clă ion. Non-metals of groups VIA and VIIA have high electron affinity and can form ionic bonds. In going across a period from left to right, electron affinity (energy released) increases and so the formation of negative ion becomes more likely. On going down a group, electron affinity decreases and so the formation of anion becomes less likely. Thus Higher the value of electron affinity of the atom, greater the ease of formation of the anion form it, i.e., other atom should have high value of electron affinity. (5) Lattice energy : Cation and anion attract each other by electrostatic force of attraction to give a molecule A+Bă. Since the electrostatic field of a charged particle extends in all ă + ă + ă + ă ă + + ă + + ă ă + Crystals of ionic compounds CHEMICAL BONDING 8 QUIZRR directions, a positive ion is surrounded by a number of negatively charged ions while each negative ion similarly surrounded by a number of positive ions. These cations and anions arrange systematically in an alternating cation-anion pattern as shown in Fig. This is called a crystal lattice. This process of clustering ions increases the force of attraction and thus potential energy decreases. The energy released when the requisite number of positive and negative ions are condensed into crystal to form one mole of the compound is called lattice energy. The value of lattice energy depends on the charges present on the two ions and the distance between them. According to CoulombÊs law, the force of attraction (F) between two oppositely charged ions in air with charges equal to q1 and q2 and separated by a distance d is given by, 1 q1 q2 F 4 0 K d 2 where d is equal to sum of ionic radii of the two ions and K is dielectric constant of medium. 1 q1 q2 F 4 0 K (r +  r ă )2 A B The value of F increases if (i) q1 and q2 are high and (ii) (rA ă  rBă ) is small. The stability of the ionic compound and the strength of the ionic bond depends on the value of F. Higher than value of F, greater shall be the stability of the ionic compound and hence greater shall be the strength of the ionic bond. For example, NaCl is more stable than CsCl as (rCs ă  rCl ă ) is less than (rA ă  rBă ). MgO is more stable than NaCl as the product q1q2 is four times more in MgO than NaCl. Conclusions : (i) An ionic bond is purely electrostatic in nature. (ii) Its formation is favoured by : (a) Low ionisation potential (I.P.) of the element that forms a cation on losing electron(s). The element should be metal, i.e., electropositive in nature. (b) High electron affinity (E.A.) of the element that forms an anion on gaining electron(s). The element should be non-metal, i.e., electronegative in nature. (c) High lattice energy (L.E.) : The energy release when isolated ions form a crystal. The value of lattice energy depends on the charges present on the two ions and distance between them. It shall be high if charges are high and ionic radii are small. (d) The summation of three energies should be negative, i.e., energy is released. I.P. + E.A. + L.E. = ăive CHEMICAL BONDING  QUIZRR 9 (iii) Greater the difference of electronegativity between two atoms, higher will be the possibility of ionic bond formation. (iv) Electrovalency The capacity of an element to form electrovalent or ionic bond is termed as electrovalency. The capacity is measured in terms of the electrons lost or accepted. Thus, electrovalency of an element is equal to the number of electrons lost by an atom of the element or gained by the atom of the element as to acquire inert gas configuration. The elements which lose electron or electrons show positive electrovalency and the elements which gain electron or electrons who negative electrovalency. Generally positive and negative signs are not used in practice and only the number is taken to represent electrovalency. No. of electrons Change in electronic Element lost or gained Electrovalency configuration by an atom Na 1 (lost ) 1 (Monovalent) 2, 8, 1 to 2, 8 (Na+) K 1 (lost) 1 (Monovalent) 2, 8, 8, 1 to 2, 8, 8 (K+) Mg 2 (lost) 2 (Divalent) 2, 8, 2 to 2, 8 (Mg++) Ca 2 (lost) 2 (Divalent) 2, 8, 8, 2 to 2, 8, 8, (Ca++) Al 3 (lost) 3 (Trivalent) 2, 8, 3 to 2, 8 (AI+++) F 1 (gained) 1 (Monovalent) 2, 7, to 2, 8 (Fă) Cl 1 (gained) 1 (Monovalent) 2, 8, 7 to 2, 8, 8 (Clă) O 2 (gained) 2 (Divalent) 2, 6 to 2, 8 (Oă ă) S 2 (gained) 2 (Divalent) 2, 8, 6 to 2, 8, 8 (S ă ă) N 3 (gained) 3 (Trivalent) 2, 5 to 2, 8 (N ă ă ă) 3. Det er minat ion of lat t ice ener gy The lattice energy of an ionic solid is determine experimentally by a process known as Born-Haber cycle. According to this theory following steps are involved : (i) The reactant are converted into gaseous state. (ii) The gaseous atoms are converted into ions. (iii) The gaseous ions are combined to form ionic compound. For example the formation of KF can occur either directly or in steps. Direct : 1 K( s)  F2 ( g) KF ( s) 2 H or heat of formation = ă 562.6 kJ/mol CHEMICAL BONDING 10 QUIZRR Steps : H = ă562.6kJ/mol K(s) + 1F (g) KF(s) 2 2 Heat of Sublimation = +89.6 kJ/mol 1/2(Heat of dissocation) = 79.1 kJ/mol (HS) (HD) ăLattice Energy (U) K(g) F(g) ăeă +eă I.P. = +419.0 kJ/mol E.A. = ă332.6 kJ/mol K+(g) + F ă(g) On the basis of above cycles (or constant heat summation rule) Heat of formation = Heat of atomization + Heat of dissociation + I.P. + E.A. + Lattice Energy HD H = HS   I.P.  (E.A.)  ( U) 2 ă 562.6 = 89.6 + 79.1 + 419.0 ă 332.6 ă U U = 817.7 kJ/mol Lattice energy of KF = ă U = ă 817.7 kJ/mol M et hod of W r it ing For mula of an I onic C omp ound In order to write the formula of an ionic compound which is made up of two ions (simple or polyatomic) having electrovalencies x and y respectively, the following points are followed: (i) Write the symbol of the ions side by side in such a way that positive ion is at the left and negative ion at the right as AB. (ii) Write their electrovalencies in figures on the top of each symbol as AxBy. (iii) Divide their valencies by H.C.F. x y (iv) Now apply crisis cross rule as i.e., formula AyBx. A B Examples : 2 1 Calcium chloride Ca Cl = CaCl2; 3 2 Aluminium oxide O = Al2O3; CHEMICAL BONDING  QUIZRR 11 1 3 Potassium phosphate PO4 = K3PO4; 2 3 Magnesium nitride N = Mg3N2; 1 1 Calcium oxide O = CaO; 1 2 Ammonium sulphate SO4 = (NH4)2SO 4 G ener al C har act er ist ics of I onic C omp ound s (i) Generally ionic compounds are hard solids. As single ions of a metal are not associated in the solid with single ions of a non-metal, separate units of ionic compounds do not exist. It is, therefore, wrong to talk of a molecule of an ionic compound. The formula only indicates the ratio of number of ions and the crystal consists of a very large number of oppositely charged ions. Thus in NaCl crystal each Na+ ion is surrounded by 6Clă ions and vice versa (in an octahedral arrangement). The attraction between Na+ and Clă ions is quite large. (ii) As a good deal of thermal energy is required to overcome the large electrostatic forces of attraction in an ionic crystal, ionic compounds have high melting and boiling points. (iii) Ionic compounds are commonly soluble in water and other polar solvents (which separate the ions). They are practically insoluble in organic solvents such as benzene, carbon tetrachloride, etc., as there is no attraction between ions and the molecules of the non- polar liquids. (iv) Ionic compounds are electrolytes. In the presence of a ionizing solvent such as water, the electrostatic forces between the ions are so greatly reduced that the ions got separated. (This is due to the electrostatic attraction between the ions and the polar molecules of the solvent.) The free ions in solution conduct electricity and on passing a current, the ionic compound undergoes chemical decomposition (called electrolysis). When an ionic compound is melted, the crystal lattice structure is broken and free ions are produced. It is the free movement of ions which makes an ionic compound a conductor and to undergo electrolysis in the molten condition. CHEMICAL BONDING  12 QUIZRR (v) When an ionic compound dissolves in water, the ions get solvated (in this case hydrated). The energy released is called solvation energy. Insoluble ionic compounds (eg., sulphates, phosphates and fluorides of Ca, Sr and Ba) have very high lattice energies and the solvation energy of the constituent ions is insufficient to counteract the high lattice energies and make them soluble. (vi) The chemical properties of an ionic compound are the properties of its constituent ions. Thus all chlorides give the characteristic reactions of the chloride ion (reactions with conc. H2SO4, AgNO3 solution, etc.). All acids which contain H+ ions give the same reactions (change blue litmus to red, effervesce with a carbonate, etc.). (vii) Reactions between solutions of ionic compounds are almost instantaneous, because they are reactions between ions (and do not involve the breaking up of bonds as in covalent compounds, q.v.). For example, when silver nitrate solution is added to sodium chloride solution, silver chloride is immediately precipitated. The reactions may be represented thus : Na+ + Clă + Ag+NOă3 AgCl + Na+ + NO3ă Var iab le E lect r ovalency Certain elements (metals) show more than one electrovalency in their electrovalent compounds. The atoms of these elements lose different number of electrons under different conditions, thereby showing variable electrovalency. The following are two reasons for variable electrovalency : (i) Instability of the core : The residue configuration left after the loss of valency electrons (electrons present in the outermost energy level) is called a core or kernel. In the case of the atoms of transition elements, ions formed after the loss of valency electrons do not possess a stable core as the configuration of outermost shell is not ns2np6 but ns2 np6 d 1 to 10. The outershell generally loses one or more electrons giving rise to metal ions of higher valencies. The variable valencies of iron can be explained on this basis. Fe 26 2, 8, 8 + 6, 2 3s2 3p6 3d 6, 4s2 Fe2+ 24 2, 8, 8 + 6 3s2 3p6 (Not stable) Fe3+ 23 2, 8, 8 + 5 3s2 3p6 3d 5 (Stable) Thus, iron shows +2 and +3 valencies. The compounds, in which iron is in trivalent state, i.e., ferric compounds are more stable than the compounds in which iron is in divalent state (ferrous compounds). (ii) Inert pair effect : some of heavier representative elements of third, fourth and fifth groups having configurations of the outermost shell ns2 np1, ns2np2 and ns2np3 show valencies with a difference of 2, i.e., (1; 3), (2; 4), (3; 5), respectively. In the case of lower valencies, only the electrons present on p-subshell are lost and ns2 electrons remain intact. The reluctance CHEMICAL BONDING  QUIZRR 13 of s-electron pair to take part in bond formation is known as the inert pair effect. A partial explanation of the inert pair effect is due to the fact that outer ns2 electrons penetrate to some extent to penultimate orit consisting 18 electrons thereby attracted strongly towards nucleus. Inert pair effect increases in a group as the atomic number increases. Tl (III group) 2, 8, 18, 32, 18, 6s2 6p1 Tl + 2, 8, 18, 32, 18, 6s2 Only 6p electron is lost Sn (IV group) 2, 8, 18, 5s2 5p2 Sn2+ 2, 8, 18, 18, 5s2 Only 5p electrons are lost Pb (IV group) 2, 8, 18, 32, 18, 6s2 6p2 Pb+ 2, 8, 18, 32, 18, 6s2 Only 6p electrons are lost 2 3 Bi (V group) 2, 8, 18, 32, 18, 6s 6p 3+ Bi 2, 8, 18, 32, 18, 6s2 Only 6p electrons are lost When sufficient energy is available, the s-electrons also enter into bond formation and higher valencies are observed. This tendency to show higher valencies is less in the case of T1, Pb and Bi but more in the case of comparatively lighter elements such as In, Sn, Sb, etc. 4. CO VAL ENT BO NDI NG Certain elements which have high ionisation energies are incapable of transferring electrons and other having low electrons affinities, fail to take up electrons. The atoms of such elements share their elections with the atoms of other elements (and sometimes among themselves) in such a manner that both the atoms form complete outer shell. In this manner they achieve stability. Such an association through sharing of electron pairs among atoms of different or of same kinds is known as Covalent Bond. This was proposed by G.N. Lewis. The covalent bonding can be achieved in two ways : Ć Sharing electrons between atoms of same kinds, formation of H2, Cl2, O2 etc. Ć sharing of electrons between atoms of different kinds, formation of HCl, CO2, H2O, CH4 etc. The bond is, non-polar bond. Covalent bond may be single, double or a triple bond. Double and triple covalent bonds are called multiple covalent bonds. Single covalent bond is formed by sharing of only one electron pair. This bond is represented by single dash (·). Double and triple covalent bonds are formed when atoms bonded together shared two or three electron pair, respectively. These bonds are represented by double dash (==) and triple dash ( ) respectively. Some examples of covalent bonding are given below : CHEMICAL BONDING 14 QUIZRR 1. Formation of hydrogen molecule : In the formation of hydrogen molecule, each hydrogen atom contributes one electron and then the pair is shared between two atoms. Both the atoms acquire stable configuration of helium. Thus, the molecule consists of one single covalent bond. ï ï H + H H H or H·H 2. For mat ion of F 2 and ot her like molecules It is formed by combination of two F atoms by sharing one electron each. It is represented by Lewis dot structure as follows : ï ï ï ï F + F F ï ïFï or FăF ï ï ï ï ï ï ï Lewis dot structure shows only the electrons of outer most shell. Note that both of F atoms now have 8 electrons each in their outer shell. Note : Pair of electrons depicted as Ć Ć or xx is called as lone pair (lp) and that by x Ć is called as shared pair or bonded pair (bp). Lone pair is the pair of electrons belonging to one atom only. 3. For mat ion of H C l molecules Both hydrogen and chlorine contribute one electron each and then the pair of electrons is equally shared. Hydrogen acquires the configuration of helium and chlorine acquires the configuration of argon. H ï Cl H ï Cl or H · Cl (1) (2, 8, 7) (2) (2, 8, 8) 4. For mat ion of wat er molecule Oxygen atom has 6 valency electrons. It can achieve configuration of neon by sharing two electrons, one with each hydrogen atom. CHEMICAL BONDING  QUIZRR 15 ï ï ï ï ï H + ï Oï + H H O H or H·O·H ï ï ï ï (1) (2, 6) (1) (2) (2, 8) (2) 5. Fo r m a t io n o f m e t h a n e Carbon has four electrons in the valency shell. It can achieve stable configuration of neon by shring four electrons with four hydrogen atoms, one with each hydrogen atom. Each hydrogen atom acquires helium configuration. H H H ï · ï H ï C ï H H C H H·C·H ï ï · ï H ï (2, 4) H H (1) (2, 8) (2) 6. Fo r m a t i o n o f O 2 m o l e c u l e s Each oxygen atom contributes two electrons and two pairs of electrons are then shared equally. Both the atoms acquire configuration of neon. ï ï ï ï ï ï O + O O O or O·O ï ï ï ï ï ï ï ï ï ï (2, 6) (2, 6) (2, 8) (2, 8) 7. For mat ion of N 2 molecules Nitrogen atom has five valency electrons. Both nitrogen atoms achieve configuration of neon by sharing 3 pairs of electrons, i.e., each atom contributes 3 electrons. N· ï ï ï ï ï ï N N N N or ·N ï ï ï ï (2, 5) (2, 5) (2, 8) (2, 8) CHEMICAL BONDING  16 QUIZRR C ov a len cy It is defined as the number of electrons contributed by an atom of the element for sharing with other atoms as to achieve noble gas configuration. It can also be defined as the number of covalent bonds formed by the atom of the element with other atoms. The usual covalency of an element except hydrogen (which has covalency 1) is equal to (8-group number of MendeleeffÊs table to which an element belongs). This is true for the elements belonging to IV, V, VI and VII groups. Element Group (8ăGroup number) Covalency C IV 8 ă 4 = 4 4 Si IV* 8 ă 4 = 4 4 N V 8 ă 5 = 3 3 P V 8 ă 5 = 3 3 O VI 8 ă 6 = 2 2 S VI 8 ă 6 = 2 2 F VII 8 ă 7 = 1 1 Cl VII 8 ă 7 = 1 1 Generally, the covalency of an element is equal to the total number of unpaired electrons in s- and p-orbitals of the valency shell. 1s Covalency Hydrogen has one unpaired orbital One Fluorine has only one unpaired orbital One Oxygen has two unpaired p-orbitals Two Nitrogen has three unpaired p-electrons Three These four elements do not possess d-orbitals in their valency shell. However, the elements having vacant d-orbitals in their valency shell like P, S, Cl, Br, I, show variable covalency by increasing the number of unpaired electrons under excited conditions, i.e., unpairing the paired orbitals and CHEMICAL BONDING  QUIZRR 17 shifting the electrons to vacant d-orbitals. [Such a shifting is not possible in the case of H, N, O and F because d-orbitals are not present in their valency shell.] Phosphorus shows 3 and 5 covalencies. 3s 3p Phosphorus atom in ground state Three unpaired electrons covalency = 3 Phosphorus atom in excited state Five unpaired electrons covalency = 5 Sulphur atoms shows 2, 4 and 6 covalencies. 3s 3p 3d Sulphur atom in ground state Two unpaired electrons covalency = 2 3s 3p 3d Sulphur atom in excited state, (a) When p-orbital is unpaired (first excited state.) Four unpaired electrons covalency = 4 (b) When s-and p-orbital is unpaired (first excited state). Six unpaired electrons covalency = 6 CHEMICAL BONDING 18 QUIZRR Chlorine shows 1, 3, 5 and 7 covalencies. 3s 3p 3d Chlorine atom in ground state One unpaired electrons covalency = 1 Chlorine atom in excited state, (a) First excited state when p-orbital is unpaired Three unpaired electrons covalency = 3 (b) Second excited state when two orbitals are unpaired Five unpaired electrons covalency = 5 (c) Third excited state when one s-and two p-orbitals are unpaired Seven unpaired electrons covalency = 7 Thus, variable covalency is shown by those elements whose atoms have vacant d-orbitals in their valency shell. G ener al C har act er ist ics of C ovalent C omp ound s (i) In a purely covalent compound the electrons in the bond are shared equally between the atoms linked by the bond; the resultant particles formed are not electrically charged. So, separate molecules of the covalent compounds exist. Covalent compounds may therefore be expected to be gases or low boiling liquids or soft, low melting solids at ordinary temperature. In the solid state they may be amorphous or present as molecular crystals the molecules being held together by what are called weak van der Waals forces of attraction. (ii) Since the molecules are held together by weak van der Waals forces, covalent compounds (except those consisting of giant molecules) have low melting and boiling points; very little thermal energy is needed to overcome these weak intermolecular forces. CHEMICAL BONDING  QUIZRR 19 (iii) They are non-electrolytes, i.e., they do not contain ions. Even in giant molecules such as diamond there are no free electrons. So they are very poor conductors of electricity. (iv) They are generally soluble in organic (non-polar) solvents such as benzene or carbon tetrachloride but are insoluble in water or other ionizing solvents. (The solubility of covalent compounds is very much dependent on the size of the molecules; giant molecules are practically insoluble in nearly all solvents.) (v) Reactions between covalent compounds are slow and often incomplete and reversible. This is so because the reaction involves breaking and making of bonds i.e., energy considerations are involved for reactants, activated complexes and products. (vi) A covalent bond is a space-directed bond and it may exhibit isomerism. Polar C ovalent Bond s – E lect r onegat ivit y The shared pair of electrons may be shared equally between two atoms; then the covalent bond is said to be non-polar. Equal sharing occurs between identical atoms, as in H ă H or Cl ă Cl (i.e., in homonuclear molecules) or between identical atoms with identical neighbours as in H3CăCH3. When the two bonded atoms are dissimilar (i.e., in heteronuclear molecules) the sharing is unequal. For example a chlorine atom has a greater electron attracting power than a hydrogen atom; so in H ă Cl, the shared pair of electrons are drawn more towards chlorine and away from hydrogen. The result is separation of charges within the molecule, the chlorine end acquiring a slight +  negative charge and the hydrogen end a slight but equal positive charge H Cl. Such covalent bonds are said to be polar (i.e. bonds formed by sharing a pair of electrons between two atoms but displaced towards the nucleus of one of the bonded atoms). The net tendency of a bonded atom in a covalent to attract the shared pair of electrons towards itself is known as electronegativity. (This word does not mean the actual content of the electric charge, but just the tendency to acquire it in a molecule). Thus F is highly electronegative, but Fă which has already an extra electron is not. To assess the tendency of an atom of a given element to attract electrons towards itself in a covalent bond, relative electronegativity values are used. (i) Electronegativity value increase across a period and decrease down a group. (ii) Smaller atoms have greater electronegativity than larger ones and so they attract electrons more towards them than larger ones. Alkali metals have low electronegativities and halogens high electronegativities. (iii) Atoms with nearly filled shells of electrons (e.g., halogens) have greater electronegativity than those with sparsely occupied shells. CHEMICAL BONDING 20 QUIZRR (iv) Elements with low electronegativity values such as Cs (0.8) and Rb (0.8) tend to form positive ions, i.e., these are metals. Elements with high electronegativity values such as F(4.0) and O (3.5) tend to form negative ions, i.e., these are non-metals. (v) Electronegativity value may be used to make rough predictions of the type of bonding to be found in a compound. The larger the difference between electronegativity values of two combining atoms, the more polar the covalent bond. If the difference is greater than 2, the greater the chance for ionic bonding (i.e., the chance of covalent bond assuming 100% ionic character). From this point of view ionic may be considered to be an extreme case of a polar bond (with total separation of charges). If the difference between the electronegativities of the combining atoms is zero or small, the bond is essentially non-polar. Let XA and XB represent the electronegativities of two atoms A and B. If XB ă XA = 1.7, the covalent bond A ă B is said to have 50% ionic character. On the basis, the % ionic character in some typical bonds are calculated (Table). These calculations are very qualitative. % Ionic Character of Bonds C ă H N ă H O ă H F ă H 4% 10% 39% 60% C ă F C ă Cl C ă Br C ăl 43% 11% 3% 0% 5. THE LEWIS THEORY The octet rule : The Lewis theory gave the first explanation of a covalent bond (in terms of electrons) that was generally accepted. If two electrons are shared between two atoms, this constitutes a bond and binds the atoms together. For many light atoms, a stable arrangement is attained when the atom is surrounded by eight electrons. This octet can be made up from some electrons which are totally owned and some electrons which are ÂsharedÊ. Thus atoms continue to form bonds until they have made up an octet of electrons. This is called the Âoctet ruleÊ. The octet rule explains the observed valencies in a large number of cases. For example nitrogen atom has 5 outer electrons and in NH3 it shares three of these, forming three bonds and thus attaining an octet, hydrogen has only one electron and by sharing one electron of two electrons. N + 3[H*] H N H * * * H CHEMICAL BONDING  QUIZRR 21 E xcep t ion t o oct et r ule It is observed that atoms in some molecules could exist with some other number of electrons in their valence shells, rather than 8 electrons without affecting the stability. We are discussing some of them. BF 3 Boron atom has only six electrons in its outershell even after making three single bonds with three F atoms (i.e. it completes only sixet). BeC l 2 Beryllium has only four electrons in its outershell even after making two single bonds with two Cl atoms. CI Be CI PC l5 Phosphorus after making five single bonds with five Cl atoms has ten electrons in its outershell. CI CI P CI CI CI SF 6 Sulphur makes six single bonds with six F atoms and thus has 12 electrons in its outershell. L ewis St r uct ur es of M olecules The formula of a molecule shows the number of atoms of each element but does not show the bonding arrangement of the atoms. To represent the bonding pattern in a molecule, the electron CHEMICAL BONDING 22 QUIZRR dot symbols of the elements are arranged such that the shared pairs and unshared pairs (called lone pairs) are shown and the octet rule (or duet for hydrogen) is satisfied. For example, a molecule of fluorine is shown as........ :F : :F : and a molecule of hydrogen fluoride is shown as H :F: or H  F:........ Arrangement of dot symbols used to represent molecules are called Lewis Structures. Lewis structures do not convey any information regarding the shape of the molecule. Usually, the shared pairs of electrons are represented by lines between atoms and any unshared pairs are shown as dot pairs. Lewis structures are written by fitting the element dot symbols together to show shared electron pairs and to satisfy the octet rule. For example,.... (i) In water (H2O), one H and two.O : complete their duet and octet respectively as : O  H. | H   (ii) In ammonia (NH3), three H and one N fit together and satisfy their duet and octet   respectively as H N H | H   (iii) In carbon tetrachloride (CCl4), four :Cl and  C  complete their octet as   : Cl: | Cl  C  Cl | : Cl:  For the given molecules, we have adopted hit & trial to fit the dot symbols together and satisfy the octet rule. But remember that hydrogen form one bond, oxygen forms two bonds, nitrogen three bonds and carbon forms four bonds. For simpler molecules, the hit & trial method works perfectly but for slightly complicated polyatomic species, this may give us more than one possible structure. Thus, a systematic approach is needed to design the Lewis structures of such polyatomic species. But before proceeding further, let us understand the limitation of this approach. CHEMICAL BONDING  QUIZRR 23 L imit at ions of L ewis T heor y of Dr awing St r uct ur e (i) This method would be applicable to only those molecules/species, which follow octet rule except hydrogen. (ii) This method would not be applicable to species, which have more than one central atom (like N2O4, N2H4 etc.). (iii) This method is also not suitable for species, which contains transition metal atom as the central atom. There are three kinds of molecules/species, which do not follow octet rule. (a) Molecules, which have contraction of octet. Such molecules are electron deficient. For example, BH3, BF3, BCl3, AlCl3, GaCl3 etc. (b) Molecules, which have expansion of octet. Such species have more than eight electrons in their outermost shell. This is possible in those molecules, which have vacant d-orbitals, thus they can expand their octet. For example, PCl5, SF6 etc. (c) Molecules containing odd number of electrons (in total) cannot satisfy octet rule. Such species are called odd electron species and are paramagnetic in nature due to presence of unpaired electron. For example, NO, NO2 and ClO2. M et hod of Dr awing L ewis St r uct ur es To draw the Lewis structures of polyatomic species, follow the given sequence. (i) First calculate n1. n1 = Sum of valence electron of all the atoms of the species  net charge on the species. For a negatively charged species, electrons are added while for positively charged species, the electrons are subtracted. For an uninegatively charged species, add 1 to the sum of valence electrons and for a dinegatively charged species, and 2 and so on. (ii) Then calculate n2. n2 = (8 ï number of atoms other than H) + (2 ï number of H atoms) (iii) Subtract n1 from n2, which gives n3. n3 = n2 ă n1 = number of electrons shared between atoms = number of bonding electrons. n3 n2  n1  = number of shared (bonding) electron pairs = number of bonds. 2 2 (iv) Subtracting n3 from n1 gives n4. n4 = n1 ă n3 = number of unshared electrons or non-bonding electrons n4 n1  n3  = number of unshared electron pairs = number of lone pairs. 2 2 CHEMICAL BONDING 24 QUIZRR (v) Identify the central atom. Generally, the central atom is the one, which is least electronegative of all the atoms, when the other atoms do not contain hydrogen. When the other atoms are hydrogen only, then the central atom would be the more electronegative atom. (vi) Now around the central atom, place the other atoms and distribute the required number of bonds (as calculated in step (iv), keeping in mind that every atom gets an octet of electrons except hydrogen. (vii) Then calculate the formal charge on each atom of the species. Formal charge is the difference between the valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis Structure. Formal charge on an atom = number of valence electrons of the atom ă (number of shared electrons of that atom + number of unshared electrons of that atom). Formal charge on an atom = number of valence electrons of the atom ă number of bonds formed by that atom ă number of unshared electrons (2 ï lone pairs) of that atom. For every electron of an atom that is shared in a bond, the „number of bonds formed by the atom‰ is one. Therefore if an atom forms only one bond (A ă B), one electron of the bond is that of A and other is that of B. So the „number of bonds‰ of A and B each is one. But if the bond were a co-ordinate bond (A B), then two electrons of A are involved in it. This makes the number of bonds of A to be 2 and that of B to be zero. (viii) When two adjacent atoms get opposite formal charges, then charges can be removed by replacing the covalent bond between the atoms by a dative (co-ordinate) bond. This bond will have the arrowhead pointing towards the atom with positive formal charge. It is not mandatory to show the dative bonds unless required to do so. (ix) The given Lewis structure should account for the factual aspects of the molecule like resonance (delocalization), bond length, p-d back bonding etc. Sometimes, there are more than one acceptable Lewis structure for a given species. In such cases, we select the most plausible Lewis structure by using formal charges and the following guidelines : Ć For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. Ć Lewis structures with large formal charges (+ 2, + 3 and/or ă 2, ă 3 and so on) are less plausible than those with small charges. Ć Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. CHEMICAL BONDING  QUIZRR 25 E xamp le 1 Determine Lewis structure of NO2. Solution : (i) n1 = 5 + (6 ï 2) ă 1 = 16 (ii) n2 = (3 ï 8) = 24 (iii) n3 = n2 ă n1 = 24 ă 16 = 8 8  Number of bonds = 4 2 (iv) n4 = n1 ă n3 = 16 ă 8 = 8 8  Number of lone pairs =  4 2 (v) nitrogen is the central atom (as it is less electronegative than O). Arranging two O atoms around it and distributing 4 bonds and 4 lone pairs as O N O (a) (b) (vi) Calculating formal change on each atom. Formal charge on N = 5ă 4 ă 0= + 1 Formal charge on O (a) = 6 ă 2 ă 4 = 0 Formal charge on O (b) = 6 ă 2 ă 4 = 0 Thus, the structures can now be shown as +1 O N O (a) (b) E xamp le 2 Determine Lewis structure of CNă ion. Solution : (i) n1 = 4 + 5 + 1 = 10 (ii) n2 = (2 ï 8) = 16 (iii) n3 = n2 ă n1 = 16 ă 10 = 6 6  Number of bonds = 3 2 CHEMICAL BONDING 26 QUIZRR (iv) n4 = n1 ă n3 = 10 ă 6 = 4 4  Number of lone pairs = 2 2 (v) Carbon is the central atom (C is less electronegative than N) and arrange N, number of bonds and number of lone pairs around it as (vi) Formal charge on C = 4 ă 3 ă 2 = ă 1 Formal charge on N = 5 ă 3 ă 2 = 0 Thus, final Lewis structure of CNă would be ă C N E xamp le 3 Draw Lewis structure for NH4+ ion. Solution : (i) n1 = 5 + (4 ï 1) ă 1 = 8 (ii) n2 = (8 ï 1) + (2 ï 4) = 16 (iii) n3 = n2 ă n1 = 16 ă 8 = 8 8  Number of bonds = 4 2 (iv) n 4 = n1 ă n3 = 8 ă 8 = 0  Number of lone pairs = 0 (v) Nitrogen being the central atom, distributing other atoms (H) around it, and 4 bonds with the 4 H atoms, the structure looks like H (b) H N H(c) (a) H (d) (vi) Formal charge on N = 5 ă 4 ă 0 = + 1 Formal charge on H(a)/H(b)/H(c)/H(d) = 1 ă 1 ă 0 = 0 Thus, final Lewis structure of NH4+ would be H + H N H H CHEMICAL BONDING  QUIZRR 27 6. COO RDINATE BOND It is a special type of covalent bond in which both the shared electrons are contributed by one atom only. It may be defined as ÂÂa covalent bond in which both electrons of the shared pair are contributed by one of the two atoms’’. Such a bond is also called as dative bond. A coordinate or a dative bond is established between two such atoms, one of which has a complete octet and possesses a pair of valency electrons while the other is short of a pair of electrons. ïï ïï A + B ïï A B or A B ïï ïï ïï This bond is represented by an arrow (). The atom which contributes electron pair is called the donar while the atom which accepts it is called acceptor. Note : Coordinate bond after formation is indistinguishable from a covalent bond. The formation of a coordinate bond can be looked upon as a combination of electrovalent and covalent bonds. The formation may be assumed to have taken place in two steps: (i) The donor atom loses one electron and transferred to acceptor atom. As a result donor atom acquires a positive charge and the acceptor atom acquires a negative charge. + ă A + B A B ï ï ï ï ï ï (ii) These two charged particles now contribute one electron each and this pair is shared by both the atoms. + ï ï ă ï ï A + B A + B ï ï ï ï ï ï ï ï As the coordinate bond is a combination of one electrovalent bond and one covalent bond, it is also termed as semi pola r bond. The compound consisting of the coordinate bond is termed coordinate compound. Some examples of coordinate bond formation are given below : (1) Formation of ammonium ion : The ammonia molecule has a lone pair of electrons i.e. an unshared pair. The hydrogen ion H+, has empty s orbital. The lone pair comes to be shared between the nitrogen and hydrogen atoms. + H H H + H N + H H N H or H N H H H H CHEMICAL BONDING  28 QUIZRR Nitrogen atom is called the donor and H , the acceptor. NH3 is a neutral molecules. H+ carries + a unit positive charge, so NH+4 ion carries positive charge. Here, all the NăH bond become identical. (2) Formation of Aluminium Chloride, Al2 Cl6 Cl Cl Cl Al Al Cl Cl Cl (3) Formation of ozone : Oxygen molecule consists of two oxygen atoms linked by a double covalent bond. Each oxygen atom has two lone pairs of electrons. When one lone pair of electrons is donated to a third oxygen atom which has six electrons, a coordinate bond is formed. ï O O + O O O O ïï ï C har act er ist ics of C oor d inat e C omp ound s The properties of coordinate compounds are intermediate between the properties of electrovalent compounds and covalent compounds. The main properties are described below : (i) Physical state : These exist as gases, liquids and solids under ordinary conditions. (ii) Melting and boiling points : Their melting and boiling points are higher than purely covalent compounds and lower than purely ionic compounds. (iii) Solubility : these are sparingly soluble in polar solvents like water but readily soluble in non-polar (organic) solvents. (iv) Stability : These are a stable as the covalent compounds. The addition compounds are, however, not every stable. It is also a strong bond because the paired electrons cannot be separated easily. (v) Conductivity : Like covalent compounds, these are also bad conductors of electricity. The solutions or fused mass do not allow the passage of electricity. (vi) Molecular reactions : These undergo molecular reactions. The reactions are slow. (vii) Isomerism : The bond is rigid and directional. Thus, coordinate compounds show isomerism. (viii) Dielectric constant : The compounds containing coordinate bond possess high values of dielectric constants. CHEMICAL BONDING  QUIZRR 29 7. DIPOLE MOMENTS A dipole consists of a positive and an equal negative charge separated by a distance within a molecule. The degree of polarity of a bond is given by the dipole moment (ø), which is the product of either charge (e) and the distance (d) between them. ø = de. ÂeÊ is of the order of magnitude of the electronic charge, i.e. about 10ă10 esu and d is the distance between the atomic centres, i.e., about 10ă8 cm. Hence dipole moment may be expected to have values around 10ă10 ï 10ă8 = 10ă18 esu-cm. It is however, general practice to express dipole moments in Debye units (D), 1 D = 10ă18 esu-cm. Electric pole Electric pole + ă d If the charge is in SI units (Coulombs) and d in metre ø will coulomb-metre (C.m) units. 1D = 3.336 ï 10ă30 C. m. Dipole moment is a vector quantity, i.e., it has both magnitude as well as direction. Thus the overall value of the dipole moment of a polar molecule depends on its geometry and shape i.e., vectorial addition of dipole moment of constituent bonds. Any covalent bond which has a certain degree of polarity will have a corresponding dipole moment, though it does not follow that compounds containing such bonds will have dipole moments, for the polarity of the molecule as a whole is the vector sum of the individual bond moments. For example, CO2 has zero dipole moment, although the C = O bond is a polar bond. This shows that CO2 is a linear molecule, O = C = O, so that the dipole moments of the two C = O bonds cancel out. The C Cl bond has a definite polarity and a definite dipole moment but carbon tetrachloride has zero dipole moment because it is a tetrahedral molecule, and the resultant of the 4C ă Cl bond moments is zero. On the contrary CH3Cl, CH2Cl2 and CHCl3 have definite dipole moments. Ap p licat ions of Dip ole M oment (i) To decide polarity of the molecule : Molecules having zero dipole moment are said to be non-polar molecules and those having R  0 are polar in nature. (ii) To determine geometry of molecules : The values of dipole moments provide valuable information about the structure of molecules. (a) CO2, CS2 molecules are linear as values of their dipole moments are zero. (b) H2O is not a linear molecule as it has dipole moment. Actually, it has V-shaped structure and the bond angle is 105Ĉ. Similarly, SO2 has a bent structure. (c) In ammonia, three hydrogen atoms do not lie symmetrically with respect to nitrogen as it has dipole moment. It has pyramidal structure. CHEMICAL BONDING 30 QUIZRR (iii) To distinguish cis and tra ns forms of geometrical isomers : Experimental values of dipole moment of the isomers (cis and trans) are determined. The trans isomer usually possesses either zero dipole moment or very low value in comparison to cis form. H C Cl H C Cl H C Cl Cl C H Cis-1, 2, dichloroethene Trans-1, 2 dicholroethene ø = 1.9 D ø 0 It should be noted that if two groups have opposite inductive character then trans isomer will have greater dipole moment of, H Cl Cl H C C C C CH3 H CH3 H Trans Cis Trans > cis (iv) To determine orientation in benzene ring : Dipole moment is useful to ascertain the orientation of substituents. The greater the dipole moment, the greater is the asymmetry. In general, dipole moment follows the order : Ortho > meta > para Cl Cl Cl Cl Cl Cl Ortho Meta Para ø = 2.54 D ø = 1.48 D ø=0 CHEMICAL BONDING  QUIZRR 31 E xamp le 4 Both CO2 and N2O are linear but dipole moment of CO2 in zero but for N2O it is non-zero, why ? Solution : The answer lies in the structure of these molecules. CO2 is a symmetrical molecule while N2O is unsymmetrical. Thus for N2O, dipoles do not cancel each other, leaving the molecule with a resultant dipole moment, while the bond moment of CO2 cancel each other, so CO2 has no net dipole moment. N N O ; O C O E xamp le 5 Compare the dipole moment of NH3 and NF3. Solution : LetÊs draw the structure of both the compounds and then analyse their dipole directions. N N H H H F F F The structure of both NH3 and NF3 are pyramidal with three bond pairs and one lone pair. In NH3, as N is more electronegative than hydrogen, so the resultant bond dipole is towards N, which means that both the lone pair and bond pair dipoles are acting in the same direction and are summed up. In case of NF3, the bond dipole (of NăF bonds) is acting towards fluorine, (as fluorine is more electronegative than N) so in NF3 the lone pair and bond pair dipoles are acting in opposition, resulting in a decreased dipole moment. Thus, NH3 has higher dipole moment than NF3. Dip ole M oment and Per cent age I onic C har act er The measured dipole moment of a substance may be used to calculate the percentage ionic character of a covalent bond in simple molecules. 1 unit charge = Magnitude of electronic charge = 4.8 ï 10ă10 e.s.u. 1 D = 1 ï 10ă18 e.s.u.ăcm Observed dipole moment  % ionic character = Theoretical dipole moment CHEMICAL BONDING 32 QUIZRR Theoretical dipole moment is confined to when we assume that the bond is 100% ionic and it is broken into ions while observed dipole moment is with respect to fractional charges on the atoms of the bond. E xamp le 6 The dipole moment of LiH is 1.964 ï 10ă29 coulomb meter, distance between Li and H is 1.596 Å. Find the % ionic character in the molecule. Solution : ø of 100% ionic molecule (Li+Hă) = 1.6 ï 10ă19 C ï 1.596 ï 10ă10 m = 2.554 ï 10ă29 Cm 1.9  10 29 % ionic character =  100 2.554  1029 = 74.4% T r ansition Fr om I onic to C ovalent Bond–FAJ ANS’ R UL E Just as a covalent bond may have partial ionic character an ionic bond may also show a certain degree of covalent character. When two oppositely charged ions approach each other closely, the cation would attract the electrons in the outer shell of the anion and simultaneously repel its nucleus. This produces distortion or polarisation of the anion, which is accompanied by some sharing of electrons between the ions, i.e., the bond acquires a certain covalent character. Cation anion Cation anion separate Polarised anion The ability of a cation to polarise the near by anion is called its polarising power and the tendency of an anion to get distorted or deformed or polarised by the cation is called its polarisability. Fact or s I nfluencing I on – Defor mat ion or I ncr easing C ovalent C har act er (i) Large charge on the ions : The greater the charge on the cation, the more strongly will it attract the electrons of the anion. For example, Al3+ can distort Clă ion more than Na+ ion. So aluminium chloride is a covalent compound whereas NaCl, AlF3, AgF are ionic. CHEMICAL BONDING  QUIZRR 33 (ii) Small cation and large anion : For a small cation, the electrostatic force with which its nucleus will attract the anion will be large. Moreover a large anion cannot hold the electrons in its outermost shell, especially when they are attracted by a neighbouring cation. Hence there will be increased covalence with a small cation and a large anion, as in AgI. (iii) Cation with a non-inert gas type of electronic configuration : A cation with a 18 electron outermost shell such as Ag+ ([Kr] 4d10) polarizes anions more strongly than a cation with a 8 electron arrangements as in K+. The ÂdÊ electrons in Ag+ do not screen the nuclear charge as effectively as the ÂsÊ and ÂpÊ electron shell in K+. Thus AgI is more covalent than KI, although Ag+ and K+ ions are nearly of the small size. Cuprous and mercurous salts are covalent. The above statements regarding the factors which influence covalent character are called FajansÊ rules. It can thus be seen easily that there is nothing like a purely ionic compound or a purely covalent compound. 8. SHAPES OF MOLECULES The shapes or geometry of a molecule is quite accurately predicted by VSEPR (valence shell electron-pair repulsion) theory. According to this theory all valence shell electron pairs surrounding the central atom arrange themselves in such a manner as to be as far away from each other as possible. By separating the electrons from each other, electrostatic repulsion, (the cause of higher energy) is minimised. As a result each molecules tends to acquire a state of lowest energy. The basic ideas can be summarized as follows : Valence shell pairs of electrons are arranged about the central atom so that repulsions among them are minimized, or so that there is maximum separation among the regions of high electron density (bond pairs) about the atom. For instance, two regions of high electron density would be most stable on opposite sides of the central atom (the linear arrangement), while three regions would be most stable when they are arranged at the corners of an equilateral triangle (the trigonal planar arrangement). The resulting arragement of these regions is referred to as the electronic geometry of the central atom. Number of Region of High Electronic Geometry Bond Angles Electron Density (bpÊs) 2 linear 180Ĉ 3 trigonal planar 120Ĉ 4 tetrahedral 109Ĉ 28 5 trigonal bipyramidal 90Ĉ, 120Ĉ, 180Ĉ 6 octahedral 90Ĉ, 180Ĉ We will discuss this table detail a little later. CHEMICAL BONDING 34 QUIZRR Some imp or t ant p oint s : (1) a covalent bond is formed by overlapping of atomic orbitals of avalency shell of the two atoms. As a result of overlapping, there is maximum electron density some where between the two atoms. (2) Greater the overlapping, higher is the strength of chemical bond. (3) Electrons which are already paired in valency shell can enter into bond formation if they can be unpaired first and shifted to vacant orbitals of slightly higher energy of the same main energy shell. (Valency shell). This point explains the trivalency of boron, tetravalency of carbon, pentavalency of phosphorus, hexavalency of sulphur inspite of the fact that these atoms have paired orbitals in the valency shell. 2s 2p 2s 2p Boron carbon Excited State One electron is shifted to p-orbital 3s 3p 3d Phosphorus One electron shifted to d-orbital (4) Between 2 orbitals of same stability (i.e. having same energy) one more directionally concentrated would form a stronger bond. Dumb-bell shaped p-orbitals will form a stronger bond as compared to spherically symmetrical s-orbital. It is formed by head on or axial overlap. CHEMICAL BONDING  QUIZRR 35 The types of bonds can be formed an account of overlapping. (1) Sigma Bond : A bond formed between two atoms by the overlap of singly occupied orbitals along their axes (end to end overlap) is called sigma () bond. The bond formed by this type of over lapping is very strong as the extent of overlapping is sufficiently high. In sigma bond, the electron density accumulates between the centres of the atoms being bounded and lies on the imaginary line joining the nuclei of bonded atoms. Sigma bonds are formed by three types of overlapping. (i) s-s overlapping : For example, hydrogen molecule. Each hydrogen atom has one electron in is orbital which is spherical. Is orbital of both the hydrogen atoms approach each other closely and when they reach a point of maximum attraction by the two nuclei, they overlap and form a sigma bond. s-s overlap 1s 1s ï Molecular axis Formation of H2 molecule by s-s overlapping The bond has two electrons which have opposite spins. The probability of finding these electrons is maximum in the region between the two nuceli on the molecular axis. The electron density of the bond is distributed symmetrically about the molecular axis. (ii) s–p overapping (Formation of HF, H2O molecules) : (a) In the formation of HF molecule the 1s-orbital of hydrogen overlaps with the p-orbital of fluorine containing unpaired electron. 1s 2p 2s 1p Hydrogen atom z y x Fluorine atom Formation of HF molecule by s-p overlapping CHEMICAL BONDING 36 QUIZRR (b) Formation of water molecule : Oxygen atom has the configuration of valency shell i.e., it has two orbitals singly occupied. These two orbitals overlap with 1 s- orbital of two hydrogen atoms forming sigma bonds. 2 ăpx Hydrogen atoms (1s1) py Oxygen atom s-p overlap H O 104.5Ĉ M.O. H Formation of water molecule by s-p overlapping Since the two orbitals of oxygen are at right angle to each other an angle of 90Ĉ is expected between two sigma bonds but actual bond angle observed is 104.5Ĉ. (iii) p-p overlapping (Formation of fluorine molecule) : This is illustrated by the formation of fluorine molecule. The electronic configuration of fluorine atom is 1s2 2s2 2 px2 2 p2y 2 p1z , i.e., one orbital is singly occupied. When p-orbitals of two fluorine atoms approach each other with their heads directly towards one another, they overlap and form a sigma bond. 2s 2p 2p 2s p-p overlap Fluorine atom Fluorine atom Head on overlap p p p-p overlap M.O. z z Formation of F2 molecule by p-p overlapping CHEMICAL BONDING  QUIZRR 37 (2) ) Bond Pi ( -bonds are formed by the sidewise or lateral overlapping of p-orbitals. The overlapping takes place at the side of two lobes and hence, the extent of overlapping is relatively smaller. Thus, -bond is a weaker bond in comparison to sigma bond. The molecular orbital is oriented above and below the plane containing nuclear axis. p p p-p overlapping  M.O. Formation of -bond Formation of oxygen molecule : Oxygen atom has two p-orbitals singly occupied in the valency shell. When two oxygen atoms approach each other, one set of p-orbitals experiences head on overlap forming a sigma bond while other set of p-orbitals overlaps sidewise to form a -bond. Thus, oxygen molecule has one -and one -bond. -bond 2s 2p 2p 2s p-p overlap (Head on) px py pz px py pz -bond Oxygen atom Oxygen atom p-p overlap (sidewise) Sidewise overlap  pz pz   M.O. py py  M.O. Head on overlap Formation of O2 molecule CHEMICAL BONDING 38 QUIZRR Similarly, the formation of nitrogen molecule can be explained. It has one sigma bond and two -bonds. All single bonds are sigma bonds. A double bond consists of one sigma and one -one while a triple bond consists of one sigma and two -bonds. (i) Bond energy increases from a single bond to a trip bond. Bond energy : Single bond < Double bond < Triple bond (ii) Bond strength increases from a single bond to a triple bond. Bond strength : Single bond < Double bond < Triple bond (iii) Bond length of a multiple bond (double or triple) is always shorter than the corresponding single bond. Bond length : Single bond > Double bond > Triple bond (iv) Reactivity of a multiple bond is always more than the single bond. This is due to the fact that -electrons are mobile in nature. Reactivity : Single bond < Double bond < Triple bond. 9. HYBRIDIZATION The valence bond theory (overlapping concept) explains satisfactorily the formation of various molecules but it fails to account the geometry and shapes of various molecules. It does not give explaination why BeCl2 is linear, BF3 is planar. CH4 is tetrahedral, NH3 is pyramidal and water is V-shaped molecule. In order to explain these cases, the valence bond theory has been supplemented by the concept of hybridization. This is a hypothetical concept and has been introduced by Pauling and Slater. According to this concept any number of atomic orbitals of an atom which differ in energy slightly may mix with each other to form new orbitals called hybrid orbitals. The process of mixing or amalgamation of atomic orbitals of nearly same energy to produce a set of entirely new orbitals of equivalent energy is known as hybridization. The following are the rules of hybridization : (i) Only orbitals (atomic) of nearly same energy belonging to same atom or ion can take part in hybridization. (ii) Number of the hybrid orbitals formed is always equal to number of atomic orbitals which have taken part in the process of hybridization. (iii) Most of the hybrid orbitals are similar but they are not necessarily identical in shape. They may differ from one another in orientation in space. (iv) Actually the orbitals which undergo hybridization and not the electrons. For example, for orbitals of nitrogen atom (2s2 2 p1x 2 p1y 2 p1z ) belonging to valency shell when hybridize, form four hybrid orbitals, one of which has two electrons (as before) and other three have one electron each. CHEMICAL BONDING  QUIZRR 39 (v) Hybrid orbitals only form sigma bonds. It is concept not only predicts the correct shapes but also explains the actual microscopic observations of the geometry of molecules. The different cases of hybridisation arises by mixing of s,pd orbitals; few of them are discussed below. Sp 3 H yb r id isat ion This type of hybridisation results from mixing of one s, three p orbitals of outer (valence) shell of an atom. Let us take the example of CH4 molecule. Carbon atom has configuration 1s2, 2s2 2px1 2py1. In ground state, it has two unpaired orbitals which can form only two covalent bonds. To get tetravalency, 2s-orbital is unpaired and the electron is shifted to 2p-orbital. Now in excited state the four unpaired orbitals undergo hybridization giving rise to four hybrid orbitals which are 109Ĉ28´ apart. The four hybrid orbitals overlap with s-orbital of each of the four hydrogen atoms forming four sigma bonds. The molecule formed is tetrahedral. 2s 2p Carbon atom in ground state Carbon atom in excited state sp3 hybridization H H C C 109 H H H Ĉ 28´ H H H Formation of CH4 molecule CHEMICAL BONDING 40 QUIZRR 2 Sp H yb r id isat ion To understand it let us take the example of BF3 molecule. BF3 molecule : Boron atom has configuration 1s2, 2s2 2p1. In ground state, it has one unpaired orbital which can form only one covalent bond. To get trivalency, the 2s-orbital is unpaired and the electron is shifted to 2p-orbital. Now in excited state the three unpaired orbitals undergo F F 120 0Ĉ B 12 Ĉ 120Ĉ B Atom F F F F