Chemical Bonding & Molecular Structure

Chemical Bonding & Molecular Structure CHEMICAL BOND (I) A force that acts between two or more atoms to hold them together as a stable molecule. (II) This process accompanied by decrease in energy. Ex.1 When two atoms combine to form a molecule (A) Energy is released (B) Energy is absorbed Exception of Octet Rule : (C) Energy is neither released nor absorbed (D) Energy may either released or absorbed (a) Incomplete octet molecules : - or (electron defficient molecules) Sol. (A) To attain stability energy is released. Compound in which octet is not complete in outer most orbit of central atom. LEWIS OCTET RULE Examples - Halides of IIIA groups, BF3, AlCl3, BCl3, hydride of III A/13th group etc. (I) Every atom has a tendency to complete its octet outermost. (II) H has the tendency to complete its duplet. (III) To acquire inert gas configuration atoms loose or gain electron or share electron. (IV) The tendency of atoms to achieve eight electrons in their outer most shell is known as Lewis octet rule. In BCl3    Boron has only 6 electron s Other examples - BeCl 2 (4e – ), ZnCl 2 (4e – ), Ga(CH3)3 (6e–) 45 (b) Expansion of octet or (electron efficient molecules) Compound in which central atom has more than 8e– in outermost orbits. Chemical Bonding & Molecular Structure Ex.3 The octet rule is not obeyed in (A) CO2 (C) PCl5 (B) BCl3 (D) SiF4 Sol. (B) & (C) As B is electron dificient, P undergoes octet expansion. Classification of bonds : - Example - In PCl5, SF6, IF7, the central atom P, S and I contain 10, 12, and 14 electrons respectively. (c) I-Pseudo inert gas configuration : (I) Cations of transition metals, which contains 18 electrons in outermost orbit Examples : Ga+3, Cu+, Ag+, Zn+2, Cd+2, Sn+4, Pb+4 etc. Electronic configuration of Ga - 1s2, 2s2 2p6, 3s23p63d10, 4s24p1 Electronic configuration of Ga+3 - 1s2, 2s2 2p6, ELECTROVALENT OR IONIC BOND (I) The chemical bond formed between two or more atoms as a result of the complete transfer of one or more electrons from one atom to another is called Ionic or electrovalent bond. (II) Electro +ve atom loses electron (group IA to IIIA) (d) Odd electron molecules : Central atom have an unpaired electron or odd no (7e–, 11e– etc) of electrons in their outer most shell. (III) Electro –ve atom gains electron (group VA to VIIA) (IV) Electrostatic force of attraction between cation and anion is called ionic bond or electrovalent bond. Electronegativity difference  nature of ionic bond. Example IA and VIIA group elements form maximum ionic compound. Examples : NO, NO2 ClO2 etc. – Ex.2 Which follows octet rule - + (A) FeCl2 (B) AgCl (C) CaCl2 (D) CuCl Sol. (C) Because in other compounds there is more than 8 e– in outermost shell of central atom. – (V) More the distance between two elements in periodic table more will be ionic character of bond. 46 (VI) Total number of electron lose or gained is called electrovalency. Chemical Bonding & Molecular Structure Higher electron affinity  Greater tendency to form anion Cl– > F– > Br– > I– e.g. Example – F– > O–2 > N–3 (c) Lattice energy - (Energy released) Amount of energy released when one mole of crystal lattice is formed Higher lattice energy  Greater will be the stability or strength of ionic compound. (a) (b) Factors affecting lattice energy : (i) Magnitude of charge  U  z+ z– (Ionic charge) Lattice energy  Magnitude of charge (c) NaCl MgCl2 AlCl3 Na + Mg+2 Al+ – Lattice energy increases – (VII) The force of attraction is equal in all direction so ionic bond is non-directional. – Size of cation decreases. (ii) Size of Cation : – Lattice energy  (VIII) Ionic compound do not have molecular formula. It has only empirical formula. eg. NaCl is empirical formula of sodium chloride. LiCl NaCl KCl Conditions for Forming Ionic Bonds : – Size of anion is constant Formation of Ionic bond depends upon these three factors – (a) Ionisation energy : Lesser Ionisation energy  Greater tendency to form cation. Mg Cs Rb 2 Al K 3 Na RbCl CsCl – Size of cation increasing – Lattice energy decreases (d) Overall lowering of energy : Energy must be released during bond formation. Amount of energy required to remove an electron from the outermost orbit of an isolated gaseous atom to form the +ve ion or cation. (energy absorbed) e.g Na 1 r  r–  Li Energy changes are involved in the following steps – A + IE = A+ + e– and B + e– = B– + EA This concludes that for lower value of IE and higher value of EA there is more ease of formation of the cation & anion respectively and consequently more chances of electrovalent bond formation. Cationformationtendency (b) Electron affinity Representation of formula of compounds : Amount of energy released when an electron is (a) Write the symbols of the ions side by side in such a added to an isolated gaseous to form –ve ion (anion) way that positive ion is at the left and negative ion is energy released. at the right as A+B– 47 (b) Write their electrovalencies in figure at the top of each symbol as AxBy (c) Now apply criss cross rule as , i.e formula Chemical Bonding & Molecular Structure (d) Conductivity – It depends on ionic mobility. In solid state - No free ions - Bad conductor of electricity. In fused state or aqueous solution Due to free ions - Good conductor of eletricity. AyBx. Examples : Calcium chloride = CaCl2 Properties of ionic compounds : conductivity order : Solid state > Fused state < Aqueous solution (e) Solubility – (a) Physical state – Highly soluble in water (Polar solvents) Ionic compounds are hard, crystalline and brittle due to strong electrostatic force of attraction. Brittleness  Example : NaCl in water (I) The Na+ ions get associates with– vely charged ‘O’ of water (II) And Cl– ions associates with +vely charged ‘H’ of water. (b) Isomorphism – (I) Two compounds are said to be isomorphous if they have similar no. of electrons i.e.similar configuration of their cation and anion. (II) They have similar crystal structure. Example – Na+ F– Mg+2 O–2 Valency + 1, –1 + 2, –2 electronic 2, 8, 2, 8 2, 8 2, 8 Ca+2 2Cl–1 2K+1 S–2 2, 8, 8, 2, 8, 8  2, 8, 8  2, 8, 8  2, 8, 8 2, 8, 8 configuration similarly (III) Thus charge on Na+ and Cl– decreases and electrostatics force of attraction also decreases which leads to free ion. (IV) The energy released due to interaction between solvent and solute is called solvation energy.If water Ionic compounds have high boiling point and melting is used as solvent it is called hydration energy. point due to strong electrostatics force of attraction among oppositely charged ions. (c) Boiling point and melting point – 48 (V) For an ionic compound to be soluble in water – Hydration energy > Lattice energy 1 Lattice energy  Solub ility Hydration energy  Solubility. Chemical Bonding & Molecular Structure Factor affecting solubility : (I) Dielectric constant – The capacity of solvent to neutralise the charge of ionic compounds is called Dielectric constant. It is represented by “” (a) Water has maximum dielectric constant ( = 80) Hydration energy (H)  1 r  1 r– {r+ & r– are radius of cation and anion} (VI) Hydration energy mainly depends on the cation radius because the value 1 r 1 r– is negligible in comparison to . (CH3OH  = 35) , (Acetone  = 21) (C2 H5 OH  = 27) , (Ether = 4.1) (Benzene  = 2.3) H2O > CH3OH > CH3CH2OH > CH3COCH3 > CH3OCH3 > C6H6 (VII) Down the group both the lattice energy & hydration energy decreases, if decreases in lattice is greater (b) Ionic compounds are more soluble in the solvents, having high dielectric constant. than hydration energy, solubility increases down the (c) H2SO4 and H2O2 have high dieletric constant but group and vice versa. these are not a good solvent due to oxidising nature (II) Size of ion : Ex.4 Out of following which has maximum value of lattice (a) Keeping size of cation constant, the lattice energy energy decreases with the increases of anionic radius. (A) NaF (B) NaCl Hence order of solubility of LiX in water is LiF < (C) NaBr (D) NaI LiCl < LiBr < LiI Sol. (A) Due to small size of anion. 1 Ex.5 The crystal lattice of electrovalent compound is As solubility  lattice energy composed of (b) in LiI covalent nature is more according to Fajan’s (A) Atoms rule but HE > LE therefore Li is more soluble in (B) Molecules water. (C) Oppositely charged ions (c) Keeping size of anion constant, the hydration energy (D) Both molecules and ions decreases with the increase of cationic radius. hence Sol. (C) It is the fact that electovalent compounds are made of ions. order of solubility of MSO4 will be – BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4 Ex.6 Ions are formed from neutral atoms by (Exception of Fajan’s rule) (A) Loss of electron (B) Gain of electrons (C) Sharing of electrons (D) Loss and gain of electrons Sol. (D) Cation and anion are formed by losing and sharing of electrons respectively. (d) If size of cation and anion is very large, solubility decreases from top to bottom. (e) Solubility decreases in a period (as ionic nature decreases and covalent nature increases) NaCl > MgCl2 > AlCl3 49 Chemical Bonding & Molecular Structure Polarizability : (III)Polarisation of anion : Polarisation capacity is maximum in pseudo inert gas (I) ability of anion to get polarised by the cation. configuration. (II) Polarisation of anion causes some sharing of electron 1 between the ions so ionic bond acquires certain solubility  polarisation covalent character. AgF > AgCl > AgBr > AgI (III) Polarisation  Covalent character (IV) Magnitude of polarisation depends upon a no. of factors, suggested by Fajan and are known as ajan’s rule. – Polarisation increases. – Covalent character increases. – Solubility decreases. (IV) Ionic reaction – Fajan’s rule : (Factors Affecting Polarisation) (a) Ionic compounds shows ionic reaction and covalent (a) Size of cation : - Polarisation of the anion increases as the size of cation decreases. compounds shows-molecular reaction. (b) Ionic reactions are faster than molecular reaction because of free ions. e.g. When NaCl is added in AgNO3 solution, white ppt of AgCl is formed at once.  Na+ NO – + AgCl  Ag+NO3– + Na+NO–  3 1 Polarisation  size of cation In a group – BeCl2 MgCl2 - Size of cation increases POLARISATION CaCl2 SrCl2 - Covalent character decreases - Ionic character increases (Fajan’s Rule) (Covalent nature in ionic bond) BaCl2 white ppt. (I) When a cation approaches an anion closely the positive charge of a cation attract the electron cloud of the anion towards itself, due to the electrostatic force of attraction between them. Greatest polarising power of Be2+, shows its  maximum covalent character In a period – Na+, Mg+2 , Al+3, Si+4 (II) At the same time cation also repel the positively charge nucleus of anion. - Size of cation decreases - Covalent charater increases (III) Due to this combined effect, cloud of anion is bulged or elongated towards the cation. This is (b) Size of anion : - If the size of the anion increases called distortion, deformation or Polarisation of the for a given cation, the covalent character increases/ anion by the cation and anion is called Polarised. Nature Polarisation  size of anion. Polarisation Power : The ability of cation to polarise a nearby anion is called Polarisation power of cation. CaF2 CaCl2 - Size of anion increases CaBr2 - Covalent character increases Cal2 - Ionic character decreases 50 Chemical Bonding & Molecular Structure COVALENT BOND (c) Charge on cation and anion : (I) Polarisation  charge on cation anion  Charge on cation  Polarisation (covalent character) eg. NaCl Na + MgCl2 Mg++ AlCl3 Al+++ SiCl4 Si++++ (I) A covalent bond is formed by the mutual sharing of electrons between two atoms of electro negative elements to complete their octet. (Except H which completes its duplet). - Charge on cation increases - Covalent character increases - Ionic character decreases (M.P. decreases) Charge on anion  polarisation  covalent nature  1 M.P. (d) Electronic configuration of cation : Polarisation capacity of cation having pseudo inert (II) The shared pair of electrons should have opposite spins, and are localised between two atoms gas configuration is high. If the size of cations is same concerned. than that of cation having inert gas configuration. CuCl (M.P. 442°C) – Cu+ 2, 8, 18 (Covalent) (III) Shairing of electrons may occurs in three ways – NaCl (M.P. 800°C) – Na+ 2, 8 (Ionic) No. of electrons Electron Bond Cu+ and Na+ both the cation (Pseudo & inert) have shared between pair same charge and size but polarising power of Cu+ is two atoms more than Na+ because – zeff of ns2p6 (inert) < zeff. of ns2p6d10 (pseudo) 2 1 Single bond + + Na < Cu 4 2 Double bond 6 3 Triple bond (Ionic) (Covalent) So CuCl has more covalent character than NaCl. Examples – Three single bonds Ex.7 Compound with maximum ionic character is formed (not triplet bond) from Triple bond. (not three bond) Double (A) Na and Cl (B) Cs and F bond (Not two single bond) (Two single (C) Cs and  (D) Na and F bonds.) Sol. (B) Using fajan’s rule, larger cation and smaller anion Lewis structure and covalent bond : will have max. Ionic character. (I) Electron dot structures, also known as Lewis Ex.8 Out of the following which one has the highest structures of covalent molecules, are written in values of covalent character accordance with octet rule. (A) ZnCl2 (B) CaCl2 (C) CdCl2 (D) CuCl Sol. (A) Because Zn+2 has smallest size among the all. (II) All atoms in a formula will have a total of eight electrons by sharing in the valence shell except the Ex.9 Compound having highest M.Pt. – H-atom which forms the largest number of bonds (A) BeCl2 (B) MgCl2 (C) CaCl2 (D) SrCl2 with other atoms placed in the centre of skeleton Sol. (D) As the size of cation increases polarisizing power structure. Other atoms surrounds it to complete the decreases hence ionic character increases octet. 51 Chemical Bonding & Molecular Structure (b) Sulphur  Ground state. (III) Structure in which valence electrons are represented by dots are called lewis structures. (IV) Structure represented by line (–) or dashes are known as couper structrue. (V) Lewis dot formulae show only the number of valency electron, the number and kinds of bonds, but do not depict the three dimensional shapes molecules and polyatomic ions. Covalency – 2 (SF2) Sulphur  Excited state 1st excited state Orbital concept of covalent bond : (I) One orbital can accomodate at the most 2 electrons with opposite spins (II) Half filled orbital or unpaired electron orbital accepts one electron from another atom, to complete its orbitals. (III) Tendency to complete orbital or to pair the electron is an essential condition of covalent bond. Completion of octet is not the essential condition of covalent bond. (IV) Covalency : It is defined as the number of electrons contributed by an atom of the element for shairing with other atoms to achieve noble gas configuration. (V) If the outermost orbit has empty orbitals then covalent bonds are formed in exicted state. 2st excited state 3s Covalency – 4 (SF4) 3p 3d Covalency – 6 (SF6) So variable covalency of S is 2, 4 & 6. (c) Iodine has three lone pair of electrons (Ground state) So it shows three excited states – Maximum number unpaired electrons = 7 Variable Valencies are 1, 3, 5, 7 Applications of Variable Valency : Variable valency in covalent bonds : (I) To explain existence of molecules – (I) Variable valencies are shown by those elements NCl3 – exists which have empty orbitals in outermost shell. NCl5 – doesn’t exists (due to absence of d-orbitals (II) Lone pair electrons gets excited in the subshell of in Nitrogen.) While PCl3 and PCl5 both exist because the same to form the maximum number of unpaired 3d orbitals are present in phosphorus. electrons. Maximum covalency is shown in excited state. OF2 –– exists, but OF4 and OF6 doesn’t exists (III) The energy required for excitation of electrons is due to absence of d-orbitals While SF4 and SF6 called promotion energy. exists due to presence of d-orbital, present in its (IV) Promotion rule – Excitation of electrons in the ame valence shell. orbit. (II) Hydrolysis of compounds – Example – CCl4 + H2O  No reaction occurs (a) Phosphorus  Ground state SiCl4 + H2O  Reaction occurs In C-atom there are no empty d-orbitals to accept Covalency 3 (PCl3) the lone pair of electrons from water. While SiCl4 has Phosphorus  Excited state empty 3d-orbitals. z Covalency – 5 (PCl5)   Si(OH) – SiO SiCl4 + 4H2O –4 HCl 4 2 Unstable 52 Chemical Bonding & Molecular Structure (d) So covalent bond has directional character. Ex.10 A covalent bond is possible between (A) Similar atom (B) Dissimilar atoms (C) Similar and dissimilar atoms (D) Similar molecules Sol. (C) It is related to electronegativity. Ex.11 Which of the following is most covalent (A) AlF3 (B) AlCl3 (C) AlBr3 (D) Al3 Sol. (D) As the size of anion increases covalent character increases. Ex.12 Among LiCl, BeCl2, BCl3 and CCl4, the covalent bond character follows the order (A) LiCl < BeCl2 > BCl3 > CCl4 (B) LiCl > BeCl2 < BCl3 < CCl4 (C) LiCl < BeCl2 < BCl3 < CCl4 (D) LiCl > BeCl2 > BCl3 > CCl4 Sol (C) The covalent character increases according to charge on cation. Wave mechanical concept of chemical bonding – (Overlapping) To explain the nature of covalent bond two theories based on quantum mechanics have been proposed. (1) Valence bond theory (VBT) (2) Molecular orbital theory (MOT) VALENCE BOND THEORY (I) It was presented by Heitler & London to explain how a covalent bond is formed. It was extended by Pauling & Slater. (II) The main points of theory are – (a) To form a covalent bond overlapping occurs between half filled valence shell orbitals of the two atoms. (b) Resulting bond acquires a pair of electrons with opposite spins to get stability. (c) Orbitals come closer to each other from the direction in which there is maximum overlapping (e) Extent of overlapping  strength of chemical bond. (f) Extent of overlapping depends on two factors. (i) Nature of orbitals – p, d and f are directional orbitals  more overlapping s-orbitals  non directional – less overlapping (ii) Nature of overlapping – Co-axial overlapping - extent o overlapping more. Collateral overlapping - extent of overlapping less Order of strength of Co - axial overlapping –p-p>s-p>s-s (g) As the value of n increases, bond strength decreases. 1-1>1-2>2-2>2-3>3-3 (h) If n is same 2p - 2p > 2s - 2p > 2s - 2s 1s - 2p > 2s - 2p > 3s - 3p (i) Electron which is already paired in valency shell can enter into bond formation, it they can be unpaired first and shifted to vacant orbitrals of slightly higher energy of the same energy shell. (j) This point can explain the trivalency of boron, tetravalency of carbon, penta valency of phosphorus etc. (k) Two types of bonds are formed on acount of overlapping. (A) Sigma () bond (B) Pi () bond 53 Chemical Bonding & Molecular Structure (d) p - p overlapping – (Coaxial) – It involves the coaxial overlapping between half filled p-orbitals of two different atoms. (A) Sigma () bond : (I) Bond formed between two atoms by the overlapping of half filled orbitals along their axis (end to end overlap) is called sigma bond. eg. Formation of Cl2, F2, Br2 (II)  bond is directional. (III)  bond do not take part in resonance. (IV) Free rotation is possible about a single  bond. (V) Maximum overlapping is possible between electron clouds and hence it is strong bond. (VI) There can be only  bond between two atoms. Sigma bond are formed by four types of overlapping (B) Pi () bond (a) s - s overlapping (H2) - Two half filled S-orbitals (I) The bond formed by sidewise (lateral) overlapping overlap along the internuclear axis. are known as  bonds. (b) s - p overlapping (Formation of HF) – When half fill s-orbital of one atom overlap with half filled p-orbital of other atom. (II) Lateral overlapping is only partial, so bonds formed are weaker and hence more reactive than  bonds (Repulsion between nucleus is more as orbitals have to come much close to each other for  bonds formation) Example – Formation of O2 molecule – (c) Bond between two hybrid orbitals – sp3 - sp3, sp2 - sp2, sp3 - sp2, sp3 - sp etc. Only py and pz of oxygen atom have unpaired e– in each orbital for bonding. Electron configuration of oxygen is – 1s22s22px2 2py1 2pz1 (III) Free rotation about a  bond is not possible. (IV)  bond is weaker than  bond (Bond energy difference is 63.5 KJ or 15 K cal/mole) (V)  bonds are non-directional, so do not determine the shape of a molecule. (VI)  bond takes part in resonance. overlapping of orbitals form  bond. (VII)  bond formed by pure or unhybrid orbitals. 54 Chemical Bonding & Molecular Structure (IV) The number of hybrid orbitals on central atom of a HYBRIDISATION molecule or ion = number of  bonds Consider an example of Be compound :+ lone pair of electron. If it is formed without hybridisation then (i) The 1st bond between two atoms will be sigma. s–p p–p (ii) The other bond between same two atoms will both the Be–Cl bonds should have different be pi bond. parameters and p-p bond strength >s-p bond (iii) Maximum two pi bonds may be present on a strength. single atom. Practically bond strength and distance of both the (iv) The electron pair of an atom which do not take Be-Cl bonds are same. part in bond formation called as lone pair of electron. This problem may overcome if hybridisation of s and (V) One element can represent many hybridisation state p-orbital occurs. depending on experimental conditions for example, C showing sp, sp2 and sp3 hybridisation in its Hybridisation : compounds. (I) It is introduced by pauling, to explain equivalent (VI) Hybrid orbitals are differentiated as sp, sp2, sp3 etc. nature of covalent bonds in a molecule. (II) Definition : Mixing of different shapes and (VII) The order of repulsion between Ip & bp is : Ip - Ip > Ip - bp > bp - bp approximate equal energy atomic orbitals, and redistribution of energy to form new orbitals, of same (VIII) The directional properties in hybrid orbital is more than atomic orbitals. Therefore hybrid orbitals form shape & same energy. These new orbitals are called stronger sigma bond. The directional property of hybrid orbitals. and the phenomenon is called different hybrid orbitals will be in following order. hybridisation. sp < sp2 < sp3 < sp3d < sp2d2 < sp3d3 Now after considering s-p hybridisation in BeCl2 Difference between hybridisation & overlapping Overlapping Hybridisation bond strength of both the bonds will be equal. It occurs between It occurs among Characteristic of Hybridisation : orbitals of two orbitals of the same (I) Hybridisation is a mixing of orbitals and not electrons. atoms atom Therefore in hybridisation full filled, half filled and Only half filled Any type of orbital can empty orbitals may take part. orbitals takes participates (II) Number of the hybrid orbitals formed is always be part in overlapping equivalent to number of atomic orbital which have taken part in the process of hybridisation. (III) Each hybrid orbital having two lobes, one is larger and other is smaller. Bond will be formed from large lobe.      It occurs during bond formation bond formed after hybridisation  Process, just before  Orbital of different energies may participates in excited states.  It may takes place in  Molecular orbitals are formed Hybridised orbitals are  formed overlapping. ground or in excited state Inground state– NH3, NCl3, PH3, PCl3 , 55 Determination of hybridisation state – Chemical Bonding & Molecular Structure Types of Hybridisation : Method (I) : sp hybridisation : Count the following pair of e– arround the central atom : (a) Count all pure  bonded electron pairs (or  bonds) (I) In this hybridisation one s– & one P–orbital of an atom are mixed to give two new hybrid orbitals which are equivalent in shape & energy known as sp hybrid orbitals. (II) These two sp hybrid orbitals are arrange in straight line & at bond angle 180°. (b) Count all lone pair of electron (c) Count Co ordinate bond (d) Count negative charge (III) s-character 50% Method (II) : Be (ground state) To predict hybridisation following formula may be used : Be (excited state) No. of hybrid orbital = 1 2 Be atom accepts two electrons from F in BeF2, [Total number of valence e– in the central atom + total number of monovalent atoms – charge on cation + charge on anion] 1 eg. NH4 [5 + 4 – 1] = 4 2 sp3 hybridisation. SF4 1 [6+4]=5 2 sp3d hybridisation. SO 24 – 1 [6+2] =4 2 sp3 hybridisation. CO2 Molecule (O = C = O ) : In CO2 molecule, C has two sp hybrid orbitals & two unhybridised p orbitals. (‘O’ is divalent so add only charge on anion) NO3– 1 [5 + 1] = 3 2 sp2 hybridisation. If such type of e– pairs are – two – sp hybridisation three – sp2 hybridisation four – sp3 hybridisation five – sp3d hybridisation six – sp3d2 hybridisation seven – sp3d3 hybridisation  Thus, CO2 molecule is a linear in shape & having 180° bond angle.  The bond length between C–O bond is reduced due to the presence of  bond. In molecules, each C atom contains two sp hybrid orbitals & two unhybridised p orbitals C(ground state) 56  C(excited state) C atom accepts four electrons from H & C, In C2H2  sp hybrid orbital of each C overlaps to give sigma bond between C – C.  The remaining one sp hybrid orbital of each C atom overlaps with s orbital of H, forming sigma bond between C – H.  The two unhybridised p orbitals of each C atom (py and px) overlap laterally to form two pi() bonds.    Chemical Bonding & Molecular Structure Therefore in H–XCA  CB–H sigma bond between CA – CB is formed sp – sp overlapping sigma bond between CA – H is formed sp – s overlapping sigma bond between CB – H is formed sp – s overlapping pi bond between CA – CB is formed : py – py, px – px overlapping Each C atom forms two sigma bonds but in C2H2, total sigma bonds are 3 Each C atom forms two  bonds. Total  bonds in C2H2 are two Total number of bonds in acetylene are : 3 + 2 bond = 5 bonds Examples on sp hybridisation Example s bond I.p.e. BeH2 2 - sp 180° Linear BeF2 2 - sp 180° Linear BeCl2 2 - sp 180° Linear BeBr2 2 - sp 180° Linear BeI2 2 - sp 180° Linear CO 2 2 - sp 180° Linear CO 1 1 sp 180° Linear NO 2 2 - sp 180° Linear C 2H 2 2 - sp 180° Linear HCN 2 - sp 180° Linear ZnCl2 2 - sp 180° Linear HgCl2 2 - sp 180° Linear CdCl2 2 - sp 180° Linear N 2O 2 - sp 180° Linear N 3– 2 - sp 180° Linear sp2 Hybridisation : (I) In this hybridisation one & two p orbitals are mixed to give three new sp2 hybrid orbitals which all are in the same shape & equivalent energies. (II) These three sp2 hybrid orbitals are at angle of 120° & giving trigonal planar shape. Hybridisation Bond angle Shape B (ground state) B (excited state) B atom accepts 3 electrons From 3 F atom  in BF3 (III) s - character 33.3% 57 Graphite :  Chemical Bonding & Molecular Structure On each carbon atom, one unhybridised. p orbital is also present, which is perpendicular to the plane of sp2 hybrid orbitals.  The distance between two layer in graphite is 3.35 Å. SnX2 (X = F, Cl, Br, I) Sn (ground state) Sn (excited state) Sn after hybridisation shares 2e– with 2 atom.  The hybridisation on each carbon atom of graphite is sp2  Three sp2 hybrid orbital are present on each carbon atom at 120° & they overlap to sp2 hybrid orbitals of adjacent C atoms forming layers of hexagons.  SnX2 having two  bonds & one LP electron therefore hybridisation is sp2  The bond angle is SnX2 will be less than 120° (due to presence of one LPe.)  The shape of SnX2 molecule is bent. Examples on sp2 hybridisation Example s bond BH3 3 BF3 3 BCl3 3 CH3 3 CH2= CH23, 3 Graphite 3 HNO3 3 – NO 3 3 HNO2 2 SO 2 2 SO 3 3 – HCO3 3 CO 3–2 3 SnCl2 2 SnBr2 2 SnI2 2 AlCl3 3 GaCl3 3 PbCl2 2 I.p.e. 1 1 1 1 Hybridisation sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 sp2 Bond angle 120° 120° 120° 120° 120° 120° 120° 120° < 120° < 120° 120° 120° 120° < 120° < 120° < 120° 120° 120° < 120° Shape Trigonal planar Trigonal planar Trigonal planar Trigonal planar Trigonal planar Trigonal planar Trigonal planar Trigonal planar Angular (V) Angular (V) Trigonal planar Trigonal planar Trigonal planar Angular (V) Angular (V) Trigonal planar Trigonal planar Trigonal planar Angular (V) 58 sp3 Hybridisation : (I) In this hybridisation one s orbital & three p orbitals of an atom of a molecule or ion, are mixed to give four new hybrid orbitals called as sp3 hybrid orbitals. (II) The angle between these four hybrid orbitals will be 109°28’ C (ground state) C (excited state) C atom share four electrons with 4 hydrogen atoms (III) The shape obtained from these hybrid orbitals would be tetrahedron. Three following examples represent this condition. Chemical Bonding & Molecular Structure (II) sp3 hybridisation, pyramidal shape & bond angle will be less than due to the presence of the lone pair electron on nirtogen repels bond pair electron more therefore bond angle is reduced from 109°28’ to 107°. The repulsion between IP – b.p > bp – bp. (c) Two sigma bonds & two lone pair lone pair electrons : (I) This condition is shown by following compounds and ions. H2 O, H2 S, H2 Se, H2Te, OCl2, OBr2, OF2, OI2 etc. (II) In all above examples, the central atom showing sp3 hybridisation, angular shape and bond angle will be either less then 109°28’ or more than 109°28’. (a) Four sigma bonds with zero lone pair electron : The following examples represent this condition. CH4, CF4, CCl4, CBr4, Cl4, NH4+, BF4–, AlF4–, BeF4–2, In above compounds, bond angle is 109°28’ & tetrahedron shape. (b) Three sigma bonds & one lone pair of electron (I) This condition is shown by following compounds & ions. In H2O the hybridisation on O atom is sp3, but due to presence of two lone pair electrons they repell each other and then repell their adjacent bond pair electron. These repulsion will be in following order. I.p. – I.p. > I.p. – b.p. > b.p. – b.p. l.p. = lone pair electron b.p. = bond pair electron NH3,PH3,AsH3,SbH3,BiH3,NF3,PF3,NCl3,PCl3 : CH3,H3O+ ClO3– 59 Chemical Bonding & Molecular Structure Examples on sp3 Hybridisation Example  bond CH4 4 I.p.e. Hybridisation Shape - sp3 Tetrahedron 3 Tetrahedron CCl4 4 - sp CBr4 4 - sp3 Tetrahedron PCl4+ 4 - sp3 Tetrahedron NH4+ 4 - sp3 Tetrahedron BF4– 4 - sp3 Tetrahedron 3 Tetrahedron – 4 AlF 4 - sp BeF4–2 4 - sp3 Tetrahedron MgF4–2 4 - sp3 Tetrahedron Ni(CO)4 4 - sp3 Tetrahedron 3 Pyramidal NF 3 3 - sp NCl3 3 1 sp3 Pyramidal N(CH3)3 3 1 sp3 Pyramidal PF3 3 1 sp3 Pyramidal 3 Pyramidal PCl3 3 1 sp AsCl3 3 1 sp3 Pyramidal SbCl3 3 1 sp3 Pyramidal BiCl3 3 1 sp3 Pyramidal NH3 3 1 sp3 Pyramidal 3 Pyramidal – : CH3 3 1 sp H3O+ 3 1 sp3 Pyramidal SO 32– 3 1 sp3 Pyramidal ClO3 3 1 sp3 Pyramidal 3 Pyramidal XeO3 3 1 sp H2O 2 2 sp3 Angular (V) H2S 2 2 sp3 Angular (V) NH2– 2 2 sp3 Angular (V) OF 2 2 2 sp3 Angular (V) 3 Angular (V) Cl2O 2 2 sp SCl2 2 2 sp3 Angular (V) Diamond 4 - sp3 Tetrahedron SiO2 4 - sp3 Tetrahedron - 3 Tetrahedron SiC 4 sp 60 Chemical Bonding & Molecular Structure sp3d Hybridisation : (I) In this hybridisation one s orbitals and one d orbital are mixed to give five new hybrid orbitals which are equivalent in shape and energy called as sp3d hybrid orbitals. (II) Out of these five orbitals, three hybrid orbitals are at 120° angle and two hybrid orbitals are perpendicular to the plane of three hybrid orbitals that is trigonal planar, the shape of molecule becomes is trigonal bipyrmaidal. For example, PF5 showing sp3d hybridisation P (ground state) 5 The shape of all above molecules is trigonal bipyramidal. (b) Four sigma bonds and one lone pair of electron The following examples represent this condition. P* (excited state) SF4, SeF4, TeF4, PoF4, PF4– , SbF4–, SCl4, SeCl4, TeCl4 etc. The shape of all above examples will be irregular tetrahedron. Examples SF4 P atom share with five e of F S ground state S excited state (III) In this hybridisation dz2 orbital is hybridised with s and p orbitals. In this way five sp3d hybrid orbitals form five sigma bond with five F atoms and give a molecule of PF5, shape of this molecule is trigonal bipyramidal. Axial two P–Cl bonds are longer than equatorial three P–Cl bond due to repulsion between 3 equatorial bp of e– and 2 axial b.p. of e– In above hybridisation there are four conditions S atom share with four e– of F (a) Five sigma bonds and zero lone pair electron : The following examples represent this conditions. PF5, PCl5, PBr5, Pl5, AsF5, AsCl5, SbCl5, SbF5 etc. 61 (c) Three sigma bonds & two lone pair of electrons : The following examples represent this condition. CIF3, BrF3, IF3, BrCl3, ICI3 etc. sp3d3 Chemical Bonding & Molecular Structure Hybridisation : (I) In this hybridisation, one s-orbitals & three d-orbitals are mixed to give seven new hybrid orbitals known as sp3d3 hybrid orbitals. (II) In this hybridisation d-orbitals used are dxy dxz & dyz orbitals. (III) These seven sp3d3 orbitals are configurated in pentagonal bipyramidal shape. (IV) Five bond angles are of 72° & two bond angles of 90°. The shape of all above compounds is ‘T’ shape. (d) Two sigma bonds & three lone pair of electrons : The following examples represent this condition. ICl2–, IBr2–, ClF2–, IF2–, BrF2–, XeF2–, I3–, Br3– The geometry of above examples will be linear (V) The following examples showing sp3d3 hybridisation –IF7 & XeF6. sp3d2 Hybridisation : (I) In this hybridisation, one s-orbitals, three p-orbitals & two d-orbitals are mixed to give six new hybrid orbitals known as sp3d2 hybrid orbitals. (II) The shape of molecule obtained from above six Shape of covalent molecules : hybrid orbitals will be symmetrical octahedral. • It was given by Sidgwick & Powel. (III) The angle between all hybrid orbitals will be 90°. • According to this concept the shape of Example : SF6, AlF6–2, PF6–, ICl5, XeF4, XeOF4, covalent compound depends on total pairs of ICl4–, electron (bonded or non bonded) present on the (IV) Two ‘d’ orbital participates in the hybridisation are central atom. Which is given as below – dx2–y2 and dz2 . S.No. Total pair Shape Hybridisation SF6 : – of e S (ground state) 1. Two Linear sp 2. Three Coplaner trigonal sp2 (Triangular) S (excited state) S (after hybridisation) 3. Four Tetrahedral sp3 4. Five Trigonal bipyramidal sp3d 5. Six Octahedral sp3d2 (Square bipyramidal) 6. Seven Pentagonal sp3d3 bipyramidal 62 Chemical Bonding & Molecular Structure Glesspie and Nyhom theory or Ex.13 The hybridization of carbon atoms ion C – C single bond of H – C  C – CH = CH2 is (A) sp3-sp3 (B) sp2-sp (C) sp-sp2 (D) sp3 -sp Sol. (C) As the first carbon has triple bond and second one has double bond with other two carbon atom. VSEPR theory : (Valence shell electron pair repulsion theory) (I) If the cenral atom possess only bonded pairs of electrons along with identical atoms then shape of the compound is symmetrical and according to Sidgwick & Powl. eg. BF3 – 120° – CH4 – 109°28’ – CO2 – 180° – triangular tetrahedral linear Ex.14 C – C bond in C2H6 undergoes heterolytic fission, the hybridisation of carbon is/are (II) If the central atom possess bonded pair of electrons as well as lone pair of electron, then (A) sp2 (B) sp3 shape of the molecule will be unsymmetrical ie. the (C) sp2, sp3 (D) sp, sp2 original bond angle will disturbed due to repulsion Sol. (C) During heterolytic fission there will be formation between lone pair of electrons. of CH 3 + and CH 3 ¯ which has sp 2 and sp 3 similarly on having different type of side atoms, hybridization respectively. molecule becomes unsymmetrical due to unequal force of repulsion between e–. Order of repulsion Ex.15 The enolic form of actone contain  bond (a),  is – bond (b) lone pair (c) .p, – .p > .p – b.p. > b.p. – b.p. (A) 9a, 1b, 2c (C) 10a, 1b, 1c (B) 8a, 2b, 2c (D) 9a, 2b, 1c 1 Bond angle  No. of lone pair of electron Sol. (A) This is clear from its structure as shown below. (III) By increasing one lone pair of electron, bond angle is decreased approx by 2.5°. H –H e.g. CH4 NH3 H2O – sp3 | | 109° 107° 105° hybridisation H—C—C=C—H (IV) In the different molecules if central atom have same | | H H Ex.16 The hybridization and geometry of BrF3 molecule are - number of lone pair of electron then bond angle will depend on electronegativities of A & B. In ABx type of molecules if side atoms are same EN of central atom increases the bond angle increases (A) sp3d and distorted trigonal bipyramidal (B) sp3d2 and Tetragonal (C) sp3d and bent (D) None Sol. (A) There is one s, three p and one d orbital while hybridization Bond angle of – NH3 > PH3 > AsH3 63 Chemical Bonding & Molecular Structure In ABx type molecules, if central atoms are same Ex.20 Shape of NH3 is very similar to and the EN of side atoms increases then bond angle (A) CH4 (B) CH3¯ decreases. (C) BH3 (D) CH3+ Sol. (B) Both have the same state of hybridization with one lone pair of e¯ each. PF3 < PCl3 < PBr3 < Pl3 (EN of side atom decrease) OF2 < Cl2O < Br2O SF2 < SCl2 < SBr2 Bond angle depends on size of side atom, On increasing size of atom bond angle increases. Cl2O > H2O Ex.17 In following which central atom has different hybridisation than other – (A) Cl2O (B) OF2 (C) H2O (D) SO2 2 Sol. (D) Because it has sp hybridisation other three have sp3 hybridisation. Ex.18 In following hydrocarbon which one has 75% p character and 25% s charater (A) C2H4 (B) C2H2 (C) CH4 (D) :CH2 3 Sol. (C) CH4 because it has sp hybridisation. Ex.21 Which of the following has pyramidal shape(A) XeO3 (B) XeF4 (C) XeF2 (D) XeF6 Sol. (A) Due to presence of lone pair of electron. Ex.22 Ether and water have same hybridization at oxygen. What angle would you expect for them Sol. In H2O bond angle is less than 109º28’ due to lone pair and bond pair repulsion. But in ether, due to strong mutual repulsion between two alkyl groups bond angle becomes greater then 109º28’ Characteristic of Covalent Compound (I) Physical state :- Covalent compounds are found in all the three states - Gas, Solid & Liquid. Separate molecules – In gaseous state Associate molecules – In liquid & solid state (Due to strong vander waal’s force and hydrogen bonding among the molecules.) As the molecule weight increases physical state changes eg. F2 and Cl2 Br2 l2, At2 gas liquid solid Top to bottom in a group. Vander waal’s force increases between the molecules. (II) Covalent solid : - Those solids in which atoms are Ex.19 Which of the following statement in incorrect linked together by covalent bonds, forms inifinite (A) Non-bonding pairs occupy more space than three dimensional giant structure. bonding pairs e.g. Diamond, Graphite, AIN, SiC, SiO2 etc. (B) The bonding orbitals in a trigonal bipyramidal 3 molecule are described as sp d hybrid Molecular solid : - Discrete (separate) molecules (C) SnCl2 has linear shape are formed by covalent bonds and then the molecules + – (D) PCl4 and AlCl4 are isoelectronic associated due to intermolecular force of attraction. Sol. (C) SnCl2 is a bent shape molecule due to presence (Vanderwaal force) of lonepair of electron other choices are correct. eg. Solid I2, dry ice (Solid CO2) etc. 64 (III)Conductivity : - Mostly covalent compounds are bad conductor of electricity. But few polar covalent compounds due to self ionisation can conduct electricity.e.g. H2O, liq. NH3 etc. H2O + H2O 2NH3 H3O+ + OH– NH4+ + NH2– Free ions are formed which can conduct electricity. Exceptions : - Graphite, HCl in water. Chemical Bonding & Molecular Structure (V) Unit of dipole moment is Debye 1 Debye = 1 × 10–18 esu cm. = 1.6 × 10–29 coulomb metre (VI) In the diatomic molecule  depends upon difference of EN i.e.   EN order of  : H–F > H–Cl > H–Br > H–I  = 0 for H–H, F–F, Cl–Cl, Br–Br, O–O (VII) For polyatomic molecules  depends on the vector sum of dipole moments of all the covalent bonds. (IV) Solubility : - Non polar compound are soluble in non polar solvents. Non polar compounds forms (VIII) For PCl and SF , etc.  = 0 due to their regular 5 6 Vander waal bond with non polar solvent molecules. geometry. (V) Isomerism : - Covalent bond is rigid and directional, (IX) Benzene, naphthalene, diphenyl have  = 0 due to planar structure. so it shows isomerism. (X) If the vector sum is zero, than compound is noneg. Organic compounds. polar compound or symmetrical compound (and it (VI) Reaction : - Reaction between covalent compounds is not essential that individual  of every bond should are slow. Because it involves breaking of odl bonds be zero). and formation of new bonds. Example : (A) BX3, CCl4 , SiCl4, CH4, CO2CS2, PCl5, SiH4 etc. DIPOLE MOMENT (I) Polarity of any polar covalent bond or moleucule is measured in terms of dipole moment. In these examples the bond B–F, C–Cl, C–H, C– O, P–Cl etc. are polar even though compounds are non-polar. (II) For measurement of extent of polarity, Pauling introduced the concept of dipole moment (). (B) (Ionic Nature in Covalent Bond) The product of positive or negative charge (q) and the distance (d) between two poles is called dipole moment. Here -  = q × d (magnitude of charge × distance) (III) Dipole moment is a vector quantity i.e. it has both magnitude as well as direction. (IV) Direction of dipole moment is represented by an arrow pointing from electro +ve to electro -ve element and from central atom to lone pair of electrons. 4 Total  = 1 + 2 + 3 + 4 = 1.47D 65 (XI) Dipole moment of H2O is 1.85 D which is resultant of two O–H bonds. Chemical Bonding & Molecular Structure (IV) To locate position of substituents in aromatic 1 compounds.   Bond angle (a) If same substituents are preset in the symmetrical position of benzene ring compounds will be zero of H2O of H2 S because electronegativity of oxygen is higher than sulphur. (XII) Angular structure of molecule have greater dipole moment. Application of dipole moment : (I) To determine polarity and geometry of molecule If = 0 compound is non polar and symmetrical eg. CO2, BF3, CCl4, CH4 . BeF2 etc. If 0 compound will be polar and unsymmetrical. H2O, SO2, NH3, Cl2 O, CH3Cl, CHCl3 etc. (a) Dipole moment of cis isomers is normally higher than trans isomers. (b) As angle between subtituents decrease value of increase Some important orders of diple momentsHF > H2O > NH3 > NF3 H2O > H2S CH3Cl > CH3F > CH3Br > CH3I BF3 < NF3 < NH3 HF > H2O > SO2 > NH3 PH3 < PCI3 eg. – NO2– > NO2 > NO 2 (II) To calculate % ionic character :% Ionic character  Experimental value of   100 Theoritical value of  (III)To distinguish cis form or trans form:- H2O < H2O2 Ex.23 Which of the following bonds is most polar (A) O – H (B) P – H (C) C – F (D) S – Cl Sol. (C) Due to maximum electronegativity difference. (b) If two groups have opposite inductive effect than Ex.24 Bond length of HCl is 1.275 Å (Charge = 4.8 × trans-isomer will have greater dipole moment \10–10 e.s.u.) if  = 1.02 D, then HCl is eg. (A) 100% ionic (B) 83% covalent (C) 50% covalent (D) 40% ionic Sol. (B) % ionic character observed  102 . = theoritica l  × 100 = × 100 1275 . x 4.8 = 17% ionic = 83% covalent 66 Ex.25 The dipole moment of the ammonia molecule is 1.48D. The length of the dipole is (A) 3.08 × 10–11 m (B) 5 × 102 m (C) 308 m (D) None Sol. (A) M q Chemical Bonding & Molecular Structure Example:   = 1.48 × 3.33 × 10–30 cm = 4.93 × 10–30 cm. (Lowry - Bronsted acid) = 1.6 × 10–19C (e– acceptor) 4.93  10 30 = q= = 3.08 × 10–11 m 1.6  10 19  = 0.0308 nm The NH3 molecule can not have the shape of an equilateral triangle because in the case its dipole moment would equal zero. It is actually constructed in the form of a triangular pyramid with the nitrogen atom at its vertex and the hydrogen atoms at the corners of its base. COORDINATE BOND (I) It is a covalent bond in which the shared e– pair come from one atom is called coordinate bond. O3 N2O Compounds in which Ionic, covalent and coordinate bonds are present, are as followsNH4Cl, CuSO4, K4[Fe(CN)6], KNC, Na3PO4, KNO3 , etc (II) Necessary conditions for the formation of co-ordinate bond are(a) Octet of donor atom should be complete and should have atleast one lone pair of electron. (b) Acceptor atom should have a defficiency of at least one pair of electron. eg. (III) Atom which provide electron pair for shairing is called donor. Ex.26 The compound which contains both covalent and co-ordinate bond is (A) C2 H5NC (C) HCN (B) C2H5CN (D) None Sol. (A) Though all compounds have covalent bond but there is coordination bond also between N and C in C2H5 NC (IV) Other atom which accepts electron pair is called Ex.27 The type of bond present in N2O5 are acceptor. That is why it is called donor-acceptor or (A) Only covalent dative bond (B) Only ionic (C) Ionic and covalent (D) Covalent and coordinate Sol. (D) The structure of N2O5 clears about it BF3 is electron defficient compound. O N —O— N O || O || O 67 Ex.28 No. of covalent and co-ordinate bond in pyrosulphuric acid are (A) 6, 4 (B) 6, 6 (C) 4, 4 (D) 4, 6 Sol. (A) Structure of H2S2O7 is as follows O O   H—O— S —O— S —OH   O O Ex.29 Which is not true according to VBT (A) A covalent bond is formed by the overlapping of orbitals with unpaired electrons of opposite spins Chemical Bonding & Molecular Structure HYDROGEN BOND Definintion : (I) It is an electrostatic attractive force between covalently bonded hydrogen atom of one molecule and an electonegative atom (F, O, N) (II) It is not formed in ionic compounds (III) H–bond forms in polar covalent compouds, (not in non-polar) (IV) It is also known as dipole-dipole attraction H — F– .... H+ — F– ...... H+ — F– Main condition for H-bonding : (B) A covalent bond is formed by the over lapping (I) H— should be covalently bonded with high electro– of orbitals with unpaired electrons of same spin ve element like F, O, N (C) The greater the extent of overlapping the (II) Atomic size of electro–ve element should be small. stronger is the bond (D) Overlapping takes place only in the direction of maximum electron density of the orbital Decreasing order of atomic size is– Sol. (B) Electrons of same spin never combine to form a bond. Other facts are true for VBT. Decreasing order of atomic size is– Ex.30 Why covalent bonds are always directional - N F (4.0) > > O O (3.5) > > F N (3.0) Sol. As the covalent bonds are formed due to overlapping (III) Strength of H–bond  Electronegativity of Z of orbitals which are directional 1 (element)  atomic size of Z Ex.31 Allyl cyanide has (IV) Hydrogen bonding occurs in HCN, due to (–C N) (A) 9 sigma bonds and 4 Pi bonds triple bond (sp hybridisation), electronegativities of (B) 9 sigma bonds, 3 pi bonds and 1 lonepairs carbon and nitrogen increases. (C) 8 sigma bonds and 5 pi bonds H — C  N.........H — C  N..........H — C  N (D) 8 sigma bonds, 3 pi bonds an 4 non-bonding electrons Sol. (B) This is clear from structure of allylcyanide H H H | | | H— C = C — C —CN: x | H 68 Chemical Bonding & Molecular Structure (A) Intermolecular H–bond H–bond formation berween two or more molecules of eithe the same or different compounds known as Inter molecular H–bonding These are two types. (i) Homointermolecular :- H–bond between molecules of same compounds. eg. (ii) Hentro intermolecular :- H–bond between molecules of different compounds. eg. alcohol, water Effect of H–bond on physical properties : (I) Solubility (A) Inter molecular H–bonding (a) Few organic compounds (Non-polar) are soluble in water (Polar solvent) due to H–bonding Other examples–Glucose, Fructose etc, dissolve in water. (b) Ketone, ether, alkane etc. are insoluble (no H– bond). Dimethyl ether is soluble in water while diethyl ether is partially soluble, due to bulky ethyl groups H–bonding interupts (c) Solubility order– CH3OCH3 < CH3OH Primary amine > secondary amine > tertiary amine (B) Intra molecular H–bonding: (a) It decreases solubility as it form chelate by H– bonding, so H–is not free for other molecule. (i) H–bonded with electronegative elements of a (b) It can not form H–bond with water molecule functional group, form H–bond with another so can not dissolves. electronegative element present on nearest position (C) Inter molecular H–bond on the same molecule. (B) Intra molecula H–bond :- It takes place within the molecule. (ii) This type of H–bond is mostly occured in organic compounds. (iii) It result in ring formation (Chelation). eg. It can form H–bond with water molecule so it can dissolvel 69 Chemical Bonding & Molecular Structure (IV) Molecular weight : (II) Viscosity: H – bond associates molecules together so viscosity increases Molecular wt. CH3COOH is double of its molecular formula, due to dimer formation occur by H–bonding (V) Physical state: H2O is liquid while H2S is gas. Water and Ice:- Both have H–bonding even then density of ice is less than water. (III)Melting point and boilling point (a) Due to intermolecular H–bond M.P. & B.P. of compounds increases. H2O > CH3OH > CH3 — O—CH3 (b) Trihydric alcohol > dihydric alcohol > monohydic alcohol Monocarboxylic acid form stronger H–bond than alcohol of comparable molecular weight. Therefore B.P. of carboxylich acid is higher than alcohol. (c) Decreasing order of M.P. & B.P. isomer amines1°–amine > 2°–amine > 3°–amine R | R — NH2 > R — NH — R > R — N — R (no hydrogen with nitrogen atom) Volume of ice is more because of open cage like crystal structure, from by association of water molecules with the help of H–bond. H2O becomes solid due to four hydrogen bond among water molecule are formed in tetrahedral manner. (VI) Base strength CH3NH2, (CH3)2 NH, (CH3)3 N form H–bond with water. So less hydrolysis i.e. it gives OH– ions. While (CH3)4 N+ OH– (ammonium compound) will give OH– ion in large amount due to no H–bonding. + 3 – 3 (d) Boiling points of VA, VIA, VIIA hydrides decreases on decreasing molecular weights. VA VIA VIIA NH3 H2O HF b.p. HF > HI > HBr > HCl PH3 H2S HCl H2O > TeH2 > SeH2 > H2 S AsH3 SeH2 HBr NH3 > SbH3 > AsH3 > PH3 SbH3 TeH2 HI Effect of intramolecular H–bonding (e) But sudden increase in boiling point of NH3, H2O (I) Strength of acid and HF is due to H–bonding (a) The formation of intramolecular H–bonding in H2O > HF > NH3 PH3 < H2S < HCl the conjugate base of an acid gives extra stability to Intramolecular H–bonding gives rise to ring formation, conjugate base and hence acid strength increases so the force of attraction among these molecules are eg. Salicylic acid is stronger than benzoic acid vander waal force. So M.P. and B.P are low. and 2, 6 – dihydroxy benzoic acid > salicylic acid. 70 Chemical Bonding & Molecular Structure Ex.32 Which one of the following does not have intermolecular H-bonding (A ) H 2O (B) o-Nitrophenol (D) CH3 (C) HF Sol. Nitrophenol has intermolecular H bonding as shown below Hence (2) (b) C 2H2SH is more acidic than C2H5OH. In C2H5OH, H–bond forms so H+ is not free (c) HF is weaker acid than HI, due to H–bond in H – F, H+ is not free (II) Stability of chloral hydrate:- Ex.33 Arrange in incresing order of their boiling pointHF, HCl, HBr, H Which order is correct – (A)HCl < HBr < H < HF (B)HCl > HBr > H > HF (C)HCl < HBr > H < HF If two or more OH group on the same atom are (D) None present it will be unstable, but chloral hydrate is stable Sol. (A) HCl < HBr < H < HF (due to H–bonding). due to strong hydrogen bonding in HF VANDER WAAL’S FORCES (a) This type of attractive forces occurs in case of non polar molecules such as H2, O2 , Cl2, CH4, CO2 (III)Maleic acid (cis) is stronger acid than fumaric acid etc. (trans). (b) The existence of weak attractive forces among the nonpolar molecule was first proposed by dutch scientist J.D. Vander Waal (c) Vander waal force  molecular weight  Atomic weight  Boiling point Types of Vander Waal’s force : (i) Ion dipole attraction - This force is between an ion such as Na+ and a polar molecule such as HCl + 71 Chemical Bonding & Molecular Structure (ii) Dipole Dipole attraction - It is again in between two polar molecules such as HF and HCl HF HCl(g) (iii) Ion induced dipole attraction - In this case a neutral molecule is induced by an ion as a dipole as shown in fig POINTS TO REMEMBER (a) Bond energy of a diatomic molecule is enthalpy change to dissociate one mole of the diatomic gaseous molecule to form the respective gaseous atoms. (b) Strong metallic bonding is favoured by smaller size of kernel. (c) Band theory is related with metallic nature  Na+ Cl2 Before induction after induction (d) The resonance structure do not have independent existence. (e) The hydration of ionic compounds involves evolution of heat, weakening of attractive forces and dissociation into ions (iv) Dipole - Induced dipole attraction : In this case a (f) The maximum covalency is equal to the actual number neutral molecule is induced as a dipole by another of s-and p-electrons in the outermost shell dipole as shown in fig. (g) Many ionic crystals dissolve in water because water decreases the interionic attraction in the crystal lattice  due to solvation. (h) Hydrogen bond stabilizes the secondary structure of proteins. e.g. HCl Cl2 (v) Induced dipole - induced dipole attraction or London dispersion force between two non polar molecules as in Cl2, He etc. (i) Isoelectronic molecular species have similar molecular orbitals and similar structure according to iso electric principle (j) The amount of energy required to break one mole of the bond and separate the bonded atoms in the gaseous state is known as bond dissociation energy (k) The average angle between the bonded orbitals is known as bond angle . (l) Bond length may be defined as the average distance between the centres of nuclei of two bond atoms. (m) Polymorphism (Allotropy) is a term used when one and the same substance can crystallise in more than Note: The relative strength of various bonds is as one forms. follows (n) In SO2 there is p - p bonding as well as p - d Ionic bond > Covalent bond > Metallic bond > Hbonding due to overlap of filled p orbitals of oxygen bond > Vander waal bond with the vacant 3d orbitals of sulphur. Cl2 Cl2 72 (o) H2SO3 exists in two forms which are always in equilibrium with one another Chemical Bonding & Molecular Structure (r) In the solid phase, SO3 polymerizes to puckered rings or the more stable layer structure as shown. (Symmetrical sulphurous acid) Molecular Orbital theory (MOT): MOT put forward by Hund & Mulliken, which can be applied to explain the properties, which the old VBT (Valence bond theory) was unable to explain (Unsymmetrical sulphurous acid) eg. Paramagniic nature Of O2 molecule, as per VBT The unsymmetrical form containing S – H bond (:O: :O:) it should be diamagnetic. explains the reducing character of H2SO3. Definition: (p) The anion of H2SO3 i.e. SO3 2– has a pyramidal The atomic orbital lose their identity during molecule structure with S atom in sp3 hybridization and all the formation (overlapping) and form new orbitals S – O bonds are of equal bond length (134 pm) termed as molecular orbitals. (q) In the gas phase, SO3 is a planar triangular molecule Characteristic of molecular orbitals: with sulphur atom in sp2 hybridization state (I) Molecular orbital formed by overlapping of atomic orbital of same energy (II) Number of melecular orbital formed = number of atomic orbital involved in overlapping (III) Half of the molecular orbital have lower energy are called Bonding molecular orbital. In order to account for the equivalency of all the S – (IV) Half are of higher energy - termed as Antibonding O bonds and short S – O distance of 143 pm, SO3 molecular orbital may be considered as a resonanchybrid involving (V) Electronic configuration in various molecular orbital p-p S – O bonding along with additional p-p are governed by same three rules. bonding. (a) Aufbau’s rule (b) Hund’s rule (c) Pauli’s exclusion principle 73 Chemical Bonding & Molecular Structure Comparision of Bonding molecular orbital & Antibonding molecular orbital : Bonding molecular orbital (BMO) Antibonding Molecular orbital (ABMO) Bonding MOis the result fo the linear combination of AO when their wave function are added Yb = YA + YB It does not have node. ABMO is result of linear combination of AO when their wave function are substracted Yb = YA + YB It always have a node between two nuclei of bonded atom. Charge density decrease in between two nuclei, leads to repulsion between two atoms. Charge density increase between two nuclei resulting attraction between two atoms Energy of BMO is less, hense stable Notation of molecular orbitals: As atomic orbitals are known by letters s, p, d and f depending on their shapes. Similarly for molecular orbital. For bonding molecular orbital- etc. For antibonding molecular orbital- etc. are used for different shapes of electron cloud. Energy of ABMO is high, hence unstable Energy Level Diagram of molecular orbital : On the basis of Aufbau’s rule - increasing order of energies of various molecular orbitals is1s < 1s < 2s < 2s < 2pz < 2px  2py < 2px 2py < 2pz Energy level diagram for homonuclear diatomic molecules like, O2, F2, Ne2 For O2 molecule- 74 Energy level diagram for B 2 , C 2 and N 2 molecules – 1s < 1s < 2s < 2s < 2px 2py <  2pz < 2px2py < 2pz For N2 molecule  Chemical Bonding & Molecular Structure The pairing in 2px and 2py or 2px and  2py will take place only when each molecular orbital of indenticle energy has one electron.  After writing the molecular orbital diagram following parameter about molecules/ion may be predicted. (i) Bond order :Bond order = ½ [No. of electron in bonding molecular orbital ie. Nb – No. of electron in antibonding molecular orbital ie. Na] = ½[Nb–Na] B.O. = ½ [Nb–Na] (ii) Bond length :- (distance between two nuclei) Bond length  If Nb  Na 1 Bond order Molecule exists Nb  Na   Molecule do not exists Nb  Na  Cause of exceptional behavior of molecular orbital in B2, C2 and N2:    (iii) Stability of molecules - stability Bond order of molecule (iv) Dissociation energy - Bond disssociation energy Bond order (v) Magnetic property (a) When electron in MO are paired – diamagnetic (b) When electron in MO are unpaired – paramagnetic Energy of 2s and 2p atomic orbitals lie fairly close Bonding in molecules :Due to small energy difference between 2s and 2pz orbitals, the interaction between them is quite large. (I) Hydrogen moleculeHaving two H atoms with one electron each (1s’) This results in loss of energy by 2s and 2s and thus 2s and 2s becomes more stable at the  cost of 2px and 2px which gets unstablised (Higher energy). Electronic configuration of molecules and their related properties :for writing electronic configuration of diatomic molecules following two rules to be followed-  Count the number of electrons present in two atoms and then fill in the appropiate energy level diagram according to Aufbau rule. M.O. configuration of H2 = (1s)2 (1s)0 Bond order = ½ [Nb – Na] = ½ [2 – 0] = 1 i.e. single bond Having paired electron so diamagnetic. Stability quite stable (having single bond) 75 Chemical Bonding & Molecular Structure (II) H2+ ion – Configuration of H2+ = (is) 1 (is)0 One electron in bonding molecular orbital. Ex.34 The bond order of N2¯ anion is (A) 1 (C) 2.5 (B) 2 (D) 3  Sol. (C) Out of 15 e¯ in N2¯, 10 are in bonding MO’s and 5e¯ are in the anti bonding MO’s Hence bond order = So paramagnetic Bond order = ½ [1– 0] = ½ Less stable (III) H2– anion M.O. configuration - (1s)2 (1s)1 Paramagnetic Bond order = ½ [2 – 1] = ½  Stability is less than [H2+] because H2– Contain ab ABMO electron 1 2 (10 - 5) = 2.5 Ex.35 Which of the following has maximum bond strength (B) O2+ (D) O2 2– (A) O2 (C) O2¯ Sol. (B) Bond order of O2 + is 2.5 which is maximum among all the species hence its bond strength will also be the maximum. Ex.36 Which of the molecules B2 and C2 has a higher energy of dissociation into atoms? compare the magnetic properties of these molecules. Sol. We draw energy level diagrams showing the formation of B2 and C2 molecules AO’s B AO’s MO’s B2 MO’s AO’s B AO’s C C2 C (IV) Helium molecule (He2) :     M.O. configuration (1s)2 (1s)2 Diamagnetic Bond order = ½ [2 – 2] = 0 (zero) Bond order zero indicates no linkage between He atoms. Hence He2 molecule does not exist  Stability (He2) Highly unstable molecule 76 Chemical Bonding & Molecular Structure Energy level diagrams showing the formation Factors affecting Bond Length :of B2 and C2 molecules. 1 The difference between the number of bonding and (a) electronegativity :- Bond lengh  ( While EN the number of antibonding electrons in the B 2 B.E. EN) molecule can be seen to be two, and in the C2 molecule, four. This corresponds to a bond H–F < H–Cl < H–Br < H–I multiplicity of 1 and 2, respectively. Consequently, (b) bond order or number of bonds :- Bond length the C2 molecule having a higher multiplicity of the bond between the atoms must be more stable. This 1  Number of bond or bond order conchosion corresponds to the experimentally established values of the energy of dissociation into Bond energy Number of bond atoms of the molecules B2 (276 KJ/mol) and C2(605 KJ/mol) e.g. C–C, C = C, C C In the B2 molecule, two electrons are arranged, Bond length 1.54 Å 1.34 Å 1.20 Å according to Hund’s rule, in two px 2py orbitals. increa sin g    The presence of two unpaired electrons imparts Bond energy 80 140 180–200K.Cal. paramagnetic properties to this molecule. In the C2 molecule, all the electrons are paired, consequently, increa sin g    this molecule is diamagnatic C—0 BOND PARAMETERS 1.43 Å 120 Å 1.13Å (I) Bond Lengthe (Bond distance) (II) Bond Angle (III) Bond Energy (I) Bond Lenght :- The average distance between the 1.47 Å 1.28 Å 1.15 Å nucleus of two atoms is known as bond lenght, (c) Resonace :- (due to resonance bond length affected) normally it is represented in Å. eg. A B eg.1. Benzene It depends mainly on electronegativilties of constituent atoms C—C bond length 1.54Å Case - I Electronegativity difference is zero thenC—C bond length 1.34 Å Bond length = rA + rB Or dA –B = rA + rB whereA is covalent radius of A rB is covalent radius of A XA is electronegativity of A but bond length is between XB is electronegativity of B single & double bond is = 1.39 Å If rA = rB then Bond length = 2rA or 2rB eg. 2. Bond length of C—O in CO is 1.15 Å  2 Case II Electronegative difference is not equal to zero thenBond length is given by shomaker & Stevenson formula is Bond length = rA + rB – 0.09 (XA – XB) Difference in electronegativities Resonance occurs in CO2 a followsBond length = 1.15 Å (Between double & triple bond) 77 Chemical Bonding & Molecular Structure (d) Hybridisation : 1 Bond length  s character •• •• Example:- NH3 PH3 Bond angle 107° 93° •• As H3 91° – Eletronegativity decreasing – Bond angle will decrease Case-III When hybridisation is same, lone pair are same, Central atom is same, bonded atoms are different. (II) Bond Angle :- The angle between any two adjacent bond is known as bond angle. It is represented in degree (°), min (‘) and second (‘’) Factors affecting the bond angle(a) Number of bond : Bond angle  Number of bonds (Bond order) 109° 120° 180° sp3 OF2 103 – 105° Electronegativity sp3 Cl2O 109 – 111° of bonded atom is sp3 Br2O 116 – 118° decreasing Here bond angle  1 electrongativity of bonded atom size of side atom (III) Bond Energy (BE) :- Bond energy may be difined as(a) Bond formation energy:- Energy released when any bond is formed is known as bond formation energy or bond energy. (b) Hybridisation :Case 1 When hybridisation is same, bonded atoms are same, central atom and lone pair are different. 1 Then bond angle  No.of lone pair Example :•• CH 4 •• N H3 H2 O •• Hybridisation sp3 sp3 sp3 Bond angle 109 > 107 > 105 (b) Bond dissociation energy :- Energy required to dissociate any bond is known as bond dissociation energy. Calculation of released energy is more diffcult than the dissociation energy therefore dissociation energy of bond is calculated and is assumed as bond energy or bond formation energy. Case-I Bond energy = bond dissociation energy eg :- N2 > O2 > H2 > F2 Case-II For polyatomic molecule :- Bond energy No 1.p. one 1.p. two 1.p. Case-II When hybridisation is same, bonded atoms are same lone pair is same but central atom is different. Then bond angle  electrongativity of central atom In diatomic molecule : eg;-  Bond dissociation energy (D) H | H  C H | H Bond energy = per CH bond is 99.5 K. Cal/mole 78 Chemical Bonding & Molecular Structure (e) Resonance :- Bond energy increases due to Theoritical values of bond dissociation energy (D) resonance of individual CH bonds CH4 are given beloweg. In benzene bond energy of CC increases D (CH3H) = 102 K Cal/mole due to electrons of C = C. D (CH2H) = 105 K Cal/mole (f) Hybridisation :- Bond energy  s-character in hybrid D (CHH) = 108 K Cal/mole orbitals. D (CH) = 83 K Cal/mole eg. spsp > sp2sp2 > sp3sp3 Hence bond energy E per CH bond in methane s. character- 50% 33.3% 25% 398  (g) Lone pair of electrons := 99.5 K Cal/mole. 4  Bond dissociation energy (D) is related to the state of hybirdisation. 1 Bond energy  lone pair of electrons Factors affecting the bond energy : (a)  Electronegativity (b) Bond order (c) Atomic size (d) Bond polarity (e) Resonance (f) Hybridisation (g) Lone pair electron (a)  Electronegativity :- Bond energy EN Size of F and O atoms small so their bond energy should be high (small atomic radius) but it is actually less due to lone pair of electrons present on F and O atoms, which repells each other in FF and OO type of bonds. RESONANCE (I) The concept of resonance was introduced by heisen (b) Bond order :- Bond energy  Bond order berg (1920) , and later developed by pauling and ingold, to explain the properties of certain molecules, eg. CC < C=C < CC 79 K. Cal, 143.3 K. Cal., 199.0 K. Cal. (II) It has been found that the observed properties of certain compounds cannot be satisfactorily 1 explained by writing a single lewis structure. The (c) Atomic size :- Bond energy  Atomic size molecule is then supposed to have many structures, each of which can explain most of the properties of eg. CC < CN < NN the molecule but none can explain all the properties •• Exception :- In case of ha AsH3 logen group, order of the molecules. The actual structure is in between of all these contributing structures and is called of bond energy isresonane hybrid and the different individual structures Cl  Cl > Br  Br > F  F > I  I are called resonating structures or canonical forms. Because of higher electron densityb and small size This phenomeni is called resonance. of F atoms, repulsion between of two F atom, (III) Let us discuss resonance in ozone, according to its weakens the bond energy. resonance structure it should have one single bond Other ex. S – S > O – O (OO = 1.48Å) but experiments show that both the bonds are same which can be proved by its C – C > Si – Si > Ge – Ge resonance hybrid as shown below. (d) Bond Polarity :- Bond energy  polarity eg. HF > HCl > HBr > HI eg. HF > HCl > HBr > HI Resonance hybrid 79 To calculate bond order in the polyatomic molecule 1. or ion use following formula : Chemical Bonding & Molecular Structure Properties of metals and their explanation by electron sea model : Bond order 1. = Total number of bonds in a molecule Re sonating Structures eg. CO 4 Bond order  3  1.33 PO Bond order  5  1.25 4 Metallic lusture : (a) It is due to presence of delocalised mobile electrons (b) These loosely bond electrons vibrate due to energy photon of incident light and radiates energy immediately 2. Electrical conductivity : (a) It is due to mobile electrons. (b) These electrons flow equaly in all direction. 7 Bond order   1.75 4 (c) While applying a potential difference across a metal, there will be a directed flow of electrons towards the positive electrode. (a) The constituent particles of metallic solids are metal atoms which are held together by metallic bond. (d) The directed flow of electrons carries the electric current from one point to another and therefore, the metals are known to be good conductors. ClO METALLIC BOND (b) In order to explain the nature of metallic bond Lorentz proposed a simple theory known as electron gas model or electron sea model. (c) A metal atom is supposed to consist of two parts, valence electrons and the remaining part (the nucleus and the inner shells) called kernel. 3. Thermal conductivity : (a) It can also be explained on the basis of electron gas model. (b) While heating a metal kinetic energy of electron increases and propagates through collisons. (d) The kernels of metal atoms occupy the lattice sites while the space between the kernel is occupied by valence electrons. 4. (e) Due to small ionisation energy the valence electrons of metal atoms are not held by the nucleus firmly. Therefore, the electrons leave the field of influence of one kernel and come under the influence of another kernel. Thus the electrons are not localised but are mobile. (b) Basically the kernels can slip over each other when a deforming force is applied. Malleability and ductility : (a) This property is due to non directional nature of metallic bond. 5. High tensile strength : It is due to positively charged kernels and the mobile valence electrons. (f) The simultaneous attraction between the kernels and the mobile electrons which hold the kernel together is known as metallic bond. 80 Chemical Bonding & Molecular Structure STRUCTURES / FORMULAE OF IMPORTANT COMPOUNDS Compound Structure 1. Sodium Sulphide (Na2S) (Na+)2S– – 2. Calcium Hydride (CaH2) Ca++(H–)2 3. Hydrogen Peroxide (H2O2) H–O–O–H 4. Phosphene (PH3) H | H –P–H 5. Carbon mono-oxide(CO) 6. Carbon di-oxide (CO2) 7. Hydrogen Cyanide (HCN) 8. Potassium Isocyanide (KNC) 9. Sodium Hydroxide (NaOH) Electron dot/Cross formula – ×× × × + × 10. Nitrous oxide (N2O) 11. Nitric oxide (NO or N2 O2) 12. Dinitrogen tri oxide (N2O3) × × ×× × × ×× × × 13. Dinitrogen tetraoxide (N2O4) 14. Nitrogen Penta oxide (N2O5) 15. Nitrous acid (HNO2) 16. Nitric acid (HNO3) ×× × × × 81 Chemical Bonding & Molecular Structure 17. Nitrate Ion (NO3–) 18. Ammonium Chloride (NH4Cl) 19. Carbonate Ion (CO3 – – ) 20. Sulphur di-oxide (SO2) 21. Sulphur Tri-oxide (SO3) –– 22. Sulphite Ion (SO3– – ) × ×× × ×× 23. Sulphate Ion (SO4– – ) 24. Thio-sulphate Ion (S2O3– – ) 82 Chemical Bonding & Molecular Structure 25. Phosphoric Acid (H3PO4) 26. Phosphorous acid (H3PO3) – × × 27. Phosphate Ion (PO4 × – – –) ×× – ×× × 28. Potassium Permagnate (KMnO4) Compound × × ×× Structure 29. Carbon di-sulphide (CS2 ) 30. Nitronium Ion (NO2+) 31. Nitrite Ion (NO2–) 32. Hydronium Ion (H3O+) 33. Sulphuryl chloride (SO2Cl2) 83 Chemical Bonding & Molecular Structure 34. Potassium chlorate (KClO3) 35. Potassium per-chlorate (KClO4) 36. Potassium dichromate (K2Cr2O7) 37. Ferry-cyanide Ion [Fe(CN)6]–3 38. Fero-cyanide Ion [Fe(CN)6]–4 39. Aluminium chloride (Al2Cl6) 40. Pyro phosphoric acid (H4P2O7) 41. Hypophosphate acid (HPO3) 42. Per-di-sulphuric acid (H2S2O8) 84 Compound Formula 43. Sulphurous acid H2SO3 44. Persulphuric acid H2SO5 Chemical Bonding & Molecular Structure Structure Peroxo monosulphuric acid (caro’s acid) 45. Thiosulphuric acid H2S2O3 46. Chlorous acid HClO2 47. Chloric acid HClO3 48. Perchloric acid HClO4 49. Perchloric anhydride Cl2O7 50. Hypophosphorous acid H3PO2 51. Perchromate CrO5 52. Chromyl Chloride CrO2Cl2 53. Carbon suboxide C3 O 2 54. Bleaching Powder CaOCl2 85 Chemical Bonding & Molecular Structure 55. Calcium carbide CaC 2 56. Ozone O3 57. Ground state of phosphorous P4 58. Ground state of sulphur S8 59. Nitrosyl chloride (Tilden reagent) NOCl 60. Sodium thiosulphate Na2 S2O3 or 86