" Due to intermolecular H-bonding, ethyl alcohol remains in the associated form andtherefore boils at a higher temperature compared to dimethyl ether.Importance of Hydrogen Bonding in Biological Systems:Hydrogen bonding plays a vital role in physiological systems. Proteins contain chains of aminoacids. The amino acid units are arranged in a spiral form somewhat like a stretched coil spring(forming a helix). The N-H group of each amino acid unit and the fourth C=O group following italong the chain, establishes the N–H---O hydrogen bonds. These bonds are partly responsible forthe stability of the spiral structure. Double helix structure of DNA also consists of two strandsforming a double helix and are joined to each other through hydrogen bond.INTERMOLECULAR FORCESWe have enough reasons to believe that a net attractive force operates between molecules of agas. Though weak in nature, this force is ultimately responsible for liquifaction and solidification ofgases. But we cannot explain the nature of this force from the ideas of ionic and covalent bonddeveloped so far, particularly when we think of saturated molecules like H , CH , He etc. The2 4 existence of intermolecular attraction in gases was first recognised by Vanderwaal‘s andaccordingly intermolecular forces have been termed as Vanderwaal‘s forces. It has beenestablished that such forces are also present in the solid and liquid states of many substances.Collectively they have also been termed London forces since their nature was first explained byLondon using wave mechanics.Nature of Vanderwaal’s Forces:The Vanderwaal‘s forces are very weak in comparison to other chemical forces. In solid NH it3 –1 –1 amount to about 39 KJ mol (bond energy N-H bond = 389 KJ mol ). The forces are nondirectional. The strength of Vanderwaal‘s force increases as the size of the units linked increases.When other factors (like H-bonding is absent), this can be appreciated by comparison of themelting or boiling points of similar compounds in a group.Origin of Intermolecular Forces: Intermolecular forces may have a wide variety of origin.?Dipole-dipole interaction: This force would exist in any molecule having a permanentdipole e.g. HF, HCl, H O etc.2 ?Ion-dipole interaction: These interactions are operative in solvation and dissolution ofionic compounds in polar solvents.?Induced dipole interaction: These generate from the polarisation of a neutral moleculeby a charge or ion.?Instantaneous dipole-induced dipole interaction: In non polar molecules dipoles maygenerate due to temporary fluctuations in electron density. These transient dipole cannow induce dipole in neighbouring molecules producing a weak temporary interaction.METALLIC BONDING26 Metals are characterised by bright, lustre, high electrical and thermal conductivity, malleability,ductility and high tensile strength. A metallic crystal consists of very large number of atoms arrangedin a regular pattern. Different model have been proposed to explain the nature of metallic bondingtwo most important modules are as followsThe electron sea Model + + + +PositiveKernles? ? ? ?+ + + +? ? ? ?Mobileelectrons+ + + +? ? ? ?In this model a metal is assumed to consist of a lattice of positive ion (or kernels) immersed in a seaof mobile valence electrons, which move freely within the boundaries of a crystal. A positive kernelconsists of the nucleus of the atom together with its core on a kernel is, therefore, equal in magnitudeto the total valence electronic charge per atom. The free electrons shield the positively charged ioncores from mutual electrostatic repulsive forces which they would otherwise exert upon one another.In a way these free electrons act as =glue‘ to hold the ion cores together. The forces that hold the atoms together in a metal as a result of the attraction between positive ionsand surrounding freely mobile electrons are known as metallic bonds. Through the electron sea predated quantum mechanics it still satisfactorily explains certainproperties of the metals. The electrical and thermal conductivity of metals for example, can beexplained by the presence of mobile electrons in metals. On applying an electron field, these mobileelectrons conduct electricity throughout the metals from one end to other. Similarly, if one part ofmetal is heated, the mobile electrons in the part of the metals acquire a large amount of kineticenergy. Being free and mobile, these electrons move rapidly throughout the metal and conduct heatto the other part of the metal. On the whole this model is not satisfactory. Example 32. Sodium metal conducts electricity because it (A) is a soft metal(B) contains only one valence electron (C) has mobile electron(D) reacts with water to form H gas2 Solution:(C)MOLECULAR ORBITAL THEORYIn Molecular Orbital Theory (MOT) the atoms in a molecule are supposed to loose their individualcontrol over the electrons. The nuclei of the bonded atoms are considered to be present atequilibrium inter-nuclear positions. The orbitals where the probability of finding the electrons ismaximum are multicentred orbitals called molecular orbitals extending over two or more nuclei. 27 In MOT the atomic orbitals loose their identity and the total number of electrons present areplaced in MO‘s according to increasing energy sequence (Auf Bau Principle) with due referenceto Pauli‘s Exclusion Principle and Hund‘s Rule of Maximum Multiplicity.When a pair of atomic orbitals combine they give rise to a pair of molecular orbitals, the bondingand the anti-bonding. The number of molecular orbitals produced must always be equal to thenumber of atomic orbitals involved. Electron density is increased for the bonding MO‘s in theinter-nuclear region but decreased for the anti-bonding MO‘s, Shielding of the nuclei by increasedelectron density in bonding MO‘s reduces inter nuclei repulsion and thus stabilizes the moleculewhereas lower electron density even as compared to the individual atom in anti-bonding MO‘sincreases the repulsion and destabilizes the system.In denotation of MO‘s, ? indicates head on overlap and? represents side ways overlap oforbitals. In simple homonuclear diatomic molecules the order of MO‘s based on increasing energyis* ?? ?? ?? 2p 2p ? ? ? ? yy * * * ? ? ? ? ? ?? ? ? ? 1s 1s 2s 2s 2p 2p xx ? * ?? 2p ?? ? ?? z 2p ?? z This order is true except B , C & N . If the molecule contains unpaired electrons in MO‘s it will be2 2 2 paramagnetic but if all the electrons are paired up then the molecule will be diamagnetic.Bond order ?? no. of e s occupying bonding MO's ?no. of e soccupy ingantibonding MO's = 2 Application of MOT to homonuclear diatomic molecules.H molecule : Total no. of electrons = 22 2 Arrangement: ?1s Bond order : ½ (2 – 0) = 1? Hmolecule : Total no. of electrons = 1 2 1 Arrangement: ?1s Bond order : ½ (1 – 0) = 1/2He molecule : Total no. of electrons = 4 2 * 2 2 Arrangement: ? ?1s 1s Bond order : ½ (2 – 2) = 0?He molecule does not exist.2 ? Hemolecule : Total no. of electrons = 3 2 2 * 1 Arrangement: ? ?1s 1s Bond order : ½ (2 – 1) = 1/2? So He exists and has been detected in discharge tubes.2 Limolecule : Total no. of electrons = 6 2 2 * 2 2 Arrangement: ? ? ?1s 1s 2s Bond order : ½ (4 – 2) = 1No unpaired e‘s so diamagneticBe molecule : Total no. of electrons = 8 22 * 2 2 *2 Arrangement: ? ? ? ?1s 1s 2s 2s Bond order : ½ (4 – 4) = 0– No unpaired e s so diamagneticBmolecule : Total no. of electrons = 10 228 2 * 2 2 * 2 2 ? ? ? Arrangement: ? ?1s 1s 2s 2s 2p x Bond order : ½ (6 – 4) = 1?diamagneticBut observed Boron is paramagneticCmolecule : Total no. of electrons = 12 2 1 ?? ? ?? 2p 2 * 2 2 * 2 2 y Arrangement: ? ? ? ? ?1s 2s 2p?? 1s 2s x 1 ? ?? 2p ?? z Bond order : ½ (4 – 0) = 2It is paramagneticBut observed C is diamagnetic 2 Nmolecule : Total no. of electrons = 14 2 2 ?? ? 2p ?? 2 * 2 2 *2 2 y Arrangement: ? ? ? ? ??? 1s 1s 2s 2s 2px 2 ? ?? 2p ?? z Bond order : ½ (6 – 0) = 3 ? It is diamagnetic Omolecule : Total no. of electrons = 16 2 *1 2 ?? ?? ? ? 2p 2p ???? y 2 * 2 2 * 2 2 y Arrangement: ? ? ? ? ????? 1s 1s 2s 2s 2p x 2 *1 ? ?? 2p ?? ? ?? z 2p ?? z Bond order : ½ (6 – 2) = 2It is paramagneticFmolecule : Total no. of electrons = 18 2 2 *2 ?? ?? ? ? 2p ???? 2py 2 *2 2 * 2 2 y * Arrangement: ? ? ? ? ? ?1s 2s 2p???? 1s 2s 2px x 2 *2 ? ???? 2p ? ?? z 2pz ?? Bond order : ½ (6 – 4) = 1It has been seen that in case of B , C & N the order of filling the e‘s is different from the normal2 2 2 sequence.1 ?? ? 2p ?? 2 *2 2 *2 y B: ? ? ? ?2 ?? 1s 1s 2s 2s 1 ? ?? 2p ?? zIt is paramagnetic2 ?? ? 2p ?? 2 *2 2 *2 y C: ? ? ? ?2 1s 2s?? 1s 2s 2 ? ?? 2p ?? zIt is diamagnetic2 ?? ? ?? 2py 2 *2 2 *2 2 N: ? ? ? ? ?2 ?? 1s 1s 2s 2s 2px 2 ? ?? 2pz ??It is diamagnetic? ? Example 33. Compare the bond energies of O , O & O2 2 2 Solution:Higher the bond order greater will be the bond energy. 29 *1 2 ?? ?? ? ? 2p 2p ???? y 2 *2 2 *2 2 yNow configuration of O = ? ? ?? ? 2 ???? 1s 1s 2s 2s 2p x 2 *1 ? ?? 2p ?? ? ?? z 2p ?? z ?Now formation of means to remove an electron from anti-bonding one, whichO 2 means increase in B.O.?B.O. of= ½ (6-1) = 2.5O 2 – ?means introduction of an e in the anti-bonding thereby reducing the bondO 2 order.?Bond order of= ½ (6 – 3) = 1.5O 2 ? ?So bond energy of > O > O O 2 2 2 – – Example34. Give MO configuration and bond orders of H , H , He and He . Which species2 2 2 2 among the above are expected to have same stabilities?2 Solution:H = ? 2 1s Bond order = 1– 2 1H = ? ?2 1s *1sBond order = 0.52 *2 He =? ? 2 1s 1s Bond order = 0– 2 * 2 1He =? ? ? 2 1s 1s 2s (3 ? 2)Bond order == 0.52 – –H and He are expected to have same stabilities.2 2 Example 35. Which of the following species have the bond order same as ?N 2 ? ? CN OH(A) (B) ? ?(C)NO (D) COSolution.In N no. of bonding electrons and anti ?bonding electrons are 10 and 42 10 ? 4 respectively. Therefore the bond order is= 3. Out of those given only2 ? ? CN CNis isoelectronic withN . Thereforehas the same bond order as N.2 2 Hence (A) is correctM.O. of Some Diatomic Heteronuclei MoleculesThe molecular orbitals of heteronuclei diatomic molecules should differ from those of homonucleispecies because of unequal contribution from the participating atomic orbitals. Let‘s take theexample of CO.The M.O. energy level diagram for CO should be similar to that of the isoelectronic molecule N .2 But C & O differ much in electronegativity and so will their corresponding atomic orbitals. But theactual MO for this species is very much complicated since it involves a hybridisation approachbetween the orbital of oxygen and carbon.HCl Molecule: Combination between the hydrogen 1s A.O‘s. and the chlorine 1s, 2s, 2p & 3s1 1 orbitals can be ruled out because their energies are too low. The combination of H 1s and 3px30 "