TYPES OF SYMMETRIES OF MO s s-s combinations of : Orbitals Molecular Orbitals s s Node s s (g) (g) Bonding orbital Antibonding orbital (u) 4 (u) s-s combinations of atomic In the bonding MO there is increased electron density between the nuclei where as in the anti-bonding MO the electron density between the nuclui is zero (since the two lobes of opposite sign cancel the function). Both in the atomic orbital pair before interaction and tthe MO s formed after interaction, there is cylindrical symmetry around the bond axis i.e, any point on the orbital surface describes a circle as the MO is rotated around the bond axis. This circle is a plane perpendicular to the bond axis. All the points on such a circle have either a positive sign or negative orbital sign. There is no change of sign i.e, no node as we move over the circle. This is term cyclindrical symmetry and with cyclindrical symmetry are designated,, if they are bonding and, if they are antibonding. s-p combinatinos of : s Molecular (g) overlap bonding orbital Node s overlap antibonding orbital (u) s-p combinatino of atomic p-p combinations of : Molecular overlap bonding orbital (g) overlap antibonding orbital (u) p-p combinations of atomic
5 Orbitals Molecular Orbitals Nodal plane overlap bonding orbital (u) Nodal plane overlap antibonding orbital (g) p-p combinations giving -bonding p-d combinations of : Molecular Orbitals Orbitals p y d yz overlap bonding orbital (u) overlap antibonding orbital p y d yz (g) p-d combinations of atomic d-d combinations of : d x y z bonding by d- (sideways overlap of two d x y ) Non-bonding combinations of : In the combinations shown in the figure below any stabilization which occurs from overlapping with is destabilized by an equal amount of overlap of with. There is no overall change in energy, and this situation is termed non-bonding. It should be noted that in all these nonbonding cases the symmetry of the two atomic is different, i.e. rotation about the axis changes the sign of one.
6 s p y d yz Some non-bonding combinations of atomic Energy level diagram for molecular : BMO s : s pz p y ABMO s : s p z px p y The energy levels of these molecular have been determined experimentally from spectroscopic data. Total bonding MO 's Total antibonding MO 's NOTE: () If total number of electrons in a molecular 4, the energy level of MO s follow this sequence. px px s, s, s, s, pz, pz py, py, Increaseing order of energy p p z p y p p y Energy p ) z s) s s s) Energy level diagram Mixing of Orbitals: However spectroscopic studies have reverse sequence of energy levels of MO s is not correct for all molecules. It has been observed that in case of homonuclear diatomic molecules of second row elements, p z MO is higher in energy than the px and py MO s. Therefore the sequence of energy levels of MO s is due to the possibility of mixing of s and p z atomic. Note: If the difference between s and p atomic orbital are small, these is a possibility of the mixing of these to form mixed atomic. Which would be comparatively more stable. pz MO s as s and s obtained from mixed atomic would be lower in energy while pz and MO s would be higher in energy than the corresponding MO s obtained from mixed atomic.
7 p z p z p y p y Energy p z p y p y s s s Molecular orbital diagrams of molecules: MO before mixing () N, N= s s p 3 = K s p 3 Total number of electrons in N molecule (7 7) = 4 s MO after mixing p z p p x p y p) z p p y s Bond order = = So, N b N a s) s) of N MO of N 0 4 6 3 s of N (triple bond between two N atoms) N N N N N non-bonding lone pair of electrons There is no unpaired electron in N molecule and hence diamagnetic. non-bonding lone pair of electrons
CHEMICAL BONDING Electronic configuration: 0 px px 0 s, s, s, s, pz,, 0 pz py, py or px s, s, s, s, pz py, Note: MO diagrams follow the same sequence of energy levels(same as N ) of molecular for H,H, He, Li, Be, B,C, N electrons are equal to 4 or less than 4. 8, etc. i.e, for those homodiatomic moleculaes in which the total number of Note: If total number of electrons in molecule > 4, then the energy sequence of energy of molecular in MO diagram will be px px s s s s p z pz p y py MO diagram of O : 4 4 O s s p s p Total number of electrons in O molecules = ( 8 8) = 6, i.e, > 4. So MO diagram follows the following sequence of energy level of molecular. 6 4 p z, then. p p y p p y p ) z s s) s of O Bond order N N B O O O A s) MO of O of O 4 8 4 (two bonds between oxygen atoms) Paramagnetic due to presence of two unpaired electrons in Electronic configuration: px px s, s, s, s, pz, py, py px and py antibonding molecular. Note: Same MO diagram for O,O, F, etc.
MO diagram for hetero diatomic molecules: AB molecule 9 A MO of AB molecule B B A, where electronegativity eg. NO N s s p 7 e 5eS 4 3 7 S 4 0 e s p 8eS p z p p x p y p y p p ) z s of N s) s) MO of NO s of O Electronic Configuration:,,,,, Bond order px px s s s s pz 0 py, py 5 0 5.5, hence paramagnetic molecule. Note: MO of NO and NO are same. NO Total e S 4 Diamagnetic molecule Bond order 6 0 4 3 ; N O 3 4 O s s p MO diagram for CO: 4 e C s s p
CHEMICAL BONDING 30 p z p p ) z p y p p y N b N s of carbon Bond order CO C O (diamegnetic) () Coulson s Treatment of CO molecules: a s) s) s of O MO of CO molecule 6 0 4 3 (Triple bond between carbon and oxygen atom) CO is ionised to give CO by removing of one electron from the reduced to p orbital then the bond order should be z and the bond lenght increased. Infact the bond lenght of CO decreases and the bond order increasess. This cannot be explained by simple CO, MO diagram. Coulson modified the MO diagram of CO by using concept of hybridization. 6C s s p 8O = 4 s s p = s p s p sp-hybridization 0 c c z y C sp sp p p 0 0 z y O sp sp p p Note: Elcetron pairs in c Non-bonding electron pairs sp and O sp hybrid are not involved in bonding and hence are called nb pairs. c sp hybrid orbital containing a lone pair of electron has higher energy and higher degree of p-character, is very reactive and hence accounts for the properties of CO. (sp) O hybrid orbital containing a lone pair of electrons, lower in energy than (sp) O hybrid orbital containing a single electron, has high degree of s-orbital character and hence is very stable (i.e. unreactive) hybrid-orbital. Now we are left with two singly filled sp c and sp O hybrid and two p-atomic ( and p y ) on both the atoms which overlap together in pair to give the following molecular. sp sp c give sp and O sp molecular
3 atomic orbital on C-atom - atomic orbital on O-atom give x and p y atomic orbital on C-atom p y atomic orbital on O-atom give p and y x MO s y MO s. The relative order of energy of different MO s in CO: sp sp O b b b s y c x y sp and hence molecular orbital configuration of CO is - sp, sp,, sp, sp Bond order O bonds b b b x y c x y 6 0 3, showing that C and O in CO molecule are linked together by a triple bond b sp molecular is mostly on oxygen atom b x and b y molecular are mostly on O-atom. Since o-atoms contributes more to the formation of these MO s than does C-atom. These MO s are leaning some what toward oxygen atom are shown in figure. The other -MO s are at right angles i.e, in front and behind the paper to the one shown below. In the formation of x and y molecular (degenerate MO s) carbon atom makes a larger contribution than O-atom. Both of these degerate MO s are also at right sides to each other. The MO s are much closer to C-atom and are, therefore concentrated mostly on C-atom. Note: Both these MO s are empty and can therefore, accept electron pairs by interacting with appropriate, filled of transitional elements. These MO s are therefore, called acceptor MO s. This acceptor property of these MO s is very important in considering the M L bond in metal carbon and hence M is the metal an L indicates - ligand such as CO.
3 sp p x y x y Energy (sp) C (sp) C ( lp) x p y (sp) C x y Catom (sp) O atomic orbital of C sp (sp) O ( l p) (sp) O atomic orbital of O Molecular of CO. Problem: Why CO molecule, co-ordinated through carbon? Electron pairs in (sp) c and (sp) O hybrid orbitls in Coulson; CO diagram, are not involved in bonding, hence called non-bonding electron pairs. Lone pair of electrons in non-bonding (sp) c hybrid orbital, has highest energy i.e. HOMO. It is concentrated on carbon atoms and easily available for bonding with other atoms/ metals. This is the reason CO coordinate with other atoms through carbon atom. Problem: Why CO has negligible dipole moment? Lone pair of electrons on carbon atom in CO molecule points away from O-atom (i.e. extends away from C O bond). This reduce the effective electronics charge between carbon and oxygen atoms. This accounting for low dipole moment of CO. C O Lone pair on carbon atom reduces charge separation and hence it has negligible dipole moment.
Molecular orbital diagram of HF: 33 An orbital interaction diagram for the formation of HF. Only the valence atomic and electrons are shown. Comparison between valence bond theory and molecular orbital theory: Common features: valence bond and molecular orbital theories have following feactures in common. (i) Both the theories are the interpretation of covalent bonds. (ii) According to both theories the of the two atoms must overlap to form a covalent bond. The overlapping each other must have the same symmetry and must be of same energy. (iii) According to both, the electronic charge reside in the region between the atomic nuclei. (iv) The basic principles involved in the distribution of electrons are similar. Thus an orbital, whether atomic or molecular cannot have more than two electrons. They are filled in order of increasing energy. (v) Both the theory postulate that a covalent bonds possesses directional properties. Differences: () Molecular are formed by LCAO approximation method. The atomic involved in the formation of these molecules are from valence shell of two atoms. () of the resulting molecule lose their individual identity. MOT VBT () Two atomic give an interatomic orbital obtained by the combination of the space function of two unpaired electrons; one being in each of the two AO s () The resulting molecule consists of atoms which although interacting retain a large measure of their individual character. (3) Molecular are polycentric. (3) are mono-centric. (4) MOT explains the paramagnetic character (4) VBT fails to explain paramagnetic of oxygen character of O. (5) Resonance play no role in MOT (5) Resonance play an important role in VBT (6) The MOT inherently gives an equal (6) However in the initial stage of weightage to the covalent and ionic development the VBT ignored the structures ionic contribution to chemical bond
SOLVED PROBLEMS 34. Which one of the following statements concerning H is incorrect? [BHU 0,5] (a) The non-degenerate LCAOMOs (without spin) must be either symmetric or anti-symmetric for inversion (b) The lowest energy MO (without spin) of the molecule is anti-symmetric for inversion (c) The ground state has a multiplicity of two (d) The MOs transform into AOs of the helium ion as the two nuclei are fused together s u H H s s s g Ground state : g spin multiplicity s The lowest energy MO (wihtout spin) of the molecule is symmetric for inversion. Correct option is (b). Which of the following combination of atoms of A and B forms bonding molecular? (a) (b) (c) A B A The additive linearly combination of wave function of atom A A and B bonding molecular orbital N C C AB A B Correct option is (b) B A B (d) A B [BHU 0] B is responsible formation of 3. Which of the following processes involves increasing in bond order? [BHU 0] (a) N N e (b) CO e CO (c) NO NO e (d) O e O According to Molecular Orbital Theory NO s s s s px py pz px py pz Bond order = 0 4 3 Correct option is (c) 4. Which molecule has zero bond order? [BHU 0] (a) H (b) H (c) HeH (d) He He s s s s Bond order 0 5. What is the bond order in NO molecule? [BHU 0,3] (a).5 (b) (c).5 (d) NO s s s s px py pz px py pz
35 0 5 Bond order.5 Correct option is (a) 6. Which molecule has zero bond order? [BHU 04] (a) H (b) H (c) HeH (d) He He s s Bond order 0 Correct option is (d) 7. What is the bond order in NO molecule? [BHU-04] (a).5 (b) (c).5 (d) Electronic configuration of NO s s s s px py pz px py pz 5 Bond order ( Nb Na ) (0 5).5 Correct answer is (a) 8. Arrange the following oxygen molecular ions and the molecule in order of increasing bond energy or in decreasing order of (O-O) bond length. The molecular ions and the molecule are : O, O, O, O and O. The possible orders are given below. Choose the correct one: [BHU 06] (a) O < O < O < O < O (b) O < O < O < O < O (c) O < O < O < O < O (d) O < O < O < O < O Bond order bond energy bond length O s s s s pz px py px py pz 4 0 6 Bond order O s s s s pz px py px py pz 5 0 5.5 Bond order O s s s s pz px py px py pz 0 4 Bond order 3 O s s s s pz px py px py pz 3 0 7.5 Bond order = O s s s s pz px py px py pz Bond order 0 8 Correct option is (b)
9. Which one of the following M, O. s of (HF ) is correct? [BHU 06] 36 (a) H F (b) F H (c) H F (d) H F Molecular orbital diagram for HF Energy H F Correct option is (b) 0. Out of the following diatomic molecules, the paramagnetic ones are (A) Be (B) B (C) C (D) N (E) O (F) F (G) C (H) C (a) Be, B and C, N (c) B, C, N and C [BHU 06] (b) C, N, O and C (d) B, O, C, C Pick the correct choice out the above four choices. B s s s s px py Number of unpaired electrons = C s s s s px py Number of unpaired electron = C s s s s px py pz Number of unpaired electron = N s s s s px py pz Number of unpaired electron = 0 Be s s s s Number of unpaired electron = 0 O s s s s pz px py px py pz Number of unpaired electron = F s s s s pz px py px py pz Number of paired electron = 0 Correct option is (d). Which of the following has a bond order.5: (a) HCl (b) CO (c) NO (d) N NO s s s s px py pz px py pz 0 5 Bond order.5 N s s s s px py pz px py pz
37 0 4 Bond order 3 N and CO both are isoelectronic hence same bond order. Correct option is (c). According to M.O. theory, the ground state of H is s. In addition to the ground state, there are following excited states of H : [BHU 07] () s s () s s (3) s s (4).. b s s The highest and lowest energy states of H will respectively be : (a) () and () (b) (3) and () (c) () and () (d) () and (3) Highest and lowest energy excited states are s s s s Highest energy state Lowest energy state Correct option is (b) 3. Which of the following is a paramagnetic molecule? [HCU-03] (a) H (b) B (c) C (d) N B s s s s px py Number of unpaired = Hence, paramagnetic H, C and N has no unpaired electron hence diamagnetic. Correct option is (b) 4. Consider four diatomic molecules M, M, M3 and M4 of same atoms with bond orders.5, 3, 4 and respectively. Which of the following arrangements describes them with decreasing bond energy? (a) M3 > M4 > M > M (b) M > M > M3 > M4 [HCU-03] (c) M3 > M > M4 > M Bond order bond energy Hence, order of bond energy will be M 3 > M > M > M 4 Thus, correct option is (d) (d) M3 > M > M > M4 5. The bond order and bond type in the C molecule are, respectively [HCU-04] (a) ; one sigma bond and one pi bond (b) 3; one sigma bond and two pi bonds (c) ; two pi bonds (d) ; two sigma bonds s s s s px py C Bond order = 4 as both the bonds are formed by -bonding hence both the bonds are -type. Correct option is (c) 6. Identify the species with the lowest bond order among the following [HCU-05] (a) C (b) NO (c) Cl (d) O s s s s pz px py px py Cl pz 3s 3s 3pz 3px 3py 3px 3py 8 5.5 Bond order
38 C, Bond order = s s s s px py NO s s s s px py pz px py 0 5.5 Bond order = 8 4 O s s s s pz px py px py pz Bond order 0 5.5 Correct option is (c) 7. The species having the shortest bond length among the following is : [HCU 07] (a) NO (b) NO (c) NO (d) NO NO s s s s px py pz pxz py 0 5.5 Bond order = NO s s s s px py pz 0 4 3 Bond order NO s s s s px py pz px py 0 6 Bond order NO s s s s px py pz 5 Bond order = 0 4.5 Correct option is (c) 8. Among the following, the most unstable molecule is (a) Li (b) Be (c) B (d) C [HCU-0] Electronic configuration of 0 Li (6) s s s s Bond order (4 ) 0 Be (8) s s s s Bond order (4 4) 0 B (0) s s s s p Bond order (6 4) x py C () s s s s p Bond order (0 4) 3 x p z py Be is most unstable because the bond order is zero. Correct answer is (b)
9. The common features among the species CN, CO and NO are [DU-04] (a) Bond order three and isoelectronic (b) Bond order three and weak field ligands (c) Bond order two and stronger field ligands (d) Iso-electronic and weak field ligand CN 4 electrons CO 4 electrons NO 4 electrons, they are iso-electronic CN s s s s pz px px py py Bond order = ( N ) (0 4) 6 b Na 3 CO s s s s p z px Bond order (0 4) 3 py NO s s s s px px pz py py 6 Bond order (0 4) 3 Correct answer is (a) 0. The HOMO in CO is? [ISM-05] (a) -bonding (b) -bonding (c) -antibonding (d) -antibonding Correct answer is (d). Among the following molecules, the shortest bond length is to be found in 39 (a) C (b) N (c) O (d) F Bond order is reciprocal to bond length, so, the molecule with higher bond order will have shorter bond length. Bond order of C = (N b N a ) = (6 ) = Bond order of N = (N b N a ) = (8 ) = 3 Bond order of O = (8 4) = Bond order of F = (8 6) = N has highest value of bond order, so it will have shortest bond length. Correct answer is (b). The bond order for N, O, N, O varies as (a) N > N > O > O (b) N > O > N > O (c) O > N > O > N (d) N > N > O > O Bond order of N = (N b N a ) = (0 4) = 3
40 Bond order of O = (0 6) = Bond order of N = (0 5) =.5 Bond order of O = (0 7) =.5 Thus, the bond order is N > N > O > O Correct answer is (a) 3. According to MO theory for the atomic species C (a) bond order is zero and paramagnetic (c) bond order is two and paramagnetic Electronic configuration of C is s s s s px py pz px py pz Bond order = 8 ( N ) 4 b Na No unpaired electrons, So, C is diamagnetic Correct answer is (d) (b) bond order is zero and diamagnetic (d) bond order is two and diamagnetic 4. On the basis of LCAO-MO theory, the magnetic characteristics of N and N are (a) both diamagnetic (b) both paramagnetic (c) N diamagnetic and N paramagnetic (d) N paramagnetic and N diamagnetic N (4) = s s s s px py pz N (3) = s s s s px py pz Nitrogen has no unpaired electron, so it is diamagnetic while N has one unpaired electron, so it is paramagnetic in nature. Correct answer is (c) 5. The highest occupied molecular orbital of HF is (a) bonding (b) anti-bonding (c) ionic (d) non-bonding Correct answer is (d) 6. Which of the following have identical bond order (a) CN (b) O (c) NO (d) CN CN and NO are isoelectronic, have the same bond order of 3. Correct answers are (a) and (c) 7. Given the species N, CO, CN and NO which of the following statements are true for these (a) All species are paramagnetic (b) All species are isoelectronic (c) All the species have dipole moment (d) All the species are linear All the species are isoelectronic (total electron is 4) N s s s s px pz py (diamagnetic) All species are diamagnetic (i.e., N is diamagnetic) and linear. Correct answers are (b) and (d)