1. Explain the formation of a chemical bond. Solution:

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1 1. Explain the formation of a chemical bond. Existence of a strong force of binding between two or many atoms is referred to as a chemical bond and it results in the formation of a stable compound with properties of its own. Molecules having two identical atoms like H 2, O 2, Cl 2, N 2 are called as homo-nuclear diatomic molecules. Molecules containing two different atoms like CO, HCl, NO, HBr etc., are called as hetero-nuclear diatomic molecules. Chemical bonds are basically classified into three types consisting of i) ionic or electrovalent bond ii) covalent bond and co-ordinate covalent bond. According to the electronic theory of valency, a chemical bond is said to be formed when atoms interact by losing, gaining or sharing of valence electrons and in doing so, a stable noble gas electronic configuration is achieved by atoms. 2. Write Lewis dot symbols for atoms of the following elements : Be, Na, B, O, N, Br. 3. Write Lewis symbols for the following atoms and ions: S and S 2 ; Al and Al 3+ ; H and H 1

2 4. Draw the Lewis structures for the following molecules and ions: H 2 S, SnCl 4, BeF 2, 3CO, HCOOH 5. Define octet rule. Write its significance and limitations. The tendency of atoms to have eight electrons in their outer-shell by interacting with other atoms through electron sharing or electron transfer is known as the octet rule of chemical bonding. Significance: 1. Used to explain chemical bonding in various compounds. 2

3 2. Able to illustrate various type of bonds like covalent, electrovalent and co-ordinate. 3. Useful for understanding the structures of most of the organic compounds. Limitations: The octet rule is not satisfied for all atoms in a compound. 1. Incomplete octet of the central atom: LiCl, BeH 2, BCl 3, AlCl 3,BF 3,LiBr, AlBr 3 2.Odd number of electrons in the outer configuration of certain atoms In some compounds like NO, NO 2, odd number of electrons are present in the outermost shell of nitrogen atom. 3.More than eight electrons in the central atom: The expanded octet In some compounds like PF 5, SF 6, H 2 SO 4 apart from s and p orbitals, d orbitals are also available for bonding. This way of bonding will lead to the presence of more than eight valence electrons around the central atom. There are 5 bonds surrounding phosphorus atom. Each bond corresponds to 2 electrons. Therefore 10 electrons are present around P atom instead of 8 electrons. Six bonds around S atom i.e 12 electrons around S atom. 4. Octet theory was not able to explain various shapes attained by molecules. 5. Octet theory is based upon the chemical inertness of rare gases instead rare gases do involve in forming compounds like XeF 2, KrF 2, XeF 4, XeOF 2, etc. 6. Write the favourable factors for the formation of ionic bond. Favourable factors for the formation of ionic bond 3

4 i) Ease of formation of cation and anion form neutral atoms. ii) Elements /Atoms with lower ionisation energies. iii) Atoms with greater negative values of electron gain enthalpy. iv) Cations and anions together form a stable crystal lattice by releasing large amount of energy. v) Lattice enthalpy of ionic solids are extremely high. 7. Discuss the shape of the following molecules using the VSEPR model: BeCl 2, BCl 3, SiCl 4, AsF 5, H 2 S, PH 3 VSEPR Theory is based upon the fact that in a polyatomic molecule, the direction of bonds around the central atom depends upon the total number of electron pairs (lone pairs as well as bond pairs) in its valence shell. These electron pairs place themselves as far apart as possible in space so as to have minimum repulsive interactions between them. The presence of lone pairs in addition to bond pairs in the molecule causes distortion in the geometry of the molecule. s Lewis Dot Structure Electron pairs around central atom bp lp Arrangement of electron pairs 2 0 Linear 3 0 Trigonal Shape R No pai bon tak opp pos due rep No pai bon tak trig geo (ma ang sep 4 0 Tetrahedral Fou pai acq tetr geo 4

5 AsF Trigonal bipyramidal H 2 S 2 2 Bent PH Pyramidal Five bond pairs acquire trigonal bipyramidal geometry. Two bond pairs and two lone pairs present. The molecule acquires angular shape instead of linear due to lp-lp, lpbp repulsion. One lp and three bps present and the molecule acquires pyramidal shape. 8. Although geometries of NH 3 and H 2 O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss. If an atom is surrounded by four bonded pairs of electrons, it must assume a tetrahedral shape due to electron repulsion. But in the case of ammonia, N is surrounded by one lone pair and three bonded pairs of electrons. The lp-bp repulsion makes the molecule to assume a pyramidal shape whose HNH angle gets reduced from 109 o 28 to 107 o. In the case of water molecule the molecule contains 2 lps and 2 bps of electrons. The lp-bp and lp-bp repulsion together exists and makes the molecule to assume a bent shape and the HOH angle is reduced from 109 o 28 to o. Since the lp-lp repulsions are much greater than the lp-bp repulsion the HOH bond angle in water is smaller than HNH bond angle in NH 3. 5

6 9. How do you express the bond strength in terms of bond order? Bond order may be defined as the half difference between number of electrons in bonding molecular orbitals and the number of electrons in anti-bonding molecular orbitals. Bond Order = [N b N a ] If the value of bond order is positive, it indicates a stable molecule and if the value of bond order is negative or zero, it means that the molecules unstable and is not formed. Bond order cannot be used for quantitative comparison of the strengths of chemical bonds. It can give us only approximate idea about the strength of the bond. Greater the bond order, greater is the strength of the bond. 10. Define the bond length. Bond length may be defined as the average equilibrium distance between the centres of the two bonded atoms. 11. Explain the important aspects of resonance with reference to the CO ion. In CO 3 2- ion all C O bonds are equivalent. This is possible only different Canonical structures of CO 3 2- undergo resonance to give resonance hybrid in which all C O bonds are equivalent. 12. H 3 PO 3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing? H 3 PO 3? If not, give reasons for the same. 6

7 The given two canonical structures of H 3 PO 3 do not have the same number of single and double bonds. 1 st structure has P = 0 and no lone pair electron on phosphorus atom while the 2 nd structure has no double bond between one of oxygen atom and P-atom and P-atom has one lone pair electrons. 13. Write the resonance structures for SO 3, NO 2 and NO 3 - SO 3 NO 2 NO 3-7

8 14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N. (a) K and S (b) Ca and O (c) Al and N 15. Although both CO 2 and H 2 O are triatomic molecules, the shape of H 2 O molecule is bent while that of CO 2 is linear. Explain this on the basis of dipole moment. A polyatomic molecule has more than two atoms bonded by covalent bonds. In such cases the idea 8

9 of a dipole can be applied to individual bonds within the molecule. The dipole moment of the individual bond in a polyatomic molecule is referred to as bond dipole. The dipole moment of the molecule depends upon the orientations of various bond dipoles. Bond dipoles Bond dipoles do not Cancel Cancel Resultant µ = 0. Resultant µ = 1.83 D. Although both CO 2 and H 2 O are both triatomic molecules, the shape of CO 2 is linear but H 2 O is bent. The shape of the molecules can be predicted by dipole moments. Experiments should that the dipole moment of CO 2 is zero. This is possible only if bond dipoles of two c = 0 bonds cancel each other. In other words, the two bond dipoles must be oriented in opposite directions. This is possible if the molecule is linear. Similarly H 2 O has a dipole moment of 1.83D. Thus its molecule cannot be linear because the bond dipoles do not cancel each other. Thus, the H 2 O molecule must have an angular shape. 16. Write the significance/applications of dipole moment. Applications of dipole moment 1. Distinction between polar and non-polar molecules. The molecules having dipole moment are called polar molecules whereas molecules having zero dipole moment are said to be non-polar molecules. For example: (i) Molecules such as H 2, N 2, O 2, Cl 2 etc., have non-polar bonds and zero value of dipole moment. (ii) Molecules such as CO 2, BF 3, CH 4, CCl 4, PCl 5, SF 6, etc., have polar bonds but zero value of dipole moment. (iii) Molecules such as HF, HCl, NH 3, H 2 O, NF 3, have polar bonds and their dipole moment is greater than zero. In other words, they are polar molecules. 2. Ionic character in a molecule. Dipole moment gives an idea about the ionic character in a bond or a molecule. For example, let us calculate the percentage of ionic character in HCl molecule. Experiments have shown that the dipole moment of HCl is 1.03 D and its bond length is A. Now for 100% ionic character the charge developed on H and Cl atoms would be e. s. u. Therefore, dipole moment in case of 100% ionic character is given as m ionic = e. s. u. cm. = e. s. u. cm = 6.12 D. The observed dipole moment m obs = D Percentage Ionic Character = = = 16.83%. 9

10 In general, larger the value of dipole moment is, more will be the ionic character. 3. Shapes of molecules. A polyatomic molecule has more than two atoms bonded by covalent bonds. In such cases the idea of a dipole can be applied to individual bonds within the molecule. The dipole moment of the individual bond in a polyatomic molecule is referred to as bond dipole. The dipole moment of the molecule depends upon the orientations of various bond dipoles. Bond dipoles Bond dipoles do not Cancel Cancel Resultant µ = 0. Resultant µ = 1.83 D. Although both CO 2 and H 2 O are both triatomic molecules, the shape of CO 2 is linear but H 2 O is bent. The shape of the molecules can be predicted by dipole moments. Experiments should that the dipole moment of CO 2 is zero. This is possible only if bond dipoles of two c = 0 bonds cancel each other. In other words, the two bond dipoles must be oriented in opposite directions. This is possible if the molecule is linear. Similarly H 2 O has a dipole moment of 1.83D. Thus its molecule cannot be linear because the bond dipoles do not cancel each other. Thus, the H 2 O molecule must have an angular shape. 17. Define electro negativity. How does it differ from electron gain enthalpy? Electronegativity Electronegativity of an element may be defined as the tendency of its atom to attract the shared pair of electrons towards it self in a covalent bond. Electron affinity 1. It is the tendency of an atom to attract outside electron to receive. 2. It is the absolute electron attracting tendency of the atom Electronegativity 1. It is the tendency of an atom to attract shared pair of electrons. 2. It is the relative electron attracting tendency of an atom. 3. It is the property of an isolated atom. 3. It is the property of bonded atom. 4. The elements with symmetrical configuration have almost zero electron affinities. 5. It has certain units i.e., KJ mol _1 and ev/atom. 4. The elements with symmetrical configuration have specific electronegativities. 5. It has no units. 18. Explain with the help of suitable example polar covalent bond. Polar covalent bond 10

11 The bond between two unlike atoms which differ in their electronegativities is said to be a polar covalent bond. In other words polar bond is due to partial ionic character of covalent bond. 19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K 2 O, N 2, SO 2 and ClF 3. N 2 < SO 2 < ClF 3 < LiF < K 2 O. 20. The skeletal structure of CH 3 COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. 21. Apart from tetrahedral geometry, another possible geometry for CH 4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH 4 is not square planar? In CH 4 the central carbon atom has no lone pair of electron. Had there been two lone pairs on carbon atom after bonding with four Hydrogen atoms, it would have been square planer in shape according to VSEPR theory. Since carbon atom is smaller and has no d-orbital space, it cannot undergo d 2 SP 3 hybridisation which is square planer in shape. Since carbon atom undergoes SP 3 hybridisation the CH 4 molecule is tetrahedral. 22. Explain why BeH 2 molecule has a zero dipole moment although the Be H bonds are polar. Be and H differ in their electronegativities. So, the Be H bond should be a polar bond. BeH 2 is sp hybridized molecule and hence it is linear. H Be-H angle is The dipole moment due to one Be H bond cancels that due to another Be H bond. So net dipole moment is zero. 11

12 23. Which out of NH 3 and NF 3 has higher dipole moment and why? In N-H bond direction of dipole moment is towards hydrogen. Resultant dipole moment is in the same direction of dipole moment due to lone pair of N: But in NF 3, resultant dipole moment is in opposite direction to that of dipole moment due to lone pair of N. Hence dipole moment of NF 3 has less dipole moment than NH What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp 2, sp 3 hybrid orbitals. Hybridisation may be defined as the phenomenon of intermixing of atomic orbitals of slightly different energies to form new orbitals of equivalent energies and identical shapes. There are many different types of hybridisation which involve the use of s and p orbitals i.e., sp 3,sp 2 and sp hybridisation. (i) sp 3 hybridisation: This type of hybridisation involves the mixing of one orbital of s-sub level and three orbitals of p-sub level of the valence shell to form four sp 3 hybrid orbitals of equivalent energies and shape. Each sp 3 hybrid orbital has 25% s character and 75% p-character. These hybridised orbitals tend to lie as apart in space as possible so that repulsive interactions between them are minimum. The four sp 3 hybrid orbitals are directed towards the four corners of a tetrahedron. The angle between the sp 3 hybrid orbitals is

13 sp 3 hybridisation is also known as tetrahedral hybridisation. The molecules in which central atom is sp 3 hybridised and is linked to four other atoms directly, have tetrahedral shape. ii)sp 2 hybridisation: This type of hybridisation involves the mixing of one orbital of s- sublevel and two orbitals of p sublevel of the valence shell to form three sp 2 hybrid orbitals. These sp 2 hybrid orbitals lie in a plane and are directed towards the corners of equilateral triangle. 13

14 Each sp 2 hybrid orbitals has one-third s- character and two-third p-character. sp 2 hybridisation is also called trigonal hybridisation. The molecule in which central atom is sp 2 hybridised and is linked to other atoms directly have triangular planar shape. (i) sp-hybridisation: This type of hybridisation involves the mixing of one orbital of s-sub-level and one orbital of p-sub level of the valence shell of the atom to form two sp-hybridised orbitals of equivalent shapes and energies. These sp hybridised orbitals are oriented in space at an angle of This hybridisation is called diagonal hybridisation. Each sp hybrid orbital has equal s and p character, i.e., 50% s-character and 50% p-character. The molecules in which the central atom is sp hybridised and is linked to two other atoms directly have linear shape. 25. Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl 3 + Cl - AlCl 4 - The phenomenon of intermixing of two or more Atomic orbitals of an atom, having nearly the same energies which result in their rearrangement to form equal number of new set orbitals of equal energy and shape, is called hybridisation. These hybrid orbitals, formed in hybridisation, orient in three dimensional space, giving specific shape, depending on the type of hybridisation. For example SP 3 hybridisation give tetrahedral shape, sp 2 gives trigonal while sp gives linear or diagonal shapes. AlCl 3 is supposed to have been formed by sp 2 hybridisation. Al - at. no Electronic configuration is 1S 2 2S 2 2S 6 3S 2 3P X 1 3P Y 0 3P Z 0 In the excited state 2s electron will go to vacant 2p y orbital. 14

15 Excited state Now 2s, 2p x and 2p y undergo sp 2 hybridisation. The three sp 2 hybrid orbitals form σ - bonds with 3 Cl- atoms. In the formation of AlCl 3, there is no change in hybridisation, because the vacant hybridized orbital can receive, one lone pair electrons from Chlorine atom. 26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF 3 + NH 3 F 3 B-NH 3 The B-atom is sp 2 hybridised. BF 3 molecule has one unhybridised orbital which can accommodate one lone pair 2S- electrons (by coordinate covalancy) from N-atom in pyramidal structure of NH 3 molecule. So there is no need for change in the hybridisation of either B-atom or N-atom. 27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C 2 H 4 and C 2 H 2 molecules. (i) C 2 H 4 (i) C 2 H 4 or H 2 C = C H 2 is formed by sp 2 hybridisation of each of two carbon atoms. The two of the three hybrid orbitals of each C-atom forms two σ -bonds with two hydrogen atoms and the third hybrid orbital of one carbon atom forms σ -bonds with third hybrid orbitals of second C-atom. Thus three σ -bonds are in one plane. The unhybridised orbitals (2p z ) of one C-atom laterally overlaps with that of 2nd C-atom forming one π -bond which is at right angles to the three σ -bonds. Ethene molecule 15

16 The Lewis structure of ethylene is represented as (ii) C 2 H 2 CH CH (Ethyne) molecule is formed by sp hybridisation of each C- atom. One hybrid bond on each C- atom forms σ -bond with one Hydrogen atom. The other hybrid bond of one C-atom forms σ -bond with another C-atom. The two unhybridised orbitals on each carbon atom form two π -bonds with those of second carbon atom. Ethyne molecule 28. What is the total number of sigma and pi bonds in the following molecules? (a) C 2 H 2 (b) C 2 H 4 (a) C 2 H 2 = (b) C 2 H 4 = 29. Considering x-axis as the inter-nuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px (c) 2py and 2py (d) 1s and 2s. a) s-s overlapping: 16

17 b) s-p overlapping: c) p-p overlapping: 2Py and 2Py form π -bond Since the p-orbitals undergo lateral overlapping it cannot form a σ - bond. d) 1S and 2S form σ - bond 30. Which hybrid orbitals are used by carbon atoms in the following molecules? (a) CH 3 CH 3 ; (b) CH 3 CH=CH 2 ; (c) CH 3 -CH 2 -OH; (d) CH 3 -CHO (e) CH 3 COOH (a) sp 3 (b) sp 3, sp 2 (c) sp 3 (d) sp 3,sp 2 (e) sp 3,sp What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type. 17

18 The pairs of electrons between two atoms in a compound, which are involved in bonding are called bond pairs. For example: Lone pairs The pairs of electron surrounding the central atom which belongs to the central atom and is not involved in bonding are called lone pairs. For Eg: NH 3 contains one lone pair of electrons. H 2 O contains two lone pair of electrons. Although CO 2 and H 2 O are both triatomic molecules, the shape of CO 2 is linear but H 2 O is bent. The shape of the molecules can be predicted by dipole moments. Experiments showed that the dipole moment of CO 2 is zero. This is possible only if bond dipoles of two C=O bonds cancel each other. In other words, the two bond dipoles must be oriented in opposite directions. This is possible if the molecule is linear. Similarly H 2 O has a dipole moment of 1.83 D. Thus its molecule cannot be linear. Because the bond dipoles do not cancel each other. Thus the H 2 O molecule must have an angular shape. 32. Distinguish between a sigma and a pi bond. The following are the differences between σ and π -bonds. σ - bond is formed by the head on (or) axial overlap of s-; s-p and p-p orbitals. 1. A π - bond is formed by the sidewise (or) lateral overlap of p-orbitals. 18

19 2. A σ- bond is stronger since there is maximum overlapping. 3. A σ - molecular orbital consists of a single electron cloud which is symmetrical along the internuclear axis. 4. There is free rotation of atoms about the σ- bond. 5. There can be only one σ - bond between two atoms. 2. It is weaker since the extent of overlapping of orbitals is minimum. 3. There are two electron clouds one above and another below the plane of the bonded atoms. 4. π- bond is rigid and no free rotation of atoms is possible about π- bond. 5. There can be one or two π - bonds between two atoms. 33. Explain the formation of H 2 molecule on the basis of valence bond theory. The formation of the simplest molecule, hydrogen (H 2 ) molecule can be explained by VB theory. H 2 molecule is a diatomic homo nuclear molecule. Each H atom has a proton as the positive nucleus and an electron in 1s orbital. According to VB theory, the H H covalent bond is formed by the overlap of the two 1s orbitals of the hydrogen atoms, thus sharing the pair of electrons in a common region of space lying between the two nuclei. Also, the two electrons overlapping orbitals exist with opposite spins only. Because of spherical and symmetrical nature of 1s orbitals, when the two 1s orbitals of the two H atoms overlap, the paired electrons are usually located at a region between two nuclei. Thus both the electrons are attracted equally by both the nuclei. The formation process of H 2 molecule can be explained by considering the potential energy changes when the two H atoms approach each other from infinity position in following manner: (i) When the two H atoms are far apart, there is no interaction between them, an the P. E of the system of two H atoms is taken as zero. (ii) As the two atoms approach each other, the inter nuclear distance of separation gets reduced and the attraction of the nuclei for the electrons increases and the repulsion between nuclei as well as between the electrons increase. (iii) As long as attraction forces are more than the repulsions the potential energy of the system decreases than zero and becomes negative. This trend continuous till the potential energy of the system reaches a minimum value. This corresponds to the most stable state of the system. (iv) Further decrease in the inter nuclear distance causes steep increase in the potential energy due to the increased inter-nuclear and inter electronic repulsion and the system becomes unstable. (v) The distance of separation between the two H atoms at which the total energy of the system is minimum is considered as the inter-nuclear distance or the bond length of the H 2 molecule. (vi) The decrease in the potential energy takes place, by the release of energy during a single bond formation. i.e. When H 2 molecule is formed heat is given off. Conversely, to break up the H H bond energy 19

20 must be supplied to the molecule. 20

21 34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. For the atomic orbitals to combine resulting in the formation of molecular orbitals the main conditions are: 1.The combining atomic orbitals should have almost same energies. For example, in case of homo nuclear diatomic molecules 1s orbital of one atom can combine with 1s orbital of the other atom but 1s orbital of one atom cannot combine with 2s orbital of the other atom. 2.The extent of overlap between the atomic orbitals of the two atoms should be large. 3.The combining atomic orbitals should have the same symmetry about the molecular axis. For example, 2p x orbital of one atom can combine with 2p x orbital of the other atom but not with 2p z orbital. It must be noted that z-axis is taken as the inter nuclear axis according to modern conventions. 35. Use molecular orbital theory to explain why the Be 2 molecule does not exist. Molecular orbital configuration of Be 2 is (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2 Bond order of Be 2 = (N b N a ) = (4-4) = 0. Since the bond order is 0, Be 2 molecule does not exist. 36. Compare the relative stability of the following species and indicate their magnetic properties:o 2,O 2 +,O 2 -,O OXYGEN MOLECULE The electronic configuration of oxygen is 1s 2, 2s 2, 2p 4. Each oxygen atom has 8 electrons, hence, in O 2 molecule there are 16 electrons. The electronic configuration of O 2 molecule, therefore, is O 2 : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (<π * 2px) 1, (π * 2py) 1. From the electronic configuration of O 2 molecule it is clear that ten electrons are present in bonding molecular orbitals and six electrons are present in anti-bonding molecular orbitals. Its bond order, therefore, is Bond order = [N b N a ] = (10-6) = 4 = 2. 21

22 So in oxygen molecule, atoms are held by a double covalent bond. The bond dissociation energy of O 2 molecule has been found to be 495 kj/mole and bond length 121 pm. Moreover, from the molecular orbital diagram of O 2 molecule, it may be noted that it contains two unpaired electrons in π 2px *, π 2py * molecular orbitals. Therefore, O 2 molecule has paramagnetic nature. Oxygen Molecule Ion (O 2 + ) This ion is formed by removal of one electron from O 2 molecule. O 2 e - O 2 + From the M. O. diagram of O 2 it is clear that this electron would be removed from one of the antibonding molecular orbitals (π 2px *, π 2py * ). Therefore, the electronic configuration of O 2 + ion is O 2 + : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (π * 2px) 1 And its bond order is (10-5) = 5 = 2.5. Since bond order of O 2 + is greater than the bond order of O 2, therefore, it will have smaller bond length and larger bond dissociation energy than that of O 2 molecule. Its bond length is 112 pm and bond dissociation energy is 625 kj/mole. Moreover it exhibits paramagnetic character because of the presence of unpaired electrons. Superoxide Ion (O 2 - ) This ion is formed by addition of one electron to O 2 molecule. O 2 + e - O 2 - This additional electron enters into one of the anti-bonding orbitals (π 2px *, π 2py * ). Therefore, electronic configuration of O 2 - ion is O 2 - : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (π * 2px) 2, (π * 2py) 1. And its bond order is = (10-7) = 3 = 1.5. Since bond order of O 2 - is less than that of O 2 molecule, therefore, it will have longer bond length and smaller bond dissociation energy than O 2 molecule. It is also paramagnetic because of the presence of an unpaired electron. Peroxide Ion (O 2 2- ) This ion is formed by addition of two electrons to O 2 molecule. O 2 + 2e - O These additional electrons enter the two half-filled p anti-bonding orbitals. Therefore, electronic configuration of O 2 2- ion is (O 2 2- ): (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (π * 2px) 2, (π * 2py) 2. 22

23 And its bond order is = (10-8) = (2) = 1. Since bond order of O 2 2- ion is less than that of O 2 molecule, it will have longer bond length and smaller bond dissociation energy than that of O 2 molecule. Since there is no unpaired electron in it, it is diamagnetic in nature. 37. Write the significance of a plus and a minus sign in representing the orbitals. The positive and negative signs in an orbital indicate whether the orbital wave function is positive or negative in a particular region. These signs do not represent the positive or negative charge. They are just analogous to the signs of amplitude in case of plane waves. 38. Describe the hybridisation in case of PCl 5. Why are the axial bonds longer as compared to equatorial bonds? Hybridisation in PCl 5 Hybridisation is the phenomenon of intermixing of atomic orbitals of slightly different energies of the atom (by redistributing their energies) to form new set of orbitals of equivalent energies and identical shape. Atomic number of phosphorus P = 15. Electronic configuration of P Ground State 1s 2 2s 2 2p 6 3s 2 3p 3 3d 0 Excited State One electron from 3s orbital has jumped to the higher 3d orbital. After attained the excited state 1s, 3p and 1d orbital (altogether 5 orbitals) are hybridised to form an equal set of equivalent five sp 3 d hybrid orbitals. These 5 sp 3 d orbitals are directed towards the five corners of a trigonal bipyramidal geometry. Cl atom is present at each corner of the trigonal bipyramid. In PCl 5, out of the five hybrid orbitals, three orbitals form P-Cl bond in one plane making an angle (P-Cl-P) 120 with each other. This plane is represented as equatorial plane and the bonds formed are equatorial bonds. Out of the remaining two hybrid orbitals, one lie perpendicularly above and the other lie perpendicularly below the equatorial plane, making an angle 90 with the plane and forms P-Cl 23

24 bonds. These two bonds are called axial bonds. Since the axial bonds suffer more repulsive interaction from the equatorial bond pairs, they are found to be slightly longer and hence slightly weaker than the equatorial bonds. 39. Define Hydrogen bond. Is it weaker or stronger than the Van der Walls forces? It is defined as the electrostatic force of attraction which exists between the covalently bonded H 2 atom of one molecule and the electronegative atom of the other molecule. For example, in case of HF, the H 2 bond exists between H atom of one molecule and F atom of other molecule.. H F. H F.. H - F. H F The attractive forces between the instantaneous dipoles and induced dipoles are called Van der Waals forces. These forces are quite weak and its strength lies below 12.5 kj mol -1. The strength of Hydrogen bond ranges between 12.5 kj mol -1 to 41.5 kj mol -1. Hence it is clear that H 2 bonding is stronger than the Van der Waals forces. 40. What is meant by the term bond order? Calculate the bond order of: N 2, O 2, O 2 +, O 2 -. Bond order cannot be used for quantitative comparison of the strengths of chemical bonds. It can give us only approximate idea about the strength of the bond. Greater the bond order, greater is the strength of the bond. The molecular orbital configuration and bond order for N 2, O 2, O 2 +, O 2 - is given as follows: N 2 : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (π 2px ) 2, (π 2py ) 2, (σ 2pz ) 2 Bond order of N 2 = (10-4) = 6 = 3. 24

25 O 2 : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (π * 2px) 1, (π * 2py) 1. Bond order of O 2 = (10-6) = 4 = 2. O 2 + : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (π * 2px) 1. Bond order of O 2 + = (10-5) = 5 = 2.5. O 2 - : (σ 1s ) 2, (σ 1s * ) 2, (σ 2s ) 2, (σ 2s * ) 2, (σ 2pz ) 2, (π 2px ) 2, (π 2py ) 2, (π * 2px) 2, (π * 2py) 1. Bond order of O 2 - = (10-7) = 3 =