Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

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1 CCB 1 CHEMICAL BONDING AND MOLECULAR STRUCTURE 41 Kossel-Lewis Approach to Chemical Bonding Q According to Kossel and Lewis why do atoms combine to form molecule Solution The atoms of different elements combine with each other in order to complete their respective octets (ie, 8 electrons in their outermost shell) or duplet (ie, outermost shell having 2 electrons) in case of H, Li and Be to attain stable nearest noble gas configuration Q Define electrovalent bond or ionic bond Solution The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as the electrovalent bond The electrovalence is thus equal to the number of unit charge(s) on the ion Q Define Octet rule Solution According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells This is known as octet rule Q Draw the Lewis Dot structure for following molecules Cl 2, O, CCl 4, CO 2, C 2 H 4, N 2, C 2 Solution Cl 2 O CCl 4 CO 2 C 2 H 4 N 2 C 2

2 CCB 2 Q Draw the Lewis representation of O 2, O 3, NF 3, CO 3 2, HNO 3 Solution Molecule/Ion Lewis Representation O 2 O 3 NF 3 CO 3 2 HNO 3 Q Write the Lewis dot structure of CO molecule [NCERT Solved Example 41] Solution Q Write the Lewis structure of the nitrile ion, NO 2 [NCERT Solved Example 42] Solution or or Q Write the method to find the formal charge of an atom in Lewis structure Solution Formal charge (FC) on an atom in a Lewis structure = [total number of valence electrons in the free atom] [total number of non bonding (lone pair) electrons] ½ [total number of bonding (shared) electrons] Q Find the formal charge on each oxygen atom in Lewis dot structure of O 3 Solution The atoms have been numbered as 1, 2 and 3 The formal charge on 1 The central O atom marked 1 = 6 2 (6) The end O atom marked 2 = 6 4 (4) The end O atom marked 3 = 6 6 (2) 1 2 Hence, we represent O 3 along with the formal charges as follows Q What are the limitation of Octet rule? Solution There are three types of exceptions to the octet rule

3 CCB 3 (a) The incomplete octet of the central atom In some compounds, the number of electrons surrounding the central atom is less than eight This is especially the case with elements having less than four valence electrons Examples are LiCl, Be and BCl 3 (b) Odd-electron molecules In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, NO 2, the octet rule is not satisfied for all the atoms (c) The expanded octet Elements in and beyond the third period of the periodic table have, apart from 3s and 3p orbitals, 3d orbitals also available for bonding In a number of compounds of these elements there are more than eight valence electrons around the central atom This is termed as the expanded octet Obviously the octet rule does not apply in such cases Some of the examples of such compounds are PF 5, SF 6, SO 4 (d) It is clear that octet rule is based upon the chemical inertness of noble gases However, some noble gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of compounds like XeF 2, KrF 2, XeOF 2 etc (e) This theory does not account for the shape of molecules (f) It does not explain the relative stability of the molecules being totally silent about the energy of a molecule 42 Ionic or Electrovalent Bond Q How the ionic bond is formed? Solution Ionic bonds will be formed more easily between elements with comparatively low ionization enthalpies and elements with comparatively high negative value of electron gain enthalpy The formation of a positive ion involves ionization, ie, removal of electron(s) from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom M(g) M + (g) + e ; Ionization enthalpy X(g) + e X (g) ; Electron gain enthalpy M + (g) + X (g) MX (s) Q Define Lattice Enthalpy Solution The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions 43 Bond Parameters Q Define Bond length Solution Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule Q Define Covalent radius Solution The covalent radius is measured approximately as the radius of an atom s core which is in contact with the core of an adjacent atom in a bonded situation The covalent radius is half of the distance between two similar atoms joined by a covalent bond in the same molecule Q Define vander Waals radius Solution The vander Waals radius represents the overall size of the atom which includes its valence shell in a nonbonded situation Further, then vander Waals radius is half of the distance between two similar atoms in separate molecules in a solid Q Define Bond angle and Bond enthalpy Solution Bond angle is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion Bond angle is expressed in degree Bond enthalpy is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state The unit of bond enthalpy is kj mol 1 for eg, (g) H(g) + H(g); a H 0 = 4358 kj mol 1 It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule

4 Q Define Bond order CCB 4 Solution In the Lewis description of covalent bond, the bond order is given by the number of bonds between the two atoms in a molecules Isoelectronic molecules and ions have identical bond orders; for example, F 2 and O 2 2 have bond order 1 N 2, CO and NO + have bond order 3 with increase of bond order, bond enthalpy increases and bond length decreases Q Define Resonance Solution According to the concept of resonance, whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately Q Draw the resonating structure of O 3 and carbonate ion Solution Q Explain the resonating structure of CO 2 molecule Solution Q Define polar and non-polar covalent bond Solution In, there is no displacement of the electric charge due to same electron affinity of both H-atoms and the bond is non-polar In HCl, the Cl atom has a more electronegativity than does the H atom Electronic charge distribution is shifted towards the Cl atom The H-Cl bond is said to be polar Q Define dipole moment Solution The magnitude of the charge displacement in a polar covalent bond is measured through a quantity called the dipole moment µ It is the product of the magnitude of charges ( ) and the distance separating them (d) (Here the symbol ( ) suggests a small magnitude of charge, less than the charge on an electron) Dipole moment is the vector quantity and direction of vector is from less electronegative to more electronegative atom, for eg, µ = d If = esu and d = 1Å = cm then µ = = esu cm In SI unit, 1 D = coulomb meter (when charge = C and d = m) In diatomic molecule, µ = d Q Why the dipole moment in case of BeF 2 and BF 3 is zero? Solution In BeF 2 two equal bond dipoles point in opposite direction and cancel the effect of each other for eg, In tetra-atomic molecule, for example in BF 3, the dipole moment is zero although the B F bonds are oriented at an angle of to one another, the three bond moments give a net sum of zero as the resultant

5 CCB 5 of any two is equal and opposite to the third, for eg, Q Which out of NH 3 and NF 3 has higher dipole moment and why? Solution NH 3 has higher dipole moment than NF 3 This is because, in case of NH 3 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N H bonds, whereas in NF 3 the orbital is in the direction opposite to the resultant dipole moment of three N F bonds The orbital dipole because of lone pair decreases the effect of the resultant N F bond moments, which results in the law dipole moment of NF 3 Q Explain briefly the Fajans rules Solution Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partial covalent character The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond The greater the charge on the cation, the greater the covalent character of the ionic bond For cations of the same size and charge, the one, with electronic configuration (n 1)d n ns 0, typical of transition metals, is more polarising than the one which a noble gas configuration, ns 2 np 6, typical of alkali and alkaline earth metal cations The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two This is precisely what happens in a covalent bond, ie, buildup of electron charge density between the nuclei The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond 44 The Valence Shell Electron Pair Repulsion (VSEPR) Theory Q Explain the main postulates of (VSEPR) theory Solution (a) The shape of a molecule depends upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom (b) Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged (c) These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them (d) The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another (e) A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair (f) Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structures The repulsive interaction of electron pairs decrease in the order Line pair (1p) Lone pair (1p) > Lone pair (1p) Bond pair (bp) > Bond pair (bp) Bond pair (bp)

6 CCB 6 Q Draw the molecular geometry of following molecules BeCl 2, BF 3, CH 4, PCl 5, SF 6, SO 2, NH 3, O, SF 4, ClF 3, BrF 5, XeF 4 Solution BeCl 2 Cl Be Cl BF 3 Linear 0 Bond angle180 CH 4 PCl 5 SF 6 SO 2 NH 3 O SF 4 ClF 3 BrF 5 XeF 4

7 CCB 7 45 Valence Bond Theory Q Explain the formation of molecule on the basis of valence bond theory Solution A labelled potential energy curve for molecule is shown in the figure The potential energy curve for the formation of molecule as a function of internuclear distance of H atoms The minimum on the curve corresponds to the most stable state of At point A when the two atoms are far apart from each other, there are no attractive or repulsive forces between the hydrogen atoms At point B when the two atoms starts approaching each other, they start interacting with each other and the system starts losing its energy because attractive forces between the two atoms become more and more dominant At point C, the two atoms acquire minimum energy since the attractive forces are balanced by the repulsive forces Beyond point C, after intermolecular distance r 0, repulsive forces predominate and molecule formed tends towards disability That is why potential energy increases abruptly Q How many types of covalent bond are possible on the basis of type of overlapping? Solution They are of two types (i) sigma( ) bond, and (ii) pi( ) bond (i) Sigma( ) bond This type of covalent bond is formed by the end to end (head-on) overlap of bonding orbitals along the internuclear axis This is called as head on overlap or axial overlap This can be formed by any one of the following types of combinations of atomic orbitals (a) s-s overlapping In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below (b) s-p overlapping This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom (c) p-p overlapping This type of overlap takes place between half filled p-orbitals of the two approaching atoms (ii) pi( ) bond In the formation of bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis The orbitals formed due to sidewise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms

8 CCB 8 Q Write the difference between sigma and pi bond? Solution Sigma bond (i) This bond is formed by overlap of orbitals along their internuclear axis (end to end overlap) (ii) This is formed by overlapping between s-s, s-p or p-p orbitals (iii) Overlapping is quite large and hence sigma bond is a strong bond (iv) Electron cloud in this case is symmetrical about the line joining the two nuclei (v) Sigma bond consists of only one electron cloud, symmetrical about the internuclear axis (vi) Free rotation about a -bond is possible Pi Bond (i) This is formed by sideway overlapping of orbitals (lateral overlapping) (ii) This is formed by the overlap of p-p orbitals only (iii) Overlapping is to a small extent Hence, -bond is a weak bond (iv) Electron cloud of -bond is unsymmetrical (v) Pi bond consists of two electron clouds, one above the plane of atomic nuclei and the other below it (vi) Free rotation about a -bond is not possible because on rotation, overlapping vanishes and so the bond breaks 46 Hybridisation Q Explain the process of hybridisation Solution Hybridisation is defined as the concept of intermixing of orbitals of same energy or of slightly different energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane New orbitals formed are called hybrid orbitals Q Write the salient feature of hybridisation and the conditions for hybridisation Solution Salient features (i) The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised (ii) The hybridised orbitals are always equivalent in energy and shape (iii) The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals (iv) These hybrid orbitals are directed in space in some preferred direction to have minimum repulsion between electron pairs and thus a stable arrangement Therefore, the type of hybridisation indicates the geometry of the molecules Conditions for hybridisation (i) The orbitals present in the valence shell of the atom are hybridised (ii) The orbitals undergoing hybridisation should have almost equal energy (iii) Promotion of electron is not essential condition prior to hybridisation (iv) It is not necessary that only half filled orbitals participate in hybridisation In some cases, even filled orbitals of valence shell take part in hybridisation Q Draw the hybridisied structure of BeCl 2, BCl 3, CH 4 Solution BeCl 2 In BeCl 2, Be is sp hybridised The shape of the molecule is linear having bond angle BCl 3 In BCl 3, B is sp 2 hybridised

9 CCB 9 CH 4 CH 4 is a tetrahedral molecule with each H C H angle equal to The four sp 3 hybrid orbitals overlap with the half-filled 1s orbitals of four H-atoms Q Draw the shape of C 2 H 4 molecule and C 2 molecule showing the and bond Solution C 2 H 4 or C 2 Q Find the hybridisation of the following molecules PF 5, PCl 5, BrF 5, SF 6 Solution Molecule Hybridisation Shape of molecule PF 5 sp 3 d Trigonal bipyramidal PCl 5 sp 3 d Trigonal bipyramidal BrF 5 sp 3 d 2 Square pyramidal SF 6 sp 3 d 2 Square bipyramidal Q Describe the hybridisation of PCl 5 and SF 6 Solution Formation of PCl 5 (sp 3 d hybridisation) The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below

10 CCB 10 Formation of SF 6 (sp 3 d 2 hybridisation) In SF 6 the central sulphur atom has the ground state outer electronic configuration 3s 2 3p 4 In the exited state the available six orbitals ie, one s, three p and two d are singly occupied by electrons These orbitals hybridise to form six new sp 3 d 2 hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF 6 These six sp 3 d 2 hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S F sigma bonds Thus SF 6 molecule has a regular octahedral geometry as shown in figure below 47 Molecular Orbital Theory Q Write the salient features of molecular orbital theory Solution (i) The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals (ii) The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals (iii) While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule Thus, an atomic orbital is monocentric while a molecular orbital is polycentric (iv) The number of molecular orbital formed is equal to the number of combining atomic orbitals When two atomic orbitals combine, two molecular orbitals are formed One is known as bonding molecular orbital while the other is called antibonding molecular orbital (v) The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital (vi) Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital (vii) The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli s exclusion principle ans the Hund s rule Q Write the formation of molecular orbital by the linear combination of atomic orbitals Solution Linear combination of Atomic orbitals (LCAO) in case of + (i) A linear combination of two atomic orbitals orbitals and A and B leads to the formation of two molecular (ii) The energy E + of molecular orbital is lowe than either of E A and E B (energies of isolated atoms) It is therefore designated as bonding molecular orbital (BMO) (iii) The energy E of molecular orbital is higher than either of E A and E B It is therefore designated as antibonding molecular orbital (ABMO) (iv) The extent of lowering of energy of the bonding molecular orbital is equal to the extent of increase of energy of antibonding molecular orbital Energies of bonding and antibonding molecular orbitals

11 Bonding MO S (2s) (2px) (2py) (2pz) Anti Bonding MO S (2s) (2px) (2py) (2pz) CCB 11 The order of energy of molecular orbital for lighter elements like boron, carbon and nitrogen are as follows (1s) < *(1s) < (2s) < *(2s) < 2py = 2pz < 2px < *2py = *2pz < *px The order of energy of molecular orbitals for heavier elements after nitrogen are (1s) < *(1s) < (2s) < *(2s) < 2px < 2py = 2pz < *2py = *2pz < *2px Q Define bond order Solution Bond order (BO) is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals ie, BO = 2 1 [Nb N a ] N b number of bonding electrons N a number of anti-bonding electrons If BO = 0, 1, 2, 3 so on it means that no bond is formed, one bond is formed, two bonds or three bonds are formed between the atoms respectively (i) BO, Bond dissociation energy (ii) BO 1 Bond Length Q What are diamagnetic and paramagnetic molecules? Solution Electronic configuration helps to predict the magnetic character of the molecule If all the electrons in a molecule are paired they are diamagnetic (repelled by magnetic field) and if unpaired electron is present they are paramagnetic (attracted by magnetic field) for eg, O 2 molecule Q Write the molecular orbital electronic configuration for He 2, Be 2, Li 2, C 2, O 2 and also write their bond order Solution Helium molecule (He 2 ) Molecular orbital electronic configuration 1s 2 ) *1s 2 ) Bond order of (He 2 ) ½(2 2) = 0 He 2 molecule is therefore unstable and does not exist Berilium molecule (Be 2 ) Electronic configuration of Be is 1s 2 2s 2 Therefore for Be 2 molecule 1s 2 ) *1s 2 ) 2s 2 ) *2s 2 ) Bond order of (He 2 ) ½(4 4) = 0 Be 2 molecule is therefore unstable and does not exist Lithium molecule (Li 2 ) Electronic configuration of Lithium is 1s 2 2s 1 Therefore molecular orbital electronic configuration is Li 2 ( 1s) 2 ( *1s) 2 ( 2s) 2 The configuration is also written as KK( 2s) 2 where KK represents the closed K shell structure ( 1s) 2 ( *1s) 2 Bond order of (Li 2 ) is ½ (4 2) = 1 It means it is stable and has no unpaired electrons Therefore it is diamagnetic Carbon molecule (C 2 ) Electronic configuration of C is 1s 2 2s 2 2p 2 Therefore for C 2 molecule 1s 2 ) *1s 2 ) 2s 2 ) *2s 2 ) ( 2p x 2 = 2p y2 ) Bond order of (C 2 ) is ½ (8 4) = 2 C 2 molecule is diamagnetic Oxygen molecule (O 2 ) Electronic configuration of O is 1s 2 2s 2 2p 4 Therefore for O 2 molecule 1s 2 ) *1s 2 ) 2s 2 ) *2s 2 ) ( 2p z ) 2 ( 2p x 2 2p y2 ) ( *2p x 1 *2p y1 ) Bond order of (O 2 ) is ½ (10 6) = 2 O 2 molecule is paramagnetic

12 49 Hydrogen Bonding Q Define hydrogen bond and what is its causes CCB 12 Solution Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule When hydrogen is bonded to strongly electronegative element X, the electron pair shared between the two atoms moves far away from hydrogen atom As a result the hydrogen atom becomes highly electropositive with respect to the other atom X Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge ( + ) while X attain fractional negative charge ( ) This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as H + X H + X H + X Q How many types of hydrogen bond are there Explain them briefly? Solution There are two types of hydrogen bonds (i) Intramolecular H-Bonding This type of H-bonding occurs when polar H and electronegative atom are present in the same molecule (a) (b) (ii) Intermolecular H-Bonding This type of H-bonding takes place between H and electronegative element present in the different molecules of the same substance (as in between O and O) or different substances (as in between O and NH 3 ) eg In water molecules Due to polar nature of O, there is association of water molecules giving a liquid state of abnormally high bp

13 CCB Explain the formation of a chemical bond NCERT EXERCISE 42 Write Lewis dot symbols for atoms of the following elements Mg, Na, B, O, N 43 Write Lewis symbols for the following atoms and ions S and S 2 ; Al and Al 3+, H and H 44 Draw the Lewis structures for the following molecules and ions S, SiCl 4, BeF 2, CO 3 2, HCOOH 45 Define octet rule Write the significance and limitations 46 Write the favourable factors for the formation of ionic bond 47 Discuss the shape of the following molecules using VSEPR model BeCl 2, BCl 3, SiCl 4, AsF 5, S, PH 3 48 Although geometries of NH 3 and O molecules are distorted tetrahedral, bond angle in water is less than of ammonia Discuss 49 How do you express the bond strength in terms of bond order? 410 Define the bond length 411 Explain the important aspects of resonance with reference to the CO 3 2 ion 412 H 3 PO 3 can be represented by structures 1 and 2 shown below Can these two structures be taken as the canocical forms of the resonance hybrid representing H 3 PO 3? If not, given reasons for the same 413 Write the resonance structures for SO 3, NO 2 and NO 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 415 Although both CO 2 and O are triatomic molecules, the shape of O molecule is bent while that of CO 2 is linear Explain this on the basis of dipole moment 416 Write the significance/applications of dipole moment 417 Define electronegativity How does it differ from electron gain enthalpy? 418 Explain with the help of suitable example polar covalent bond 419 Arrange the bonds in order of increasing ionic character of the molecules LiF, K 2 O, N 2, SO 2 and ClF 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 421 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 atoms at its centre Explain why CH 4 is not square planar? 422 Explain, why Be molecule has a zero dipole moment although the Be H bonds are polar 423 Which out of NH 3 and NF 3 has higher dipole moment and why? 424 What is meany by hybridisation of atomic orbitals? Describe the shapes of sp, sp 2, sp 3 hybrid orbitals

14 425 Describe the change in hybridisation if any, of the Al atom in the following reaction AlCl 3 + Cl AlCl 4 CCB 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 BNH Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C 2 H 4 and C 2 molecules 428 What is the total number of sigma and pi bonds in the following molecules? (a) C 2 (b) C 2 H Considering x-axis as the intermolecular axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2p x (c) 2p y and 2p y (d) 1s and 2s 430 Which hybrid orbitals are used by carbon atoms in the following molecules? (a) CH 3 CH 3 (b) CH 3 CH = C (c) CH 3 C OH (d) CH 3 CHO (e) CH 3 COOH 431 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type 432 Distinghish between a sigma and pi bond 433 Explain the formation of molecule on the basis of valence bond theory 434 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals 435 Use molecular orbital theory to explain why the Be 2 molecule does not exist 436 Compare the relative stability of the following species and indicate their magnetic properties O 2, O 2+, O 2 (superoxide), O 2 2 (peroxide) 437 Write the significance of a plus and a minus sign shown in representing the orbitals 438 Describe the hybridisation in case of PCl 5 Why are the axial bonds longer as compared to equatorial bonds? 439 Define hydrogen bond Is the weaker or stronger than the van der Waals forces? 440 What is meant by the term bond order? Calculate the bond order of N 2, O 2, O 2 + and O 2

15 CCB 15 ANSWERS 42,,,, 43,, 44,,,, 45 Lewis postulated that atoms achieve the stable octet when they are linked by chemical bonds and in doing so each atom attains a stable outer octet of electrons Significance of octet rule It helps to explain why different atoms combine with each other to form ionic compounds or covalent compounds Limitations of octet rule (i) In the formation of hydrogen molecule duplet is completed and not the octet (ii) Also in the formation of BeCl 2, BF 3, AlCl 3 etc octet is not completed 46 (i) Low ionization enthalpy of the metal atom (ii) High electron gain enthalpy of the non-metal atom (iii) High lattice enthalpy of the compound formed 49 Greater the bond order, shorter the bond length and greater the bond strength 410 Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule ,, 415 Studies have shown that the net dipole moment of CO 2 molecule is zero This is possible only when CO 2 molecule is linear ie, O = C = O, such that dipole moments of C O are equal and opposite and hence cancel out On the other hand, it is found that O molecule has a net dipole moment of 184 D, though it contains 2O H bonds This suggests that it has a bent structure 416 (i) In determining the polarities of bonds As µ = q d, greater is the magnitude of dipole moment higher will be the polarity of bonds H F > H Cl > H Br > HI Further in case of non-polar molecules like O 2 and N 2, the dipole is found to be zero (ii) In calculation of percentage ionic character 417 Electronegativity (i) It is the tendency of an atom to attract shared pair of electrons (ii) It is the property of bonded atom (iii) The elements with symmetrical configuration have specific electronegativities (iv) It has no units Electron gain enthalpy (i) It provides a measure of the ease with which an atom adds an electron to form an anion (ii) It is the property of an isolated atom (iii) The elements with symmetrical configuration have almost zero electron gain enthalpy (iv) It has units of kj mol Polar character of covalent bonds when two unlike atoms form a covalent bond, the electrons may not be shared equally by both the atoms In such a case, the electrons cloud is closer to the one atom than to the other One end of the bond, therefore, develops partial positive charge and the other

16 CCB 16 partial negative charge Such a bond is called a partial polar bond This is represented by symbols + and, where d stands for a partial charge Example H Cl 419 N 2 < SO 2 < ClF 3 < K 2 O < LiF 422 (3) In Be two dipoles act in opposite direction and hence the two dipoles cancel each other completely and hence has a zero dipole moment 423 NH 3 has higher dipole moment than NF 3 This is because, in case of NH 3 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N H bonds, whereas in NF 3 the orbital is in the direction opposite to the resultant dipole moment of three N F bonds The orbital dipole because of lone pair decreases the effect of the resultant N F bond moments, which results in the law dipole moment of NF Initially BF 3 is sp 2 hybridised and NH 3 is sp 3 hybridised F 3 BNH 3 is sp 3 hybridised Thus, there is a change in hybridisation of boron 428 (a) (3, 2 ) (b) (5, 1 ) 429 Only (c) will not give a -bond because taking x-axis as the internuclear axis, there will be lateral (sideway) overlap between two 2p y orbitals forming a -bond 430 (a) both carbon atoms are sp 3 hybridised (b) C 1 = sp 3, C 2 = sp 2, C 3 = sp 3 (c) both carbon atoms are sp 3 hybridised (d) C 1 = sp 3, C 2 = sp 2 (e) C 1 = sp 3, C 2 = sp A labelled potential energy curve for molecule is shown in the figure The potential energy curve for the formation of molecule as a function of internuclear distance of H atoms The minimum on the curve corresponds to the most stable state of At point A when the two atoms are far apart from each other, there are no attractive or repulsive forces between the hydrogen atoms At point B when the two atoms starts approaching each other, they start interacting with each other and the system starts losing its energy because attractive forces between the two atoms become more and more dominant At point C, the two atoms acquire minimum energy since the attractive forces are balanced by the repulsive forces Beyond point C, after intermolecular distance r 0, repulsive forces predominate and molecule formed tends towards disability That is why potential energy increases abruptly 434 (a) The combining atomic orbitals should have comparable energies (b) The combining atomic orbitals must have proper orientation eg, same symmetry, so that they are able to overlap to a considerable extent (c) The extent of overlaping should be large Greater the overlap, greater will be the electron density between the nuclei 435 Molecular orbital electronic configuration of Be 2 = 1s 2, *1s 2, 2s 2, *2s 2 Bond order = 1 = (4 4) 0 From molecular orbital theory bond order of Be 2 2 is 0, which shows that Be 2 does not exist Bond order of O 2 = 2, O 2 = 2, O 2 2 = 1, O 2 2 = 1 We know that higher the bond order higher the 2 + stability ie, O 2 < O 2 < O 2 < O they undergo sp 3 d hybridization to form five equivalent orbitals 439 Hydrogen bond is stronger than ver der Waals forces N b 2 N a

17 CCB 17 ADDITIONAL QUESTIONS AND PROBLEMS Q Identify the compounds which do not obey the octet rule from among SO 2, SF 2, SF 4, SF 6, AlCl 3, MgCl 2, PCl 3, PCl 5 and XeF 2 Q Why is it easier to remove an electron from an O 2 molecule than from an N 2 molecule? Q Using Lewis electron-dot symbols, show the formation of bonds in (a) O (b) C 2 H 4 Q What is meant by formal charge? Calculate the formal charges of all the atoms in the nitrite and the carbonate ion Q In the following pairs, which compound will be more ionic and why? (a) NaCl and KCl (b) NaCl and NaI (c) NaCl and MgCl 2 Q Arrange the following in order of increasing bond length, giving reasons (a) H F, H Cl, H Br and H I (b) Carbon-carbon bond length in C 2 H 6, C 2 H 4 and C 2 Q Draw the resonance structures for (a) 2 SO 4 (b) NO 3 (c) ClO 4 Q Both CO 2 and SO 2 contain polar covalent bonds, yet the former is nonpolar, while the latter is polar Give reasons Q The N F bond is more polar than the N H bond, yet NH 3 has a higher dipole moment than NF 3 Justify Q Using the VSEPR theory, predict and draw the structures of (a) IF 7 (b) SF 4 (c) PCl 5 Q How does the VSEPR theory explain the (a) bent structure of water (b) pyramidal structure of ammonia Q Why is a sigma bond stronger than a pi bond? Explain Q Consider the following reaction BF 3 + F BF 4 What is the hybridisation of boron in the reactant and the product? Q + Is there any change in the hybridisation of nitrogen in NH 3 and NH 4? Justify your answer Q Describe the hybridisation of PCl 5 and SF 6 Q Explain why OF 6 is not known but SF 6 is known Q Using the molecular orbital theory, explain why Be 2 does not exist Q Explain why oxygen is paramagnetic but the peroxide ion is diamagnetic Q Which is more stable the peroxide ion or the superoxide ion? Justify your answer by giving a suitable example Q Define bond order How does it affect the stability of a molecule? Q Which of three +, and are paramagnetic and why? Q Which will have a higher boiling point and why NH 3 or PH 3? Q Give an example of each of the following (a) A molecule with a trigonal bipyramidal shape (b) A molecule with a pyramidal shape (c) A molecule containing a coordinate bond Q Using the VSEPR theory, explain why (a) CF 4 and XeF 4 differ in structures, and (b) the shapes of PF 5 and BrF 5 are different Q What is meant by sigma and pi bonds Taking a suitable example, show under what conditions and bonds are formed Q (a) What is meant by bonding and antibonding molecular orbitals? Explain by taking the example of (b) Using the molecular orbital theory, explain the magnetic behaviour of O 2, B 2 and C 2

Class XI Chapter 4 Chemical Bonding and Molecular Structure Chemistry

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