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TPU CHEMISTRY 1M € 2 M QUESTIONS pprr>sT > nn > > 2 > 11. CHEMICAL BONpyy 7 ACT Cl crystals? What are the coordination numbers of NaCI Ena $ Coordination number of NaClis 6, Coordination of Ñ the central atom? Predict the shape of a molecule with four bond pairs e hape is tetrahedral. Eg: CH, Tf a molecule with four bond pairs on the central atom then its shap 4 Explain what is meant by localised orbitals? : ati led localised orb; Localised orbitals: The orbitals which are involved in be eE are ca itals, mutually shared by the nuelei of both bonded atoms in molecules. CP has greater stability than Clatom. Explain why? st noble gas co; pe CT has argon configuration. Henoe itjs stable, but Cl has one electron short to nearest noble g: "Berto Hence it is unstable, i ns Why argon is not represented by Ar,? Argonatomhas only paired electrons with stable octet configuration. It cannot share ¡ts electrons with another argon atom and does not form (Ar,) diatomic molecule. l What is the valeney of the element £(A) Z=15) ? Asthe Z=15 ofelement A, its electronic configuration is [Ne] 35? 3p*, It can show valency 3 due to thres unpaired electrons in ground state. but in excited state it can exhibit 5 due to the presence of 5 unpaired electrons, A Write the consequences of hydrogen bonding? Consequences of Hydro gen bonding. : 1) Thecovalent compounds are usually insoluble in water. But ifthe compound is able to form hydrogen: bonds with water it can dissolve in water. B y the same reason glucose and alcohol are soluble in water 2) Dueto hydrogen bonding the M.P. and B.P of compounds increases anomalously than expected. Die to the same reason HO has high B.P than any other hydride ofthe same group elements. What type of bonds are present in 'NH,CT? Writeits structure? H e | NACI contains ionic, covalent and dative bonds. Structure: | H-N>H | Cr Define hydrogen bond. ls it weaker or stron; ger than vander Waals forces ? Hydrogen bond : The electrostatic forces ofattrací : attraction between a Partially positively charged hydrogen atom ofa polar molecule and highly electronegative atom is knownas hydrogen bond. It is stronger than vandet | Waals forces. a HO, LT. Correct the given structure. H= C -C-0-H .. "mo 0 y . H-C-C-0-8H 1 .. H 22, [PU CHE STRY CHEMICAL BONDING Which of the two ions (7 (or) 2" is more stable and why? Ca” is more stable (han Zn?" because Ca?' las octet contiguration in valenee shell where as zp has pscudo octet configuration ín velenos shell (18 electreon cofiguration). Octet configuration is more stable than pseudo octet configuratoín CI Son is more stable than Cf atom. CL ion poss: Why? $ses stable octet configuration but Cl atom doesnot have octet configuration Why Argon does not form Ar, molecule? Argon is monoatomie pas as it possesses stable vctet confuguration in ¡ts valence shell What is the best possible Arrangement of four bond paírs in the valency shell of an atom to minimise repulsion? The best possible arrangement to minimiso repulsión with four bond pairs in the valency shell of an atom is Tetrahedral shape with bond angle 109%8' IA and B are two different atoms when docs AB moclcule become covalent? When the electro negativity difference between the atoms A and Bis less than 1.7, then AB molecule becomes covalent What is meant by localised orbitals? The molecular orbital with bonded electron cloud localised between the two nuelei of bonded atoms is called localised orbtial. How many sigma and Pi bonds are presentin 2) C,H,andb) CH, In C,/1, number of sigma bonds are three and pi bonds are two in C,H, number of sigma bonds are five and pi bond is one Is there any change in the hybridisation of Boron and Nitrogen atoms as a result of the following reaction BF + NH, > F,BNH,? Before the reaction in the hybridisation of Boron is sp? and after the reaction it is sp". But for nitrogen it is sp' before and after the reaction. What is octet rule? An atom must possess 8 electrons in its valency shell for stability by losing, gaining or sharing electrons. ¿ " Write Lewis dot structures for S and s? The lewis dot structure are .. .. .. CA .S, . s. Z ; S s? S Ñ Write the possible resonance structures for 50,7 lo) l EN, 1050 bond angle 120" S—0Obond distance =],434 I PU CHEMISTRY CHEMICAL BONDING - More than 2 M Questions 1... Explain the formation ofionic bond in terms ofatomic orbitals. Discuss the factors that are favour the ionic bond formation 2 A: — Theelectrostatic force that binds the Oppositely charged ions which are formed by the transfer ofelectrons from one atom with low ionization Potential to the other with high electron affinity is called ionic bond or electrovalent bond. It is formed when the electronegativity difference between the two atoms is 1.7 or more. For example : Formation ofsodium chloride in terms oforbital concept, Na (2-11) Theelectronio configuration is 152522 . K,L CH (Z=17). The electronic configurationis 1525? 2p35%3p* «LAA Theconfigurationsaferthetransferofelectrons forming jonscan beexpressedas: Na” K,L35* 3p" 38 LK Na” ionattained the electronic CI” ionattained the electronic configuration ofNeon. configuration of Argon. In the formation of sodium chloride the 3s electron of. sodium atomis transferred to the 3 ¡p, orbital of chlorine atom. The Na” lonand C/” ionso formed are now bound by strong coulombic electrostatic forces ofattraction forming sodium chloride. Conditions : a) Cation formation : 1) Lowerionization energy: E Lower is the jonization energy ofan atom greater is the case of. formation of cation. Eg: Theionization energy of sodium is less than that of. Potassium. So K * ioncan readily formed than ii) Large size of the atom : R Large atoms can easily lose the valence electrons. Ifthe size js large, the distance between the nucleus and the valence electrons is more and so the force ofattraction is less. Therefore the electron can be removed easily from the atom forming cation. Li“ M?> ap iv) Cations withinert gas configuration: Cations possessing inert gas configuration are more stable than those. cations which have pseudo inert gas configuration. Ca? jonismore stable. Because Cy”? ion has electronic configuration ofinert gas (Argon), where as z,*2 has pseudo inert gas configuration, E 5) Anion formation : : E D High electron afíinity and high electronegativity: z lfthe electron affinity of an element is high its anion can be easily formed. NARAYANA STUDY MATERIAL 61 Há, , EEN CHEMICAL BONDING - PU CHEMISTRY Eg: (ly +0 > Cl ñi) Smaller size ofatom: Smaller the size of the atom lesser is the distance the nuclear attraction on incoming electron is moro. ii) LowerCharge: lons with lower negati E: cr 0 > nucleus and the valence orbit, Hero is readily formed. q between the So the anion with higher negative charge, charge are more readily formed than those 2.7 Describe the energy changes occuring in the formation of sodium chloride according to Born iS Maber Cycle. A: — Theencreyreleased when gaseous positive mole ofionic erystal is called lattice encrgy. Energy Changes in NaC/ according to Born - Haber Cyele ; e Born - aber eyele method is an indirect method for the calculation of lattice energies ofionic compounds Icis based on Hess law of constant heat summation. "The energy released in the formation ofan ionig substance is he same, whether the reaction proceeds in one step or in several steps”. 5 and negative ions are brought together from infinity to formorg +1 a Na(g) Mag) A “E a s ; a CE ud | a EN El 4 1 -3H, Ne + Zi ERICA EST DEE dE Na Ca a) In dírect method : The amount of heat evolved when.one mole of sodium chloride is formed from its elements is called heat of formation ( AH 1) «Its value is - 410,5 kj/mole. Na, y EVOL yy > Na Cl 3 5 (al LE A10S Imola, 5) Inanindi lrcet ' method : The above reaction can take place in several steps and the energy changes | in the various steps of formation of. NaCT can be calculated according to Born- Haber cycle. (1) Solid sodium sublimes to gaseous sodium. In this energy is absorbed. Itis sublimation. A Itísequalto 108.7 kJ/mole. Na, > Na, ; +S=-+108.7 kJ/mole, A (1) Gaseoussodiumionis formed by the lossofan electron fromsodiumatom. For this first jonizati energy is needed ¡.e 492,82 kJ/mole. Na, >) NO" ¿te "+ 1=+ 492,82 kJ/mole. (ii) Chlorine molecule dissociates into chlorine atoms. In the gascous state the energy absorbed pb called dissociation energy. (D). It is 239.1 kJ/mole for chlorine molecules. For one mole ofthe atom it is('/,D) =+ 119.55 kJ/mole, YO ¿y > CK(g); +1, D= +119,55 kJ/mole, (iv) Chloride ion is formed from chlorine atomin the e: gaseous state by taking an electron. Ts enel released is 361.57 kJ/mole is called electron affinity (E) Oli HO Cl E =-361.57 Kj/ mole (v) One mole of sodium chloride crystal is formed from Na” and CT: gaseous ions. The en NARAYANA STUDY MATERIAL 62 TPU CHEMISTRY electrons. They can attain ns np" configuration 4 le: NH, A Significance: The molecules with 8 electrons inits valence shell are more stable example: NH, H,O have 8 E electrons in its valence shell they are more stable, figurallon in lts outerinost orbít and to have 4 6. CA: NARAYANA STUDY MATERIAL 64 CHEMICAL BONDING Evety atom, therefore tries to acquire the eight electrons con £onfiguration ofthe zero group elements of the periodic table. ] r es. Discuss the phenomenon of hybridisation of atomic orbitals with suitable examp Hybridisation is defined as the process of mixing ofatomic orbitals of nearly equal energy ofanatomto give the same number ofnew set oforbitals ofequal eneray and shapes. Depending on the number and nature o orbitals involving in hybridisation ít is elassified into different types, If's' and 'p' atomic orbitals are involveg + three types are possible namely sp',sp*and sp. sp* hybridisation : Ñ In this hybridisation one 's' and three 'p' atomic orbitals of the excited atom combine to form fourequivalent sp* hybridised orbitals. This hybridisation is known as 'sp* hybridisation. Each sp* hybridised orbital possess 25% 's' character and 75% of 'p' character. The shape of the molecule a istetrahedral with a bond angle 109128" : eg: Formation of Methane molecule : . The central atom of methane is carbon. . : + Theelectronic configuration ofcarbonin ground state is 1525? 2p", 2p',2p",andon excitationitis 1525! 2p*, 2p',2p", + Thels'2s'2p! 2p!, 2p',. undergo sp* hybridisation giving four equivalent s* hybridised orbitals. . Each sp” hybrid orbitalof carbon overlaps with the 1s orbitals ofhydrogen forming four O, pss Dond. The bonds are directed towards the fowr corners ofa regular tetrahedron. The shape of methane molecule is tetrahedral witha bond angle 10928. y Hybrid orbitals Methane sp'* hybridisation : Ñ E In this hybridisation one 's' and two 7 atomic orbitals ofthe excited atom combine to formthres equivalen! sp? hybridised orbitals, N This hybridisation isalso known as 'sp” hybridisation, In sp? hybridisation each sp" hybrid orbital has 3333 E 's character and 66.66% 'p' character, The shape of the molecule is trigonal planar with a bond angle 120 eg: Boron trichloride molccule formation : , *:.. . The electronic configuration of'B'in the pround state is 15? 95? 2p.2p*2p".. $. z ¡PU CHEMISTRY CHEMICAL BONDING + Onexcitation the configuration is 15? 25! 2p.! 2p.!2p" ¡Now there are three half filled orbitals are available for hybridisation, E + Nowsp*hybridisation takes place at boron atom giving three sp?hybrid orbitals, + Eachsp* hybrid orbital ofboron overlaps with 27, orbitals ofchlorine forming three C.,,»_, bonds. The bonds are directed towards three corners of! triangle. So, the shape of molecule is trigonal planar with bond angle 1200. ek sp hybridisation ; In this hybridisation one 's' and one P' atomic orbitals ofthe excited atom combine to form two equivalent sp hybridised orbitals. This hybridisation is also known as 'sp' hybridisation. In sp hybridisation each sp hybrid orbital has 50% 's character and 50% 'p' character. The shape of the molecule is linear with a bond angle 1800. eg: Beryllium chloride molecule formation ; e Beatomhas 1525 2p,22p,'2p", electronic configuration. . In ground state it has no half filled orbitals.On excitation the configuration becomes 15? 251 2p 2p, a 2p", . + | Nowsphybridisation takes place at beryllium atom giving two sp hybrid orbitals.. . Each sp hybrid orbital of. Beryllium overlaps with 2, -p, orbitals of chlorine forming two Gp. p DOnds. The bonds are directed linearly. So, the shape of the molecule is linear with bond angle 180". ER ISO *p-p E Write an essay on valence shell electron pair repulsion (VSEPR) theory. VSEPR theory was proposed by sidgwick and Powell and later extended by Gillespie and Nyholm. This theory explains the shapes of simple molecules having electron pairs bonded or non-bonded. The repulsions among the electron pairs present in the valence shell of the central atom decides the shape of the molecules. According to this theory : + Theshape ofthe molecule is determined by repulsions between all of the electrón pairs present in the valency shell of central atom. : + Theelectron pairs orient in space so as to have minimum repulsions among them. + Themagnitude ofrepulsions between bonding pairs of electrons depends on the electronegativity NARAYANA STUDY MATERIAL 65 IPU CHEMISTRY CHEMICAL BONDING O P,-Py.The p, orbital inthe two atoms will be atright angles to the internuelear axis, These two can have lateral overlap to forma 7 bond, The electron density of the bonded pair is distributed in two banana like regions lying on cithersido of he internuclcar axis. Thus the oxygen molecule has a double bond. The molecule hasone 0, -», andone ¿y P.= p, Dotween the two atoms, E) 8 3 0- so O Formation of 0, molecule (To El tr bonds) (0=0) o Formation ofH,O molecule: 0 ol + A Ñ al , e 1040 311 H e H 4 H H In water molecule thete are two 'O-H bonds which aré formed due to overlap ofs-p orbitals. These bonds ate sigma bonds, The bond angle between these s-p orbitals should be 90" because the overlapping orbitals ofoxygen (p' and p,!) are at right angles to each other. But the experimental results shows that bond angle in 104.30 ; ; Ñ e and describe Hydrogen bonding. Hydrogen bond is a weak electrostatic bond formed between partially positively charged hydrogen atomanda highly electronegative atomofthe same molecule or another molecule, : + - Hydrogen bondis formed when the hydrogen is bonded to small size and highly electronegative atoms like F, O and N, A partial positive charge will be on hydrogen atom and partial negative charge on the electronegative atom. + Thebond dissociationenergy ofhydrogen bond is 40 kj/mole, . Hydrogen bond is represented with dotted lines (--=). Hydrogen bond is stronger thian vander Waal's forces and weaker than covalent bond. Hydrogen bonding is o£two types. 1) Intermolecular hydrogen bond and 2) Intramolecular hydrogen bond. a) Intermolecularhydrogen bond: If the hydrogen bond is formed between two polar molecules itis called intermolecular hydrogen bond. l.e,, the hydrogen bond is formed between hydrogen atomofone molecule and highly electronegative atom of another molecule is known as intermolecular hydrogen bond. : eg: Water (41,0); HF: NH,; p - nitrophenol. water: e NARAYANA STUDY MATERIAL 67 1PU CHEMISTRY 5 -6 E m7 + -6 Ma Vie e ? O in O HO ke K X N H NARAYANASTUDY MATERIAL 68 7 Write an essay on "Co-ordinate covalent bond". CHEMICAL BONDING; m $" H i a , Due to Water molecule formsan associated molecule through intermolecular hydrogen bond, molecular association water possess high boiling point. b) Intramolecular hydrogen bond : the hydrogen bond is formed within the molecule it is kno eg: o-nitrophenol; o-hydroxy benzaldehyde. oye ==O 1 H Oo - hydroxy benzaldehyde (Salicvlaldehyde) wn as intramolecular hydrogen bond, Abnormal behaviour due to hydrogen bond : 1) Thephysical state of substance may alter. They have high melting and boiling points, 2) Ammonia has higherboiling point than Z/C! eventhough nitrogen and chlorine have same electronegativity values (3.0). Ammonia forms an associated molecule through intermolecular hydrogen' bond. 3) p-hydroxy benzaldehyde have higher boiling point than o-| Ed benzaldehyde. This is due to intermolecular hydrogen bonding in para isomer. 4) Ethylalcohol is highly soluble in water due to association and to association through intermolecular hydrogen bonding. 5)- HO is liquid but H/,Sisa gas, since hydrogenbond is present in 1,0. Co-ordinate covalent bond (dative bond) is special type of covalent bond. It is formed by the sharing of electrons between two atoms in which both the clectrons of the shared electron pair are contributed by one atomand the other atom nearly participates in sharing. 5 The bond is represented as ("> *) an arrow starting from the donar atom and directed towards the acceptor atom. Examiples : 1) Ammonia- Boron trifluoride (H,N: > BF) Ammonía combines with boron trifluoride to give ammonia boron trifluoride. A ed HN OBP HN > BF H Fo TF Donar Acceptor Ammonia boron trifluoride In ammonia nitrogen has a complete octet and also it has a lone pair of electrons. In BF, the boron atomhas a total of six electrons after sharing wit; fluorine, Nitrogen donates the electron pairto. 'boronto formaco- CHEMICAL BONDING IPU CHEMISTRY ns in to of O, is written as the electron y AÑ Write the MORD's of (2) NM) Oy prelectro A. MOEDOofN, and O, written as follows. The distribu MOED ofO, molecule is 220 ap 10.013.028 RT ÓN 2p!=1"2p, ols, O *1S. Na pj Ed os. 0*ls caño t0sS m2 =12p,,02P: MOED of N, molecule is Calculate the number of sodium ions and chloride ions per unit 14. Define "lattice energy" oferystal. cell of NaCL . The energy released when gaseous positive and negative jons are brought together from infinity to form + Aa e one mole of jonic crystal is called lattice energy. Calculation ofions in NaC7 erystal per unit cell : In NaC 1 crystal lattice, (a) Theno. of body centred Na' ¡ons=1 The no. of Na' ions on the 12 edges= 12 Contribution of Na” jon on the edge ofthe unit celi= 1/4 Effective no. of Na” ions per unit cell= ( 2x Y) +1=4 The no. of 7” ions present at the 8 corners= 8 - (0) Contribution of 7” ion at the corner to the unit cell= 1/8 The no. of (7 ¡onsat the 6 face centres= 6 Contribution of (/” ions at face center = 1/2 Effective no.of £'7* ion per unitcell = (8 x Y) + (s x Y) =4 Four Na” and four (7 ¡ions present per unit cell of NaC] NARAYANA STUDY MATERIAL 70 < Az (1) Theionsinionic bond are held by strong coulombic electrostati CHEMICAL BONDING ¡PU CHEMISTRY hich is th ES 15. Which of the ions, Ca'? and Zn'? is more stable? Why? For ionie substances WIlE Pico is term to use "Formula weight' or'Molecular weight"? s A. (a) Ca? ionismorestable, Because Ca * ion has electronic configuration of inert gas Argon, whereas rtgas configuration. 2p 3 3p 3d E gas Argon. The outermost orbit has a stable ¿ion all the orbitals are completely stable than Ca? ion. ion pairs are produced. ts. In case of covalent Zn * docs not have such inert gas configuration. Zn? has pseudo incl Ca” = 18? 29? 2938 3p", Zn = 1 According to this Ca'? ion has the electronic configuration of inert inert gas configuration (ns? np*) with 2, 8. 8 configuration. Though in ] filled with electrons(2. 8, 18) ¡has no octet electronic configuration hence it 'S (0) Inioniccrystals only ions are present. On vaporisation of ionic erystals onl In sucha condition it is better to express their quantitics in terms O fformula wcigh compounds molecular weight concept can be used because they dissociale into molecules. . The concentrations ofionic compounds can be better expressed in terms of formalit y rather than molarity. ES NaClis anionic compound, The formula weight ofitis 58.5 gm. The formula weight ofMgC1,is9S gm. 16. Explain the properties of lonic compounds. a le forces ofattraction. (2) - Jonic compounds have high melting and boiling points. . , (3) Generally ionic compounds are soluble in polar solvents like water but insoluble in the solvents like benzene, chloroform, CCI, ete, (4) - Tonic compounds in solid state are bad conductors due to absence of free ions. (3) Inmoltenstate and in aqueous solution, free ions are present. So they are good conductors of heat and electricity, They act as good electrolytes. : (6) — Lonic reactions in solution state are very fast. eg: NaCl] (ag) + A ZNO (ag) > AgC! + NANO, (aq) White precipitate (7) Tonic bondisnon-directional. Hence ionic compounds do not exhibit isomerism. non-polar 17. Even though nitrogen in ammonia undergoessp* hybridisation, the bond angle is not 109 28'. Explain. Az (1) Inammonia the central atom nitrogen has the electronic configuration 15725? 2", 2p'2P", (2) Inammonia the expected hybridisation is sp? because four outer orbitals ofnitrogen are hybridised. (3) Three sp” hybridised orbitals contain shared pairs ofelectrons and one orbital contains one lone pair. The expected shape is tetrahedral with a bond angle 109928”. (4) Butthe actual shape is pyramidal with a bond angle 107%. (3) The decrease in bond angle is due to repulsions between lone pair-bond pair and bond pair - bond pair. 18. Explain the formation of "a ' bonds in molecules with atleast two examples. A: ASigma(a) bondis a covalent bond in which the electron cloud is concentrated mainly along the internuclear axis of the bonded atoms. It has cylindrical symmetry. A sigma bond is formed by overlapping of pure orbitals or hybrid orbitals of two atoms along their axial line, Sigma bond is a strong bond. It is also sometimes called as localised bond. The bond indicates the distance between the two atomic nuelei and also its direction, All single bonds are sigma bonds, NARAYANASTUDY MATERIAL 71 FPU CHEMISTRY CHEMICAL BONDING In case of hydrogen atom the strength ofs=s bonds is more. This is because of the small size of Is orbitals involved in Overlapping, Even thovgh they are spherical because of their small size they can overlap strongly. according to Linus Pauling the inercasing order of bond energies for hybrid orbitals ¡ssp HBr>HC?, A: (1) HCIand HBrare gases. Among themas molecular weight increases, the vander Waal's, forces of attractions increases. Since the mol. wt. of HBr is more than that of HCT, the boiling point of HBris more than ACJ. il (2) HBrandHCI do not have any hydrogen bonds. But hydrogen fluoride in liquid state exists as hexamer (HF), due to hydrogen bonds. Hence the B.P. ofHF is higher than that of H Br and HC/. (3) HCl is more volatile gaseous compound. The order boiling points is HF > HBr> HC]. 23. Explain Fajan's rules? Discuss. A: — Fajan's rules: * (1) Tonic nature of the bond increases with increase in the size of cation. eg :. The ionic nature increases in the order. Li 
