C1-Allens Made Chemistry Theory {PART-1}

2111R(1)49 194R36   15 R36 = 365R  3615R  = 36912 Å5= 6566.4 ÅEx.Calculate the frequency of the last line of the lyman series in hydrogen spectrum.Sol. For last line of Lyman series n = 1, n = 121 =RZ2 221211nn1=R 10 1  1=R1=109700 cm–1=C=C ×1=C R×=3 10 m sec×8 –1 × 109700 cm—1=3 10×10 cm sec–1 × 109700 cm–1=3.29 10×15 sec–1Ex.Calculate wavelength of 3 line of Bracket series in hydrogen spectrum.rd Sol. For 3 line of Bracket series n = 4, n =7rd121=RZ2   2211471=R1116491=R49 1616 491=R 337841=33R784Therefore,=78433R78491221666 Å33

Ex.What will be the shortest and longest wavelength of absorption lines of hydrogen gas containing atoms inground state ? Give Z = 1, R = 109737.5Sol.22212111RZnnFor shortest wavelength E should be maximum for that n = 1, n = 1211109737.5cm ×   21111 = 109737.5 cm–1 = 911 × 10 cm–8For largest wavelength E should be minimum so n = 1, n = 212 1 = 109737.5 cm × –1 221112 = 1215 ÅEx.A series of lines in the spectrum of atomic hydrogen lies at wavelengths 656.46, 482.7, 434.17, 410.29 nm.What is the wavelength of next line in this series.Sol. The give series of lines are in the visible region and thus appears to be Balmer seriesTherefore, n = 2 and n =? for next line12If = 410.29 × 10 cm and n = 2–71n may be calculated for the last line2 1 = R 221211nn71410.29 10 = 109737 222112nn = 62Thus next line will be obtained during the jump of electron from 7 to 2 shell, i.e.thnd22111R27 = 109737  11449 = 397.2 × 10 cm = 397.2 nm–7Ex.The wave number of 1 line of Balmer series of hydrogen spectrum is 15200 cmSt–1 . The wave number of 1Stline of Balmer series of Li spectrum will be ?+2Sol. Wave number of 1 line of Balmer series of hydrogen spectrum.st22212111RZnn or2221211ZRnn  for H, Z = 1221211Rnn =15200 cm–1Wave number of 1 line of Balmer series of Li ion is.st+22221211ZRnn  {Z = 3 for Li }+2  = 3 2× 15200 = 9 × 15200 = 136800 cm–1

Ex.Calculate the ratio of maximum of Lyman & Balmer series ?Sol.E1ststMaximum of Lyman series 1 line of Lyman series Maximum of Balmer series1 line of Balmer seriesLymanBalmerLB11 222211R1211R23RR11141419       =3 R45 R36   BL275LB527Ex.A certain electronic transition from an excited state to ground state of the Hydrogen atom in one or more stepsgives rise of 5 lines in the ultra violet region of the spectrum.How many lines does this transition produce in theinfra red region of the spectrum?Sol. (Lyman Series) ultra violet region : 5 Lines i.e. e is coming from 6 to 1 –thst Orbitn –1 = 52 n = 62 Infrared region line(i)Paschen series = (6 – 3) = 3(ii)Bracket = (6 – 4) = 2(iii)Pfund = (6 – 5) = 1Total Number of lines are = 6Limitation of the Bohr's model :1.Bohr's theory does not explain the spectrum of multi electron atom.2.Why the Angular momentum of the revolving electron is equal to nh2 , has not been explained byBohr's theory.3.Bohr interrelate quantum theory of radiation and classical law of physics with out any theoriticalexplanation.This was the biggest drawback of this model.4.Bohr's theory does not explain the fine structure of the spectral lines. Fine structure of the spectralline is obtained when spectrum is viewed by spectroscope of more resolution power.5.Bohr theory does not explain the spiliting of spectral lines in the presence of magnetic field (Zemman'seffect) or electric field (Stark's effect)

SOMMERFELD EXTENSION OF THE BOHR'S MODELAccording to sommerfeld electron revolve around the nucleus in the Elliptical Orbits.Circular orbit is a special case of elliptical orbit when the length of major axis becomes equal to thelength of minor axis then the shape of orbit will be circular.If electrons revolve in elliptical orbit then its angular momentum shows two components1 .Radial component :J = r rn h2 where n = radial quantum number.r[n = (n – 1).................0]r n = Shell number2 .Azimuthal Components:J =n h2 n =Azimuthal quantum number[ n = 1, 2, 3, 4.............n]n = Shell numberSo total Angular momentum = JnJ n=J + Jr nh2 =rn h2  + nh2 n=n + nr where n = principal quantum numberEx.Letn = 4n=n + nrThen4=3 + 14=2 + 24=1 + 34=0 + 4The length of major axis indicates by n + n i.e. n and length of minor axis indicates by nr The path of electronK = r nnn=nn = Length of major axisLength of minor axisIf n = 4 then n = 1, 2, 3, 4 K = 41 ,42 ,43 ,44 3 Elliptical path circular pathverFocusr = constant= VariableFocusr = Variable= VariableMajor axisr1r2Minor axis

If n = 5then n = 1, 2, 3, 4, 5 K = 51, 52, 53, 54 ,554 Elliptical path Circular pathIf n = 1 Then Elliptical path=( n – 1) = ( 1 – 1) = 0Circulars path=1In nth orbit :Number of elliptical path = (n – 1)Number of circular path = 1In every atom, 1 orbit is always circular.st THE DUAL NATURE OF MATTER (THE WAVE NATURE OF ELECTRON)1.In 1924. a French physicist, Louis De Broglie suggested that if the nature of light is both that of aparticle and of a wave, then this dual behavior should be true for the matter also.2.According to De Broglie, the wavelength of an electron is inversely proportional to its momentum p.1p   or1mv  hp  Here h = Planck's constantp = momentum of electronMomentum (p) = Mass (m) × Velocity (c)hmv   = h2m(K.E.)From the de-Broglie equation it follows that wavelength of a particle decrease with increase in velocityof the particle. Moreover, lighter particles would have longer wavelength than heavier particles, providedvelocity is equal.If a charged particle Q is accelerated through potential difference V from rest then de-broglie wavelengthish2mQV  de-Broglie concept is more significant for microscopic or sub-microscopic particles whose wavelengthcan be measured.The circumference of the n orbit is equal to n times the wavelength of the electron.thn 2 rn Wavelength of electron is always calculated using De-broglie calculation.Ex.Two particles X and Y are in motion. If the wavelength associated with particle X is 4 × 10 m, calculate–8the wavelength associated with particle Y if its momentum is half of X.Sol.According to de Broglie equation x =xhpand  y =yhpyxyxpp But p = ½ p (given)yxxxyx1 / 2pp  = ½ B = 2 A = 2 × 4 × 10 m = 8 × 10 m–8–8

Ex.Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving with a speed of 30 ms .–1Sol. = 34h6.6 10mv0.1 30 = 2.2 × 10–34 mThis is apparent that this wavelength is too small for ordinary observation.Although the de Broglie equation is applicable to all material objects but it has significance only in case ofmicroscopic particles.Since, we come across macroscopic objects in our everyday life, de Broglie relationship has no significancein everyday life.HEISENBERG UNCERTAINTY PRINCIPLEBohr's theory considers an electron as a material particle. Its position and momentum can be determinedwith accuracy. But, when an electron is considered in the form of wave as suggested by de-Broglie, it is notpossible to ascertain simultaneously the exact position and velocity of the electron more precisely at a giveninstant since the wave is extending throughout a region of space.In 1927, Werner Heisenberg presented a principle known as Heisenberg uncertainty principle which statesas : \"It is impossible to measure simultaneously the exact position and exact momentum of a body as small asan electron.\"The uncertainty of measurement of position, x, and the uncertainty of momentum p or m v, are relatedby Heisenberg's relationship as : ( p = mv, p = m v)x . p > h4 orx . m v > h4  orhx. v4 m   where h is Planck's constant.  x v = uncertainty productFor an electron of mass m (9.10 × 10–28 g), the product of uncertainty is quite large.x . v >  276.624 104 m> 27286.624 104 3.14 9.10 10= 0.57 erg sec per gram approximatelyWhen x = 0, v =  and vice-versa.In the case of bigger particles (having considerable mass), the value of uncertainty product is negligible. If theposition is known quite accurately, i.e., x is very small, v becomes large and vice-versa. In terms of uncertainty in energy E, and uncertainty in time t, this principle is written as,hE. t4   Heisenberg replaced the concept of definite orbits by the concept of probability.Ex.Why electron cannot exist inside the nucleus according to Heisenberg's uncertainty principle ?Sol.Diameter of the atomic nucleus is of the order of 10–15 mThe maximum uncertainty in the position of electron is 10–15 m.Mass of electron = 9.1 × 10–31 kg.x. p =  h4x × (m. v) = h/4

v = h14x.m = 346.63 1022 47 × 15311109.1 10v = 5.80 × 10 ms10–1This value is much higher than the velocity of light and hence not possible.DE BROGLIE RELATIONSHIP & HEISENBERG'S UNCERTAINTY PRINCIPLEEx.The mass of a particle is 1 mg and its velocity is 4.5 × 10 cm per second. What should be the wavelength of5this particle if h = 6.652 × 10–27 erg second.(1) 1.4722 × 10–24 cm(2) 1.4722 × 10–29 cm(3) 1.4722 × 10–32 cm(4) 1.4722 × 10–34 cmSol.Given thatm = 1 mg = 1 × 10 g–3c = 4.5 × 10 cm/sec.5h = 6.652 × 10–27 erg sec.hmc   = 27356.652 101 104.5 10 = 296.652 104.5cm = 1.4722 × 10–29 cmEx.Which of the following should be the wavelength of an electron if its mass is 9.1 10× –31 kg and its velocity is1/10 of that of light and the value of h is 6.6252 10× –24 joule second?(1) 2.446 × 10 metre (2) 2.246 –7× 10 metre (3) 2.246 –9× 10–11 metre (4) 2.246 × 10–13 metreSol.Given thatm = 9.1 × 10–31 kg1c10  of velocity of lightor1c10 × 3 × 10 metre/second i.e. 3 8× 10 metre/second7h = 6.6252 × 10–34 joule secondhmc  = 343176.6252 109.1 103 10  = 3424 6.6252 1027.3 10or0.2426 × 10–10 metreor2.426 × 10–11 metreEx.What should be the momentum (in gram cm per second) of a particle if its De Broglie wavelength is 1 Å andthe value of h is 6.6252 × 10–27 erg second ?(1) 6.6252 × 10–19(2) 6.6252 × 10–21(3) 6.6252 × 10–24(4) 6.6252 × 10––27Sol.Given that= 1 Å = 1 × 10 cm–8 h = 6.6252 × 10–27 erg secondor2786.6252 10 p1 10  = 6.6252 × 10–19 gram cm/sec.

Ex.What should be the mass of the sodium photon if its wavelength is 5894Å, the velocity of light is 3 × 108metre/second and the value of h is 6.6252 × 10–34 kg m /sec.?2(1) 3.746 × 10–26(2) 3.746 × 10–30(3) 3.746 × 10–34(4) 3.746 × 10–36Sol.hm c  m = hc (= 5894Å = 5894 × 10–10 m)348106.652 10m3 105894 10 or 326.6521017682= 0.0003746 × 10–32or3.746 × 10–36 kgEx.What should be the uncertainty in the velocity of an electron if the uncertainty in its position is 0.005 nm, themass of electron is 9.109 × 10–31 kg and the value of h is 6.6252 × 10–34 joule/second?(1) 2.316 × 105(2) 1.158 × 107(3) 2.316 × 109(4) 2.316 × 1011Sol.Uncertainty in position ( x) = 0.005 nm = 0.005 × 10–9 m= 5 × 10–12 mMass of electron (m) = 9.109 × 10–31 kg. v= h4 mx   = 3412316.6252 104 3.14 5 109.109 10  m/sec.or v= 34436.6252 104 3.14 5 9.109 10  v= 1.15816 × 10 m/sec.7 Ex.What should be the uncertainty in velocity of a particle of 1 kg mass if uncertainty in position is 1Å and thevalue of h is 6.6252 × 10–34 Joule sec.?(1) 1.055 × 10–22(2) 1.055 × 1022(3) 5.25 × 10–25(4) 1.055 × 1024Sol.Given thatx = 1Å = 1 × 10–10 mm = 1 kgh = 6.6252 × 10–34 Joule sec.hv4mx   = 34106.6252 104 3.14 1 10 or34106.6252 10 v12.56 10  m/sec.= 0.52525 × 10–24 m/sec.= 5.25 × 10–25 m/sec.Ex.What should be the uncertainty in position if uncertainty in momentum is 1 × 10–2 g cm/sec. and value of his 6.6252 × 10–34 Joule sec. ?(1) 1.054 × 10–22 m(2) 1.054 × 10–25 m(3) 0.525 × 10–27 m(4 ) 1.054 × 10–32 mSol.Given thatp = 1 × 10–2 g cm/sec. = 1 × 10–7 kg m/sec.h = 6.6252 × 10–34 Joule sec.

 x × p = h4 x = h4p or3476.6252 10 x4 3.14 10  = 0.525 × 10–27 mEx.A ball weighs 25 g moves with a velocity of 6.6 10 cm/sec then find out the De Broglie ×4 associated withit.Sol.= hmv=34746.6 1010erg sec25 6.6 10 cm / sec= 1381025=0.04 10×–38 10×7=0.04 10×–31 cm=4 10×–33 cmEx.Which of the following has least De Broglie if they have same velocity.1 .e –2 .p3 .CO24 .SO2Sol.= hmvmass of SO is greater than the mass of e , p, CO2– 2= h constant= v Sameleast will be SO21mEx.If uncertainty in position of an e is same as the x of He atom. If p of e is 32 10 then find p in He– – ×5 atom.Sol.x p = ×h4 Since x is same for both.therefore p will be same byhee4(He)HexphxP4  HePeP 1 Pe P He32 10 = 32 10×5×5PHe= 32 10×5Ex.Calculate the uncertainty in the position of a particle when the uncertainty in momentum is(a) 1 × 10 g cm sec–3–1(b) Zero.Sol.Givenp = 1 × 10 g cm sec–3–1h = 6.62 × 10–27erg sec.= 3.142According to uncertainty principle

  x ph.4 So,h1x.4p  2736.62 1014 3.14210=0.527 × 10–24 cm(b)When the value of p = 0, the value of x will be infinity.Ex.The uncertainty in position and velocity of a particle are 10–10 m and 5.27 × 10–24 ms respectively. Calculate–1the mass of the particle (h = 6.625 × 10–34 Joule Sec.)Sol.According to Heisenberg's uncertainty principle,hx.m v4 or mh4x. v = 3410246.625 104 3.143 105.27 10= 0.099 kgEx.Calculate the uncertainty in velocity of a cricket ball of mass 150 g if the uncertainty in its position isof the order of 1Å (h= 6.6 × 10–34 kg m s ).2–1Sol.x . m v =h4  v= h4x.m= 3410 6.6 104 3.143 100.150= 3.499 × 10–24 ms–1QUANTUM NUMBERS :The set of four intergers required to define an electron completely in an atom are called quantum numbers.Thefirst three have been derived from Schordinger wave equation.(i)Principle quantum number (n) :It describes the size of the electron wave and the total energy of the electron. It has integral values1,2,3,4......., etc, and is denoted by K,L,M,N..........,etc.The maximum number of electrons which can be present in a principal energy shell is equal to 2n .2No energy shell in the atoms of known elements possesses more than 32 electrons.(ii)Azimuthal quantum number ( ) :It describes the shape of electron cloud and the number of subshells in a shell. It can have valuesfrom 0 to (n – 1), i.e., = 0 (s-subshell), = 1 (p-subshell), = 2 (d-subshell), = 3 (f-subshell).(iii)Magnetic quantum number (m) :It describes the orientations of the subshells. It can have values from – to + including zero, i.e.,total (2 + 1) values. Each value corresponds to an orbital. s-subshell has one orbital, p-subshell threeorbitals (p , p and p ), d-subshell five orbitals (d , d , d , dxyzxyyzzxx – y 22, d ) and f-subshell has seven orbitals.z2The total number of orbitals present in a main energy level is 'n '.2

(iv)Spin quantum number (s) :It describes the spin of the electron. It has values +1/2 and –1/2. (+) signifies clockwise spinningand (–) signifies anticlockwise spinning.RULES FOR FILLING OF ORBITALS1 .Aufbau Principle :Aufbau is a German word and its meaning 'Building up'Aufbau principle gives a sequence in which various subshell are filled up depending on the relative order ofthe Energies of various subshell.Principle : The subshell with minimum energy is filled up first and when this subshell obtained maximum quota of electrons then the next subshell of higher energy starts filling.The sequence in which various subshell are filled is the following.1s , 2s , 2p , 3s , 3p , 4s , 3d , 4p , 5s , 4d , 5p , 6s , 4f , 5d , 6p , 7s , 5f , 6d , 7p22626210621062141062141062 .(n + ) rule :According to it the sequence in which various subshell are filled up can also be determined with the help of( n + ) value for a given subshell. PRINCIPLE OF (n + ) RULE :The subshell with lowest( n + ) value is filled up first, when two or more subshell have same (n+ ) value then the subshell with lowest value of n is filled up first. Sub Shellnn +1s1012s2022p213(1)3s303(2)3p314(1)4s404(2)3d325(1)4p415(2)5s505(3)4d426(1)5p516(2)6s606(3)1s2s3s4s5s6s7sStartingpoint2p3p5p7p6p3d6d4p4d5d4f5f

3 .Pauli's Exclusion principle :In 1925 Pauli stated that no two electron in an atom can have same values of all four quantum numbers.An orbital can accomodates maximum 2 electrons with opposite spin.4 .Hund's Maximum Multiplicity Rule :(Multiplicity : Many of the same kind)According to Hund's rule electrons are distributed among the orbitals of subshell in such a way as togive maximum number of unpaired electron with parallel spin. i.e. in a subshell pairing of electronwill not start until and unless all the orbitals of that subshell will get one electron each with same spin.SPIN MULTIPLICITYIt is given by 2S + 1 where S is the total spin.(a)(b)For (a), S = 1122 = 0Spin multiplicity = 2S + 1 = 0 + 1 = 1 (singlet)For (b), S =  1122 = 1Spin multiplicity = 2S + 1 = 2 × 1 + 1 = 3 (triplet)Ex.Find out the angular momentum of an electron in(a)4s orbital(b)3p orbital(c)4 orbitalthSol.Angular momentum in an orbital = h12 (a) = 0 for 4s orbital, hence orbital angular momentum = 0(b) = 1 for 3p orbitalAngular momentum =  hh1 1122(c)Angular momentum in 4 orbitth= nh4h2h22Ex.Given below are the sets of quantum numbers for given orbitals. Name these orbitals.(i)n = 4, = 2, m = 0(ii)n = 3, = 1, m = ±1(iii)n = 4, = 0, m = 0(iv)n = 3, = 2, m = ±2Sol.(i)4dz2(ii)3p or 3pxy(iii)4s(iv) 22xy3d or 3dxyELECTRONIC CONFIGURATION OF ELEMENTSBased on the rules, we can easily determine the electronic configurations of most element. We just needto know the atomic number of an element, the order in which orbitals are to be filled and the maximumnumber of electrons in a shell, sub-shell or orbital. The configurtion so obtained can be represented intwo ways. As an illustration, let us consider fluorine (Z = 9) :F(Z = 9) = 1s , 2s , 2p , 2p , 2p or22x2y2z11s2s2p x2p y2p zImportance of knowing the exact electronic configuration of an element lies in the fact that the chemicalproperties of an element are dependent on the behaviour and relative arrangement of its electrons.

Electronic configurations of heavier elements (beyond Z = 56) deviate a little from the order mentionedpreviously. These are listed below :LanthanidesLa (Z = 57):[Xe]6s 5d (not 4f )211Ce (Z = 58):[Xe]6s 5d 4f211Pr (Z = 59):[Xe]6s 5d 4f212ActinidesAc (Z = 89):[Rn]7s 6d (not 5f )211Th (Z = 90):[Rn]7s 6d 5f211Pa (Z = 91):[Rn]7s 6d 5f212Beyond Z = 103Z = 104:[Rn]5f 6d 7s1422Z = 105:[Rn]5f 6d 7s1432Z = 106:[Rn]5f 6d 7s1442Z = 112:[Rn]5f 6d 7s14102EXCEPTIONAL CONFIGURATIONSStability of Half Filled and Completely Filled OrbitalsCu has 29 electrons. its expected electronic configuration is 1s ,2s , 2p , 3s , 3p , 4s , 3d .2262629But a shift of one electron from lower energy 4s orbital to higher energy 3d orbital will make the distributionof electron symmetrical and hence will impart more stability.Thus the electronic configuration of Cu is 1s , 2s , 2p , 3s , 2p , 4s , d22626110Fully filled and half filled orbital are more stable.Ex.We know that fully filled and half filled orbital are more stable. Can you write the electronic configurationof Cr(Z = 24) ?Sol.Cr (Z = 24)1s , 2s , 2p , 3s , 3p , 4s , 3d .2262615Since half filled orbital is more stable, one 4s electron is shifted to 3d orbital.Ex.A compound of vanadium has a magnetic moment of 1.73 BM work out the electronic configuration ofthe vanadium in the compound.Sol.Magnetic moment =  n n 2Where n is number of unpaired electrons1.73 =  n n 2 or (1.73) = n + 2n, n = 122Vanadium atom must have the unpaired electron and thus its configuration is :23V : 1s 2s 2p 3s 3p 3d4+226261WAVE MECHANICAL MODEL OF ATOMSchrodinger wave equation :General wave equationNodes y = A sin twhere, y = displacement A = amplitude t = time

Developed by schrodinger, this model is based on the particle and wave nature of electron is known asWAVE MECHANICAL MODEL of atom. The motion of electron around nucleus is round motion and maybe considered to be analogous to the STANDING WAVES, the waves which are generated by plucking thestretched string. The amplitude of the standing wave is independent of time and is a function of the distancefrom one fixed end. The derived eq. by schrodinger isSchrodinger wave equation 222222228m(EV)0xyzh    where = Amplitude of e wave (or wave function)–m = mass of e–E = Total energyV = Potential energyor2228m(EV)0h    where  2 = Laplacian operator = 222xyz        22h8m   2+ (E – V) = 0222hV8m     = EHE  222hHV8m    Hemiltonion operatorSCHRODINGER EQUATION IN CARTESIAN COORDINATE :Z = rcosy = rsin sinyxrsinrcos rsincoszRrsin sinx = rsin cosThe schrodinger equation can be written in terms of cartesiancoordinates (x, y, z) or in terms of spherical polar coordinates (r, , ). However for most calculations it is simpler to solve the wave equationin polar coordinates. When Schrodinger wave equation in polarcoordinates is solved for hydrogen atom the solution obtained can befactorized into 2 separate parts, one being the function of r and otherthe function of and . (r, ) = R(r) f ( , ) R(r) = Radial functionf ( , ) = Angular function 

SIGNIFICANCE OF ( ) :The wave function may be regarded as amplitude of electron wave expressed in terms of coordinates(x, y, z) or (r,  , ). The wave function may have +ve or –ve values depending on the values of coordinates.As such there is no physical significance of .SIGNIFICANCE OF ( ) : 2 In classical theory of electromagnetic radiation, the square of amplitude is proportional to the intensity oflight. A very similar concept was suggested by MAX BORN in QUANTUM MECHANICS according towhich the square of function at any point is proportional to the probability of finding an electron at thatpoint  2 is known as PROBABILITY DENSITY and is always +ve.The region of space in which there is maximum probability of finding an electron (say 90%) is termed as anorbital.QUANTUM NUMBERS :In an atom, a large no of orbitals are permissible. These orbitals are designated by a set of 3 numbersknown as QUANTUM NUMBERS (principle, azimuthal, magnetic) which arise as a natural consequence inthe solution of schrondinger wave equation. These quantum numbers describe energies of electron in anatom, information about shapes and orientation of orbitals. In order to designate the electron an additionalquantum number called as SPIN QUANTUM NUMBER is needed to specify spin of the electron.GRAPHICAL REPRESENTATION OF : = Amplitude of wave 2 = probability density  It gives us probability of finding an electron at a point or per unit volume.2Probabilitydv   = (x, y, z) = (r, , )  = R(r) ~ ()R = radial probability density, r = radius2 R = radial probability density of finding an electron at a distance R from the nucleus in any direction.2 (i)rRHyperbola21s0012rR(r)2ewhereana a = 0.529 Å0n = principal quantum number03r2a1s01R(r)2earR 2 Graph between R and r will be same hyperbolic.2

(ii)3222s011R(r)(2)ea 2 2 After reaching 2a now curve would start decreasing and again goes up to cut at .0 R(r)2sr2a 0R (r)22sr(iii)R(r) vs r3s R(r)3srR (r)23srRADIAL NODES : Points at which the probability of finding an electron is zero is known as radial nodes.No. of radial nodes = n – – 1Electron cannot be present at 2a distance from nucleus. If we join all 2a points 0 0 2a 02a 02a 02a 0to form a sphere, we can say that electron cannot be present on surface ofsphere, however it may be present inside or outside. At 2a distance probability0 of presence of electron is zero. Since a sphere is formed radial nodes are alsocalled SPHERICAL NODES.Radial nodes are spherical in shape also known as spherical nodes or nodal sphere.GRAPHS FOR P-ORBITALS :3222p011R(r)ea 2 6R(r)2pr2pR(r)3pr3prR (r)23pR(r)4pr4pR(r)3dr3d

RADIAL PROBABLITY DISTRIBUTION FUNCTION :dr is very very lessvolume = 334((rdr)r ) shell3rdr= 322334((r3r dr3r drdrr )3dr and dr are neglected23= 24(3r dr) = 4 r dr3 2Radial probability in given shell = 4 r drR 22Radial probability function = 4 r R 22The probability of finding an electron at a distance r from the nucleus in all the direction is called radialprobability function (RPF).GRAPHS BETWEEN RPF AND r :R (r) = 1s 2Ce4 R r = 4 r C 22 2  222 e4 R r 22 2r(max)1s = C'r202ra e 4 R r = C'r 22 2 202ra e For s-orbital, R and R at nucleus is not zero but probability is almost equal to zero because of very small size.2R dv 0 2 as dv 0 for nucleusrmax distance at which the probability of finding an electron is max.221sP4 R r 0r3 / 2a1s01R2eanow differentiate eq. by putting value of R1sP = 4 r × 4 202r3a01eaP = 02ra2Cr e002r2raa2dPC(2r er e)dr02ra0dP2rCr e20dra0ra  Maxima4 R r2 2r2sPeaks of the curves are increasing4 R r2 2rrmax3sHere 1 peak is smaller than 2 and 2 smaller than 3 .st ndndrd

4 R r2 2rrmax3pANTINODE POINT - Point at which probability of finding an electron is max.COMPARISON OF rma x. & ravg FOR DIFFERENT ORBITALS :Note : ravg is always greater than rmaxCase I - when is same but n is different.As value of n increases rmax increases.2s(r) (r1s(max)4 R r2 24 R r2 2)2s(max)r1srPeaks are numbered according to value of (n – ).As n increases ravg increases if ravg is more, electron will be more away from nucleus.PENETRATION POWER : Penetration power of orbital is a measure of its closeness to the nucleus.1s > 2s > 3sCase II - When 'n' is same but ' ' different.As value of ' ' increases, r increasesavgAs value of ' ' increases, rmax decreases rmax. 3d3p3sr4 R r2 24 R r2 24 R r2 2 ravg. Closeness to nucleus - 3s > 3p > 3dENERGY COMPARISON :For energy comparison Aufbau rule should be used. But for hydrogen atom, subshell belonging to particularshell possess equal energy.2p = 2s, 3s = 3p = 3dANGULAR FUNCTION :It gives us an idea about the shape, orientation of an orbitaleg. = 1  m = –1, 0, +1PxPyPzFor s orbital, angular part is independent of and .There would be zero angular nodes for s-orbital.No.of angular nodes for any orbital 

SHAPE OF ANGULAR NODE :For P :x Angular Node or Nodal plane Angular node or nodal planeP x yz planeP y xz planeP z xy planeFor d :xy xyd xy xz and yzd yz xz and xyd zx xy and yz45°dx –y 22We cannot predict the designation of angular nodes but can be said that at an angle of 45° with axis.PHOTOELECTRIC EFFECTSir J.J. Thomson observed that when a light of certain frequency strikes the surface of a metal, electronsare ejected from the metal. This phenomenon is known as photoelectric effect and the ejected electrons arecalled photoelectrons.A few metals, which are having low ionisation energy like Cesium, show this effect under the action of visiblelight but many more show it under the action of more energetic ultraviolet light.LightelectronsEvacuated quartz tube– +AV+–

An evacuated tube contains two electrodes connected to a source of variable voltage, with the metal platewhose surface is irradiated as the anode. Some of the photoelectrons that emerge from this surface haveenough energy to reach the cathode despite its negative polarity, and they constitute the measured current.The slower photoelectrons are repelled before they get to the cathode. When the voltage is increased toa certain value V , of the order of several volts, no more photoelectrons arrive, as indicated by the current0dropping to zero. This extinction voltage (or also referred as stopping potential) corresponds to the maximumphotoelectron kinetic energy i.e. eV = ½ mv02The experimental findings are summarized as below :Electrons come out as soon as the light (of sufficient energy) strikes the metal surface.The light of any frequency will not be able to cause ejection of electrons from a metal surface. Thereis a minimum frequency, called the threshold (or critical) frequency, which can just cause the ejection.This frequency varies with the nature of the metal. The higher the frequency of the light, the moreenergy the photoelectrons have. Blue light results in faster electrons than red light.Photoelectric current is increased with increase intensity of light of same frequency, if emission ispermitted, i.e. a bright light yields more photoelectrons than a dim one of the same frequency, butthe electron energies remain the same.Light must have stream of energy particles or quanta of energy (h ). Suppose, the threshold frequencyof light required ejecting electrons from a metal is , when a photon of light of this frequency strikes ametal it imparts its entire energy (h 0) to the electron.E=h 0E>h 0K.E.=0MetalK.E = h – hmax 0\"This energy enables the electron to break away from the atom by overcoming the attractive influenceof the nucleus\". Thus each photon can eject one electron. If the frequency of light is less than  0 thereis no ejection of electron. If the frequency of light is higher than  0 (let it be ), the photon of this lighthaving higher energy (h ), will impart some energy to the electron that is needed to remove it away fromthe atom. The excess energy would give a certain velocity (i.e. kinetic energy) to the electron.h = h 0 + K.E.h = h + ½ m 0 2½ m = h – h 2 0where = frequency of the incident light 0 = threshould frequencyh is the threshold energy (or) the work function denoted by = h (minimum energy of the photon to liberate00electron). It is constant for particular metal and is also equal to the ionization potential of gaseous atoms.The kinetic energy of the photoelectrons increases linearly with the frequency of incident light. This, ifthe energy of the ejected electrons is plotted as a function of frequency, it result in a straight line whoseslope is equal to Planck's constant 'h' and whose intercept is h 0 .K.E.ofPhotoelectrons

Ex.A photon of wavelength 3000 Å strikes a metal surface, the work function of the metal being 2.20 eV.Calculate (i) the energy of the photon in eV (ii) the kinetic energy of the emitted photo electron and (iii)the velocity of the photo electron.Sol.(i)Energy of the photonE = h = hc=348176.6 10Js 3 10 ms3 10 m = 6.6 × 10–19 J1eV = 1.6 × 10–19 JTherefore E = 19196.6 10J1.6 10J / ev = 4.125 eV(ii)Kinetic energy of the emitted photo electronWork function = 2.20 eVTherefore, KE = 2.475 – 2.20 = 1.925 eV = 3.08 × 10–19 J(iii)Velocity of the photo electronKE =12mv = 3.08 × 102 –19 JTherefore, velocity (v) = 19312 3.08 109.1 10 = 8.22 × 10 ms5–1Ex.Photoelectrons are liberated by ultra violet light of wavelength 2000 Å from a metallic surface for whichthe photoelectric threshold is 4000 Å. Calculate the de Broglie wavelength of electrons emitted with maximumkinetic energy.Sol.K.E. = Quantum Energy – Threshold energy= 348106.626 103 102000 10  – 348106.626 103 104000 10= 348106.626 103 10101120004000= 4.969 × 10–19 Joule.21mv = 4.969 × 102–19m v = 2 × 4.969 × 102 2–19 × 9.1 × 10–31mv = 9.51 × 10–25 = hmv = 34256.626 109.51 10 = 0.696 × 10 m–9

MEMORY TIPS1 .Frequency, = c2 .Energy/photon,E = h = hcAlso, E = 12375eV , if is in Å3 .Electronic energy change during transition, E = 21nnEEn > n , emission spectra if electron jumps from n to n shell and absorption spectra if electron excites2121from n to n shell.124 .Radius of n Bohr orbit of H atom, r =thn2222n h4me K (where K = 9 × 10 )9r for H = 0.529 Å ; r for H like atom r = 0.529 × 1nn2nÅZ5 .Velocity of electron in n Bohr orbit of H atom, v =th22 KZenhv = 2.18 × 10 8Zcm / sec .n6 .Energy of electron in n Bohr orbit of H atom, E =th 2242222mZ e Kn hwhere n = 1, 2, 3..........[E = –13.6 × 22Znkcal/mole (1 cal = 4.18 J)]E for H = – 21.72 × 101–12 erg = – 13.6 eV, E for H like atom = E for H × Z1127 .Wavelength emitted during transition in H atom,H2212111Rnn=24322122me11chnn (in C.G.S.)8 .Photoelectric effect hv = w + 21mu2orhv = I.E. + K.E.9 .Possible transitions for a jump from n to n = 21 21(nn )1 0 .Angular momentum of electron in an orbit = n. (h/2 )1 1 .Angular momentum of electron in an orbital = (nh/2 ) 11 2 .Total spin = ±  1n ; where n is no. of unpaired electrons.21 3 .Magnetic moment of an atom  n n 2B.M.; where n is no. of unpaired electrons.1 4 .Nodal planes : Radial nodes = n – – 1, Angular nodes = 1, Total nodes = (n – )1 5 .de Broglie equation : =2hhmu2K.E. mwhere is wavelength, m is mass and u is velocity of particle.1 6 .Heisenberg uncertainty principle : p. x  h4 u. x  h4 mwhere p, u and x are uncertainties in momentum, velocity and position respectively. Planck's constantis h and m is mass of subatomic particle.

SOLVED PROBLEMS (SUBJECTIVE)Ex 1.Find the wavelengths of the first line of He ion spectral series whose interval with extreme lines is+1211= 2.7451 × 10 cm4–1Sol.Extreme lines means first and last1211=RZ 222111n   – RZ 2221111n(n1)or 1211= 221RZ(n1)2.7451 × 10 = 4221109677.76 2(n1)(n + 1) = 41 n = 31Wavelength of first line,1= 109677.76 × 2 × 2221134 = 4689 × 10 cm = 4689 Å–8Ex 2.Find the wavelength emitted during the transition of electron in between two levels of Li ion whose sum2+is 5 and difference is 3.Sol.Let the transition occurs between the level n and n and n > n1221Given that n + n = 512n – n = 321n = 1 and n = 412Therefore, 1= R × Zh2  221114= 109678 × (3)21516  = 1.08 × 10 cm–6Ex 3.The Lyman series of the hydrogen spectrum can be represented by the equation.v = 3.2881 × 10 s15–1  22111n(where n = 2, 3,.....)Calculate the maximum and minimum wavelength of lines in this series.Sol.1  = c = 1583.2881 103 10 m –1 2211n1Wavelength is maximum min  when n is minimum so that 21n is maximumminmax1=1583.2881 103 10  221112

max = 8153 10433.2881 10= 1.2165 × 10 m = 121.67 nm–7Wavelength is minimum max  when n is i.e. series convergemax =158min13.2881 103 10min = 0.9124 × 10 m 91.24 nm–7Ex 4.When certain metal was irradiated with light frequency 0.4 × 10 Hz the photo electrons emitted had13twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency1.0 × 10 Hz. Calculate threshold frequency (13 0) for the metal.Sol.hv = h 0 + KEKE = h(1 1 –  0 )KE = h(2 2 –  0) = 1 KE22010vv   = 121301301.0 10120.4 10   0 = 1.6 × 10 Hz13Ex 5.Iodine molecule dissociates into atoms after absorbing light of 3000 Å. If one quantum of radiation is absorbedby each molecule, calculate the kinetic energy of iodine atoms. (Bond energy of I = 240 kJ (mol).2Sol.Energy given to iodine molecule34810hc6.62 103 103000 10  = 6.62 × 10–19 JAlso energy used for breaking upI molecule = 231923240 103.984 10J6.023 10Energy used in imparting kinetic to two atoms = (6.62 – 3.984) × 10–19 JKE of iodine atom =196.623.984102= 1.318 × 10–19 JEx 6.Two hydrogen atoms collide head on and end up with zero kinetic energy. Each atom then emits a photonof wavelength 121.6 nm. Which transition leads to this wavelength ? How fast were the hydrogen atomstravelling before collision ?Sol.Wavelength is emitted in UV region and thus n = 11For H atom =H22111R1n 91121.6 10=1.097 × 10722111n n = 2Also the energy released is due to collision and all the kinetic energy is released in form of photon.21hcmv2

12 × 1.67 × 10–27 × v = 234896.625 103 10121.6 10v = 4.43 × 10 m/sec4Ex 7.Find the energy in kJ per mole of electronic charge accelerated by a potential of 2 volt.Sol.Energy in joules = charge in coulombs × potential difference in volt= 1.6 × 10–19 × 6.02 × 10 × 2 = 19.264 × 10 J or 192.264 kJ234Ex 8.Which hydrogen like ionic species has wavelength difference between the first line of Balmer and first lineof Lyman series equal to 59.3 × 10 m ? Neglect the reduced mass effect.–9Sol.Wave number of first Balmer line of an species with atomic number Z is given by22211v 'RZ23= 25RZ36Similarly wave number of v of first Lyman line is given byv = RZ 2221112  = 23RZ ; v41 and v '1 '' – =223645RZ3RZ =2136453RZ=28815RZZ =2978859.3 1015 1.097 10 = 9 or Z = 3Ionic species is Li2+Ex 9.(i)What is highest frequency photon that can be emitted from hydrogen atom ? What is wavelengthof this photon ?(ii)Find the longest wavelength transition in the Paschen series of Be .3+(iii)Find the ratio of the wavelength of first and the ultimate line of Balmer series of He ?+Sol.(i)Highest frequency photon is emitted when electron comes from infinity to 1 energy level.stE = 2213.6Z13.6 eV1 or,13.6 × 1.6 × 10–19 Joule = 2.176 × 10–18 JouleE = h = Eh = 18342.176 10J6.626 10Js = 0.328 × 10 Hz16 = c = 8163 100.328 10 = 9.146 × 10 m–8(ii)2H221211RZnn For He ; Z = 4 ; For Paschen series n = 31For longest wavelength n = 42

1 = 109678 × (4) × 2221134 = 109678 × 16 × 11916  = 109678 × 16 × 7144 = 1172.20 Å(iii)Wave number of first line of Balmer,212211RZ23  = 5 4R5R369Wavelength of first line of Balmer = 95RWave number of ultimate line of Balmer, 22211RZ2   =4R4=RWavelength of ultimate line of Balmer = 1RRatio = 95Ex 10. An electron beam can undergo difraction by crystals. Through what potential should a beam of electronsbe accelerated so that its wavelength becomes equal to 1.0 Å.Sol.For an electron21mveV2where V is accelerating potential =hmv21hmeV2m V =221h2me= 2 343110 21916.625 102 9.108 10(1.0 10)1.602 10= 150.40 voltEx 11. The angular momentum of an electron in a Bohr's orbit of H-atom is 4.2178 × 10–34 kgm /sec. Calculate2the wavelength of the spectral line emitted when electrons falls from this level to next lower level.Sol.mvr = nh2 nh2  = 4.2178 × 10–34n = 34344.2178 1023.146.625 10 = 42212111RnnThe wavelength for transition from n = 4 to n = 32211110967834 = 1.8 × 10 cm.–4

Ex 12. The kinetic energy of an electron in H like atom is 6.04 eV. Find the area of the third Bohr orbit to whichthis electron belongs. Also report the atom.Sol.K.E. = 6.04 in 3 orbitrdEtotal = K.E. + P.E. = K.E. – 2 × K.E.–K.E. = – 6.04 eVE for H = –13.6 eV and not for any orbit E = – 6.04 eV for H atom. Thus, atom for which K.E. is1given is other than H.E H like atom = E × ZnnH2212EZ n6.04 =2213.6Z3Z = 3.99 24 Z = 2The atom is He +r = 0.529 × n2nZ= 0.529 × 232= 2.3805 ÅArea, r =  22 8222.3805 107 = 17.8 × 10–16 cm2Ex 13. O undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom 1.967 eV2more energetic than normal. The dissociation of O into two normal atom of oxygen2requires 498 kJmol . What is the maximum wavelength effective for photo chemical dissociation of–1O ?2Sol.We knowP 2h ONormal + OExcitedO 2ONormal + ONormalEnergy required for simple dissociation of O into two normal atoms = 498 × 10 Jmol23–1= 8123498 10Jmol6.023 10If one atom in excited state has more energy, i.e.. 1.967 eV= 1.967 × 1.602 × 10–19 JThe energy required for photochemical dissociation of O2= 31923498 101.967 1.602 106.023 10= 82.68 × 10–20 + 31.51 × 10–20 = 114.19 × 10–20 JouleE = hc114.19 × 10–20 =3486.625 103 10 = 1740.2 × 10–10 m = 1740.2 Å.

SOLVED PROBLEMS (OBJECTIVE)Ex 1.The wave-mechanical model of atom is based upon :-(A) de Broglie concept of dual character of matter(B) Heisenberg's uncertainty principle(C) Schrodinger wave equation(D) All the above threeSol.( D )Ex 2.An orbital is correctly described by :-(A)  2(B) (C) |2|(D) noneSol.( A )Ex 3.The orbital angular momentum of a d-electron is :-(A) 6 (B) 2 (C) (D) 2Sol.For d–electron, = 2, orbital angular momentum =1 =2 2 1=6 So, (A) is the correct answerEx 4.The following electron configuration of an atom in the ground state is not correct because :-3s 3p 3d(A) the energy of the atom is not minimum(B) Pauli's exclusion principle is violated(C) Hund's rule is violated(D) Aufbau principle is not followedSol.(C) is the correct answer.Ex 5.In the first bohr orbit of H atom the energy of an electron is –13.6 eV. The possible energy value (s) ofexcited state (s) for electron in Bohr orbit of hydrogen is/are :-(A) –3.4 eV(B) –4.2 eV(C) 6.8 eV(D) +6.8 eVSol.E = n213.6eVn For n = 2, E = 213.63.4eV4  So, (A) is the correct answer.Ex 6.The electronic configuration of an element is 1s , 2s , 2p , 3s , 3p , 3d , 4s . This represents its :-2262651(A) excited state(B) ground state(C) cationic form(D) anionic formSol.The given electronic configuration is ground state for chromium.So, (B) is the correct answerEx 7.Which of the following sets of quantum number is/are incorrect ?(A) n = 3, = 3, m = 0, s = 12(B) n = 3, = 2, m = 2, s = –12(C) n = 3, = 1, m = 2, s = –12(D) n = 3, = 0, m = 0, s = +12Sol.When n = 3, cannot be 3 so (A) is incorrect when l = 1, m cannot be = +2.So, (C) is incorrectSo, (A) and (C) is the correct answer.

Ex 8.Select the pairs of ions which have same electronic configuration ?(A) Cr , Fe3+3+(B) Fe , Mn3+2+(C) Fe , Co3+3+(D) Se , Cr3+3+Sol.Fe and Mn have same electronic configuration3+2+So (B) is the correct answer.Ex 9.If an electron in H atom has an energy of –78.4 kcal/mol. The orbit in which the electron is presentis :-(A) 1st(B) 2nd(C) 3rd(D) 4thSol.E = n2313.6kcal / mol n –78.4 =2313.6n  n = 2Ex 10. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition,n = 4 to n = 2 in the He spectrum ?+(A) n = 4 to n = 2(B) n = 3 to n = 2(C) n = 3 to n = 1 (D) n = 2 to n = 1Sol.22211RZ = 24 3R4In H–spectrum for the same  or as Z = 1, n = 1, n = 22So, (D) is the correct answer.Ex 11. Principal, azimuthal and magnetic quantum numbers are respectively related to :-(A) size, orientation and shape(B) size, shape and orientation(C) shape, size and orientation(D) none of theseSol.Principal gives size, i.e. azimuthal gives shape and magnetic quantum number gives the orientation.So, (B) is the correct answer.Ex 12. If the radius of 2 Bohr orbit of hydrogen atom is r . The radius of third Bohr orbit will be :-nd2(A) 24 r9(B) 4r2(C) 29 r4(D) 9r2Sol.r = 2222n h4mZe 2223r2r3  r = 3 29 r4So, (C) is the correct answer.Ex 13. Difference between n and (n + 1) Bohr's radius of H–atom is equal to its (n – 1) Bohr's radius. Thethththvalue of n is :-(A) 1(B) 2(C) 3(D) 4Sol.r nn2But r + 1 – r = r – 1nnn(n + 1) – n = (n – 1)222n = 4So (D) is the correct answerEx 14. The dissociation energy of H is 430.53 kJ mol . If H is dissociated by illumination with radiation of2–12wavelength 253.7 nm. The fraction of the radiant energy which will be converted into kinetic energy isgiven by :-(A) 8.86%(B) 2.33%(C) 1.3%(D) 90%

Sol.323hc430.53 10K.E.6.023 10K.E. = 34896.626 103 10253.7 10 – 323430.53 106.023 10 = 6.9 × 10–20 Fraction = 20196.9 107.83 10 = 0.088 = 8.86%Ex 15. No. of wave in third Bohr's orbit of hydrogen is :-(A) 3(B) 6(C) 9(D) 12Sol.Number of waves = CircumferenceWavelength2 r2 rh / mv  = 2(mvr) = h2nhh2 n = 3So, (A) is the correct answer.Ex 16. In the hydrogen atoms, the electrons are excited to the 5 energy level. The number of the lines that maythappear in the spectrum will be :-(A) 4(B) 8(C) 10(D) 12Sol.No. of lines produced for a jump from fifth orbit to 1 orbit is given byst= n n 12 = 5 5 12 = 10So, (C) is the correct answer.Ex 17. Light of wavelength shines on a metal surface with intensity x and the metal emits Y electrons per secondof average energy, Z. What will happen to Y and Z if x is doubled ?(A) Y will be double and Z will become half(B) Y will remain same and Z will be doubled(C) Both Y and Z will be doubled(D) Y will be doubled but Z will remain sameSol.When intensity is doubled, number of electrons emitted per second is also doubled but average energyof photoelectrons emitted remains the same.So, (D) is the correct answer.Ex 18. Which of the following is the ground state electronic configuration of nitrogen :-(A) (B) (C) (D) Sol.In (A) and (D), the unpaired electrons have spin in the same direction.So, (A) and (D) are the correct answer.Ex 19. Select the wrong statement (s) from the following ?(A) If the value of = 0, the electron distribution is spherical(B) The shape of the orbital is given by magnetic quantum number(C) Angular momentum of 1s, 2s, 3s electrons are equal(D) In an atom, all electrons travel with the same velocitySol.(B) is wrong because shape is given by azimuthal quantum number and magnetic quantum number tellsthe orientation. (D) is wrong because electrons in different shells travel with different velocities.So, (A) and (C) are the correct answer.

Ex 20. For the energy levels in an atom, which one of the following statement/s is/are correct ?(A) There are seven principal electron energy levels(B) The second principal energy level can have four sub-energy levels and contain a maximum of eight electrons(C) The M energy level can have a maximum of 32 electrons.(D) The 4s sub-energy level is at a lower energy than the 3d sub-energy level.Sol.(A) and (D) are true. (B) is wrong because for n = 2, = 0, 1 (two sub-energy levels). (C) is wrong becauseM shell means n = 3. Maximum electrons it can have = 2n = 2 × 3 = 1822So, (A) and (D) is the correct answer.Ex 21. Which of the following statement (s) is (are) correct ?(A) The electronic configuration of Cr is [Ar]3d , 4s (Atomic No. of Cr = 24)51(B) The magnetic quantum number may have a negative value(C) In silver atom 23 electrons have spin of one type and 24 of the opposite type (Atomic No. of Ag = 47)(D) The oxidation state of nitrogen in HN is –33Sol.Only (D) is wrong because oxidation state of N in HN is –1/3.3So, (A), (B) and (C) are the correct answer.Ex 22. Many elements have non-integral atomic masses because :-(A) they have isotopes(B) their isotopes have non-integral masses(C) their isotopes have different masses(D) the constituents, neutrons, protons and electrons combine to give rational massesSol.Non-integral atomic masses are due to isotopes which have different masses.So, (A) and (C) are the correct answer.



CARBOHYDRATES, AMINO ACIDS & POLYMERSIntroduction :Complex organic compound which governs the common activities of the living organism are called biomolecules.Living systems are made up of various complex biomolecules like carbohydrates, proteins, nucleic acids, lipids,etc.In addition, some simple molecules like vitamins and mineral salts also play an important role in the functionsof organisms.CARBOHYDR ATESCarbohydrates are primarily produced by plants and form a very large group of naturally occurring organiccompounds. Some common examples are cane sugar, glucose, starch, etc. Most of them have a generalformula, C (H O) , and were considered as hydrates of carbon from where the name carbohydrate was derived.X2yFor example, the molecular formula of glucose (C H O ) fits into this general formula, C (H O) . But all the6126626compounds which fit into this formula may not be classified as carbohydrates. Rhamnose, C H O is a6125carbohydrate but does not fit in this definition. Chemically, the carbohydrates may be defined as opticallyactive polyhydroxy aldehydes or ketones or the compounds which produce such units onhydrolysis. Some of the carbohydrates, which are sweet in taste, are also called sugars. The most commonsugar, used in our homes is named as sucrose whereas the sugar present in milk is known as lactose.Classification of Carbohydrates :Carbohydrates are classified on the basis of their behaviour on hydrolysis. They have been broadly divided intofollowing three groups.1 .Monosaccharides :A carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or ketone is calleda monosaccharide. Some common examples are glucose, fructose, ribose, etc.Monosaccharides are further classified on the basis of number of carbon atoms and the functional grouppresent in them. If a monosaccharide contains an aldehyde group, it is known as an aldose and if it contains aketo group, it is known as a ketose. Number of carbon atoms constituting the monosaccharide is also introducedin the name as is evident from the examples given in TableDifferent Types of MonosaccharidesCarbon AtomGeneral termAldehydeKetone3 Triose Aldotriose Ketotriose4 Tetrose Aldotetrose Ketotetrose5 Pentose Aldopentose Ketopentose6 Hexose Aldohexose Ketohexose7 Heptose Aldoheptose Ketoheptose2 .Oligosaccharides :Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides. They arefurther classified as disaccharides, trisaccharides, tetrasaccharides, etc., depending upon the number ofmonosaccharides, they provide on hydrolysis. Amongst these the most common are disaccharides. The twomonosaccharide units obtained on hydrolysis of a disaccharide may be same or different.

For example, sucrose on hydrolysis gives one molecule each of glucose and fructose whereas maltose gives twomolecules of glucose only.3 .Polysaccharides :Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Somecommon examples are starch, cellulose, glycogen, gums, etc. Polysaccharides are not sweet in taste, hence theyare also called non-sugars.The carbohydrates may also be classified as either reducing or nonreducing sugars. All those carbohydrates whichreduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides whetheraldose or ketose are reducing sugars.In disaccharides, if the reducing groups of monosaccharides i.e., aldehydic or ketonic groups are bonded, theseare non-reducing sugars e.g. sucrose. On the other hand, sugars in which these functional groups are free,are called reducing sugars, for example, maltose and lactose.Configuration in monosaccharides :Glucose is correctly named as D(+)-glucose. ‘D’ before the name of glucose represents the configurationwhereas ‘(+)’ represents dextrorotatory nature of the molecule. It may be remembered that ‘D’ and ‘L’ have norelation with the optical activity of the compound. The meaning of D– and L– notations is given as follows. Theletters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer.This refers to their relation with a particular isomer of glyceraldehyde. Glyceraldehyde contains one asymmetriccarbon atom and exists in two enantiomeric forms as shown below.CHOCHOH2OHH(+)-GlyceraldehydeCHOCHOH2OHH(–)-GlyceraldehydeAll those compounds which can be chemically correlated to (+) isomer of glyceraldehyde are said to haveD-configuration whereas those which can be correlated to (–) isomer of glyceraldehyde are said to haveL- configuration. For assigning the configuration of monosaccharides, it is the lowest asymmetric carbon atom(as shown below) which is compared. As in (+) glucose, —OH on the lowest asymmetric carbon is on the rightside which is comparable to (+) glyceraldehyde, so it is assigned D-configuration. For this comparison, thestructure is written in a way that most oxidised carbon is at the top.CHOCHOH2OHHD–(+)-GlyceraldehydeCHOH–C–OHHO–C–HH–C–OHH–C–OHCHOH2D-(+) – Glucose

GLUCOSE (ALDOHEXOSE)Structure of Glucose :Its molecular formula was found to be C H O .6126CHOH–C–OHHO–C–HH–C–OHH–C–OHCHOH2GlucoseCyclic structure of Glucose :-Glucose is found to exist in two different crystalline forms which are named as and . The -form of glucose(m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the -form(m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in ring formation. Thisexplains the absence of —CHO group and also existence of glucose in two forms as shown below. These twocyclic forms exist in equilibrium with open chain structure.H–C–OHH OHHO HH OHH 1O23456CHOH 2-D-(+) – GlucoseH–CH OHHO HH OH123456CHOH 2OH OHHO–C–HH OHHO HH OHH 1O23456CHOH 2-D-(+) – GlucoseThe two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C , called1anomeric carbon (the aldehyde carbon before cyclisation). Such isomers, i.e., -form and -form, are calledanomers. The six membered cyclic structure of glucose is called pyranose structure ( – or –), in analogywith pyran. Pyran is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring. Thecyclic structure of glucose is more correctly represented by Haworth structure as given below.OPyranCHOH2H HHOOHHHOHOHOH643215D – (+) – Glucopyranose– CHOH2H HHOOHHHOHOHOH643215D – (+) – Glucopyranose– Mutarotation :The ring structures for glucose also explains the mutarotation by the fact that the change in specific rotation isdue to the interconversion of the -form [( ) = +113°] of glucose to -[(x) = + 19.5) and vice verse through D Dthe aldehydrol structure (hydrate of aldehyde) till an equilibrium is reached between two structures. The specificrotation value of this equilibrium mixture consisting of 36% -glucose and 64% -is +52.50.

H—C—OHH C —OH2(1)H—C—OH(2)H—CHO—C—H(3)(4)H—C—OH(5)(6)O-D-glucose [ ] =+113° D+H O2–H O2CH(OH)2H C —OH2H—C—OHH—C—OHHO—C—HH—C—OH–H O2+H O2HO—C—HH C —OH2(1)H—C—OH(2)H—CHO—C—H(3)(4)H—C—OH(5)(6)O-D-glucose [ ] =+19.5° Dintermediate aldehydrol form(36%)[ ] =+52.5° D(64%)Epimers :Many common sugars are closely related, differing only by the stereochemistry at a single carbon atom. Forexample, glucose and mannose differ only at C-2, the first asymmetric carbon atom. Sugars that differ only bythe stereochemistry at a single carbon are called epimers, and the carbon atom where they differ is generallystated. If the number of a carbon atom is not specified, it is assumed to be C-2. Therefore, glucose andmannose are 'C-2 epimers' or simply 'epimers'. The C-4 epimer of glucose is galactose and the C-2 epimer oferythrose is threose.CHOCHOCHOCHOCHOH C —OH2(1)(1)(1)(1)(1)HO—C—HH—C—OHH—C—OHH—C—OHHO—C—H(2)(2)(2)(2)(2)H—C—OHHO—C—HHO—C—HHO—C—HH—C—OHH—C—OH(3)(3)(3)(3)(3)(4)(4)(4)(4)(4)H—C—OHH—C—OHH—C—OH(5)(5)(5)(6)(6)(6)D-mannoseD-glucoseD-galactoseD-erythroseD-threoseC-2 epimersH—C—OHH C —OH2H C —OH2H C —OH2H C —OH2HO—C—H C-4 epimersC-2 epimersFRUCTOSE (KETOHEXOSE)Structure of Fructose :Fructose also has the molecular formula C H O and on the basis of its reactions6126C=OCHOH2HO CHH CH COHOHCHOH2Fructoseit was found to contain a ketonic functional group at carbon number 2 and sixcarbons in straight chain as in the case of glucose. It belongs to D-series and is alaevorotatory compound. It is appropriately written as D-(–)-fructose. Its openchain structure is as shown.It also exists in two cyclic forms which are obtained by the addition of —OH atC5 to the (C=O ) group. The ring, thus formed is a five membered ring and isnamed as furanose with analogy to the compound furan. Furan is a five memberedcyclic compound with one oxygen and four carbon atoms.

HOHC–C–OH2HO HH OHH 1 2O3456CHOH 2– D – (–) – FructofuranoseHO–C–CHOH2HO HH OHH 1O23456CHOH 2– D – (–) – FructofuranoseOFuranThe cyclic structures of two anomers of fructose are represented by Haworth structures as given.OHHOHC265HOH H 4OH32CHOH21OH– D – (–) – FructofuranoseOHHOHC265HOH H 4OH32CHOH2 1OH– D – (–) – FructofuranoseGlucose (Aldohexose) :Preparation of Glucose :( a )From sucrose (Cane sugar): If sucrose is boiled with dilute HCl or H SO in alcoholic solution,24glucose and fructose are obtained in equal amounts.C H O + HO1222112SucroseH C H O + C H O61266126GlucoseFructose( b )From starch : Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute H SO24at 393 K under pressure.(CH O) +nHO6105 n2Starch or celluloseH393K : 2 3 atmnC H O6126GlucoseReactions of glucose :(i)On prolonged heating with Red P/HI, it forms n-hexane, suggesting that all the six carbon atoms arelinked in a straight chain.CHO(CHOH)4CHOH2Re d P HICH–CH–CH–CH–CH–CH322223(n-Hexane)(ii)Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to givecyanohydrin. These reactions confirm the presence of a carbonyl group (>C = O) in glucose.CHO(CHOH)4CHOH22NH OHCH=N–OH(CHOH)4CHOH2

CHO(CHOH)4CHOH2HCNCH(CHOH)4CHOH2CNOH(iii)Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agentlike bromine water. This indicates that the carbonyl group is present as an aldehydic group. (Aldohexose)COOH(CHOH)4CHOH2Gluconic acidCHO(CHOH)4CHOH2Br water2(iv)Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence offive –OH groups. Since it exists as a stable compound, five –OH groups should be attached to differentcarbon atoms.CHO(CHOH)4CHOH2Acetic anhydrideCHO(CH–O–C–CH)3 4CH–O–C–CH23OO(v)On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharicacid. This indicates the presence of a primary alcoholic (–OH) group in glucose.CHO(CHOH)4CHOH23OxidationHNOCOOH(CHOH)4COOHSaccharic acidOxidationCOOH(CHOH)4CHOH2Gluconicacid(vi)Glucoside formation : Glucose reacts with methyl alcohol in the presence of hydrochloric acid gasforming two isomeric ( and ) methyl glucosides. This reaction indicates the presence of a ring structurein glucose in which free —CHO group is not present but it is converted into (—CHOH) group.CH OH2OHOCH3HOHOHHHHOH 5Methyl -D-glucopyranoside+Methyl -D-GlucopyranosideD(+) glucose CH OH(pyranose form)3HClCarbohydrate acetals, generally are called glycosides and an acetal of glucose is called glucoside.(vii)Action of phenylhydrazine : Like the normal aldehydes, glucose reacts with phenylhydrazine inequimolecular proporation to form phenylhydrazone but unlike the normal aldehydes, glucose reactswith excess of phenylhydrazine (three molecular proporations) to form glucosazone.

According to Fischer, glucosazone formation takes place in the following steps :(a)One molecule of phenylhydrazine reacts with the aldehydic group of glucose to form glucosephenylhydrazone.HC —OHHC(CHOH)3H C —OH2(Glucose)O+ H NNHC H265Phenylhydrazine (Ist molecule)HC —OHHC(CHOH)3H C —OH2(Glucose Phenylhydrazone)NNHC H65(b)The second molecule of phenylhydrazine oxidizes the secondary alcoholic group (—CHOH—) adjacentto the aldehydic group to a ketonic group and itself is reduced to aniline and ammonia.+ C H NHNH652Phenylhydrazine (2nd molecule)HC —OHHC(CHOH)3H C —OH2(Glucose phenylhydrazone)NNHC H65CO HC(CHOH)3H C —OH2(Phenylhydrazone of glucosone)NNHC H65+ C H NH + NH6523Aniline(c)The new keto group reacts with the third molecule of phenylhydrazine to form glucosazone.+ C H NHNH652Phenylhydrazine (3rd molecule)HC(CHOH)3H C —OH2(Glucosazone)NNHC H65CO HC(CHOH)3H C —OH2(Phenylhydrazone of glucosone)NNHC H65CNNHC H65Glucosazone is a yellow crystalline compound, sparingly soluble in water and has a sharp melting point.On account of these properties, it is used in the identification of glucose.Disaccharides :The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Sucha linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.( i )Sucrose : One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixtureof D- (+)-glucose and D-(-) fructose.122211261266126SucroseD ( ) Glu cos eD ( ) FructoseC H OH OC H OC H O   These two monosaccharides are held together by a glycosidic linkage between C1 of -glucose and C2of -fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation,sucrose is a non reducing sugar.~zCHOH2H HHOOHHHOHOH643215OHOH H4HO 35CHOH2 6HOHOHC21Glycosidiclinkage2

Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+)to laevo (–) and the product is named as invert sugar.Sucrose is dextrorotatory, its specific rotation being +66.5%, D-glucose is also dextrorotatory, [ ] =+53°, Dbut D-fructose has a large negative rotation, [ ] =–92°. Since D-fructose has a greater specific rotation Dthan D-Glucose, the resulting mixture is laevorotatory. Because of this the hydrolysis of sucrose is knownas the inversion of sucrose and the equimolecular mixture of glucose and fructose is known as invertsugar or invertose.(ii)Maltose : Another disaccharide, maltose is composed of two -D-glucose units in which C-1 of oneglucose (I) is linked to C-4 of another glucose unit (II). The free aldehyde group can be produced at C-1of second glucose in solution and it shows reducing properties so it is a reducing sugar.CHOH2H HHOOHHHOHOH643215D – Glucose– – D – FructoseOMaltoseH(I)(II)CHOH2HOHHHOHO643215OHH(iii)Lactose : It is more commonly known as milk sugar since this disaccharide is found in milk. It iscomposed of -D-galactose and -D-glucose. The linkage is between C-1 of galactose and C-4 of glucose.Hence it is also a reducing sugar.CHOH2H HOOHHHOHO643215D – Galactose– – D – GlucoseLactoseHCHOH2HOHHHOHO643215OHHHO HPolysaccharidesPolysaccharides contain a large number of monosaccharide units joined together by glycosidic linkages. Theymainly act as the food storage or structural materials.(i)Starch : Starch is the main storage polysaccharide of plants. It is the most important dietary source forhuman beings. High content of starch is found in cereals, roots, tubers and some vegetables. It is apolymer of -glucose and consists of two components :Amylose and Amylopectin Amylose is water soluble component which constitutes about 15-20% of. starch. Chemically amylose is a long unbranched chain with 200-1000 -D-(+)-glucose units held byC1– C4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80-85% of starch. Itis a branched chain polymerof -D-glucose

CHOH2CHOH2CHOH2HHHHOHOHOHHHHHHHOHOHOHOOO411AmyloseHH–OOOHH446532HO– 1-Link-LinkCHOH2CHOH2HHHOHOHHHHHOHOHOO41H–OOH4HO1-LinkBranch at C6CHOH2CH 2HHHHOHOHOHHHHHHHOHOHOHOOO411HH–OOOHH446CHOH2O–H-Link-LinkAmylopectin1units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidiclinkage.(ii)Cellulose : Cellulose occurs exclusively in plants and it is the most abundant organic substance in plantkingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain polysaccharidecomposed only of -D-glucose units which are joined by glycosidic linkage between C1 of one glucoseunit and C4 of the next glucose unit.OOOOOHOHOHOHOHOHHOHC2HOHC2HOHC2OOO-linksCellulose

(iii)Glycogen : The carbohydrates are stored in animal body as glycogen. It is also known as animalstarch because its structure is similar to amylopectin and is rather more highly branched. It is present inliver, muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose.Glycogen is also found in yeast and fungi.AMINO ACIDSAmino acids contain amino (–NH ) and carboxyl (–COOH) functional groups. Depending upon the relative2position of amino group with respect to carboxyl group, the amino acids can be classified as    , , , and soon. Only -amino acids are obtained on hydrolysis of proteins. They may contain other functional groups also.R–CH–COOHNH 2– amino acid (R=side chain)All -amino acids have trivial names, which usually reflect the property of that compound or its source.Glycine is so named since it has sweet taste (in Greek glykos means sweet) and tyrosine was first obtained fromcheese (in Greek, tyros means cheese.).Natural Amino acids :COOHHN2HRName of theCharacteristic featureThree letterOne letteramino acidsof side chain. Rs ym bo lof code1. GlycineHGlyG2. Alanine– CH3AlaA3. Valine*(H C) CH–32ValV4. Leucine*(H C) CH-CH –322LeuL5. Isoleucine*HC–CH–CH–3CH 3IleI6. Arginine*HN=C–NH–(CH) –2 3NH 2ArgR7. Lysine*H N–(CH ) –22 4LysK8. Glutamic acidHOOC–CH –CH –22GluE9. Aspartic acidHOOC–CH –2AspD10. GlutamineHN–C–CH–CH–222OGlnQ11. AsparagineHN–C–CH–22OAsnN12. Threonine*H C–CHOH–3ThrT

13. SerineHO–CH –2SerS14. CysteineHS–CH –2CysC15. Methionine*H C–S–CH –CH –322MetM16. Phenylalanine*C H –CH –652PheF17. Tyrosine(p) HO–C H –CH –642TyrY18. Tryptophan*NH–CH2TrpW19. Histidine*NHC2NHHisH20. ProlineHNHCOOHCH 2ProP essential amino acid, a = entire structureClassification of Amino Acids :Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxylgroups in their molecule.The amino acids, which can be synthesised in the body, are known as non-essential amino acids. Onthe other hand, those which cannot be synthesised in the body and must be obtained through diet, areknown as essential amino acids.Amino acids are usually colourless, crystalline solids. These are water-soluble, high melting solids and behavelike salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic(carboxyl group) and basic (amino group) groups in the same molecule.R–CH–C–O–HR–CH–C–OOO:NH2NH 3(Zwitter ion)–+In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise toa dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges.In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids andbases.Except glycine, all other naturally occurring -amino acids are optically active, since the  -carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms. Most naturally occurringamino acids have L-configuration. L-Aminoacids are represented by writing the –NH group on2left hand side.

Methods of Preparation of -amino acid :1 .Amination of -halo acids :-Chloro or -bromo acids on treatment with excess of liquid or concentrated ammonia forms the respectiveamino acids.Cl––CH COOH + 2NH 23 H NCH COOH + NH Cl224-Chloroacetic acid Glycine33342CH CHCOOH2NHCH CHCOOHNH Cl||BrNH-Bromopropionic acid -Amino propionic acid (Alanine)Lab preparation of glycine :ClCH COOH + 3NH 2350 C H NCH COONH + NH Cl2244-Chloroacetic liquid Amm. salt of glycine acidThe ammonium salt so obtained is boiled with copper carbonate and cooled when blue colour needles of coppersalt of glycine are obtained.2[H N––CH COONH ] + CuCO 2243 (H NCH COO) Cu + (NH ) CO2224 23 Copper salt of glycineIt is now dissolved in water and H S is passed till whole of the copper precipitates as copper sulphide leaving2glycine as the aqueous solution.(H N––CH COO) Cu + H S22222H NCH COOH + CuS 22Glycine Black ppt.2 .Gabriel phthalimide synthesis :-Halogenated acid or ester combines with potassium phthalimide. The product on hydrolysis gives -aminoacid.COCOCOCONK + ClCHCOOC H 225NCHCOOC H225Pot. phthalimideChloroethyl acetate–KCl2H O2HClCOOHCOOH+ CHNHCOOH + C HOH2225GlycinePhthalic acid3 .Strecker synthesis :An aldehyde reacts with HCN and ammonia or NH CN and the product on hydrolysis yields -amino acid.4R––C=OR––C––OHR––C––NH2R––C––NH2AldehydeHCNNH 3H OH 2+CNCNCOOHHHHHCyanohydrinAminonitrile-Amino acid

4 .From natural proteins :Natural proteins are hydrolysed with dil. HCl or H SO at 250°C in an autoclave when a mixture of -amino24acids is obtained. This mixture is esterified and the various esters are separated by fractional distillation. Theesters are then hydrolysed into respective -amino acids.Properties :( i )Amino acids are colourless, crystalline substances having sweet taste. They melt with decompositionat higher temperature (more than 200°C). They are soluble in water but insoluble in organic solvents.(ii)Except glycine, all the -amino acids are optically active and have an asymmetric carbon atom ( -carbonatom). Hence, each of these amino acids can exist in two optical isomers. In proteins, however, onlyone isomer of each is commonly involved.(iii)Zwitter ion and isoelectric point :Since the –NH group is basic and –COOH group is acidic, in neutral solution, it exists in an internal2ionic form called a Zwitter ion where the proton of –COOH group is transferred to the –NH group2to form inner salt, also known as dipolar ion.HN––CHCOOH2HN––CH––COO + H2HN––CH––COO3-Amino acidIn waterRRR..––++Zwitter ion (Dipolar ion)The Zwitter ion is dipolar, charged but overall electrically neutral and contains both a positive and negativecharge.Therefore, amino acids are high melting crystalline solids and amphoteric in nature. Depending on the pHof the solution, the amino acid can donate or accept proton.HN––C––C––OH3HN––C––C––O3HN––C––C––O2H +OH–HO2RRRH OH OH OLow pH (Acidic soln.)Positive form (II)(Cation)Zwitter ion (I) Neutral formHigh pH (Basic soln.)Negative form (III)(Anion)(Proton removed)–––When an ionised form of amino acid is placed in an electric field, it will migrate towards the opposite electrode.Depending on the pH of the medium, following three things may happen :(i)In acidic solution (low pH), the positive ion moves towards cathode [exist as cation, structure (II)].(ii)In basic solution (high pH), the negative ion moves towards anode [exist as anion, structure (III)](iii)The Zwitter ion does not move towards any of the electrodes [neutral dipolar ion, structure (I)].The intermediate pH at which the amino acid shows no tendency to migrate towards any of the electrodesand exists the equilibrium when placed in an electric field is known as isoelectric point. This is characteristicof a given amino acid and depends on the nature of R-linked to -carbon atom.(A) As regards the chemical properties, -amino acids show the reactions of –NH group, –COOH group2and in which both the groups are involved. A summary of chemical properties is given below :

NaOHCHOHDry HCl25Decarboxylation Ba(OH) , 2LiAlH4H4AlkylationCHBr3AcetylationCHCOCl3HClPCl5HNO2RCHNHCOONa2HNCHCOOCH325HN––CH22HNCHCOOCH225Sodium saltAgOHRRREthyl esterAmineR––CHRCHRCHRCHRCHRCHRCHOHCOOHNH 2NH 2NHCH3NHCOCH3NHCl3NH(HCl)2CHOH2COOHCOOHCOOHCOOHCOClAmino alcoholN-Methyl derivativeN-Acetyl derivativeSalt formationAmino acid chloride-Hydroxy acid-Amino acid– Action of heat : -Amino acids lose two molecules of water and form cyclic amides.R––CHR––CHCH––RCH––R + 2HO2NH H OH OCNH––CCO OH H HNC– NH–+OOCH––NH H HO OC2CH––NH––CO2CO OH H HN––CH2CO––NH––CH2+Heat(-2H O)2Glycine (2 molecules)Diketo piperazine-Amino acid lose a molecule of ammonia per molecule of amino acid to yield , -unsaturated acids. 

CH–––CH–––COOH2CH=CHCOOH2Acrylic acid(-unsaturated acid)NH H2Heat–NH3-Amino propionic acidCH–––CH–––CH–––C3OOHCH––CH=CH COOH3– –Crotonic acidNH H2Heat(–NH )3-Amino butyric acid-Amino acids and -amino acids undergo intramolecular dehydration to form cyclic amides called Lactams.CH––CH––CH––CO222CH––C ––CH––COH 222NHButyro lactumNH H H–OHeat(-HO)2Amino butyric acidCHCHCHCHCO2222CH HCHCHCO22 C22NH-ValerolactamNH H H–OHeat(-HO)2-Amino valeric acidThese lactams have stable five or six membered rings.PROTEINSThe word protein is derived from Greek word, “proteios” which means primary or of prime importance.All proteins are polymers of -amino acids.Structure of Proteins :Proteins are the polymers of -amino acids and they are connected to each other by peptide bond orpeptide linkage. Chemically, peptide linkage is an amide formed between –COOH group and –NH group.2HN–CH–COOH + HN–CH–COOH222CH 3–HO2HN–CH–CO–NH–CH–COOH22CH 3Peptide linkageGlycylalanine [Gly-Ala]The reaction between two molecules of similar or different amino acids, proceeds through the combination ofthe amino group of one molecule with the carboxyl group of the other. This results in the elimination of a watermolecule and formation of a peptide bond –CO–NH–.Proteins can be classified into two types on the basis of their molecular shape.( a )Fibrous proteins : When the polypeptide chains run parallel and are held together by hydrogen anddisulphide bonds, then fibre– like structure is formed. Such proteins are generally insoluble in water.Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc.

( b )Globular proteins : This structure results when the chains of polypeptides coil around to give aspherical shape. These are usually soluble in water. Insulin and albumins are the common examplesof globular proteins. Structure and shape of proteins can be studied at four different levels, i.e., primary,secondary, tertiary and quaternary.Primary structure of proteins : Proteins may have one or more polypeptide chains. Each polypeptidein a protein has amino acids linked with each other in a specific sequence and it is this sequence ofamino acids that is said to be the primary structure of that protein. Any change in this primary structurei.e., the sequence of amino acids creates a different protein.Secondary structure of proteins : The secondary structure of protein refers to the shape in which along polypeptide chain can exist. They are found to exist in two different types of structures viz. -helixand -pleated sheet structure. These structures arise due to the regular folding of the backbone of thepolypeptide chain due to hydrogen bonding between C=O and –NH– groups of the peptide bond.-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogenbonds by twisting into a right handed screw (helix) with the –NH group of each amino acid residuehydrogen bonded to the C=O of an adjacent turn of the helix.In -structure all peptide chains are stretched out to nearly maximum extension and then laid side by sidewhich are held together by intermolecular hydrogen bonds.Tertiary structure of proteins : The tertiary structure of proteins represents overall folding ofthe polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecularshapes viz. fibrous and globular. The main forces which stabilise the 2 and 3 structures of proteins arehydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction.Quaternary structure of proteins : Some of the proteins are composed of two or more polypeptidechains referred to as sub-units. Subunits with respect to each other is known as quaternary structure.Formation of proteins–Peptide bond :Proteins are formed by joining the carboxyl group of one amino acid to the -amino group of another aminoacid. The bond formed between two amino acids by the elimination of water molecule is called a peptide linkageor bond. The peptide bond is simply another name for amide bond.––C—OH + H ––N––––C––N––+HO2OO HHCarboxyl groupof one amino acidPeptide bondAmine group of other amino acidThe product formed by linking amino acid molecules through peptide linkages, ––CO––NH––, is called a peptide.Peptides are further designated as di, tri, tetra or penta peptides accordingly as they contain two, three, fouror five amino acid molecules, same or different, joined together in the following fashions.HN––CH––C––OH + H––N––CH––C––OH2HN––CH––C––N––CH––C––OH (Dipeptide)2OO HOORR(2 molecules)RRH(–H O)2PeptidelinkageWhen the number of amino acid molecules is large, the product is termed polypeptide which may be rep-resented as,HN––CH––C––2––NH––CH––C––––NH––CH––COOHOORRR\"n

The formation of a protein linking together different amino acids is not a random process. Each moleculeof a given protein has the same sequence of the amino acids along its polypeptide chain. In fact, it is thisvery sequence that imparts to a protein its own specific properties. Even a change of just one amino acidcan drastically change the properties of entire protein molecule. Haemoglobin consists of 574 amino acidunits in its molecule in a definite sequence. When only one specific amino acid in the sequence is changed,it become a defective haemoglobin which loses its specific property of carrying oxygen in blood stream andresults in a disease called sickle cell anaemia.Normal haemoglobin,––Val––His––Leu––Thr––Pro––Glu––Glu––Lys––Sickle cell haemoglobin––Val––His––Leu––Thr––Pro––Val––Glu––Lys––Composition of proteins : Composition of a protein varies with source. An approximate composition isas follows :Carbon 50–53%; hydrogen 6–7%; oxygen 23–25%; nitrogen 16–17%; sulphur about 1%. Other elements mayalso be present, e.g., phosphorus (in nucleoproteins), iodine (in thyroid proteins) and iron (in haemoglobin).Denaturation of Proteins :Protein found in a biological system with a unique three-dimensional structure and biological activity is called anative protein. When a protein in its native form, is subjected to physical change like change in temperature orchemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helixget uncoiled and protein loses its biological activity. This is called denaturation of protein. The coagulation ofegg white on boiling is a common example of denaturation. Another example is curdling of milk which iscaused due to the formation of lactic acid by the bacteria present in milk.NUCLEIC ACIDSIntroduction :Every generation of each and every species resembles its ancestors in many ways. How are these characteristicstransmitted from one generation to the next? It has been observed that nucleus of a living cell is responsible forthis transmission of inherent characters, also called heredity. The particles in nucleus of the cell, responsiblefor heredity, are called chromosomes which are made up of proteins and another type of biomolecules callednucleic acids. These are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid(RNA). Since nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides.Chemical composition of Nucleic Acids :Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric acid and nitrogen containingheterocyclic compounds (called bases). In DNA molecules, the sugar moiety is -D-2-deoxyribose whereas inRNA molecule, it is -D-ribose.OHH3214OHOHHOHC25OHHH-D-riboseOHH 3214OHHHOHC25OHHH-D-2-deoxyribose

DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also containsfour bases, the first three bases are same as in DNA but the fourth one is uracil (U).NH 2CCNNHCCNHCHNAdenine (A)OCCNHNHCCNHC–NH2NGuanine (G)NH 2CHCNCNCytosine (C)HCHOCCNHCNThymine (T)HCHOHC3OOCHCNHCNUracil (U)HCHOStructure of Nucleic Acids :A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside. In nucleosides, thesugar carbons are numbered as 1', 2', 3', etc. in order to distinguish these from the bases. When nucleoside islinked to phosphoric acid at 5'-position of sugar moiety, we get a nucleotide.O(a)HH 3'2'1'4'OHOHHO–HC25'BaseHHO(b)HH 3'2'1'4'OHOHO–P–O–HC25'BaseHHOOStructure of (a) a nucleoside and (b) a nucleotideNucleotides are joined together by phosphodiester linkage between 5' and 3' carbon atoms of the pentosesugar. Biomolecules A simplified version of nucleic acid chain is as shown below. Fig.BaseSugarPhosphateSugarPhosphateSugarBaseBasenInformation regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure.Nucleic acids have a secondary structure also. James Watson and Francis Crick gave a double strand helixstructure for DNA (Fig.).