Tamilnadu Board Class 11 Chemistry Chapter 2
2UnitQuantum Mechanical Model of AtomLearning Objectives:After studying this unit, students will be able to Recognise various atomic models Explain the dual behaviour of matter Derive de Broglie equation and solve numericalproblemsErwin Schrödinger(1887 - 1961)ErwinSchrödingerwasawarded the Nobel Prizein physics in 1933 for “thediscovery of new productiveforms of atomic emistry, physics, maths andbotany. He was not satisfiedwith the quantum conditionin Bohr's orbit theory andbelieved that atomic spectrashould really be determinedby some kind of eigenvalueproblem and proposed thewave equation, now namedafter him. Explain Heisenberg’s uncertainty principle andsolve related problems Appreciate the significance of quantum numbers Summarise important featuresmechanical model of atomofquantum Draw the shapes of various atomic orbitals Explain the Aufbau principle Describe Hund's rule and Pauli’s exclusion principle Apply the relevant rules for filling electrons inatoms and write the electronic configuration ofvarious atoms38
Based on these observations,he proposed that in an atom there isa tiny positively charged nucleus andthe electrons are moving around thenucleus with high speed. The theoryof electromagnetic radiation statesthat a moving charged particle shouldcontinuously loose its energy in theform of radiation. Therefore, the movingelectron in an atom should continuouslyloose its energy and finally collide withnucleus resulting in the collapse of theatom. However, this doesn't happenand the atoms are stable. Moreover, thismodel does not explain the distributionof electrons around the nucleus and theirenergies.2.1 Introduction to atom models:Let us recall the history of thedevelopment of atomic models fromthe previous classes. We know that allthings are made of matter. The basicunit that makes up all matter is atom.The word ‘atom’ has been derived fromthe Greek word ‘a-tomio’ meaning nondivisible. Atom was considered as nondivisible until the discovery of subatomic particles such as electron, protonand neutron. J. J. Thomson’s cathode rayexperiment revealed that atoms consistof negatively charged particles calledelectrons. He proposed that atom is apositively charged sphere in which theelectrons are embedded like the seeds inthe watermelon. Later, Rutherford’s α-rayscattering experiment results proved thatThomson’s model was wrong. Rutherfordbombarded a thin gold foil with a streamof fast moving α–particles. It was observedthat2.1.1 Bohr atom model:The work of Planck and Einsteinshowed that the energy of electromagneticradiation is quantised in units of hν(where ν is the frequency of radiation andh is Planck's constant 6.626 10-34 Js).Extending Planck’s quantum hypothesisto the energies of atoms, Niels Bohrproposed a new atomic model for thehydrogen atom. This model is based onthe following assumptions:(i) most of the α–particles passed throughthe foil(ii) some of them were deflected througha small angle and(iii) very few α–particles were reflectedback by 180 1.The energies of electrons are quantised2.The electron is revolving around thenucleus in a certain fixed circularpath called stationary orbit.3.Electron can revolve only in thoseorbits in which the angular momentum(mvr) of the electron must be equal toan integral multiple of h/2π.i.e. mvr nh/2π -------- (2.1)Figure. 2.1 Rutherford's α-ray scatteringexperimentwhere n 1,2,3,.etc.,39
4.As long as an electron revolves in thefixed stationary orbit, it doesn’t loseits energy. However, when an electronjumps from higher energy state (E2)to a lower energy state (E1), the excessenergy is emitted as radiation. Thefrequency of the emitted radiation is2.1.2 Limitation of Bohr's atom model:The Bohr's atom model is applicableonly to species having one electron suchas hydrogen, Li2 etc. and not applicableto multi electron atoms. It was unable toexplain the splitting of spectral lines inthe presence of magnetic field (Zeemaneffect) or an electric field (Stark effect).Bohr’s theory was unable to explain whythe electron is restricted to revolve aroundthe nucleus in a fixed orbit in which theangular momentum of the electron isequal to nh/2π and a logical answer forthis, was provided by Louis de Broglie.E2 – E1 hνand(E2 – E1)ν -------- (2.2)hConversely, when suitable energy issupplied to an electron, it will jump fromlower energy orbit to a higher energyorbit.2.2 Wave particle duality of matterAlbert Einstein proposed that lighthas dual nature. i.e. light photons behaveboth like a particle and as a wave. Louisde Broglie extended this concept andproposed that all forms of matter showeddual character. To quantify this relation, hederived an equation for the wavelength ofa matter wave. He combined the followingtwo equations of energy of which onerepresents wave character (hυ) and theother represents the particle nature (mc2).Applying Bohr’s postulates to ahydrogen like atom (one electron speciessuch as H, He and Li2 etc.) the radius ofthe nth orbit and the energy of the electronrevolving in the nth orbit were derived.The results are as follows:rn(0.529)n2A ---------------- (2.3)ZEn ( 13.6) 2eV atom 1 --------(2.4)n2(i)(or)En Planck’s quantum hypothesis:E hν( 1312.8) 2 kJ mol 1 --------(2.5)n2(ii)Einsteins mass-energy relationshipE mc2The detailed derivation of rn and Enwill be discussed in 12th standardatomic physics unit.-------- (2.6)---------(2.7)From (2.6) and (2.7)hν mc2hc/λ mc240
λ h / mc ------------(2.8)6.626 10 34 kgm2s 1 9.11 10 31 kg 72.73 ms 16.626 10 3 m 1 105 m662.6The equation 2.8 represents thewavelength of photons whose momentumis given by mc (Photons have zero restmass)For the electron, the de Brogliewavelength is significant and measurablewhile for the iron ball it is too small tomeasure, hence it becomes insignificant.For a particle of matter with massm and moving with a velocity v, theequation 2.8 can be written asEvaluate Yourselfλ h / mv ------------ (2.9)?1. Calculate the de-Broglie wavelengthof an electron that has been acceleratedfrom rest through a potential differenceof 1 keV.This is valid only when the particletravels at speeds much less than the speedof Light.This equation implies that a movingparticle can be considered as a wave anda wave can exhibit the properties (i.emomentum) of a particle. For a particlewith high linear momentum (mv) thewavelength will be so small and cannot beobserved. For a microscopic particle suchas an electron, the mass is of the order of10-31 kg, hence the wavelength is muchlarger than the size of atom and it tum and de Broglie concept:According to the de Broglieconcept, the electron that revolves aroundthe nucleus exhibits both particle andwave character. In order for the electronwave to exist in phase, the circumferenceof the orbit should be an integral multipleof the wavelength of the electron wave.Otherwise, the electron wave is out ofphase.Let us understand this bycalculating de Broglie wavelength in thefollowing two cases:Circumference of the orbit(i) A 6.626 kg iron ball moving with10 ms-12πr nλ ------------(2.10)2πr nh/mv(ii) An electron moving at 72.73 ms-1Rearranging,---(2.1)λiron ball h/mvAngular momentum nλmvr nh/2π ------ nh/2πThe above equation was alreadypredicted by Bohr. Hence, De Broglie andBohr’s concepts are in agreement witheach other.6.626 10 34 kgm2s 1 1 x 10 35 m 16.626 kg x 10 msλelectron h/mv41
AllowedNot allowed -nucleusdetermination of position and velocityof a microscopic particle. Based on this,Heisenberg arrived at his uncertaintyprinciple, which states that ‘It is impossibleto accurately determine both the positionas well as the momentum of a microscopicparticle simultaneously’. The product ofuncertainty (error) in the measurement isexpressed as follows. -nucleusΔx.Δp h/4π -------- (2.11)n 3n 4where, Δx and Δp are uncertaintiesin determining the position andmomentum, respectively.nucleusnucleusn 5n 6The uncertainty principle hasnegligible effect for macroscopicobjects and becomes significant only formicroscopic particles such as electrons.Let us understand this by calculatingthe uncertainty in the velocity of theelectron in hydrogen atom. (Bohr radiusof 1st orbit is 0.529 Ǻ) Assuming that theposition of the electron in this orbit isdetermined with the accuracy of 0.5 % ofthe radius.Figure. 2.2 Wave nature of electrons inallowed Bohr orbitsDavison and Germer experiment :The wave nature of electron wasexperimentally confirmed by Davissonand Germer. They allowed the acceleratedbeam of electrons to fall on a nickel crystaland recorded the diffraction pattern. Theresultant diffraction pattern is similar tothe x-ray diffraction pattern. The findingof wave nature of electron leads to thedevelopment of various experimentaltechniques such as electron microscope,low energy electron diffraction etc Uncertainity in position Δx0. 5 % x 0.529 Α100 %0.5 x 0.529 x 10 10 m100Δx 2.645 x 10 13 mFrom the Heisenberg’s uncertainityprinciple,hΔx.Δp 4π2.3 Heisenberg’s uncertainty principleaThe dual nature of matter imposeslimitation on the simultaneous42
Δx.(m.Δv ) Δv Δv the Heisenberg's principle and the dualnature of the microscopic particles, a newmechanics called quantum mechanics wasdeveloped.h4πh4 π. m . Δx6.626 x 104 x 3.14 x 9.11 x10 34 312 -1kg m sErwin Schrödinger expressedthe wave nature of electron in terms ofa differential equation. This equationdetermines the change of wave functionin space depending on the field of forcein which the electron moves. The timeindependent Schrödinger equation can beexpressed as, 13kg x 2.645 x 10 mΔv 2.189 x 108 mTherefore, the uncertainty in thevelocity of the electron is comparable withthe velocity of light. At this high level ofuncertainty it is very difficult to find outthe exact velocity. Evaluate YourselfH Ψ EΨ (2.12)? Where H is called Hamiltonian operator ,Ψ is the wave function and is a funcitonof position co-ordinates of the particleand is denoted as Ψ(x, y , z) E is the energyof the system2. Calculate the uncertainty in the positionof an electron, if the uncertainty in itsvelocity is 5.7 105 ms-1.2.4 Quantum mechanical model of atom– Schrödinger Equation: h 2 2 2 2H 2 2 2 2 V z 8π m x y (2.12) can be written as, The motion of objects that we comeacross in our daily life can be well describedusing classical mechanics which is basedon the Newton’s laws of motion. In classicalmechanics the physical state of the particleis defined by its position and momentum.If we know both these properties, wecan predict the future state of the systembased on the force acting on it usingclassical mechanics. However, accordingto Heisenberg’s uncertainty principleboth these properties cannot be measuredsimultaneously with absolute accuracy fora microscopic particle such as an electron.The classical mechanics does not considerthe dual nature of the matter which issignificant for microscopic particles.As a consequence, it fails to explain themotion of microscopic particles. Based on h 2 2 ψ 2 ψ 2 ψ 2 2 2 2 Vψ Eψ y z 8 π m x 28π mMultiply by 2 and rearrangingh222 ψ ψ ψ 8π 2 m 2 (E V)ψ 0 x 2 y 2 z 2h (2.13)The above schrÖdinger waveequation does not contain time as a variableand is referred to as time independentSchrödinger wave equation. This equationcan be solved only for certain values ofE, the total energy. i.e. the energy of thesystem is quantised. The permitted totalenergy values are called eigen values andcorresponding wave functions representthe atomic orbitals.43
2.4.1 Main features of the quantummechanical model of atomand spin quantum number (s). WhenSchrödinger equation is solved for a wavefunction Ψ, the solution contains the firstthree quantum numbers n, l and m. Thefourth quantum number arises due to thespinning of the electron about its ownaxis. However, classical pictures of speciesspinning around themselves are incorrect.1. The energy of electrons in atoms isquantised2. The existence of quantized electronicenergy levels is a direct result of thewave like properties of electrons. Thesolutions of Schrödinger wave equationgives the allowed energy levels (orbits).Principal quantum number (n):This quantum number representsthe energy level in which electron revolvesaround the nucleus and is denoted by thesymbol 'n'.3. According to Heisenberg uncertaintyprinciple, the exact position andmomentum of an electron can not bedetermined with absolute accuracy. Asa consequence, quantum mechanicsintroduced the concept of orbital.Orbital is a three dimensional spacein which the probability of finding theelectron is maximum.4. The solution of SchrÖdinger waveequation for the allowed energies of anatom gives the wave function ψ, whichrepresents an atomic orbital. Thewave nature of electron present in anorbital can be well defined by the wavefunction ψ.1.The 'n' can have the values 1, 2, 3, n 1 represents K shell; n 2 representsL shell and n 3, 4, 5 represent theM, N, O shells, respectively.2.The maximum number of electronsthat can be accommodated in a givenshell is 2n2.3.'n' gives the energy of the electron,En ( 1312.8) 2 kJ mol 1n2andthedistance of the electron from the2nucleus is given by rn (0.529)n AZ5. The wave function ψ itself has no physicalmeaning. However, the probabilityof finding the electron in a smallvolume dxdydz around a point (x,y,z)is proportional to ψ(x,y,z) 2 dxdydz ψ(x,y,z) 2 is known as probabilitydensity and is always positive.Azimuthal Quantum number (l) orsubsidiary quantum number :1.It is represented by the letter 'l', andcan take integral values from zero ton-1, where n is the principal quantumnumber2.5 Quantum numbers2.The electron in an atom can becharacterised by a set of four quantumnumbers, namely principal quantumnumber (n), azimuthal quantum number(l), magnetic quantum number (m)Each l value represents a subshell(orbital). l 0, 1, 2, 3 and 4 representsthe s, p, d, f and g orbitals respectively.3.The maximum number of electronsthat can be accommodated in a givensubshell (orbital) is 2(2l 1).44
4.It is used to calculate the orbital angular momentum using the expressionAngular momentum l(l 1)h (2.14)2πMagnetic quantum number (ml):1.It is denoted by the letter 'ml'. It takes integral values ranging from -l to l through 0.i.e. if l 1; m -1, 0 and 12.Different values of m for a given l value, represent different orientation of orbitals inspace.3.The Zeeman Effect (the splitting of spectral lines in a magnetic field) provides theexperimental justification for this quantum number.4.The magnitude of the angular momentum is determined by the quantum number lwhile its direction is given by magnetic quantum number.Spin quantum number (ms):1.The spin quantum number represents the spin of the electron and is denoted by theletter 'ms'2.The electron in an atom revolves not only around the nucleus but also spins. It isusual to write this as electron spins about its own axis either in a clockwise directionor in anti-clockwise direction. The visualisation is not true. However spin is to beunderstood as representing a property that revealed itself in magnetic fields.3.Corresponding to the clockwise and anti-clockwise spinning of the electron, maximumtwo values are possible for this quantum number.4.The values of 'ms' is equal to -½ and ½Table 2.1 Quantum numbers and its significanceShellKLmaximumPrincipalnumber ofquantumelectronnumberin a shell(n)(orbital) 2n2122(1)2 22(2)2 8Azimuthalquantumnumber(l) 0,1.(n-1)MagneticMaximumquantumno. ofnumber (m)electron indifferenta orbitalpossible2 (2l 1) orientation oforbitalDesignation oforbitals in a givenshell02[2(0) 1] 201s0202s12[2(1) 1] 6-1, 0, 12py, 2pz, 2px45
ShellMNmaximumPrincipalnumber ofquantumelectronnumberin a shell(n)(orbital) 2n2342(3)2MagneticMaximumquantumno. ofnumber (m)electron indifferenta orbitalpossible2 (2l 1) orientation oforbitalAzimuthalquantumnumber(l) 0,1.(n-1)0203s16-1, 0, 13py, 3pz,3px,22[2(2) 1] 10-2, -1, 0, 1, 20204s16-1, 0, 14py, 4pz, 4px210-2, -1, 0, 1, 24dx2-y2, 4dxy, 4dz2,32[2(3) 1] 14-3, -2, -1, 0, 1, 1, 2, 3 182(4)2 32Designation oforbitals in a givenshell3dx2-y2, 3dyz, 3dz2,3dzx, 3dxy4dyz, 4dzxf y(3x2 y2),fz(x2 y2),fyz2, fz3, fxz2,fxyz, fx(x2 3y2),The labels on the orbitals, such as px, dz2, fxyz etc. are not associated with specific 'm' valuesEvaluate Yourself(where R(r) is called radial wavefunction, other two functions are calledangular wave functions)?3. How many orbitals are possible in the4th energy level? (n 4)As we know, the Ψ itself has nophysical meaning and the square ofthe wave function Ψ 2 is related to theprobability of finding the electrons withina given volume of space. Let us analyse how Ψ 2 varies with the distance from nucleus(radial distribution of the probability) andthe direction from the nucleus (angulardistribution of the probability).2.5.1 Shapes of atomic orbitals:The solution to SchrÖdingerequation gives the permitted energy valuescalled eigen values and the wave functionscorresponding to the eigen values arecalled atomic orbitals. The solution (Ψ)of the SchrÖdinger wave equation forone electron system like hydrogen canbe represented in the following form inspherical polar coordinates r, θ, φ as,Radial distribution function:Consider a single electron ofhydrogen atom in the ground state forwhich the quantum numbers are n 1 andΨ (r, θ, φ) R(r).f(θ).g(φ) ------ (2.15)46
The plot of 4πr2Ψ2 versus r is givenl 0. i.e. it occupies 1s orbital. The plotof R(r)2 versus r for 1s orbital is given inFigure 2.3below.0.50.60.41s4 r Y20.4R(r)20.30.20.20.1005101153579r (Å)r (a0)Figure. 2.3 Plot of R(r)2 versus r for 1sorbital of hydrogenFigure 2.5 Plot of 4πr2ψ2 versus r for 1sorbital of hydrogenThe graph shows that as the distancebetween the electron and the nucleusdecreases, the probability of finding theelectron increases. At r 0, the quantityR(r)2 is maximum i.e. The maximumvalue for Ψ 2 is at the nucleus. However,probability of finding the electron in agiven spherical shell around the nucleusis important. Let us consider the volume(dV) bounded by two spheres of radii rand r dr.The above plot shows that themaximum probability occurs at distanceof 0.52 Å from the nucleus. This is equalto the Bohr radius. It indicates that themaximum probability of finding theelectron around the nucleus is at thisdistance. However, there is a probabilityto find the electron at other distances also.The radial distribution function of 2s, 3s,3p and 3d orbitals of the hydrogen atomare represented as follows.2s orbital0.1rr dr0.054 r Y2Figure 2.4Volume of the sphere,dVdr V 2s2s4πr3304π(3r2)30510152025r (units of Bohr radius / Z)Figure 2.6 (a) - Plot of 4πr2ψ2 versus rfor 2s orbitals of hydrogendV 4πr2drΨ2dV 4πr2Ψ2dr ---------- (2.16)47
For 2s orbital, as the distance fromnucleus r increases, the probability densityfirst increases, reaches a small maximumfollowed by a sharp decrease to zero andthen increases to another maximum,after that decreases to zero. The regionwhere this probability density functionreduces to zero is called nodal surface or aradial node. In general, it has been foundthat ns-orbital has (n–1) nodes. In otherwords, number of radial nodes for 2sorbital is one, for 3s orbital it is two and soon. The plot of 4πr2ψ2 versus r for 3p and3d orbitals shows similar pattern but thenumber of radial nodes are equal to(n-l-1)(where n is principal quantum numberand l is azimuthal quantum number of theorbital).3s orbital0.13s4 r Y20.0500510152025r (units of Bohr radius / Z)Figure 2.6 (b) - Plot of 4πr2ψ2 versus rfor 3s orbitals of hydrogen3p3p orbital0.14 r Y20.050Angular distribution function:0510152025The variation of the probabilityof locating the electron on a spherewith nucleus at its centre depends onthe azimuthal quantum number of theorbital in which the electron is present.For 1s orbital, l 0 and m 0. f(θ) 1/ 2and g(φ) 1/ 2π. Therefore, the angulardistribution function is equal to 1/2 π.i.e. it is independent of the angle θ andφ. Hence, the probability of finding theelectron is independent of the directionfrom the nucleus. The shape of thes orbital is spherical as shown in thefigure 2.7r (units of Bohr radius / Z)Figure 2.6 (c) - Plot of 4πr2ψ2 versus rfor 3p orbitals of hydrogen3d3d orbital0.14 r Y20.0500510152025r (units of Bohr radius / Z)Figure 2.6 (d) - Plot of 4πr2ψ2 versus rfor 3d orbitals of hydrogen48
zzzzz1s1syyxyxNodal plane yzyxpypxxyxNodal plane xzpzNodal plane xy(b) Cartoon representations of 2p orbitalszz2syxz3syxFor ‘d’ orbital l 2 and thecorresponding m values are -2, -1, 0 1, 2. The shape of the d orbital lookslike a 'clover leaf '. The five m values giverise to five d orbitals namely dxy , dyz, dzx,dx2-y2 and dz2. The 3d orbitals contain twonodal planes as shown in Figure 2.9.yx3sFigure 2.8 Shapes of 2p orbitalsNodezNodesyzxFigure 2.7 Shapes of 1s, 2s and 3sorbitalsxFor p orbitals l 1 and thecorresponding m values are -1, 0 and 1. The angular distribution functionsare quite complex and are not discussedhere. The shape of the p orbital is shownin Figure 2.8. The three different m valuesindicates that there are three differentorientations possible for p orbitals. Theseorbitals are designated as px, py and pz andthe angular distribution for these orbitalsshows that the lobes are along the x, y andz axis respectively. As seen in the Figure2.8 the 2p orbitals have one nodal plane.y3dxyzyx3dyz49
zzzxxNodalplanesyyzyzxxy3dxzyzzzxxyyzxyxyFigure 2.10 shapes of f-orbitals3dx2-y2zEvaluate Yourself?4. Calculate the total number of angularnodes and radial nodes present in 3dand 4f orbitals.yx2.5.2 Energies of orbitalsIn hydrogen atom, only one electronis present. For such one electron system,the energy of the electron in the nth orbitis given by the expression3dz2Figure 2.9 shapes of d orbitalsEn For 'f ' orbital, l 3 and the m valuesare -3, -2,-1, 0, 1, 2, 3 correspondingto seven f orbitals fz3, fxz2, f yz2, fxyz, fz(x2 y2), fx(x2 3y2), f y(3x2 y2), which are shown inFigure 2.10. There are 3 nodal planes inthe f-orbitals.( 1312.8) 2 kJ mol 1n2From this equation, we know thatthe energy depends only on the value ofprincipal quantum number. As the n valueincreases the energy of the orbital alsoincreases. The energies of various orbitals50
will be in the following order:5p5d5f6s6p6d7s1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7sThe electron in the hydrogen atomoccupies the 1s orbital that has the lowestenergy. This state is called ground state.When this electron gains some energy, itmoves to the higher energy orbitals suchas 2s, 2p etc These states are calledexcited states.ln 5456756786787As we know there are three differentorientations in space that are possible fora p orbital. All the three p orbitals, namely,px, py and pz have same energies and arecalled degenerate orbitals. However, inthe presence of magnetic or electric fieldthe degeneracy is lost.In a multi-electron atom, inaddition to the electrostatic attractiveforce between the electron and nucleus,there exists a repulsive force among theelectrons. These two forces are operatingin the opposite direction. This resultsin the decrease in the nuclear force ofattraction on electron. The net chargeexperienced by the electron is calledeffective nuclear charge. The effectivenuclear charge depends on the shape ofthe orbitals and it decreases with increasein azimuthal quantum number l. Theorder of the effective nuclear charge feltby a electron in an orbital within the givenshell is s p d f. Greater the effectivenuclear charge, greater is the stability ofthe orbital. Hence, within a given energylevel, the energy of the orbitals are in thefollowing order. s p d f.Table 2.2 n l values of different orbitalsn1230120Based on the (n l) rule, theincreasing order of energies of orbitals isas follows:1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6dHowever, the above order is nottrue for atoms other than hydrogen(multi-electron systems). For suchsystems the Schrödinger equation is quitecomplex. For these systems the relativeorder of energies of various orbitals isgiven approximately by the (n l) rule. Itstates that, the lower the value of (n l)for an orbital, the lower is its energy. Iftwo orbitals have the same value of (n l), the orbital with lower value of n willhave the lower energy. Using this rule theorder of energies of various orbitals can beexpressed as follows.Orbital555666751
Hund's rule. These rules are describedbelow.7p 6d7s 5f2.6.1 Aufbau principle:4fThe word Aufbau in German means'building up'. In the ground state of theatoms, the orbitals are filled in the orderof their increasing energies. That is theelectrons first occupy the lowest energyorbital available to them.Energy, E5p 4d5s4p3d4s3p3sOnce the lower energy orbitals arecompletely filled, then the electrons enterthe next higher energy orbitals. The orderof filling of various orbitals as per theAufbau principle is given in the figure 2.12which is in accordance with (n l) rule.2p2sn 11sn 2Figure. 2.11 Energy levels of atomicorbitalsEvaluate Yourselfn 4n 3The energies of same orbitaldecrease with an increase in the atomicnumber. For example, the energy of the 2sorbital of hydrogen atom is greater thanthat of 2s orbital of lithium and that oflithium is greater than that of sodium andso on, that is, E2s(H) E2s(Li) E2s(Na) E2s(K).?2s2p3s3p 3d4s4p 4d4f5s 5p5d5f6s6p6d7s7pn 75. Energy of an electron in hydrogenatom in ground state is -13.6 eV. Whatis the energy of the electron in thesecond excited state?1sn 56s5dn 66pFigure. 2.12 Aufbau principle2.6 Filling of orbitals:2.6.2 Pauli Exclusion Principle :In an atom, the electrons are filledin various orbitals according to aufbauprinciple, Pauli exclusion principle andPauli formulated the exclusionprinciple which states that "No twoelectrons in an atom can have the same set52
of values of all four quantum numbers."It means that, each electron must haveunique values for the four quantumnumbers (n, l, m and s).the electrons are filled in various orbitals.But the rule does not deal with the fillingof electrons in the degenerate orbitals (i.e.orbitals having same energy) such as px,py and pz. In what order these orbitals tobe filled? The answer is provided by theHund's rule of maximum multiplicity.It states that electron pairing in thedegenerate orbitals does not take placeuntil all the available orbitals contains oneelectron each.For the lone electron present inhydrogen atom, the four quantum numbersare: n 1; l 0; m 0 and s ½. Forthe two electrons present in helium, oneelectron has the quantum numbers sameas the electron of hydrogen atom, n 1,l 0, m 0 and s ½. For other electron,the fourth quantum number is differenti.e., n 1, l 0, m 0 and s –½.We know that there are three porbitals, five d orbitals and seven f orbitals.According to this rule, pairing of electronsin these orbitals starts only when the 4th,6th and 8th electron enters the p, d and forbitals respectively.As we know that the spin quantumnumber can have only two values ½and –½, only two electrons can beaccommodated in a given orbital inaccordance with Pauli exclusion principle.Let us understand this by writing allthe four quantum numbers for the eightelectron in L shell.For example, consider the carbonatom which has six electrons. Accordingto Aufbau principle, the electronicconfiguration is 1s2, 2s2, 2p2It can be represented as below,Table 2.3 Quantum numbers ofelectrons in L shellຯElectronnlms1st200 ½2nd200-½3rd21-1 ½4th21-1-½5th210 ½6th210-½7th21 1 ½8th21 1-½1s2ຯ2s2։2px1։2py12pz0In this case, in order to minimisethe electron-electron repulsion, the sixthelectron enters the unoccupied 2py orbitalas per Hunds rule. i.e. it does not get pairedwith the fifth electron already present inthe 2Px orbital.Evaluate Yourself?6. How many unpaired electrons are2.6.3 Hund'smultiplicityruleofpresent in the ground state of Fe3 maximum(z 26), Mn2 (z 25) and argon (z 18)?The Aufbau principle describes how53
Table 2. 4 Electronic configuration and2.6.4 Electronic configuration of atomsorbital diagrams for first 10 elementsThe distribution of electrons intoby applying the aufbau principle, PauliElectronicConfigurationelectronic configuration. It can be writtenElementvarious orbitals of an atom is called itsH11s1։1s1He21s2։ 1s2Li31s2 2s1։ 1s2։2s1Be41s2 2s2։ 1s2։ 2s2B51s2 2s2 2p1։ 1s2։ 2s2։2px1 2py 2pzC61s2 2s2 2p2։ 1s2։ 2s2։։2px1 2py1 2pzN71s2 2s2 2p3O81s2 2s2 2p4F91s2 2s2 2p5exclusion principle and Hund's rule. Theelectronic configuration is written asnlx , where n represents the principle ofOrbital diagramquantum number, 'l' represents the letterdesignation of the orbital [s(l 0), p (l 1),d(l 2) and f(l 3)] and 'x' represents thenumber of electron present in that orbital.Let us consider the hydrogenatom which has only one electron and itoccupies the lowest energy orbital i.e. 1saccording to aufbau principle. In this casen 1; l s; x 1.Hence the electronic configurationis1s1.(spoken as one-ess-one).Theorbitaldiagramforthisconfiguration is,։ s1։ ։ ։։։1s22s22px12py12pz1։։։ ։ 1s22s22px22py։ ։ ։ ։ 1s22s22px22py։ ։ ։ ։ ։ 1s22s22py22pz2The electronic configuration andorbital diagram for the elements uptoNe10 1s2 2s2 2p6atomic number 10 are given below :։ 2px2122pz1։2pz1The actual electronic configurationof some elements such as chromium andcopper slightly differ from the expectedelectronic configuration in accordancewith the Aufbau principle.54
For chromium - 24degenerate orbitals. This can be explainedon the basis of symmetry and exchangeenergy. For example chromium has
chemistry, physics, maths and botany. He was not satisfied with the quantum condition in Bohr's orbit theory and believed that atomic spectra should really be determined by some kind of eigenvalue problem and proposed the wave equat
Chemistry ORU CH 210 Organic Chemistry I CHE 211 1,3 Chemistry OSU-OKC CH 210 Organic Chemistry I CHEM 2055 1,3,5 Chemistry OU CH 210 Organic Chemistry I CHEM 3064 1 Chemistry RCC CH 210 Organic Chemistry I CHEM 2115 1,3,5 Chemistry RSC CH 210 Organic Chemistry I CHEM 2103 1,3 Chemistry RSC CH 210 Organic Chemistry I CHEM 2112 1,3
Distribution of Free Textbook Programme Untouchability is a sin Untouchability is a crime Untouchability is a inhuman DIRECTORATE OF TECHNICAL EDUCATION GOVERNMENT OF TAMILNADU A Publication under Government of Tamilnadu (NOT FOR SALE) ii Government of Tamilnadu First Edition – 2015
Physical chemistry: Equilibria Physical chemistry: Reaction kinetics Inorganic chemistry: The Periodic Table: chemical periodicity Inorganic chemistry: Group 2 Inorganic chemistry: Group 17 Inorganic chemistry: An introduction to the chemistry of transition elements Inorganic chemistry: Nitrogen and sulfur Organic chemistry: Introductory topics
Accelerated Chemistry I and Accelerated Chemistry Lab I and Accelerated Chemistry II and Accelerated Chemistry Lab II (preferred sequence) CHEM 102 & CHEM 103 & CHEM 104 & CHEM 105 General Chemistry I and General Chemistry Lab I and General Chemistry II and General Chemistry Lab II (with advisor approval) Organic chemistry, select from: 9-10
CHEM 0350 Organic Chemistry 1 CHEM 0360 Organic Chemistry 1 CHEM 0500 Inorganic Chemistry 1 CHEM 1140 Physical Chemistry: Quantum Chemistry 1 1 . Chemistry at Brown equivalent or greater in scope and scale to work the studen
Chemistry is the science that describes matter, its properties, the changes it undergoes, and the energy changes that accompany those processes. Inorganic chemistry Organic chemistry Physical chemistry Biochemistry Applied Chemistry: Analytical chemistry, Pharmaceutical Chemistry, . Istv an Szalai (E otv os University) Lecture 1 6 / 45
Chemistry of Cycloalkanes 13. Chemistry of Alkyl halides 14. Alcohols 15. Chemistry of Ethers and Epoxides 16. Chemistry of Benzene and Aromaticity 17. Chemistry of Aryl Halides 18. Aromatic Sulphonic Acids 19. Chemistry of Aldehydes and Ketones 20. Carboxylic Acids 21. Chemistry of Carboxylic Acid Derivativ
ADVANCED DIPLOMA Diploma in Chemistry 60% in Analytical Chemistry 3 Theory & Practical, Chemical Quality Assurance, Mathematics 2 Chemical Industrial 1 or S5 Subjects and Chemistry project II. Semester 1 Analytical Chemistry IV Physical Chemistry IV Research Methodology in Chemistry Semester 2 Inorganic Chemistry IV Organic Chemistry IV .
