ATOMIC STRUCTURE

FUNDAMENTAL PARTICLES

An atom consists of three fundamental particles namely (i)Electron, (ii)Proton and (iii)Neutron.We usually define atom as the smallest unit of an element which cannot be further subdivided.But this is not entirely correct. Atom can be divided into positive ions by removing electrons.But the ions thus formed lose the property of the element. Hydrogen atom consists of oneelectron and one proton and we can remove the electron from the hydrogen atom to producea proton, which we call hydrogen ion. But hydrogen ion exhibit properties which are differentfrom hydrogen atom. So atom is the smallest unit of an element which retains thecharacteristic properties of the element. Can you guess how small an atom is!!! It isunbelievably small in size and the radius of an atom is roughly 10-8 cm. We were not ableto see an atom by any powerful microscope till the last part of twentieth century. Butrecently a very powerful microscope called scanning tunnelling microscope (STM) hasbeen invented by which we are able to see an atom of an element. Let us know the importantproperties of the fundamental particles.Electron:

It is a negatively charged particle having almost negligible mass.Charge(q) = - 1.602 X 10-19 coulomb(This is regarded as one unit of negative charge

i.e -1)Mass(m)=9.1X10-28gm=9.1X10-31 kg

Proton:It is a positively charged particle having mass nearly 1840times heavier than the

mass of an electron. The magnitude of charge is same as that of an electron.Charge(q)=+ 1.602 X 10-19 coulomb(This is regarded as one unit of positive charge

i.e +1)Mass(m)= 1.66X10-24gm = 1.66X10-27kg

Neutron:Neutron is a neutral particle having no charge. But its mass is nearly same as that

of a proton.Charge(q) = 0; Mass(m)= 1.66X10-24gm = 1.66X10-27kg

SAQ 1:(i) How many times a neutron is heavier than an electron?(ii) Find the charge of three electrons in coulomb.(iii) Find the charge of two units of positive charge.(iv) A certain number of protons carried a total charge of +1.602X10-18 coulomb. How

many protons were there?(v) Find the magnitude of charge of Avogadro's number of electrons.

RUTHERFORD'S MODEL OF ATOM

After many unsuccessful attempts by several scientists to explain how electrons, protons andneutrons are arranged in an atom, Rutherford gave an elegant and convincing model of atom.He gave his model after conducting his famous gold foil experiment. First let us knowRutherford's model of atom and after that we know his experiment which led him to suggestthe model.Atom consists of two parts: (i)Nucleus (ii)Extra-nuclear part

Nucleus:Nucleus is the extremely minute spherical body situated at the centre of the atom. Nucleuscarries negligible volume compared to the total volume of the atom. It carries all protons andall neutrons. This means that it carries all the positive charge and almost all the mass of theatom because electrons carry negligible mass. Its radius was calculated by Rutherford to be2.72 X 10-12cm. In other words, almost the whole mass of the atom is concentrated in a verytiny spot called nucleus.Extra-nuclear Part:The part of the atom other than nucleus is called the extra-nuclear part. Most of the spaceoccupied by this part is empty or vacant. Electrons are situated far away from the nucleusand revolve continuously round the nucleus just like planets revolve round the sun in the solarsystem. Therefore electrons were designated as planetary electrons. The radius of an atomi.e the size of the extra-nuclear part (note that the extranuclear part embeds the nucleusinside it) is approximately 10-8 cm. Thus nucleus is nearly 10,000 times smaller in size thanthe atom.So nucleus of an atom is responsible for the mass of the atom while the extra-nuclearpart is responsible for the volume of the atom.

Rutherford's Gold Foil Experiment:This is the experiment which led Rutherford to discover nucleus and give the model of atomthat we have discussed above.A radioactive metal polonium was taken in a lead box with a narrow hole. Polonium is anemitter of alpha particles. Alpha particles are nothing but the dipositive helium ions(He++)which are spontaneously emitted from radioactive elements such as polonium. A thin beamof very high speed alpha particles were allowed to come out from the lead box through thefine hole and that is then allowed to strike on a very thin gold foil of 0.00004cm thickness.A movable fluorescent screen (ZnS) was placed behind the gold foil to monitor the path ofthe alpha rays after the beam strikes on the gold foil. Note that when alpha particles strikeon a fluorescent screen, the screen produces bright light at the spot where the particlesstrike. The following observations were made.

(i) Most of the alpha particles penetrated straight through the gold foil without anydeviation in its direction of motion. The thin beam of the particles almost fell straight ontothe ZnS screen where it produced a bright glow. This was not a new observation byRutherford as it was known before that radioactive rays can penetrate through thin metallicfoils and sheets.

(ii) The remarkable observation was that a few alpha particles were deviated from theiroriginal direction of motion and deflected at different angles. This was known by moving theZnS screen to different positions and verifying whether any glow appears or not. Faint glowappeared at different angles in the ZnS screen which confirmed the above observation.(iii) The most remarkable observation was that a very few alpha particles (about 1 inabout 20,000) rebounded back after hitting the gold foil. This was known by moving the ZnSscreen to the other side(same side as the polonium source and placing it in between thepolonium and gold foil) and finding a very faint glow on the spot where the rebounded alphaparticles were hitting.

Lead Box

Polonium

rays

thin gold foil

rays

rays

rays

rays

rays

rays

rays

rays

rays

rays

(Rutherford's Gold Foil Experiment)

******SINGH: incomplete

Explanation:(i) Since most of the alpha particles passed through the gold foil straight without sufferingany deviation in their path, most of the space occupied by the atom must be empty. Mostalpha particles passed through the empty space undeviated.(ii) Since a few alpha particles were deviated at different angles, there must be a minutebut very heavy body at the centre of the atom which carries all the positive charges andalmost all the mass of the atom. Those few alpha particles which pass near this heavy +velycharged body, suffer repulsion and deviate at different angles with respect to their originaldirection. You already know that alpha particles are dipositive helium ions for which theysuffer repulsion when they pass through the heavy +vely charged body(nucleus of gold atom)located at the centre of the atom. This heavy positively charged body was named asnucleus.(iii) Those very few alpha particles which directly hit on the heavy positively chargedbody(nucleus) rebounded back without passing through the gold foil. Thus the presence ofthe nucleus is further confirmed by this observation.

Since nucleus carries all the positive charge at a very small space at the centre, the electronsmust be present farther away from the nucleus and must be revolving round the nucleus.Thus there must be a large gap or vacant space between the minute nucleus and theelectrons through which most of the alpha particles could pass through the gold foil withoutsuffering any deviation. Note that, at the time of the discovery of nucleus, the fundamental

particle, neutron was not discovered. Long after that neutron was discovered by Chadwickand was believed to be present in the nucleus alongwith the protons.SAQ 2:(i) Why is polonium taken in a lead box?(ii) Why can the alpha particles not dislodge and remove the nucleus from the atom byhitting onto it rather themselves get rebounded?(iii) Why gold was chosen for the experiment?

ATOMIC SPECTROSCOPYWhat is spectrum? When a beam of sunlight is passed through a prism and allowed to fallon a screen, we find seven colours VIBGYOR(V-violet, I-Indigo, B-blue, G-green, Y-yellow,O-orange, R-range). This is called a spectrum. Spectrum is the pattern of colours viewed ona screen when a mixture of colours (white sunlight) is passed through a prism. The whitelight is dispersed in the prism and get separated according to colour. We shall know moreabout spectrum later and now let us study more about different types of radiation like light,heat etc.

Light is a form of radiation coming out from sun or a glowing electric bulb, tube light etc.and are called visible radiation. The visible radiation consist of infinite number of individualradiation or waves. Unlike light radiation which are visible to our eyes there are many moreradiations which are invisible. Some of them also come from sun and others are producedby other sources. Radiations like cosmic rays, ultraviolet rays (UV) come from sun. Radiationslike gamma rays are produced by radioactive elements like uranium, polonium etc. Do youknow that the deadly disease cancer is treated by allowing the gamma rays produced fromthe radioactive isotope Co60 to strike on the cancer affected area? This is called radiationtherapy. Radiations like x-rays which are used to take photographs of our bones are preparedwhen strong electron beam(cathode rays) are allowed to strike on metals like copper. Similarlythe heat is another form of radiation called infrared radiation (IR radiation). There are a fewmore radiations also which are invisible like the radiations we discussed above. Excepting thelight radiation (visible radiation), all other radiations are invisible.

Electromagnetic Radiations:All the radiations that we have discussed above are called electromagnetic radiations.

All of them are forms of energy of different magnitudes which travel in the form of transversewaves. Have you seen a transverse wave? Yes, when you throw a stone onto a still pond,you find ripples originate from the point at which the stone strikes the water and the wavespropagate from this centre in all directions to large distances. This is a transverse wave. Alltransverse waves are like this. Electromagnetic waves are simply energy waves and do notcarry any material particles like the water waves. But like water waves, they are transversewaves. A wave has three parameters to describe it. These are called wavelength (l),

frequency (n) and wave-number( ). Only one of the parameters is required to identifya particular wave and distinguish it from others. The other two parameters are automaticallyknown. Just like you have three names, one good name(say Sujata) and two nick names(sayMuni and Mama). All the three names refer to same person, i.e you. In the similar mannerif wavelength of a wave is known, then the other two parameters namely frequency and

wavenumber are automatically known. If frequency is known, then the remaining twoparameters i.e wavelength and wavenumber are known. Similarly if wavenumber is known,wavelength and frequency are also known. Let us first know what are these three parameters.

Crest and Trough: The top(maximum) position of a wave is called crest and thebottom(minimum) position is called trough.

(i) Wave length(λλλλλ): It is the distance between two adjacent crests or two adjacenttroughs of a wave. It is expressed in distance units i.e cm(CGS) and m(MKS). But usuallythe wavelengths of electromagnetic waves are so small that they are expressed by anothersmall unit called angstrom represented by the symbol Å.

1 Å = 10-8cm = 10-10m

Another smaller unit of wavelength which is also used is nanometer(nm).1nm= 10-9m = 10-7cm

SAQ 3: The wavelength of a wave is 0.001cm. Express in Å as well as nm units.SAQ 4: The wavelength of a wave is 1000Å. Express it in cm and m units.

(ii) Frequency(ννννν): Look to the above diagram. A full wave or cycle which consists ofone crest and one trough. If you stand on the orgiin of the above diagram and count howmany full waves are passing through that point in one second, it is called the frequency ofthat wave. The number of full waves passing through a particular point per second iscalled frequency. The unit of frequency is waves/sec or cycles/sec (Hertz) or more commonly1/sec or sec-1. Note that as a matter of convention in chemistry we shall use sec-1 as theunit of frequency rather than Hertz.

(iii) Wave-number( ): It is the reciprocal of wavelength. = 1

The unit of wave-number is reciprocal of distance unit i.e 1/cm or cm-1 and 1/m or m-1. Notethat the unit cannot be 1/Å or Å-1. So if the wavelength is given in Å unit, it is first convertedto cm or m unit and then its reciprocal is found out to get wavenumber. Look to the followingexample.

Example: Calculate the wave-number of a wave which has a wavelength of 3000Å.Solution: Wavelength= λ= 3000Å = 3000 X 10-8cm (Since 1Å = 10-8cm)

Wave-number= = 1

= 13000 X 10 cm-8 =

10 3

3000 X 10 X 10 cm3-8 = 3.33 X 105

cm-1

So we had to first change the wavelength from angstrom unit to cm unit and then we foundthe wavenumber in cm-1 unit.Example : Calculate the wavelength in angstrom unit if its wave-number 4000cm-

1.

Solution: Since = 1

, ⇒ = 1 i.e wavelength is reciprocal of wavenumber.

λ = 1

4000=4000 cm-1

1cm

To covert cm to angstrom unit, you have to multiply by a factor 108, since 1 Å = 10-8cm ⇒1cm=108 Å.

0

A4

2.5 X 10=4000

4X 1010,000=

0

A4000

810cm1

4000=4000 cm-1

1 = =

(Note that students often get confused in the two closely related symbols(ν and )givento frequency and wave-number respectively. They are two different parameters althoughthe symbols are very closely related. A bar over ν converts frequency into wave number.The students are advised to to be careful about their use.)

ELECTROMAGNETIC RADIATIONS OR WAVES:Already we have discussed before that the radiations like ultraviolet rays(UV) coming fromsun, gamma(γ) rays coming from radioactive elements, infrared(IR) radiations present in heatradiations etc. are called electromagnetic waves. Each kind of wave is made up of anelectric field and a magnetic field and hence is called electromagnetic wave. These aresimply energy radiations and execute wave nature. These are not material particles and donot carry any mass. They do not need any medium to travel unlike sound waves which needair medium to travel from one place to the other. It means that these waves can pass througheven vacuum. The sun rays carrying visible radiations, ultraviolet radiations etc. come throughlarge distances in vacuum and then reach on the earth's atmosphere. A medium kept on theway of electromagnetic radiations rather obstructs the movement of these radiations. Forexample, light waves can pass through a transparent medium but will be obstructed andreflected by a opaque medium. Various electromagnetic waves are obstructed in differentways by different media which we shall not discuss here.Electromagnetic Spectrum:Different electromagnetic radiations are arranged in a systematic order of increasingwavelengths. This systematic arrangement of all the forms of electromagnetic radiations arecalled electromagnetic spectrum. The following table gives all the electromagnetic radiationsand their wavelength ranges. Note that each type of radiation has a range of wavelengths.For example, all the radiations whose wavelengths lie between 4000Å to 8000Å belong tothe category of visible radiation (light radiation). In fact, the visible radiations that are presentin the sun light contains infinite number of electromagnetic waves having wavelengths inbetween 3800 to 7600Å.

Name of Radiation Wavelength Range in Å

Cosmic Rays 0.001 - 0.01Gamma Rays(g) 0.01 - 0.1x-rays 0.1 - 150ultraviolet(UV) Rays 150 - 3800Visible Rays 3800 - 7600Infrared(IR) Rays 7600- 6 X 106

Microwave 6 X 106 - 6 X 109

Radiowaves 6 X 109 - 3 X 1013

SAQ 5: Indicate to which type of electromagnetic radiations the following wavesbelong. The wavelengths of the radiations are given below.

(i)3000Å (ii) 10Å (iii)104 Å (iv)0.05Å (v)120Å(vi)108Å (vii)0.005Å (viiii)6X 103Å (ix)1011Å

RELATIONSHIP BETWEEN WAVELENGTH AND FREQUENCY:

In the above diagram you find electromagnetic waves having smaller and longer wavelengths.One can well understand that if the wavelength of a wave is smaller, then larger number ofsuch waves can pass through a point in one second and hence its frequency is more.Similarly, if the wavelength of a wave is large, fewer number of such waves can passthrough a point in one second, hence its frequency is small. Thus we can say that wavelengthis inversely proportional to frequency. If wavelength of a wave is small, its frequency is largeand vice versa. Mathematically,

ν ∝ 1

⇒ 1= k (where k is the proportionality constant)

Let us find the unit of k first. From the above relation,k= νλ ⇒ (1/sec) X cm = cm/sec

So the unit of k becomes the unit of velocity. This k is designated as c for electromagneticwaves, which is conventionally called the velocity of light. As a matter of fact this is aconstant for all electromagnetic radiations such as ultraviolet, infrared, x-rays etc., not merelyit is for velocity of light or visible radiations. It is wrong to call it as velocity of light becauseit is not only the velocity of light but also the velocity of very short wavelength radiations suchas gamma rays, x-rays and also the velocity of very long wavelength radiations such asinfrared, microwave and radiowaves. So we write again the mathematical relations,

1= c ⇒ = c ⇒ = c

c= 3 X 108cm/sec = 3 x 1010 cm/sec

From the above relations we find that the product of wavelength(λ) and frequency(ν) of allelectromagnetic waves is constant and is equal to c, which we usually call the velocity of

light. Since wave-number( ) is reciprocal of wavelength(λ), we find the relation between

frequency(ν) and wavenumber( ). Frequency(ν) is obtained by multiplying wavenumber( )with velocity of light(c). Go through the following numerical examples.Example: The wavelength of an ultraviolet radiation is 2000Å. Calculate itsfrequency and wave-number.

Solution: λ = 2000Å= 2000 X 10-8cm. We know that 1= c

So 1 = 3 X 10

8cm/secX

2000 X 10-8 =

2000 X 10-83000 X 10

5

= 1.5 X 1013

sec-1

So the frequency of the wave is 1.5 X 1013 sec-1 or Hertz.

Again = 1

⇒ cm4X 10-82000 X 10

410=

cm-82000 X 10= 1

= 5 X 104cm

-1

So the wave-number of the radiation is 5 X 104 cm-1.Example: The frequency of a certain electromagnetic wave is 1017 sec-1. Find itswavelength in angstrom unit. Also find its wave number. Indicate to which region of theelectromagnetic spectrum does it belong.Solution: ν = 1017 sec-1. So λ = c/ν = (3 X 1010)/1017

= 3 X 10-7 cm = 3 X 10-7 X 108 Å = 30 Å.This wave belongs to x-rays of the electromagnetic spectrum(Refer table).

Wave-number = = 1

= 1/(3 X 10-7cm) = 3.33 X 106 cm-1.

Note that, we used the cm unit (not angstrom unit) for finding wave number.Example: The wave-number of a certain electromagnetic spectrum is 4000 cm-1.Find its wavelength in angstrom unit. Also find its frequency. To which region of theelectromagnetic spectrum does this wave belong.

Solution: λ = 1/ = 1/4000 cm = (1/4000) X 108 Å = 2.5 X 104 .This wave falls in the infrared(IR) region of the electromagnetic spectrum.

ν = c X = 3 X 1010 X 4000 = 1.2 X 1014 sec-1.SAQ 6:(i) What is the velocity of x-rays? How it is related with the velocity of infrared rays.(ii) A certain electromagnetic radiation has a wavelength of 500 nm. What is its frequency

and wavenumber? To which region does it belong?(iii) Arrange the following radiations in the order of increasing wavelengths.

microwave, x-rays, uv rays, cosmic rays and radiowaves.(iv) Arrange all the electromagnetic radiations in the decreasing order of frequencies.(v) Wavelength is inversely proportional to frequency. Indicate in which way thewavenumber varies with frequency? Arrange the following in the increasing orderof wave number.

visible rays, ultraviolet rays, gamma rays, microwaves.(vi) Give one application of x-rays, gamma rays, microwave and radiowaves.

(vii) How many different waves are present in sunlight?(viii) Where can we get ultraviolet radiations? What is its harmful effect?PLANCK'S QUANTUM THEORY

Now we shall discuss about energy of the electromagnetic radiations. Planck suggested thatthe electromagnetic radiations do not move continuously as we expect of a transverse wave.They travel in the form of discontinuous energy packets or bundles. Imagine that thereis a heap of bricks in one place of a field and you want to transfer all the bricks to a distantplace of the field. You can do so with the help of a large number of your friends. You shallmake a queue of all the persons starting from the heap of bricks and ending at the placewhere the bricks are to be transferred. The first person takes one brick from the heap andtransfers it to the second person and the second person transfers to the third and so on tillit is kept at the other end to make another heap. If this process of transfer of bricks is carriedout very rapidly for some time, it will appear from a distance that the movement of bricksis continuous, while in reality it is a discontinuous process. Moreover if the transfer of bricksis done in a wavy manner and not in a linear manner, then the movement of bricks will appearas a wave motion. Such is the case with all the electromagnetic radiations. These radiationsmove in the form of discontinuous energy packets called quanta (singular-quantum)orphotons. Although the term photon was originally used to represent a packet of lightenergy(quantum), it is now conventionally used to denote all types of electromagnetic radiations.Imagine each energy packet is like a brick of the example described before. These packetsor quanta were believed to be hypothetical particles having no rest mass. Thus the dualtheory or corpuscular theory of light was put forth by Einstein which says that light and infact all electromagnetic radiations can be considered both as waves as well as particles(corpuscles). The particles in this case are not real particles but are the massless photonsor quanta. For some phenomena of light, its wave nature is taken into consideration and forsome other phenomena the particle nature is considered. These photons are emitted from itssource and propagated in space so fast that the discontinuity between successive photonscannot be detected.

(Rapid Wave Like Movement of Bricks)

Schematic View of Dual Theory of Light

Bricks light photons

Features of Planck's quantium theory:• Substances radiate or absorb energy discontinuously in the form of small packets orbundles of energy.• Each packet of energy(small or big) is called quantum or photon.• The energy of each packet of electromagnetic radiation(quantum) is directlyproportional to the frequency of that wave(ννννν). Mathematically

E ∝ ν ⇒ E = h ν(where h= the proportionality constant called Planck's constant whose value ish = 6.627 X 10-27 erg.sec or 6.627 X 10-34 joule.sec)But we know the relationship between frequency(ν), wavelength(λ) and wave-

number( ) before. 1= c and = c . So substituting these values in the energy

equation, we get.

E = hc

= hch =

• A body can radiate or absorb energy in whole number multiples of a quantum hν,2hν, 3hν………..nhν. Where ‘n’ is the positive integer.• Neils Bohr used this theory to explain the structure of atom.Let us take up some numerical problems to calculate the value of an energy packet(photon).Example: Calculate the energy of a light photon having frequency 1014 sec-1(Hertz).Solution: According to Planck's quantum theory, E = hν =6.627 X 10-27erg.secX1014sec-1

= 6.627 X 10-13 erg (CGS).In MKS unit, E = hν = 6.627 X 10-34 joule.sec X 1014 sec-1= 6.627 X 10-20joule.Example: Calculate the energy of an ultraviolet photon whose wavelength is 300nm.Solution: l = 300 nm = 300 X 10-7cm. We know that

hc=E , So E = 6.627 X 10 erg.sec X 3 X 10 cm/sec

300 X 10 cm-7

-27 10

= 6.627 X 10-12 erg

Example: Calculate the energy of a photon having wave-number 107 cm-1.

Solution: We know that hcE = ,So E = 6.627 X 10-27 erg.sec X 3 X 1010 cm/secX 107 cm-1 = 19.88 X 10-10 erg

= 1.988 X 10-9 erg.Thus we can calculate the energy of a single photon or quantum(packet) of any electromagneticradiation, if we know any one of the three parameters such as wavelength, frequency orwave-number.SAQ 7:(i) Energy of a photon of electromagnetic radiation is inversely proportional to thewavelength. Is the statement true? Indicate their relation.(ii) Between x-ray photon and IR photon, which is more energetic and why?(iii) In the electromagnetic spectrum, which radiation is most energetic and which least?

Indicate its cause.(iv) How does energy of a photon vary with wave-number?(v) "A radioactive element emits gamma rays continuously i.e the energy is emittedcontinuously without any break". Is the statement true? If not correctly justify.(vi) Why do we not observe the discontinuity of any electromagnetic radiations: sayvisible radiations.(vii) Calculate the energy of light photon having wavelength of 7000Å.(viii) What will be the energy of a photon having frequency of 1015 sec-1?(ix) Calculate the energy of a microwave photon having wavenumber 1 cm-1.

(x) The energy of a photon is 1.98 X 10-18 erg. Calculate its wavelength, frequency andwavenumber. To which region of the electromagnetic spectrum does it belong.

Atomic SpectroscopyWhen a mixture of radiations of different wavelengths is allowed to pass through a prism,the individual radiation gets separated from lower wavelength to higher wavelength frombottom to top respectively. When white beam of sunlight is passed through a prism weobserve seven colours starting from violet at the lowermost part followed by indigo, blue,green, yellow, orange in the intermediate part and red at the uppermost part. Violet colourhas the smallest wavelength and red colour has the greatest wavelength and the wavelengthregularly increases from violet to red. Spectrum is a pattern of different colours (VIBGYOR)that we see on a screen or on photographic plate when white light is allowed to passthrough a prism. A schematic diagram of the solar spectrum has been shown in the figurebelow. Note that this is not the actual spectrum. The white sunlight has been shown by athick black line and the band of seven colours has been shown by a black patch. The actualcoloured spectrum has been shown in the next page. Have a look now.Spectrum is not only restricted to visible spectrum or the solar spectrum that we have justnow discussed. It is extended to all the electromagnetic radiations such as UV spectrum,IR spectrum, Microwave spectrum and so on. When a beam of ultraviolet rays will passthrough a prism, similar phenomenon occurs like the visible spectrum, the mixture getsseparated and disperse into a band of radiations with increasing wavelength from 150 to3800Å as we move from bottom to top. The difference is that there is no colour in this caselike that we found in visible spectrum, so we cannot observe them in our eyes on a screen.The very large number(in fact infinite) of UV radiations in the range of 150 to 3800Å thatare present in the beam get split and disperse into a wide band which can only be seen ona photographic plate. This is the UV spectrum. Same is the case with IR spectrum and infact with the spectra of all electromagnetic radiations. So now what we learnt about aspectrum? Spectrum is a regular pattern or arrangement of radiations of increasing wavelengthfrom lower part to higher part that is observed either on a screen or photographic plate.Excepting the visible(light)spectrum, no other spectrum can be observed on a screen. All thespetra(singular: spectrum) can be recorded on a photographic plate. This is becausephotographic plate is affected by all types of electromagnetic radiations.

(white light)

(prism)

(photographic plate or screen)

(spectrum: VIBGYOR)

(A Schematic View of a Spectrum in Black and White)

SAQ 8:(i) What colour will you get if you mix all the seven colours, VIBGYOR?(ii) Which colour produces photons of the greatest energy which of least energy?(iii) Green colour in the VIBGYOR spectrum refers to a single wavelength. Is thisstatement true? Give reason.(iv) Why violet light appears at the lowermost part of the visible spectrum?(v) UV radiations fall in the range of 3800 Å to 7600Å of the electromagnetic

spectrum.Which other region of the electromagnetic spectrum has wavelength greaterthan 3800Å(and just above it) and which region has wavelength smaller than 150Å(just below it)?

(vi) Will you get different colours in ultraviolet or infrared spectra as we find in visiblespectrum?

TYPES OF SPECTRA:(i) Discontinuous or Line Spectrum:Hold a crystal of sodium chloride (common salt) in a flame of a bunsen burner with the helpof a holder, what will you observe? You will find that the flame is emitting golden yellow light which is characteristic of sodium. If you conduct this experiment in a dark room and allowthe yellow light emitted from the flame to pass through a prism and put a white screen toview the spectrum, you will find a few discrete (well separated) lines out of which two areprominent(intense) in the yellow region of the spectrum. These yellow lines are called D

1 and

D2 lines of sodium at 5890 and 5896Å respectively. This type of spectrum is called a

discontinuous spectrum or a line spectrum. Note that sodium emits these two radiationsat 5890 and 5896Å and this is the identity mark of sodium. In other words if you find thetwo lines in a spectrum, you will be sure that it belongs to sodium and none else. From thisyou can identify the unknown element by looking to its spectrum.If a little KCl is held in a bunsen flame, you will observe it emitting violet (lilac) colour. Ifyou pass the violet light through a prism and the spectrum is recorded on a screen, you willfind a few discrete lines unique to potassium. Like Na and K, calcium will emit brick redcolour and barium will emit pea green colour. Each of the colour on analysis through a prismgives a few lines well separated from each other in the spectrum. These are the examplesof discontinuous spectra. Spectrum of each element is unique and gives fixed number of lineswhich are different from the spectra of another element. Therefore spectrum is calledthe fingerprint of the element. Just like the fingerprint of one person is unique and isdifferent from any other person, in the same manner the spectrum of each element is unique

and different from others. From the atomic spectrum of an element it is possible to detectthe element if it is not known to us.

(ii) Continuous or Band Spectrum:If you compare the discontinuous spectrum of sodium that we just learnt with the spectrumof sunlight (VIBGYOR) also called the solar spectrum, what difference you mark? Insunlight spectrum you will find a continuous band of colours and there is no gap from startto finish. The violet colour consists of a short range of wavelengths in which the deep violetcolour fades gradually and finally changes to the new colour, indigo. Like violet, the indigocolour is intense at the beginning and gradually fades to change finally to another colour, blue.Similarly deep blue colour fades and changes to green and in the same manner to yellow thento orange and finally to red. The spectrum gives a band of colours and not single andseparated colours. In sodium spectrum, on the other hand, there was only two yellow linesD

1 and D

2 with a gap in between. No other colour or line was present there. So the solar

spectrum is an example of continuous or band spectrum while sodium spectrum an exampleof line or discontinuous spectrum. As a matter of fact when a large number of lines areplaced very close to each other and overlap with each other, we get a band or continuousspectrum. Only when we get few lines with a detectable separation between them, we callit a line spectrum.SAQ 9:(i) What is the nature of spectrum of elements: continuous or discontinuous?(ii) "In line spectrum, we get a continuous band or patch of colours." Is the statement

true or false? Give reason.(iii) "The seven colours in the solar spectrum are well separated from each other having

distinct gaps between them". State whether the statement is true or false. Givereason.

Continuous or Band Spectrum (Example : Solar spectrum)

Discontinous or Line Spectrum (Example 1 : Sodium sepctrum)

Line Spectrum (Example 2 : Helim Spectrum)

Line Spectrum : (Example 2 : Hydrogen Spectrum - Balmer Series)

Types of Spectra on the basis of emission and absorption:

1. Emission SpectraWhen the spectrum of an element or substance is taken while it is emitting radiations

it is called emission spectrum. Solar spectrum(VIBGYOR), sodium spectrum(D1 and D

2

lines), the blue lines for the Cs spectrum, a few lines of hydrogen and helium spectra andother examples that we have discussed so far are examples of emission spectra. Most ofthe spectra studied come under this category. When the substance is heated or subjected toelectric discharge, emission spectra are formed.2. Absorption Spectra

When the spectrum of the substance is taken when it is absorbing radiation fromsunlight or any other continous spectral region, it is called absorption spectrum.

When a beam of sunlight is first passed through NaCl solution and then through aprism then the spectrum is observed on a white screen the spectrum is found to contain theentire VIBGYOR pattern expect two lines in the yellow region i.e D

1(5890 Å) and D

2(5896

Å) which appear as dark lines or missing lines. This is because Na+ ions present in thesolution have absorbed the two radiations(D

1 and D

2) from the sunlight and that is why they

are missing in the spectrum. In the diagram above, the VIBGYOR pattern has been shownfor a continuous solar spectrum follwed by a certain element's heated vapour(or underelectric discharge) giving emission line spectrum(bright lines) in a white screen in a darkroom and the last one is the absorption line spectrum(dark lines)of the same gaseous elementin the cold state. In the last spectrum all the VIBGYOR is present except at the dark linesat which the gas has absorbed. The bright lines in the emission spectrum of an element fallexactly in the same positions as the dark lines in the absorption spectrum for the sameelement. Hence emission and absorption spectra of a substance give the same information,one directly and the other in the indirect manner. Out of the two types of spectra, emissionspectra are most common and widely used. We shall henceforth discuss only on emissionspecra.

DO YOU KNOW WHAT ARE FRAUNHOFER LINES ?

Under very careful examination using powerful spectrometers, the “continuous” spectrum ofsunlight(emisison spectrum) turns out to be an absorption spectrum. In order to reach earth,sunlight needs to pass through the sun’s atmosphere, which is a lot cooler than the part ofthe sun where light is emitted. Gases in the atmosphere thus absorb certain frequencies,creating nearly 600 dark lines that Fraunhofer observed. They are now called Fraunhoferlines in his honor. In fact these are absoption spectrum of all substances present betweensun and earth.

HYDROGEN SPECTRUMWhen hydrogen gas is taken in a glass tube and is subjected to high voltage and very lowpressure, it glows by emitting bluish white light. When this bluish white light is analysed ina dark room by passing it through a prism and the spectrum observed in a photographic plate,we see only few lines. Hence hydrogen spectrum is a line or discontinuous spectrum. Thelines approach closer and closer towards the lower wavelength region and after few linesmerge into each other and stops giving further lines. See the hydrogen spectrum given in thecoloured photograph. This is called a series. Balmer was the first scientist to observe thisseries in the visible region and the series is called Balmer series. In this series the first lineappears in the orange(6560Å) region, the second in the blue(4860Å), third in the violetregion(4340Å) and further lines start merging onto one another in the same violetregion(4100Å). So it is not a continuous spectrum like the sunlight spectrum which containsall the colours(VIBGYOR) without any gap in between them. See the solar spectrum andhydrogen spectrum in its actual coloured pattern in the picture given and compare.After the discovery of Balmer series in the visible region, on careful analysis of the hydrogenspectrum in regions other than visible region, other series of spectra were discovered bydifferent scientists. These are Lyman series, Paschen series, Bracket series and Pfundseries. Each series shows the same pattern i.e consist of few lines which approach closerand closer and finally merge onto one another towards the lower wavelength region. Lymanseries falls in the ultraviolet region while the remaining series of lines fall in the infraredregion of the electromagnetic spectrum. We shall discuss more about hydrogen spectrumlater.SAQ 10:(i) Hydrogen spectrum is discontinuous: State true or false(ii) Solar spectrum and hydrogen spectrum belong to the same the category of linespectrum. State true or false.(iii) What is a spectrometer and what are its functions?

BOHR'S MODEL OF ATOMDemerits of Rutherford's Model of atom:You remember that Rutherford was the first person who suggested that electrons are revolvinground the very tiny but heavy nucleus like planets revolve round the sun. That is undoubtedlythe crucial discovery which suggested the presence and nature of nucleus and the behaviourof electrons with respect to the nucleus. Unfortunately he did not mention whether the

electrons are revolving in fixed orbits about the nucleus like the planets revolve round the sunor are moving in any orbit. This was a subject matter of criticism against his theory by somephysicists, the first among them was his own student, Bohr. Let us see what was thecriticism.1. According to the theory of electrodynamics, if a charge particle like electron(-ve)revolves round another charged particle(nucleus:+ve), the electron should continuously loseor emit energy and therefore should gradually be drawn nearer to the nucleus and finallyshould merge onto the nucleus. If that would have really happened, then there would not haveexisted any proton or electron separately as they would have destroyed each other. Such athing really does not take place.

To explain this, it was believed that the law of electrodynamics failed in case of anatom and the electrons move in different fixed orbits and do not emit energy as long as theyrevolve in those fixed orbits. That was precisely what Bohr proposed in his model.Postulates of Bohr's model of atom

1. Electron revolves round the nucleus in fixed circular orbits called energy levels orshells designated by the letters K, L, M, N, O ........ The first shell is the K shell, the secondshell the L shell, the third M shell and so on. The electron possesses the minimum energywhen revolves in the K shell. The energy of the electron goes on increasing as it revolvesin higher and higher shells such as L, M, N, O....2. As long as electron revolves round a particular orbit, the energy possessed by it isfixed. It neither loses or gains energy.3. Electron can jump from one orbit to another. If the electron jumps from a lower orbitto a higher orbit, it has to absorb energy. If the electron jumps from a higher orbit to lowerorbit, it emits energy.Hydrogen atom:There is one electron in hydrogen atom which usually revolves in the K shell. This is calledthe ground state of hydrogen atom. All the other shells such as L, M, N, O... are alsopresent in hydrogen atom. They are vacant or empty. If the electron likes to jump from theK shell to the L shell, how it can do so? It can do so by absorbing appropriate amount ofenergy. Similarly if the electron wants to jump from the K shell directly to M shell whilebypassing the L shell, it can also do so by absorbing still greater but appropriate amount ofenergy. Likewise when the electron wants to jump back from M shell to K shell, what willhappen? It will emit or evolve energy. By what amount? The amount of energy emitted isexactly same as it had absorbed while jumping from K shell to M shell. When electron isrevolving in higher shells such as L, M, N, O etc., it is said to remain in the excited state.As long as the electron remains in a particular orbit(shell), it neither emits nor absorbs energy.We say this as the energy levels are quantized which means that electron takes fixed orpermitted energy values.4. The energy absorption and emission that takes place when electron jumps from oneshell to another occurs in the form of single packet or photon(quantum) according to Planck'sQuantum theory.Explanation: Let the energy of the electron when revolves in first shell(K shell)=E

1,

in the 2nd shell(L)= E2, in the 3rd shell(M)= E

3 and so on.

KL

MN

E1

E2

E3

E4

(E2-E1) absorbed

(E3-E1) emitted

ENERGY LEVELS(SHELLS) OF HYDROGEN ATOM

E1< E

2 < E

3 < ........ the energy values go on increasing as we go further away from

the nucleus.If the electron jumps from 1st(K) to 2nd shell(L), the energy difference is equal toΔE=E

2-E

1. This differential amount of energy has to be absorbed by the electron in order

to jump from K to L shell. According to the postulate of Bohr this absorption will take placein the form of a single photon of energy selecting the particular frequency (hence wavelengthand wavenumber) from the electromagnetic radiation following Planck's quantum theory.

11

1 =h hc=hcE = E - E = 12

Suppose the frequency of the radiation necessary for it is ν1(hence wavelength λ

1 and wave-

number 1 ). So when electron revolving in the K shell gets a photon having frequency ν1,

then only it will jump to the L shell. Note that electron cannot absorb any energy less thanthat and thus cannot jump to any other orbit between the K and L shell. Thus we say thatenergy levels are quantized. This means that electron can possess fixed energy values suchas E

1, E

2, E

3 and so on and not any energy value in between these values. Now when the

electron will like to jump from L shell back to the K shell, it will emit the same amount ofenergy in the form of a single photon having same frequency ν

1(hence wavelength λ

1 and

wave-number 1 ). Similarly when electron jumps from 1st shell(K) to 3rds shell(M) directly,

then it has to absorb the differential amount of energy i.e (E3-E

1)in form of a single photon

having a fixed frequency(ν2) (hence fixed wavelength λ

2 and wavenumber 2 )

'3 2

221E - E = E = hc

= hch =

So when electron absorbs a photon having frequency ν2, it jumps from K shell to M shell.

Now you can well guess how much of energy, the electron will evolve when it jumps fromM shell back to the K shell? It is same in magnitude that it had absorbed while jumping fromK to M shell. A photon having frequency ν

2 will be evolved. So we conclude by saying that

electron absorbs a photon of a fixed energy (hence fixed frequency, wavelength andwavenumber) when it jumps from any lower shell to any higher shell. Likewise it will emitphoton of fixed energy when it jumps from a higher shell to a lower shell.

SAQ 11:(i) The electron in H atom jumps from 4th shell to 2nd shell. Will energy be absorbedor emitted?(ii) The electron in H atom jumps from 3rd shell to 7th shell. Will energy be absorbedor emitted?(iii) The energy of the electron in the different shells is as follows.

K shell= p joule, L shell = q joule, M shell= r joule, N shell = s jouleAnswer the following questions.(a)Can the electron in the hydrogen atom lose energy when it is revolving in the 2ndshell(L)? If so by what amount and where will it go?(b)Can the electron in the H atom lose energy when it revolves in the K shell? Givereason.(c)The electron is in the M shell. Can it lose energy? If so, by what amounts andwhere it can go after losing energy?(d)In which shell, the electron can absorb energy only but cannot lose energy?Justify.(e)The electron is present in the 4th shell(N) of H atom. How many possible photonsit can emit? Which jump will emit greatest energy photon and which the least energyphoton?

5. By virtue of the electron's motion in a circular orbit, it possess angular momentumwhich is equal to the product of mass of the electron, velocity of the electron and the radiusof the orbit.

Angular Momentum = m X v X rAccording to Bohr's postulate, the angular momentum of an electron takes fixed values likeenergy levels. Thus the angular momentum of the electron is quantized and is expressed bya whole number multiple of h/2π.

Angular momentum = n X (h/2π)Where n= 1, 2, 3, and so on.

If n=1, the angular momentum = h/2π = angular momentum of electron in K shellIf n=2, the angular momentum = 2 X h/2π = angular momentum of electron in L shellIf n=3, the angular momentum = 3 X h/2π = angular momentum of electron in M shell andso on.

Energy of Electron in H atom:Bohr derived mathematically the energy of electron in different shells of hydrogen atom. Theenergy equation is presented below in its simplest form.

E = kn2 (where k is a constant whose value is 2.11 X 10 -11 erg per atom and

n= shell number or the principal quantum number i.e 1 for K shell, 2 for L shell, 3 for Mshell and so on)From the above relation, we can find the energy values of electron in the different shells.

E1 = - k, ( n=1) E

2 = - k/4, (n=2)

E3 = - k/9, (n=3) E

4 = - k/16, (n=4) and so on.

From this we can calculate the energy difference between any two shells and then usingPlanck's quantum theory, we can calculate the frequency (or the wavelength or wavenumber)of the photon absorbed or emitted for a particular jump. Read the following example.Example: Calculate the wavelength of the photon to be absorbed by the electronin H atom to jump from 1st to 2nd shell. To which region of the electromagneticspectrum does this photon belong?Solution: The energy of the electron in the 1st shell= E

1 = - k

Energy of the electron in the 2nd shell =E2 = -k/4

So the energy difference =DE=E2 - E

1= -k/4 - (-k) = 3k/4

According to Planck's Quantum theory, 2 1E - E = E = hc ⇒ 4

3khc

=

We know the values of h(Planck's constant), c(velocity of light) and k(the constant for Hatom= 2.11 X 10-11erg)

⇒ = 4 hc3 k

=4 X 6.627 X 10 erg.sec X 3 X 10 cm/sec

3 X 2.11 X 10 erg-11

10-27

= 1.256 X 10-5

cm

= 1.256 X 10-5 X 108 Å = 1.256 X 103 Å = 1256Å.The electron will absorb a photon of wavelength 1256Å in order to jump from K shell to Mshell. This wave belong to the ultraviolet region of the electromagnetic spectrum as UVradiations have wavelength range between 150 to 3800Å. Once you find the wavelength, youcan very well find its frequency and wavenumber.SAQ 12:(i) Why is the energy of the electron is negative?(ii) What will be the wavelength of the photon emitted when electron in H atom jumpsfrom 2nd shell to 1st shell?(iii) Calculate the wavelength of the photon absorbed when electron in H atom jumpsfrom 1st shell to 3rd shell. In which region of the electromagnetic spectrum does this belong.(iv) Calculate the wavelength of the photon emitted when electron jumps from 3rd shellto 1st shell in H atom.(v) Calculate the wavelength of the photon absorbed when electron jumps from 2ndshell to 3rd shell. To which region does it belong? What photon will be emitted when theelectron jumps from 3rd shell back to the 2nd shell?(vi) What is the value of energy of electron in the 7th shell and 3rd shell in H atom?Which is of higher energy? Answer in terms of the constant k(2.11X 10-11erg).(vii) When the electron jumps from 1st shell to infinite distance, how much energy isrequired? What is the energy of the electron at the infinite distance?(viii) Suppose in one H atom, the electron jumps from 2nd shell to 1st shell, in another Hatom electron jumps from 3rd shell to 1st shell and in the still another atom, electron jumpsfrom 4th to 1st shell. How does the wavelength vary in respect of the three photons emittedin the three jumps? Does it increase or decrease in the sequence of the jump given above?

Origin of Hydrogen Spectrum:You already know that when hydrogen gas is subjected to high voltage and low pressure ina tube and the bluish white light that it emits is analysed through a prism, we find severalseries of lines such as Balmer series, Lyman series etc. in the spectrum on a photographic

plate. This is called hydrogen spectrum. We also know that each series is a discontinuousspectrum consisting of discrete and few lines which gradually come closer to each other andtowards lower wavelength region merge onto one another after which it stops giving furtherlines. Now we shall know how the spectral lines arise in spectrum. To understand this, letus look to the energy profile in H atom. The energy values of electrons go on increasingas we move to higher shells. The 1st shell has energy -k, the 2nd shell -k/4, the 3rd shell-k/9 and so on.

-k

k4

-

k9-

k16

--

4

k25

k-36

L

K

M

NO

3k

5k36

7k144

P

(Energy Profile for the different shells in H atom

If we calculate the gaps between the successive energy levels, we find that the energy gapbetween two successive shells gradually decreases as we move to higher and higher shellsand after a few shells the gap becomes practically equal to zero. That means, the electronsacquire nearly equal energy values as we move to higher shells. Now let us know the originof hydrogen spectrum.A glass tube containing hydrogen gas contains thousands and thousands of H atoms. Whenthe gas is subjected to high voltage at low pressure, the only one electron present in the Kshell(ground state) jumps to different higher shells in different atoms. In some atom, theelectron jumps from 1st to 2nd, in some other atom from 1st to 3rd, and in other atom from1st to 4th and so on. This takes place by absorbing photons of different energy availableto it when it is subjected to high voltage. These are called the excited H atoms in whichelectrons are present in different higher energy levels in different atoms. Note that each atomcontains one electron(not more than one) but present in different higher shells in differentatoms. After some time the electrons will like to come back from different higher energylevels present in different atoms to different lower energy levels in the same atoms. In theseprocesses photons of different energies are emitted. When this mixture of photons emittedis passed through a prism, the radiations get dispersed to give the spectrum on the photographicplate. The spectrum consists of several series of lines that have been named as Lymanseries, Balmer series, Paschen series, Bracket series and Pfund series after the scientistswho had invented them. Say for example, all the lines observed in the Lyman series are dueto transition of electrons from different higher shells to the first shell and the lines of Balmerseries are due to transitions from different higher shells to the 2nd shell and so on. Let usknow more about each series. Note that only because the electrons can take up fixed energy

levels(K,L,M,N....) and not any energy level or all possible energy levels, we get discrete andfixed lines in each series of H spectrum and get a discontinuous spectrum. Each particulartransition(say 3rd shell to 2nd shell) gives rise to one line in the spectrum. Had electronassumed all possible energy levels, it would have jumped to infinite number of energy levelsand hence emitted infinite number of radiations which would have resulted in a band orcontinuous spectrum like solar spectrum. But such a thing does not really take place. We getdiscontinuous spectrum consisting of a few discrete lines in each series. This clearly provedthe validity of the Bohr's postulate of atomic structure that the atom consists of some fixednumber of shells having fixed energy values. In fact, Bohr took the idea of fixed orbits onlyfrom the experimental finding of the H spectrum by Balmer, the first scientist to observe theH spectrum. Another point may be discussed here. Why the lines in each series of spectrumcome close to each other and finally merge onto one other and stops further growth aftersome distance? This is because of the following reason.Since the gaps between successiveenergy levels diminish(as shown before in the energy profile) and gradually tend to becomezero, the lines in each spectral series also approach closer as we move on to lower wavelengthregion and after a few discrete lines, they merge onto one another and stop furtherexetension.Further discussion on the hydrogen spectrum, we postpone to higher classes.

123

4

5

6

Lyman Series

Balmer Series

Paschen Series

2

34

5

6

Bracket Series 4

56

Transitions of Electrons giving Different Series in H-Spectrum

1. LYMAN SERIES:(i) This series of lines falls in the ultraviolet region of the electromagnetic spectrum.(ii) The different lines appear due to the transition of electrons from various higherenergy levels in various H atoms to the 1st shell.

The transition : 2nd shell to 1st shell gives the → 1st line3rd shell to 1st shell gives the → 2nd line4th shell to 1st shell gives the → 3rd line and so on.

If n1= any lower shell and n

2 = any higher shell, then we can say that for Lyman

series: n1 = 1 and n

2 = 2, 3, 4, 5..........

2. BALMER SERIES:(i) This series falls in the visible region of the electromagnetic spectrum.(ii) The different lines in this series are due to the transition of electrons from various

higher energy levels in various H atoms to the 2nd shell.For Balmer series: n

1= 2 and n

2 = 3, 4, 5, 6..........

So 1st line → the transition from 3rd shell to 2nd shell2nd line → the transition from 4th shell to 2nd shell3rd line → the transition from 5th shell to 2nd shell and so on.

3. PASCHEN SERIES:(i) This series appears in the first part of the infrared(IR) region of the electromagneticspectrum.(ii) The different lines in this series are due to the transitions of electrons from varioushigher energy levels in various H atoms to the 3rd shell.

For Paschen series: n1= 3 and n

2 = 4, 5, 6, 7 .......

So 1st line → the transition from 4th shell to 3rd shell.2nd line → the transition from 5th shell to 3rd shell.3rd line → the transition from 6th shell to 3rd shell and so on.

4. BRACKET SERIES:(i) The lines appear also in the IR region but in higher wavelength region.(ii) The different lines in this series appear due to the transitions of electrons fromvarious higher energy levels in various H atoms to the 4th shell.

For Paschen series: n1= 4 and n

2 = 5, 6, 7, 8 .......

5. PFUND SERIES:(i) The lines appear also in the IR region but still at higher wavelength region.(ii) The different lines in this series are due to the transitions of electrons from varioushigher energy levels in various H atoms to the 5th shell.

For Paschen series: n1= 5 and n

2 = 6, 7, 8, 9 .......

6. HUMPHREYS SERIES :(i) The lines appear also in the IR region but still at higher wavelength region.(ii) The different lines in this series are due to the transitions of electrons from varioushigher energy levels in various H atoms to the 6th shell.

For Paschen series: n1= 6 and n

2 = 7, 8, 9, 10.....

SAQ 13:(i) Indicate to which transitions the following lines of each series in H spectrum refers.

(a)) 2nd line of Balmer series (b)1st line of Lyman series (c)3rd line of thePaschen series (d)3rd line of the Lyman series(e)3rd line of Balmer series

(f)2nd line of the Bracket series (g)1st line of the Pfund series(ii) Calculate the wavelength of the 2nd line of the Lyman series. Check whether itbelongs to UV region or not.(iii) Calculate the wavelength of the 3rd line of Balmer series. Check whether it belongsto visible region or not.(iv) Calculate the wavelength of the 2nd line of Paschen series and check whether it fallsin the IR region or not.(v) Predict what would have been the nature of H spectrum, if the electrons would have

taken all possible orbits(infinite number) instead of some fixed energy levels. What is theactual nature of H spectrum.Advantage of Bohr's Model of Atom:Bohr's postulates and mathematical derivation of energy of electron could satisfy theexperimental finding of the hydrogen spectrum. The actual wavelengths of the spectral linescould match well with the mathematical calculation based on Bohr's theory. That became thegreatest success of the Bohr's theory.Demerits of Bohr's Model of Atom:1. The entire Bohr's postulates that we studied are applicable to H atom and H likespecies which contain one electron such as He+, Li2+ etc. It failed to explain the spectra ofspecies containing more than one electron such as He, Li etc.2. When powerful spectrometers were discovered, it was found that each line of thehydrogen spectrum was found to split into two finer lines called a doublet. This splitting ofspectral lines could not be explained by Bohr's theory.Note that it was subsequently revealed that each shell consists of a fixed number of subshells.Each subshell contains a few orbital where electrons reside. In each orbital, two electronscan at best exisit with opposite spins- one spinning clockwise and the other anticlockwise.This different spin states is responsibe for splitting the spectral line. Further details on thiswill not be given here.3. When the spectrum was taken in the presence of external magnetic field or electricfield, it was found that each spectral line is split into some finer lines analogous to theprevious case. These were called the Zeeman effect and Stark effect respectively. Bohr'stheory could not explain these effects. We shall not also discuss about the cause of theseeffects here as it is beyond the scope of this book.4. It was suggested by de Broglie that electron moves in the form of waves and notin the conventional rectilinear motion as was commonly believed. He proved beyond anydoubt that electron in each energy level moves like a standing wave. It has a wavelength,frequency and a wavenumber. This is called electron wave which is a matter wave and isdifferent from electromagnetic waves. Bohr's theory could not explain this wave nature ofelectron.5. Heisenberg proved that it is not possible to accurately know the position and velocityof very tiny(microscopic)particles like electron. So it is wrong to say that electrons arerotating in fixed orbits about the nucleus. This was a major challenge to Bohr's theory. WhileBohr suggested that the electrons move in fixed orbits(K,L,M,N...) having fixed radii andenergy levels, it was found subsequently to be untrue. The electrons do move in fixed energylevels in agreement with Bohr's theory, but not in fixed orbits. The term 'orbit' therefore wasreplaced by another term 'orbital'. Orbital is a three dimensional space(and not a twodimensional circular path which is called the orbit)about the nucleus which possessesfixed energy and the electron moves around the nucleus anywhere inside this space.The electron does not repeat its motion again and again in the same track or orbit assuggested by Bohr's theory. It remains within the boundary of an orbital and rotates insideanywhere and everywhere as a wave. Have you seen the event of motor cycling inside aspherical cage in a circus? The motor cyclist rides the motor cycle inside a spherical cagein all possible tracks in very high speed. He does not repeat a fixed track but moves in allpossible tracks but inside the cage. In the same manner, an electron moves inside an orbitaland possesses the same fixed energy, moves anywhere and everywhere.

SAQ 14:(i) What is the difference between electromagnetic waves and electron wave(matterwaves).(ii) Define the term, 'orbital' and explain clearly what it is.

Wave-mechanical Approach to the Structure of Atom:It has been already mentioned that electron moves in the form of waves. This was provedby de Broglie. This was the beginning of a new approach called wave-mechanical approachto the structure of atom. Subsequently Heisenberg proved that electrons do not move in fixedorbits, rather they move in fixed orbitals, which are the three dimensional spaces inside whichelectrons move in all possible places. Later, another scientist Schrodinger derived a mathematicalequation called electron wave equation. While solving this wave equation three indexnumbers were obtained. These index numbers are called quantum numbers which weredesignated as n(principal quantum number), l (azimuthal quantum number) and m(magneticquantum number). The details of all these are beyond the scope of this book. However wediscuss below the significance of the quantum numbers.

Quantum Numbers:There are four index numbers which give the complete information about an electron, suchas: in what shell the electron is present, in what sub-shell it is present, it what orbital it ispresent, what is its energy value etc. Three of the quantum numbers(n,l and m) have resultedfrom the solution of the Schrodinger's wave equation as mentioned above. The forth quantumnumber called spin quantum number(s) has come from spectroscopic studies. When all thefour quantum numbers n, l, m and s are assigned to any electron, we know the completeidentity of that electron. Let us discuss the importance of each quantum number one by one.1. Principal Quantum Number(n):

(i)It is the shell number that we studied in Bohr's theory. It takes whole numbervalues 1, 2, 3...... representing the main energy levels.

n=1 stands for the 1st or the K shelln=2 stands for the 2nd or the L shelln=3 stands for the 3rd or the M shelln=4 stands for the 4th or the N shell and so on.

(ii)It tells about the distance of the electron with respect to nucleus. If n is larger,the electron is situated farther away from the nucleus and if n is smaller it is located closerto the nucleus.

(iii)It determines the average energy value of the electron according to Bohr's theory,E= - k/n2

2. Azimuthal Quantum Number(l):This is also called the subsidiary quantum number or angular momentum quantum

number or orbital quantum number.(i)It suggests that each shell is divided into a fixed number sub-shells.(ii)It takes values 0,1,2...... upto (n-1), where n= principal quantum number.The number of subshells present in a particular shell is equal to the number of l

values that the shell can have. This is dependent on the principal quantum number of theshell. The number of l values and hence number of subshells is equal to the n value, theprincipal quantum number.

1st shell(K shell):For n = 1

l = 0 (s-subshell)(there is only one l value and thus K shell has one subshell called the s-subshell)2nd shell(L shell):For n = 2

l = 0 and (s-subshell)= 1 (p- subshell)

Thus there are two values of l and the 2nd shell has two subshells namely s-subshell(l=0) and p-subshell(l=1).

3rd shell(M shell):For n = 3

l = 0 (s-subshell)= 1 (p-subshell)= 2 (d-subshell)

Thus there are three values of l and 3rd shell has three subshells namely s-subshell(l=0),p-subshell(l=1) and d-subshell(l=2).

4th shell (N shell):For n = 4

l = 0 (s-subshell)= 1 (p-subshell)= 2 (d-subshell)= 3 (f-subshell)

Thus there are four values of l and the 4th shell has four subshells namely s-subshell(l=0),p-subshell(l=1), d-subshell(l-2) and f-subshell(l=3).Similarly the 5th shell must have five subshells, 6th shell must have six subshells and so on.SAQ15:(i) What is the value of l for a p-subshell ?(ii) How many subshells are there for the 3rd shell. What are their l values and whatsubshell does each represent?(iii) An electron is present in the 2nd shell and in the p-subshell. Assign the n(principalquantum number) and l(azimuthal quantum number) to the electron.(iv) An electron has quantum numbers as follows: n= 4 and l=3, In which shell and whichsubshell it is present.(v) How will you distinguish between a p subshell of 2nd shell and p subshell of the 3rdshell? Are they same? Explain.(vi) An electron is present in the 5th shell and s-subshell. Assign the n and l values.(vii) One electron is present in the f-subshell of 4th shell and another electron is presentin the f-subshell of 5th shell. How can you distinguish between the two electrons?(viii) How many subshells does the 5th shell(n=5) have? What are their l values andsymbols representing them._____________________________________________________________________(iii) The value of l tells us the shape of an orbital. So far, we have not studied aboutorbitals. We shall know about orbitals in our next quantum number(m). But, first let us knowtheir shapes.Every orbital has a fixed shape. The shape of the different orbitals are as follows.

s-orbital: spherically symmetricalp-orbital: dumbbell shapedd-orbital: double dumbbell shaped.f-orbital: complex

p p px y z

xp

py

zp

(dumbbell shaped p-orbitals shown separtely)

(p-orbitals shown together(spherically symmetrical s-orbital)

When l value of electron is known to be 0, immediately we know that it is s-orbital and itsshape is spherically symmetrical. Similarly, if the l value is 1, it is a p-orbital and its shapeis dumbbell shaped. Similarly when l=2, it is d-orbital and is double dumbbell shaped.(iv) In a given shell, the order of energy among the subshells is as follows:

s<p<d<fFor example, in the 4th shell, which contains all the four subshells s, p, d and f, the 4ssubshell possesses the least energy while 4f subshell possesses the greatest energy and the

order is: 4s<4p<4d<4f.Note that we should not compare a 4s subshell with a 3p subshell or a 3d subshell with a4p subshell and so on. For comparison, the subshells of the same shell is considered. Aquestion arises at this point why s- subshell possesses the least energy and f- subshellpossesses the greatest energy? It is because, s-subshell is spherically symmetrical in shapeand it is most tightly bound to the nucleus. In other words, the nucleus attracts the s-subshellto the maximum extent and for that reason, it possesses the minimum energy. This is whatis called the pentration power of s- orbital is maximum. As we move from s to p and thento d and finally to f- subshells, the subshells become less and less tightly bound to the nucleusi.e become more and more diffused and penetration power decreases. The nuclear attraction,thus, gradually falls in the same order. Hence the energy of subshell decreases in the ordercited above.SAQ 16:(i) Give the decreasing order of energy for the sub-shells present in the 2nd shell.(ii) Is the order correct. 3s>3p>3d? Explain.(iii) Between 4p and 4f which has higher energy?(iv) An electron has n=4 and l=1. Where is it located?(v) Can an elenctron have n=2 and l=2? Justify.(vi) Between 2s and 2p subshell which is of higher energy?

3. Magnetic Quantum Number(m):(i) It suggests that each subshell is further subdivided into a fixed number of orbitals.From the l value we know the presence of subshells in a shell while from m value we knowthe presence of orbitals in a subshell.(ii) The number of m values depends on l value, that means the subshell determines howmany orbitals it can possess. As many m values are possible for a given subshell, samenumber of orbitals it will have. An orbital can have (2l+1) values of m, and so it will contain(2l+1) number of orbitals. For example

s-subshell: l=0: (2l+1)= 0+1 = 1, So s-subshell has one orbital which is calledthe s-orbital. This means that the s-subshell itself is the orbital

p-subshell: l=1: (2l+1)=2+1=3, So p-subshell has three orbitals which are designatedas p

x, p

y and p

z orbitals.

d-subshell: l=2: (2l+1)=4+1=5, So d-subshell has five orbitals.f-subshell: l=3: (2l+1)=6+1=7, So f-subshell has seven orbitals.(The designations of the d and f- orbitals are beyond the scope of this book)

(iii) m takes values -l, -l+1, -l+2.......0, 1, 2.........+lIn the previous point we knew how many values of m are possible for a given l value. Herewe know what values, m can take. m can take values from -l to +l through 0. Eachsuccessive value is obtained by adding 1 to the previous value.

l=0(s-subshell):m=0 (one value for the one s-orbital)l=1(p-subshell)m=-1, 0, +1 (three values for the three p orbitals p

x, p

y and p

z)

l=2(d-subshell)m=-2, -1, 0, +1, +2 (five values for the five d orbitals)l=3(f-subshell)m=-3, -2, -1, 0, +1, +2, +3 (seven values for the seven f orbitals)

(iv) Magnetic quantum number(m) tells about the orientation of an orbital in space relativeto other orbitals. For example, the three dumbbell shaped p orbitals(p

x, p

y and p

z) are lying

along three mutually perpependicular axes(x, y and z). These p-orbitals are represented bythe m values -1, 0 and +1. Each value will represent one p orbital lying along a particularaxis. So from the value of m, we can know along which direction the orbital is oriented.(v) All the orbitals of a subshell possess same energy. For example, all the three p-orbitals(p

x, p

y and p

z) possess equal energy. Similarly all the five d orbitals possess equal

energy and all the seven f orbitals of the f-subshell possess equal energy. Orbitals possessingidentical energy are called the degenerate orbitals.The three p orbitals are degenerate, thefive d orbitals are degenerate and also seven f orbitals are degenerate.SAQ 17:(i) Give the magnetic quantum numbers for the p- subshell. How many orbitals does thep-subshell contain?(ii) How many m values are possible for l=2? What does it mean? Also give the absolutem values for all these.(iii) How many orbitals can a s-subshell have? How can you justify through l and mvalue.(iv) How can you prove through quantum numbers that a f-subshell has seven orbitals.

4. Spin Quantum Number(s)(i) Just like earth, while revolving around the sun also spins(rotates) about its own axis,in the same way, electron spins about its own axis besides revolving around the nucleus. Thespinning of the electron may be clockwise or anticlockwise. Spin quantum number determinesthe spin of an electron in an orbital, whether clockwise or anticlockwise.(ii) It takes values +½ or -½. If +½ will indicate the clockwise spin of the electron then-½ will indicate the anticlockwise spin of the electron. The clockwise spin(+½) is alsorepresented by an upward arrow mark(↑) while the anticlockwise spin(-½) is also representedby a downward arrow mark(↓). Note that these values(+½ and -½) do not have any actualsignificance. These are merely quantum mechanical values assigned to the two differentspins of an electron. Also note that these values can be interchangeably used for clockwiseand anticlockwise spins. That means clockwise spin can also be represented by -½ and ifthis is so, anticlockwise spin will be represented by +½. The importance of this quantumnumber will be made more clear when we study Pauli's exclusion principle just after a while.SAQ 18:(i) An electron has the following quantum numbers. n=3, l=1, m=-1 and s=+½,Give the identity of the electron.(ii) An electron has the following quantum numbers: n=2, l=0, m=0 and s=+½. Designateit.(iii) Can an electron have the following set of quantum numbers: n=2, l=0, m=+1 ands=-½. If not give reason and indicate which value is not possible?(iv) Can an electron have the following set of quantum numbers: n=4, l=2, m=-3 ands=+½. Give reason.(v) Give all possible quantum numbers to an electron present in a 4s orbital.(vi) Give all possible quantum numbers to an electron present in a 2p orbital.(vii) Give all possible quantum numbers to an electron present in a 3d orbital.

PAULI'S EXCLUSION PRINCIPLE:From the discussion on quantum numbers, we know now, that a principal shell consists ofa fixed number of subshells and each subshell further consists of a fixed number of orbitals.Each orbital carries electrons. Let us take an example. Suppose there is a multistorey buildingwhich can be thought to be analogous to the main shell, say N(4th)shell. The building consistsof several flats. These flats are considered as different subshells present in that shell. If themain building is the 4th shell, then it has four flats(subshells) namely s-, p- d- and f subshells.Again each flat consists of a fixed number of rooms. These rooms are analogous to orbitals.We know that the s- flat(s-subshell) has one room(orbital) called s-orbital, the p-flat(p-subshell) contains three rooms(p

x, p

y and p

z orbitals), d-flat(d-subshell) contains five rooms(the

five d orbitals), while the f-flat(f-subshell) will have seven rooms(the seven f orbitals). Themain buiding(main shell) is known from the n(principal quantum number)value. Then theflats(subshell)of that building(shell) will be known from the l(azimuthal quantum number)valuewhile the rooms(orbitals) present in each flat and their orientation can be known from them(magnetic quantum number)values. In each room(orbital), people like us are present. Peopleis analogous to electrons, because electrons are present in orbitals. But the question is howmany electrons at the maximum can be accommodated in one orbital and in which way? Thisis known from Pauli's Exclusion Principle which states that: An orbital at the most canaccommodate two electrons with opposite spins, i.e if one electron spins clockwise(s=+½),

the other electron has to spin anticlockwise (s=-½). Look to the following boxdiagram. The upward arrow mark indicates clockwise spin of one electron and the downwardarrow mark for the other electron present in the same orbital spinning anticlockwise.

Explanation: You know that electrons are negatively charged particles. There will be greatamount of repulsion between them. So how can they stay in one single room(orbital)?Theydo live happily by spinning in opposite directions. You know that a spinning charge is just likea bar magnet, and the direction of the magnetic field can be known from the right hand rule.If all the fingers except the thumb of your right hand curl to give the direction of the spinningof the electron, the thumb will give the direction of the magnetic field. So if the electron spinsclockwise, its magnetic field points upwards and if it spins anticlockwise, its magnetic fieldpoints downwards. When two electrons in one room spin in opposite directions, their magneticfields lie in opposite directions and cancel each other. This results in some attraction betweenthem which reduces the electrostatic repulsion to a large extent. So the two electrons remaintogether in one orbital.

ALTERNATIVE DEFINITION OF THE PRINCIPLE:No two electron can have all the four quantum numbers same.At the most three quantum numbers namely n, l and m can be same for two electrons. Inother words the two electrons may be present in the same shell(n), same subshell(l) andsame orbital(m), but they will not have the same spin. Thus the forth quantum number i.espin quantum number(s) have to be different. If one spins clockwise(s=+½), the other spinsanticlockwise(s=-½). Let us take the example of two electrons.

1st electron 2nd electronn = 3 n = 3l = 2 l = 2m = -2 m = -2s = +½ s = -½

You found in this example that the two electrons have the same n, l and m values i.e bothremain in the 3rd shell(n=3), d-subshell(l=2) and the same orbital(m=-2), but their spin quantumnumbers(s) are different. If one spins clockwise(s=+½), the other spins anticlockwise(s=-½).The reason for this has already been discussed before.SAQ 19:(i) Explain why after filling one electron in an orbital, second electron does not go toa higher orbital rather it remains in the same orbital with opposite spin?(ii) Supposing an orbital has two electrons with opposite(antiparallel) spin. Can the thirdelectron stay with them in the same orbital? If not, why?

Calculation of electrons in different Shells, Sub-shells and Orbitals:Let us now calculate total number of electrons that can remain in different subshells andshells on the basis of quantum numbers and Pauli's exclusion principle.K shell (1st shell):

n=1, l =0, m=0, s = +½ and -½ ---------> 2 electronsThe 1st shell has one subshell called the s-subshell represented by the l value 0. This s-subshell has only one orbital, represented by the m value of 0. You know from Pauli's

exclusion principle that one orbital can contain at the most two electrons with opposite spins(s=+½ and -½). So can you say how many electrons can K shell accommodate? Twoelectrons. The two electrons are present in the s-subshell which is also the s-orbital. HereK shell itself is the s-subshell which is also the s-orbital.

L shell (2nd shell):n=2, l= 0 (s-subshell) m= 0 (s-orbital) s=+½ and -½ (two electrons)

So total electrons in s-subshell -----> 2 electrons= 1 (p-subshell) m= -1, 0, +1(three p-orbitals)

For each orbital s= +½ and -½(two electrons)Total electrons in p-subshell=3X2 --->6 electrons

________________________________Total number of electrons in L shell ------> 8 electrons

In the 2nd shell there are two subshells namely s and p. s-subshell has one orbital namelys-orbital which can accommodate two electrons (s=+½ and -½). Thus s-subshell canaccommodate two electrons. The p-subshell has three orbitals and each of them can containtwo electrons (s=+½ and -½) and so the total electrons that the p-subshell can contain is3X2=6. Thus taking s and p subshells together, the 2nd shell can contain 2+6= 8 electrons.

M shell(3rd shell):n=3 l= 0 (s-subshell) m= 0 (s-orbital) s=+½ and -½(two

electrons)So total electrons in s-subshell -----> 2 electrons

= 1 (p-subshell) m= -1, 0, +1(three p-orbitals)For each orbital s= +½ and -½(two electrons)

Total electrons in p-subshell=3X2 --> 6 electrons = 2 (d-subshell) m= -2, -1, 0, +1, +2(five d-orbitals)

For each orbital s= +½ and -½(two electrons)Total electrons in d-subshell=5X2 -->10 electrons

________________________________________Total number of electrons in M shell ---> 18 electrons

3rd shell has three subshells namely s, p and d. s and p subshells contain 2 and 8 electronsrespectively in the same way as we have discussed for 2nd shell. The additional d-subshellhas five orbitals each of which can contain two electrons(s=+½ and -½). So the total numberof electrons in the d-subshell=5X2=10. The sum of all the electrons present in all the threesubshells in the 3rd shell=2+6+10=18.

N-Shell (4th shell):n=4 l= 0 (s-subshell) m= 0 (s-orbital) s=+½ and -½(two electrons)

So total electrons in s-subshell -----> 2 electrons

= 1 (p-subshell) m= -1, 0, +1(three p-orbitals)

For each orbital s= +½ and -½(two electrons)Total electrons in p-subshell=3X2 ---------> 6 electrons

= 2 (d-subshell) m= -2, -1, 0, +1, +2(five d-orbitals)For each orbital s= +½ and -½(two electrons)Total electrons in d-subshell=5X2 ----> 10 electrons

= 3 (f-subshell) m=-3, -2, -1, 0, +1, +2, +3 (seven f-orbitals)For each orbital s= +½ and -½(two electrons)

Total electrons in f-subshell=7X2 --------> 14 electrons________________________________________

Total number of electrons in N shell ----> 32 electrons4th shell has four subshells namely s, p, d and f. Like the 3rd shell, s will contain 2 electrons,p- will contain 6 electrons and d-will contain 10 electrons. The additional f-subshell has sevenorbitals each of which will accommodate two electrons. So the f-subshell, in total, willcontain 7X2=14 electrons. Thus the sum of the electrons in s, p, d and f subshells presentin the 4th shell=2+6+10+14=32.To sum up, the maximum capacity of accommodating electrons of different shells is asfollows:

1st shell: 2, 2nd shell:8, 3rd shell:18, 4th shell:32and so on.From this a general rule can be framed to know the total number of electrons in differentshells.

The total number of electrons in a shell = 2n2

(where n=principal quantum number or the shell number):1st shell: 2 X 12 = 2 electrons; 2nd shell: 2X 22= 8 electrons3rd shell: 2 X 32=18 electrons; 4rd shell: 2 X 42 = 32 electrons

SAQ 20:(i) Calculate the maximum number of electrons that can be accommodated in the 5th

shell.(ii) Show by calculations how many electrons can a p-subshell can accommodate.(iii) How many electrons can a d-subshell accommodate? How can you justify that.

HUND'S RULEThis rule speaks about how shall we distribute and fill electrons in a set of degenerateorbitals. Orbitals having same energy are called degenerate orbitals. The three p-orbitals aredegenerate, the five d orbitals are degenerate and the seven f orbitals are degenerate.

p-subshell:Supposing we are going to fill the p-orbitals with a total of 4 electrons. There are twopossibilities.

(i)The four electrons are filled in two orbitals such that each orbital contains twoelectrons with opposite spins.

2p4(wrong)

Note that although orbitals have different shapes, we represent the filling of orbitals in boxlike diagram as drawn above. This is called electron filling diagram or the box diagram.

Don't think that the actual shape of orbitals are like rectangular boxes. The actual shapesof different orbitals, you already know.

(ii)First, each of the orbital is filled with one electron and then the forth electron willpair with the first one.

2p4(correct)

The second possibility is correct while the first one is wrong according to Hund's rule. Letus now define Hund's rule.Electron pairing occurs only after filling one electron each with parallel (same)spin toa set of degenerate orbitals.Let us take another example. Supposing we have three electrons to fill the three degeneratep orbitals. Look to the following three possibilities.

(wrong) (wrong) (correct)

(i) (ii) (iii)

(i) is wrong because electron pairing starts from the first orbital and one orbital isleft vacant without containing any electron. This has been explained before.

(ii) is also wrong even though each orbital is singly occupied. This is becauseelectrons have not the same spin. In second and third orbitals the electrons spin clockwisewhile in the first orbital electron spins anticlockwise. According to Hund's rule all the unpairedelectrons filled in the degenerate orbitals should have the same or parallel spin, i.e either allclockwise or all anticlockwise. Conventionally we show all clockwise.

(iii) is thus the correct filling diagram.Explanation of Hund's Rule:Supposing you with two other friends go to a building which contains three adjacent rooms.Each room has one plate of your most favourite dish, say ice-cream. How will three of youdistribute in these three rooms of same nature? Will two of you crowd in one room and theother go to the second room or three of you go to different rooms and remain single in eachroom and comfortably eat your favourite dish, ice-cream? Undoubtedly, the second possibilityis most favourable. If the fourth friend of yours comes to the same place, then he will haveno other way than to share with one of you. Similarly, when there are three electrons to fillthree p orbitals of same energy, they will spread in the three rooms first with single occupancyand then if the fourth electron comes, it will pair in one of the singly occupied orbitals.Usually we show it in the first orbital of the electron box diagram. Actually all the threeelectron boxes representing orbitals are same and pairing can start from any orbital. Forconvenience, we start from the LHS.SAQ 21: Give the electron filling diagram for the following electrons. Calculate thenumber of unpaired electrons in each case.

(i) d7 (ii) p3 (iii) d4 (iv) f10 (v) p2 (vi) d6

SAQ 22: Explain why the unpaired electrons cannot spin in opposite directions whenoccupy the degenerate orbitals? They spin always with parallel spin, why?

AUFBAU PRINCIPLENormally we expect that when electrons are filled in different orbitals, subshells and shells,it is filled first in the lowermost orbital then gradually the electrons are filled in orbitals ofhigher and higher energy. The electron cannot enter to a higher energy orbital bypassing anorbital of lower energy. This is an unstable and unfavourable situation. A lower energyconfiguration is more stable than a higher energy system. So first the electron will enter intoa lower energy orbital and then to a higher energy orbital. This is called Aufbau principle.The German word, Aufbau means build up process, which means the electrons are to beadded up systematically in orbitals with increasing energy values. We know that shell-wisethe increasing order of energy values is

1(K)<2(L)<3(M)<4(N)<5(O)<......Subshell-wise the increasing order of energy values is

s<p<d<fSo if we consider both the subshells and shells into account the expected order of the energylevels with increasing order is

1s< 2s<2p <3s<3p<3d <4s<4p<4d<4f <5s<5p<5d<5f.......... (wrong)This means that the s subshell of the 1st shell is to be filled first, then comes the 2nd shell.In the 2nd shell first the s-subshell is to be filled then the p-subshell. Then comes 3rd shell.In the third shell, first s, then p and finally d subshell are to be filled. Then comes the 4thshell. In the 4th shell, first s, then p, followed by d and finally f-subshell. Like this highershells are to be filled. Although this is the expected order of filling the electrons, severalfactors complicate the situation and change the order at places. The complicating factorswhich creep in as more and more electrons are added in the different orbitals will not bediscussed in this book as they are beyond its scope. However, a general rule which determinesthe actual order of energy levels will be discussed. This is called the (n+l) rule. The Aufbaurule is based on this (n+l) rule.(n+l) Rule:(i) The orbital having greater value of the sum, (n+l) will have greater energy andconversely the orbital having less (n+l) value will possess lower energy. n and l are theprincipal and azimuthal quantum numbers respectively.Example: 2s and 2p: For 2s = n+l = 2+0=2 and for 2p= 2+l =3Since 2p orbital has greater n+l value than 2s, the order of energy is 2s<2p

3d and 4s: For 3d = n+ l = 3+2=5 and for 4s = 4+0=4,4s<3d

Note that although we expected that 3d should be of lower energy than 4s, in reality it isthe opposite.

4d and 5s: For 4d = n+l = 4+2=6 and for 5s = n+l = 5+0=5, so5s<4d

(ii) When two orbitals have the same (n+l) value, then the one having higher n valuepossesses greater energy.

2p and 3s: For 2p= n+l = 2+1=3 and for 3s=n+l = 3+0=3, In thiscase the (n+l) values are same for the two orbitals. According to rule, 3s has greater nvalue(3) than 2p(2), so 3s possess more energy than 2p. Conversely 2p has less energy than3s.

2p<3s

SAQ 23: Indicate the increasing order of energy in each pair by applying the (n+l)rule (i) 4p, 4d (ii)6s, 4f (iii)5s, 4f(iv)5s, 4p (v)4f, 5d

Using the (n+l) rule the new and correct order of energy levels is as follows:

1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p

Look to the orbitals which have been underline. These are the places where deviations fromthe expected order have taken place. 4s orbital has come before 3d. Similarly 5s orbital hascome before 4d and likewise 6s orbital has come before both 4f and 5d. This order of orbitalsstrictly follows the (n+l) rule. To remember this order, there is another very simple andinteresting method. You have to make a diagram as given below. First write the similarorbitals in the same columns. All the s orbitals will fall in one column, all the p orbitals willfall in the adjacent column followed by all the d orbitals and then the f orbitals. Then drawparallel and slanting lines passing through the orbitals. Then put downward arrow marks toeach line. Finally join the head of each line(arrow)to the tail of the next line as drawn in thisdiagram below.

Filling Arrow Diagram:

We have to fill the different orbitals by following the arrows. First, the 1s orbital is filled up.Then the arrow moves on to the 2s orbital, followed by 2p and 3s. Then we go to the next

arrow. The 3p orbital is filled first and after that 4s(not 3d as expected). Then we move ontothe next arrow. 3d is filled first followed by 4p and then comes 5s(not 4d as expected). Thenwe move onto the next arrow. First 4d, then 5p followed by 6s. Then we move on to thenext arrow. 4f is filled first followed by 5d, then 6p and finally 7s. We then proceed to thenext arrow. 5f is filled first followed by 6d and so on. Thus when we have some electronsto fill up in the different orbitals according to Aufbau principle, then we have to follow thisorder. Note that while going to the next higher orbital, the lower orbitals should be completelyfilled. To remind again, the maximum capacities of s, p, d and f subshells are 2, 6, 10 and14 electrons respectively. So these number of electrons are to be filled in the respectivesubshells before moving onto the next subshell. Examples given below will make you understandstill better.

ELECTRONIC CONFIGURATION:We know that the number of electrons present in an atom of an element is equal to its atomicnumber. Say for example, Na has the atomic number 11, so it contains 11 electrons arrangedin different orbitals from lower to higher energy according to the Aufbau principle. Thisarrangement of electrons is called electronic configuration of the atom. We shall write theelectronic configuration of an atom as follows.

Example1 Na(11): 1s2, 2s2, 2p6, 3s1

Note that the number of electrons in each subshell is to appear as superscript on the symbolof the subshell. While filling the subshells we always have to look to the filling arrowdiagram. We start filling first 1s subshell to its full capacity(2), then to 2s to its full capacity(2),followed by 2p subshell to its full capacity(6). From time to time you have to add all theelectrons you have filled so far and check with the total electrons that have to be filled. Inthis case we find that the total electrons filled so far is 10, and one more electron has to befilled to make 11 for sodium. This electron will be filled in the next higher subshell. It goesto 3s, so we write 3s1.Shell wise Configuration:If we add all the electrons present in each shell, the configuration will be

Na = 2,8,1The 1st shell will contain 2 electrons, the 2nd shell 8 electrons(2+6) and the 3rd shell 1electron. This is the electronic configuration that students use to write commonly. But youare advised to write the electronic configuration in the way we discussed before by fillingthe electrons in different subshells of different shells according to the Aufbau order.Valence Shell:The last shell in an atom is called the valence shell. In case of sodium that we discussedbefore, the 3rd shell is the valence shell. It has only one electron(3s1). To know the numberof unpaired electrons we have to draw the box diagram(filling diagram) of all the subshellsof the valence shell. In this case we have to draw the box for the 3s orbital.

3s1 So that the number of unpaired electrons is one in sodium.

Note that the valence shell can only contain one or more unpaired electrons provided it isnot completely filled. The inner shells contain only electron pairs.

Example 2: N(7): 1s2, 2s2, 2p3.Note that although the maximum capacity of p-subshell is 6, we are to fill only 3 electronsas we have to fill a total of 7 electrons, not more. The electronic configuration shell-wisewould be N(7)=2,5.Two electrons in 1st shell and 5 electrons in 2nd shell(2+3).

In this case the 2nd shell is the valence shell and that contains two subshells s and p. Thereare 5 electrons in the valence shell(2 in s and 3 in p subshells).We have to draw the boxdiagram of each subshell and then find how many unpaired electrons are present. Note thatwhile filling electrons in p, d or f subshells, we have to make use of the Hund's rule becauseeach of these subshells contain more than one degenerate orbitals.

2s 2p2 3

Here we find the two s-elctrons are paired according to Pauli's exclusion principle. But whilefilling the three degenerate orbitals of the p-subshell we have to apply Hund's rule. Eachorbital has to be filled with one electron with parallel(same) spin. So there are three unpairedelectrons in nitrogen atom. Let us take another example.Example 3: Cl(17): 1s2, 2s2, 2p6, 3s2, 3p5.and shell wise it will be, 2, 8, 7The valence shell in this case is the 3rd shell and that contains 7 electrons, out of which twoare in s and remaining 5 are in p-subshells. Let us draw the box diagram of this subshells.

23s 3p5

The s-subshell contains one pair of electrons. But while filling the p-subshell according toHund's rule, we find that there is only one unpaired electrons.LONE PAIRS:The pairs of electrons present in the valence shell are called the lone pairs. In the aboveexample, there are three lone pairs in chlorine atom.Example 4: Ca(20): 1s2, 2s2, 2p6, 3s2, 3p6, 4s2

Shell-wise the configuration is: 2, 8, 8, 2.In this case the valence shell is the 4th shell and we have to draw its box diagram.

24sSo we find that there is no unpaired electron. But there is one lone pair.

Example 5: Fe(26): 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6

Shell-wise configuration: 2, 8, 14, 2In this case, we have to draw both 4s and 3d subshells. Since 3d subshell has 6 electronsand is incomplete, it must contain unpaired electrons. Note that although the valence shell isthe 4th shell, we have to draw the box diagram of 3d as it contains unpaired electron.

24s 3d6

When filling six electrons in the five d orbitals following Hund's rule, we find that there arefour unpaired electrons.SAQ 24: Write the electronic configuration both subshell wise and shell-wise for thefollowing. Also draw the box diagram of the valence shell and other incomplete subshell, ifany, and find the number of unpaired electrons in each case.The atomic numbers are givenwithin brackets.He(2), Be(4), B(5), C(6), O(8), F(9), Ne(10), Mg(12), Al(13), Si(14), P(15), S(16), Ar(18),K(19), Sc(21), Ti(22), V(23), Mn(25), Co(27), Ni(28), Zn(30), Br(35), I(53), Sn(50), Hg(80)

Stability of Half-filled and Full-filled Subshell configuration:(Configuration of Cr and Cu)We know from (n+l) rule that 4s<3d, but their energy difference is very small. Exchange ofelectron from 4s-subshell to 3d subshell sometimes occurs particularly when there is extrastability achieved due to full-filled and half filled subshell configuration. Look to the followingcase to get a better understanding. The expected electronic configuration of Cr(24) is givenbelow.

Cr(24): 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d4 (wrong)

43d4s2

In this configuration there are four unpaired electrons in Cr. But actually Cr contains sixunpaired electrons which has been revealed from experiment. How can we account for this?Actually one electron slips from the 4s subshell to the vacant 3d subshell, then both thesubshells become half filled. The maximum capacity of s-subshell is 2 and that of d-subshellis 10. So 1 and 5 electrons make them half-filled respectively. Since half-filled or full-filledsubshell configuration are more stable, such a shifting of electron occurs from 4s to 3d. Sothe correct configuration is:

Cr(24): 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 (correct)

3d4s2 5

In this configuration of Cr, we find a total of six unpaired electrons.Cu(29): 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d9 (wrong)

In this case too, one electron from 4s subshell goes to the 3d subshell to make it full-filled(3d10) and itself becomes half filled. Since this is more stable configuration, the correctconfiguration is

Cu(29):1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10 (correct)

3d4s2 10

In this configuration of Cu, the 3d subshell is complete(10) and 4s subshell contains oneunpaired electron(half filled).Electronic Configuration of Ions:Example: Na+: Sodium has 11 electrons, when Na is converted to Na+ ion, one electron

has been lost. So the number of electrons in Na+ is 10. We have to fill these 10 electronsin different shells and subshells.

Na+(10): 1s2, 2s2, 2p6: (2, 8). Thus we find that 1st and 2nd shell are complete andthere is no unpaired electrons.

Cl-: Cl has 17 electrons, but when Cl is converted to Cl- ion, it gains one additionalelectron. So the total number of electrons is 18.

Cl-(18): 1s2, 2s2, 2p6, 3s2, 3p6 (2, 8, 8). Thus we find that 1st and 2nd shells arecompletely filled. The 3rd shell however is not completely filled but the two subshells s andp are completely filled and there is no unpaired electrons.SAQ 25: Write the electronic configurations of the following. Find the number of unpairedelectrons.

(i)Fe2+ (ii)S2- (iii)Cr3+ (iv)Cu+ (v)Zn2+

Isotopes, Isobars and IsotonesWe have studied in details about isotopes in the chapter, Atomic Mass.Isobars are different elements which have the same mass number. For example:

18Ar40,

19K40 and

20Ca40 have same mass numbers(40), but they have different atomic numbers.

They differ both in the number of their protons and neutrons.Another pair of isobars is 22

Ti50

and 24

Cr50.Isotones are different elements but have same number of neutrons in their nuclei.

19K39 and

20Ca40 have same number of neutrons(20) and are isotones.

SAQ 26: Indicate which pair of elements are isotopes, isobars and isotones.(i)

6C14,

7N14 (ii)

6C12,

6C13 (iii)

1H1,

1D2 (iv)

5B11,

6C12

(v)27

Co60, 28

Ni60 (vi)8O18,

9F19

Do you know why the shells of an atom are designated as K, L, M. N......., not A,B, C, D.......?A: A girl was given a 18th century old child garment to mend for exhibition purpose. To herutter surprise she found a 7" wide tucks of extra fabric in the shoulder and side seams. Then sherealised the reason for it. Since in those times material was valuable and never wasted and thedress was meant to last a lifetime, and the extra cloth was added to allow the dress to grow withits wearer over the years. Unfortunately the child wearing that dress might not have grownfurther or might have died and it remained as such without expanding. Same thing happened tothe designation of K, L, M..... to the shells.

In early 20th century, x-rays were discovered by the bombardment of cathode rayson metals. Barkla found two types of x-rays from metals- one was of more penetrating typecalled as K- radiation and the other less penetrating type called the L-radiation. He anticipatedthat there might be still more penetrating x-rays which will be discoverd later and so hestarted naming from the middle of alphabets. But unfortunately that never happened. So theletters K, L remained as such and the shells were designated as K, L, M, N.....In fact themost intese K- radiaon (x-rays) are produced when electron from a metal is knocked fromthe 1st shell by electron beam and is again recaptured and L radiation(less intense) areproduced when electron is knocked out from 2nd shell and is then recaptured.

RESPONSE TO SAQs(Atomic Structure)

SAQ 1:(i) Nearly 1840 times because a neutron has almost same mass as a proton.(ii) - 3 X 1.602 X 10-19 coul. (iii) 2 X 1.602 X 10-19 coul.(iv) Ten (v) 6.023 X 1023 X 1.602 X 10-19=96,488 coul. This is called one faraday(1F ≈ 96,500coul) of charge.SAQ 2:(i) The alpha particles emitted out of the radioactive element, polonium are dangerousto environment and human health. Thick lead sheet is a good absorber of alpha particles. Itdoes not allow the alpha particles to pass through it. Therefore the radioactive metal, poloniumis kept inside a lead box and a thin beam of alpha rays is allowed to pass through a fine hole.(ii) It is because alpha particle is much lighter than the nucleus of gold atom. Thereforethe impact of the alpha particle onto the massive gold nucleus is not able to dislocate anddislodge the nucleus from its original position.(iii) Gold is a very malleable metal. Malleable means it can be hammered to form veryvery thin sheet to extent of 0.0004cm thickness. Imagine how thin such a foil would be!!!Other metals do not have such malleability and for that reason gold was preferred.SAQ 3: 1 cm = 108 Å, So 0.001cm = 0.001X 108 Å = 105 Å

Again 1 cm = 107nm, So 0.001cm = 0.001 X 107 = 104 nmSAQ 4: 1 Å = 10-8cm, So 1000 Å = 1000 X 10-8cm = 10-5 cm.

Again 1 Å = 10-10m, So 1000 Å = 1000 X 10-10= 10-7m.SAQ 5: (i)UV rays (ii)x-rays (iii)IR rays(iv)Gamma rays

(v)x-rays (vi)Microwave (vii)Cosmic rays(viii)Visible rays

(ix)Radiowaves.SAQ 6:(i) 3 X 1010cm/sec. It is same as the velocity of infrared rays and in fact any other

electromagnetic wave.(ii) ν = (3 X 1010)/(500 X 10-7)=6 X 1014 sec-1;

Wave number = 1/(500 X 10-7)=2X 104 cm-1.This wave belongs to visible radiations( 500nm=5000 Å)

(iii) cosmic rays < x-rays < uv rays < microwaves < radiowaves.(iv) cosmic > gamma > x-rays > uv rays > visible > IR > microwave > radiowaves(v) Wave number is directly proportional to frequency.

microwaves < visible rays < ultraviolet rays < gamma rays(vi) x-rays: picture of bones; gamma rays: treatment ofcancer

microwave: telephone communications; radiowaves: radio broadcasting(vii) Infinite(viii) We can get UV radiations from a mercury vapour lamp. Sun also produces UV rays

which is largely absorbed by the ozone layer. UV rays can cause damage to the skin.SAQ 7:

(i) The statement is true. Their relation is hc=E

(ii) x-ray photon is more energetic than IR photon, because the wavelengths of x-rayradiations are smaller than IR radiations(Refer the electromagnetic spectrum) and energy isinversely proportional to wavelength.(iii) In electromagnetic spectrum, cosmic rays are most energetic as its wavelength is theshortest and radiowaves are least energetic as its wavelength is the greatest.

(iv) Energy of a photon is directly proportional to wave-number. hcE = (v) The statement is not true, because according to Planck's quantum theory,electromagnetic radiations are not emitted or propagated in space continuously but take placein the form of discontinuous energy packets called quanta or photons.(vi) Because the photons are emitted and propagated so fast that the discontinuity is notnoticed.

(vii) E-12

2.84 X 10=-87000 X 10

-27 106.627 X 10 X 3 X 10

= erg

(viii) 6.627 X 10=15

6.627 X 10 X 10E-12-27

= erg

(ix)-16

1.98 X 106.627 X 10 X 3 X 10 X 1 = E -27 10= erg

(x) hc=E ⇒

10-276.627 X 10 X 3 X 10 1.98 X 10

-18=

⇒ = 100 cmIt belongs to the radiowaves region of the electromagnetic spectrum.E = hν = 6.627 X 10-27 X ν = 1.98 X 10-18 erg.⇒ ν = 2.98 X 108 sec-1

hcE = ⇒ 6.627 x 10-27 X 3 X 1010 X = 1.98 X 10-18 erg.

⇒ = 0.01 cm-1

SAQ 8:(i) White (ii)Violet and red colours respectively(iii) No, this statement is incorrect. Green colour consists of a large number of waveshaving approximate wavelength range of 5000-5500 Å. In the vicinity of 5000 Å, the greenis mixed with faint blue as the green colour comes just after blue colour. In the vicinity of5500 Å green colour is mixed with faint yellow colour as yellow colour comes just aftergreen. Thus there is a continuity of colour throughout the spectrum(VIBGYOR). Refer thepicture of solar spectrum in the text.(iv) Violet colour radiations have the shortest wavelengths and travel with minimumvelocity in the prism and hence is deviated to the maximum.(v) Visible radiations have wavelengths greater than UV radiations and come just afterit while x-rays radiations have wavelengths smaller than UV radiations and come just beforeit.

(vi) No, we do not get any colour in any spectrum other than visible spectrum.SAQ 9:(i) Discontinuous(ii) FALSE. We get individual lines which are well separated from each other.(iii) FALSE. We get a band of seven colours having no gap in between.SAQ 10:(i) True(ii) False: Solar spectrum is continuous while H-spectrum is discontinuous (line) spectrum.(iii) Spectrometer is the instrument which separates a mixture of waves and disperses

them into pattern of waves with increasing wavelengths from bottom top. The simplespectrometer consists of a prism and a photographic plate(screen).

SAQ 11:(i) Emitted. (ii) Absorbed(iii) (a)Yes. It will lose by an amount (q-p) joule and it will go to the first shell(K).

(b)No. It cannot go closer to nucleus than the K shell.(c)Yes. It can lose (r-q)joule and go to L shell(2nd) and lose (r-p) joule to go to Kshell.Thus there are two possibilities.(d)K shell. By absorbing it will go to L,M,N..... It cannot go closer to the nucleus,

so will not emit energy.(e)There are six possible transitions: These are 4---->3, 4---->2, 4---->1, 3---->2,3---->1 and 2---->1. The jump 4---->1 will emit greatest energy photon and 2---->1

will emit the least energy photon.SAQ 12:(i) When the electron is at infinite distance i.e when the electron is far away from thenucleus, the energy of the electron is assumed to be zero. When it comes closer to thenucleus, it is more and more attracted by the nucleus. Remember that whenever there isattraction, there is a net decrease or loss of energy. The loss is associated with a negativesign. As the electron comes nearer to the nucleus its energy becomes more and morenegative as more and more energy is lost. Let us take one example. You have some moneywith you which you declare to us to be zero rupee. Supposing you lose 500 rupees out ofthat. What will be your wealth amount? It will be -500. Similarly if you lose 1000 rupees,your wealth value will be -1000 and so on. Similarly as electron goes closer to nucleus itloses more and more energy and takes more and more negative values. Numerically thevalue increases but as a whole the value decreases. When electron reaches in the 1st shellfrom infinite distance, it has lost k(2.11X 10-11erg) amount of energy, so that its energy in the1st shell is -k. Similarly when electron comes from infinite distance to 2nd shell, it has lostk/4 amount of energy, so that its energy in the 2nd is -k/4. Similarly, the energy in the thirdshell is -k/9 and so on. I guess now you understand why the energy of the electron isnegative.(ii) In the example solved just before in the text, the photon absorbed when electronjumps from 1st to 2nd shell has a wavelength of 1256Å. So when it jumps from 2nd shellback to 1st shell, photon of same value i.e 1256Å will be emitted.(iii) Energy of electron in 3rd shell = E

3 = - k/9 and in the first shell = E

1= -k

So the difference = hcE = E - E = 13 = -k/9 - (-k) = 8k/9

⇒ hc

= 8k9

⇒ =8 k9 hc -5

1.060 X 10=-27 10

-118 X 2.11 X 10 erg

9 X 6.627 X 10 erg.sec X 3 X 10 cm/sec= cm

= 1.060 X 10-5 X 108 Å = 1060 Å.So the photon absorbed for the jump is 1060Å and it falls in the ultraviolet region of theelectromagnetic spectrum.(iv) When electron jumps from 3rd shell back to 1st shell, photon of same wavelengththat was absorbed while jumping from 1st to 3rd shell i.e1060Å(SAQ iii) will beemitted.(v) Energy of the electron in the 3rd shell = E

3 = -k/9 and in the 2nd shell=E

2= -k/4

So the difference = 4 hcE = E - E = 2 = -k/9 - (-k/4) = 5k/36

⇒ hc

= 5k36

⇒ =5 k36 hc -5

6.784 X 10=-27 10

-115 X 2.11 X 10 erg

36 X 6.627 X 10 erg.sec X 3 X 10 cm/sec= cm

= 6.784 X 10-5 X 108 Å = 6784 Å. So the photon having wavelength of 6784Å isto be absorbed when electron jumps from 2nd shell to 3rd shell. This photon belongs to thevisible region(3800-7600Å)of the electromagnetic spectrum.If the electron jumps from 3rd shell back again to 2nd shell, then the photon of samewavelength i.e 6784Å will be emitted out.(vi) Energy of the electron in 6th shell= E

6 = -k/36 and in the 3rd shell = E

3 = -k/9.

E6 is greater than E

3. As electron goes farther away from the nucleus to higher and higher

shells, its energy value goes on increasing.(vii) The energy of the electron is zero at infinite distance and -k in the 1st shell. So theenergy difference = E∞ − Ε1 = 0 - (-k)= kSo k(2.11 X 10-11erg per atom) is required for the electron to jump from 1st shell to infinitedistance.(viii) The energy difference between E

2 and E

1 is less than between E

3 and E

1 which in

turn less than the energy difference between E4 and E

1. As the energy of the photons go

on increasing in the three jumps mentioned, the wavelength will go on decreasing. This isbecause the energy is inversely proportional to wavelength.SAQ 13:(i) (a)n=4 to n=2 (b) n=2 to n=1 (c) n=6 to n=3 (d) n=4 to n=1

(e)n=5 to n=2 (f)n=6 to n=4 (g)n=6 to n=5(ii) ΔE = E

3-E

1 = -k/9 -(-k) = 8k/9 = hc/λ ⇒ λ = 1060 Å (UV)

(iii) ΔE = E5-E

2 = -k/25 - (-k/4) = 21k/100 = hc/λ ⇒ λ = 4485 Å (Visible)

(iv) ΔE = E5-E

3 = -k/25 - (-k/9) = 16k/225 = hc/λ ⇒ λ = 13246 Å (IR)

(v) Continuous or band spectrum. Actually the spectrum is a line spectrum.

SAQ 14:(i) Electromagnetic waves do not carry any real material particles. They carry onlyenergy packets(photons); eg. UV, Visible, IR etc. Electron waves carry real material particles,like electron(microscopic), bullet(macroscopic) etc. All electromagnetic waves travel with thesame speed( 3 X 1010cm/sec) while electron wave or any other matter waves can travel inany speed.(ii) Orbital is the three dimensional space in which the electron possessing a particularenergy value can remain anywhere within it.SAQ15:(i) l value for any p-subshell is 1. (l=1)(ii) 3rd shell has 3 subshells. The three l values are 0, 1 and 2 for s- , p- and d- subshellsrespectively.(iii) n=2, for p-subshell l=1. Note that p-subshell is denoted by l=1 irrespective to which

shell it belongs.(iv) The electron is present in the 4th shell(n=4) and f-subshell(l=3) of the 4th shell. Note

that the 4th shell has four subshells namely s-, p- , d- and f-subshells.(v) p-subshell of second shell has n=2 and l=1 but the p-subshell of the third shell hasn=3 and l=1. Although their l values are same, their n values are different. Fromthis the two can be distinguished.(vi) n=5 and l=0. Note that for any s-orbital irrespective of the shell, the l value is 0.(vii) The first electron will have n=4 and l=3 while the second electron will have n=5 and

l=3. Their principal quantum numbers(n) are different.(viii) The 5th shell will have five subshells given by the l values 0, 1, 3 and 4 represented

by the symbols, s, p, d, f and g respectively.SAQ 16:(i) 2p>2s (ii) No, The actual order is 3s<3p<3d(iii) 4f (iv)In the 4th shell(N) and in its p subshell.(v) No, the second shell(n=2) has two subshells represented by l=0(s) and l=1(p).(vi) 2p has higher energy.SAQ 17:(i) m= -1, 0, +1. p-subshell contains three p orbitals(ii) For l=2, five(2l+1) values of m are possible. They are: -2, -1, 0, +1, +2. There arefive d- orbitals.(iii) s-subshell has only one orbital. The subshell itself is the orbital. For this, l=0 and m=0(iv) For f-subshell, l=3 and for l=3, the values of m are: -3, -2, -1, 0, +1, +2 and +3. These

seven values represent seven f-orbitals.SAQ 18:(i) It is 3p orbital. We cannot say which orbital out of the three orbitals(p

x, p

y or p

z) at

this level. The elctron is either spinning clockwise or anticlockwise.(ii) It is in the 2s orbital.(iii) It is the 2s subshell. An s-subshell has the l value=0, hence the m value given is not

possible.(iv) NO. It is the 4d subshell which has m= -2, -1, 0, +1, +2. So the m value given isnot possible.(v) n=4, l=0, m=0 and s = +½ or -½.

(vi) n=2, l=1, m = -1, 0 or +1 and s= +½ or -½.(vii) n=3, l=2, m=-2, -1, 0, +1 or +2 and s=+½ or -½.SAQ 19:(i) The second electron has two options. One, it may go to a higher energy orbital,second it may remain with the first electron in the same orbital with opposite spin withrespect to the first electron. The first requires more energy than the second option. So itremains in the same orbital.(ii) The repulsion between the third electron with the electron pair already present in theorbital is unfavourably high and requires more energy than if the third electron would havegone to a higher energy orbital. Thus the third electron cannot stay in the same orbital andit goes to the next higher energy orbital.SAQ 20:(i) 2 X 52 = 50 elctrons(ii) For p-subshell: l=1 and for this there are three m values: m= -1,0,+1 for threeorbitals(p

x, p

y and p

z). Each orbital contain two electrons with s= +½ and -½.

So in total the p- subshell will contain 3X2=6 electrons.(iii) A d-subshell can accommodate 10 electrons. The l value for the d -subshell = 2 and

there are five m- values(-2,-1,0,+1,+2) possible for it. The five m- values representfive orbitals. Each orbital contains two electrons with s= +½ and -½. So the totalelectrons in the d-subshell= 2 X 5 =10SAQ 21:

(i)3d7

The number of unpaired electrons: 3

(ii)3p

The number of unpaired electrons: 3

(iii) Four d orbitals will be singly occupied. The number of unpaired electrons:4(iv) After filling seven f-orbitals with one electrons, the 8th, 9th and 10th electrons willpair the three orbitals. Another four f-orbitals will contain 4 unpaired electrons.Do the box diagram for yourself.(v) Two unpaired electrons (vi) 4 unpaired electrons.SAQ 22: Let us hypothetically imagine the following situation. If the two electrons inthe neighbouring degenerate orbitals would spin in opposite directions, they would developsome affinity towards each other. This is due to the equal and opposite magnetic fields theyproduce. It may so happen that one electron may leak or slip into the other room and themoment the two remain in one orbital rendering the other one vacant, that becomes anunfavourable high energy situation as one orbital would lie vacant while the other orbitalhaving same energy would be occupied by two electrons. Thus one of the electrons fromthe pair would immediately come back to the vacant orbital from which it slipped and nowit would spin in the same direction as the other one and not in the opposite spin as it didbefore. Thus we logically proved that all the unpaired electrons in the degenerate orbitals spinin the same directions.SAQ 23: The reader is advised to find the (n+l) values of the orbitals and check withthe answers.

(i)4p<4d (ii)6s<4f (iii)5s<4f (iv)4p<5s(here n+l values are same)

(v)4f<5d(here also n+l values are same).SAQ 24: The abbreviation UE has been used for Unpaired Electrons.He(2):1s2(No UE) Be(4):1s2,2s2(No UE);

B(5):1s2,2s2,2p1(UE-1); C(6):1s2,2s22p2(UE-2)O(8): 1s2, 2s2,2p4(UE-2) F(9):1s22s22p5(UE-1);Ne(10): 1s2,2s2,2p6(No UE) Mg(12):1s2,2s2,2p6,3s2(No UE)Al(13): 1s2,2s2,2p6,3s2,3p1(UE-1) Si(14): 1s2,2s2,2p6,3s2,3p2(UE-2)P(15): 1s2,2s2,2p6,3s2,3p3(UE-3) S(16): 1s2,2s2,2p6,3s2,3p4(UE-2)Ar(18): 1s2,2s2,2p6,3s2,3p6(No UE) K(19): 1s2,2s2,2p6,3s2,3p6,4s1(UE-1)Sc(21): 1s2,2s2,2p6,3s2,3p6,4s2, 3d1 (UE-1) Ti(22): 1s2,2s2,2p6,3s2,3p6,4s2,3d2(UE-2)V(23): 1s2,2s2,2p6,3s2,3p6,4s2,3d3(UE-3) Mn(25):1s2,2s2,2p6,3s2,3p6,4s2,3d5((UE-5)Co(27): 1s2,2s2,2p6,3s2,3p6,4s2,3d7(UE-3) Ni(28): 1s2,2s2,2p6,3s2,3p6,4s2,3d8(UE-2)Zn(30): 1s2,2s2,2p6,3s2,3p6,4s2,3d10(No UE)Br(35): 1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p5(UE-1)I(53): 1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,5s2,4d10,5p5 (UE-1)Sn(50): 1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,5s2,4d10,5p2(UE-2)Hg(80): 1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,5s2,4d10,5p6,6s2,4f14,5d10(No UE)The reader is advised to add the electrons present in all the subshells of each shell and writedown the shell-wise configuration. Let us take the case of Hg(80)Hg(80): 2, 8, 18, 32, 18, 2SAQ 25:

(i)No. of electrons=26-2=24: 1s2,2s2,2p6,3s23p64s03d6: 4 unpaired electrons.Note that the two electrons lost from Fe atom to form Fe2+ goes from 4s, the valenceshell(not 3d). So in the configuration of Fe2+, 4s0 has been written. You can also delete 4sfrom the configuration.In all the elements in the transition series given below(exceptingS2-, which is not a transition element), first the electronic configuration of the neutral atomis written, then the required number of electrons are removed first from the 4s and then ifrequired from 3d. Hund's rule is applied to the incomplete 3d subshell to find the number ofunpaired electrons. One thing you bear in mind that while filling electron, the energy orderis 4s<3d [ ns < (n-1)d), but after filling 3d(at least with one electron), the energy order isreversed i.e 4s>3d [ns>(n-1)d]. Since 4s is of higher energy, the electron is lost first fromthat orbital and subsequently from 3d. Why this is so, we will not discuss here.

(ii)No of electrons=16+2=18: 1s2,2s2,2p6,3s2,3p6: 0 unpaired electron(iii)No. of electrons=24-3=21: 1s2,2s2,2p6,3s23p64s03d3 3 unpaired electrons(iv)No. of electrons=29-1=28: 1s2,2s2,2p6,3s23p64s03d10 0 unpaired electron(v)No. of electrons =30-2=28: 1s2,2s2,2p6,3s23p64s03d10 0 unpaired electron

SAQ 26: (i)isobars (ii)isotopes (iii)isotopes (iv)isotones (v)isobars(vi)isotones