Atomic Structure

Reconstruct Your Chemistry With Prince Sir

BASIC KNOWLEDGE Atom & Molecules (a)

The samllest particle a matter that takes part in a chemical reaction and maintains its identity through out the chemical reaction is called an atom. The atom of all gases of all except those of noble gases, cannot exist in free state. These exist in molecular form. The molecules of hydogen, nitrogen, oxygen and halogens are diatomic (H2, N2). Phosphorus molecule is tetratomic and that of sulphur is octa atomic.

(b)

The smallest paticle of matter that can exist in free state in nature, is known as a molecule.

(c)

Some molecules are composed of homoatomic atom, e.g., H2, O2, N2 Cl2, O3 etc., while the molecules of some compounds are made up of two or more hetroatomic atoms e.g., HCl, NaOH, HNO3, CaCO3, etc.

Dalton’s Atomic Theory

r i S e c n i r P By

The concepts put forward John Dalton regarding the composition of matter are known as Dalton’s atomic theory. Its important points are as follows. (a)

Every matter is composed of very minute particles, called atoms that take part in chemicel reactions.

(b)

Atoms cannot be further subdivided.

(c)

The atoms of different elements differ from each other in their properties and masses, while the atoms of the same element are identical in all respects.

(d)

The atoms of different elements can combine in simple ratio to form compounds. The masses of combining elements represent the masses of combining atoms.

(e)

Atom can neither be created nor destroyed.

Modern Concept : Many of the concepts of Dalton’s atomic theory cannot be explained. Therefore, foundation of modern atomic theory was laid down by the end of nineteenth century. The modern theory is substantiated by the existence of isotopes, radioactive disintegration, etc. The imnprotant points of the modern atomic theory are as foilows. (a)

An atom is mainly composed of three fundamental particles, viz. electron, proton and neutron.

(b)

Apart from the aforesaid three fundamental particles, many others have also been identified, viz. positron, meson, neutrino, antiproton, etc.

(c)

Soddy discovered the existence of isotopes, which were atom of the same element having different masses. For example, protium, deuterium and tritium are atoms of hydrogen having atomic masses 1, 2 and 3 a.m.u. respectively.

(d)

Atoms having same mass have different atomic numbers. These are known as isobars. For example,14 80 Ar 40

and Ca. 20

(e)

Atoms of elements combines to form molecules.

(1)


Cathode Rays (Discovery of Electron)

Production of Cathode Rays


The salient features of cathode ray are as follows. (a)

Cathode rays travel in a straight line. This indicates that the formation of a shadow when an opaque object is placed in its path.

(b)

If a light metal pinwheel is placed in the path of cathode rays, the wheel starts revolving. This proves that is cathode rays consits of tiny particles having momentum.

(c)

Cathode rays get deviated in electrical and magnetic fields. This proves that they are composed of charged particles. Their derivation towards anode indicates their negatively charged nature. The direction of their deviation in magnetic field depends on pole of the magnet which has been placed near the cathode ray tube.

(d)

Cathode rays produce green fluorescence on the walls of the glass tube.

(e)

Cathode rays produce incandescence in at thin metal foil.

(f)

Cathode rays effect the photographic plate.

(g)

Cathode rays ionize gasses proving that they are charged.

(h)

Cathode rays pernetrates across a thin metal foil.

(i)

Cathode rays produce X-rays when hit a piece of tungsten or any other metal having high melting point.

Nature of Cathode Rays J.J. Thompson (1897) proved through experiments that. (a)

Cathode rays are composed of ex tremely tiny negatively charged particles (electrons).

(b)

The ratio of negative charge (e) and mass (m) for cathode ray particle (electrons) is a constant. This ratio is independent of the material used in the preparation of the electrodes of the discharge tube or the gas filled in it. Thus, e/m of an electron is a universal constant. e ch arg e on electron = = 1.76 × 108 Coulomb/gm m mass of electron

In addition to the above proofs, photoelectric effect, thermionic effect and emisson of beta particles from radioactive elements also confirm that electron is an essential constituent of matter. These negatively charged tiny particles discovered by Thomson. It is denoted by e– or –1e0. (2)


Electron (e– or –1e0) (a)

Electron is a fundamental particle of an atom, which is an essential constituent of every matter.

(b)

The value of

(c)

R.A. Mulliken calculated the charge on an electron by his famous Oil Drop Experiment. The value came out to be 1.6012 × 10–19 coulomb or 4.803 × 10–10 e.s.u.

(d)

The value of e/m of an electron is known as its specific charge. With the help of this specific charge and the charge on the electron (determined by Mulliken), the mass of the electron could be calculated as follws.

e was found to be 1.76 × 108 coulomb/gram by Thomson. m

e 1.6012  1019 coulomb   mass of electron(m e )  9.1091  1028 gram 8 e / m 1.76  10 coulomb / gram (e)

Molar mass of electron is obtained on multiplying mass of electron by Avogadro number (6.023×1023). Therefore gram molecular mass of electron is as follows.

r i S e c n i r P By = 9.1091 × 10–28 × 6.023 × 1023 = 5.483 × 10–4g

(f)

The mass of an electron in motion is calculated with the help of the following expression. Mass of electron in motion (m) =

Re st mass of electron   v 2  1     c  

where v is velocity of electron and c velocity of light.

When v = c, the mass of the electron in motion becomes infinity.

Therefore the mass of an electron increases with increases in its velocity due to which specific charge e/m on it decreases. (g)

The physical and chemical properties of an element depend on the distribution of electrons in its outermost energy level.

Anode Rays or Canal Rays : Discovery of proton

Production of Cathode Rays

(3)


Properties of Anode Rays (a)

Positive rays travel in straight line.

(b)

Positive rays affect photographic plate.

(c)

Positive rays are deviated in the electron and magnetic fields. The direction of their deviation proves the presence pf positive charge on their particles.

(d)

Positive rays pass across a very thin sheet of metal. But their penetrating power is less than that of cathode rays.

(e)

Positive rays produce fluorescence and phosphorescence.

Nature of Anode Rays Thomson and Wein studied the nature of positive rays and proved with the help of experiments that. (a)

Positive rays are composed of positively charged particles.

(b)

The ratio (e/m), of positive charge (e) and mass (m) for the particles of positive rays depends on the nature of the gas filled in the discharge tube. The value of e/m for the particles of positive rays obtained from different gases is different. The e/m value for positive rays is not a universal constant. Thomson and Wein found out theough experiments that the maximum value of e/m is for particles of positive rays of hydrogen gas.

r i S e c n i r P By Proton

(a)

Proton is a fundamental particle of an atom. It is an essential constituent of every matter.

(b)

The credit for the discovery of proton goes to Goldstein.

(c)

Proton bears one unit positive charge.

(d)

Thomson and Wein estimated the value of e/m as 9.578 × 104 coulomb per gram for the positively charged particle proton.

(e)

The amount of positive charge (e) on proton is 1.602 × 10–19 coulomb or 4.8 × 10–10 e.s.u.

(f)

Mass of proton (m) = 1.6725 × 10–24 gram ; = 1.6775 × 10–17 kilogram = 1.6725 × 10–29gram ; = 1837 times that of electron = 1.00757 a.m.u.

; = Mass of hydrogen atom

1.6725  1024 Mass of proton (m) a.m.u = = 1.00757 a.m.u. 1.66  1024 (g)

Mass of proton (m) multiplied by Avogadro number (6.023 × 1023) gives molar mass of proton. Thus Gram molecular mass of proton = 1.6725 × 10–24 × 6.023 × 1023 = 1.008 (Approx)

(h)

Proton is present in the nucleus of an atom.

(i)

The number of electrons is equal to the number of protons in a neutral atom.

(j)

The atomic number of an atom is equal to the number of protons present in the nucleus of that atom.

(k)

Proton is the nucleus of protium i.e. the common hydrogen atom.

(i)

Proton is ionized hydrogen atom, i.e. (H+)

(m) Proton is obtained when the only one electron present in hydrogen atom is removed. Hydrogen atom consists of anly one electron and proton.

(4)


Discovery of Neutron Penetrating rays are emitted on bombarding-particles on the elements like beryllium, boron and aluminium. James Chadwick in 1932 studied the nature of these radiation and came to the conclusion that these rays are composed of very tiny electro neutral particles. The mass of these particles is almost equal to that of the hydrogen atom. This particle is called neutron and is denoted by the symbol, 0n1.

Neutron(0n1) (a) It is a fundamental particle of atom that is present in the nuclei of all atoms except hydrogen or protium. (b) It was discovered by James Chadwick in 1932. (c)

It is an electro neutral particle, i.e. it does not have any positive or negative charge on it.

(d) The mass of a neutron is almost equal to that of a proton. Actually it is a little bitheavier than proton. Its mass (m) is as fallows : Mass (m) of a neutron = 1.6748 × 10–24 gram = Approximately mass of a proton


(e) Neutron is relatively heavier out of the three fundamental particles of an atom.

Other Particles of Atom

(a)

Positive : It was discovered by C.D. Anderson in 1932. It beards a unit positive charge and its mass is equal to that of an electron. Thus its mass regarded as negligible. It merges with an electron and emit electromagnetic radiations. It is denoted by e+.

(b)

Meson : Yukawa in 1935 discovered this particle. Different types of meson particles are possible in the atom. These are called meson family.

(c)

Neutrino : Pauling discovered this particles in 1927. They do not bear any charge, i.e. they are electro neutral particle.

(d)

Antiproton : Segre discovered this paticle in 1956. It bears a unit negative charge and its mass is equal to that of a proton.

Nuclear Charge and Atomic Number Positive charge on the nucleus of an atom is equal to the atomic number of that atom. A scientist named Mosley studied the frequency of X-rays emitted by showering high velocity electrons on a metal and established the following relationship.

v = a (z – b) where v = frequency of X-rays

z = atomic number or nuclear charge

a and b are constants. Thus nuclear charge of an atom is equal to the atomic number of that atom. Sincc an atom is electron neutral, the number of positively charged proton in its nucleus is equal to the negatively charged electrons moving around the nucleus in the atom. Thus Atomic number = number of protons in the atom or number of electrons in the atom Atomic Weight or Mass Number The value of mass number of an atom (in amu) is always a whole number. Mass number of an atom is the sum of number of protons and number of neutrons present in that atom. Mass number = Number of protons (Z) + Number of neutrons (n) = Atomic number + Number of neutrons (5)

Reconstruct Your Chemistry With Prince Sir (a)

The protons and neutrons present in the nucleus are known as nucleons.

(b)

The weight of electrons is neglected during calculation of the atomic weight, be because the mass of an electron is negligible in comparison to that of a proton or a neutron.

(c)

In the nucleus of anutral atom, the number of positively charged protons is equal to that of negatively charged electrons.

(d)

The number of protons present in an atom is called atomic number of that atom.

(e)

Kernel : The group of all the electrons except those of the outermost energy level, is called that kernel of that atom and the electrons present in the kernel are known as electron of the kernel. For example, if the electroinc configuration of an atom is 2, 6 then the number of kernel electrons is 2. If the electronic configuration of an atom is 2, 8, 8, then the number of kernel electrons is 10. If the electronic configuration of an atom is 2, 8, 8, 8, then the number of kernel electrons is 18.

Isotops


(a)

The atoms of the same element having same atomic number but different atomic weights, are called isotopes.

(b)

Isotopes of an element have same number of protons but different number of neutrons in their atoms. Hence their atomic weight are different. For example, oxygen has the following three isotopes. 8

(c)

O 16

8

O 17

8

O 18

Proton

8

8

8

Neutrons

8

9

10

Atomic weights

16

17

18

Hydrogen has the following three isotopes. H1(Protium)

D2 (Deuterium)

T3 (Tritium)

1

1

1

Proton

1

1

1

Neutrons

0

1

2

Atomic weights

1

2

3

Isobars (a)

Isobars are the atoms of different elements having same atomic weight.

(b)

Isobars have different numbers of proton as well as neutrons.

(c)

The sum of number of protons and neutrons in isobars is same. For example Atomic weight of three elements 18Ar40, 19K40, and 20Ca40 is 40. (i)

Ar40

K 40

Ca40

Protons

18

19

20

Neutrons

22

21

20

Isotones The atoms having same number of neutrons are called isoneutronic or isotones. For example 14

Si30

15

P 31

14

S 32

Protons

14

15

16

Neutrons

16

16

16

Atomic weight

30

31

32 (6)


Isoelectronic The chemical species in which number of electrons is same are called isoelectronic. For example (a)

Li+

Be+2

B+3

Electrons

2

2

2

(b)

Na+

Mg+2

Al+3

F–

O –2

Electrons

10

10

10

10

10

(c)

K

+

Ca

Electrons

18

18

+2

Ar 18

THOMSON’S MODEL OF AN ATOM


(i)

Atom is a very minute, spherical, electro neutral particle that consists of positively and negatively charged matter.

(ii)

The positively charged matter is uniformly distributed in the atom and the negatively charged electrons are embedded in it just as the seeds in water melon. Therefore, Thomson model of an atom is also called “water melon model”.

(iii) Thomson’s model of an atom failed to explain the production of the atomic spectrum.

RUTHERFORD’S MODEL OF AN ATOM

Ernest Rutherford in 1911 put forward the “nuclear model” of atom on the basis of  particle scattering experiment. In this experiment, Rutherford bombarded  - particles (Helium nuclei, He+2) on a thin gold foil and observed that most of the  -particles travelled straight without deviation in the direction of their path, some of them deviate form their path by different angles, while very few get rebounded after colliding with the foil. Rutherford gave the following nuclear model on the basis of the experiment. (i)

Atom is a very m inute, spherical, electro neutral particle composed of the following two parts : (a) Positively charged nucleus and

(ii)

(b) a vast extranuclear space in which electrons are present.

Whole of the positive charge and almost all the mass of atom is confined to a very mintue part at the center of the atom, called the nucleus of the atom. The radius of nucleus is about 10–13 to 10–12 cm (or10–15 to 10– 14 meter), while the radius of atom is in the order of 10–8 cm.

(iii) The number of electrons in an atom is equal to the number of protons present in the nucleus. That is way an atom is electroneutral. (iv) This model of an atom is also called “solar model” of “planetary model”. This is because, the movement of electrons around the nucleus in this model has been compared to that of planets moving around the sun in the solar system.

DEMERITS OR RUTHERFORD’S MODEL OFAN ATOM (i)

According to Clark Maxwell’s theory of electrodynamics, an electrically charged particle in motion continuously emits energy. This results in regular decrease in the energy of that particle. On the basis of this principle, it can be conciuded that an electron moving around the nucleus will contnuously emits the energy. This will result in decrease in the radius of the electron orbit, due to which the electron would ultimately fall into the nucleus. (7)


(ii)

Plunging of an electron into the nucleus would definitely mean destruction of the atom or end of the existence of the atom. But we know that it never happens. Atom is a stable system. Therefore Rutherford model failed in explaining the stability of an atomic system. (iii) If an electron moving around the nucleus continuously emits energy, then the atomic spectrum must be continuous, i.e. the spectrum should not have lines of definite frequency. However, the atomic spectrum is actually not continuous and possesses so many lines of definite frequency. Therefore, Rutherford model failed to explain the line spectrum of an atom.

BOHR’S MODEL OF AN ATOM (i)

(ii)

Neil Bohr in 1913 presented a quantum mechanical model of atom. An electron moves around the nucleus in constant circular paths.


Ou t of the infinite possible orbits electrons can revolve only in those orbits in which the angular momentum (mvr) is integral multiple of i.e.

mvr =

h . This is called the condition of quantization. 2

nh , 2

n 0, n = 1, 2, 3,.............

(iii) The centrifugal force required for the circular motion of electron around the nucles is provided by the electrostatic force of attraction between the negatively charged elctron and positively charged nucles.

m2 KZe2  2 r r

(8)


(iv) An electron when present in an orbit neither loses gains energy, these orbit are called stationary orbit. When electron jumps from lower to higher orbit it absorbs energy and when it jumps from higher to lower orbit it emitts energy. (v) The energy emitted or absorbed is equal to the difference in energies between the two energy levles. If the electron transits from n2 to n1 then energy absorbed or emitted is E = E n 2  E n1 

hc = hv 

Radius of nth Bohr’s orbit According to Bohr’s hypothesis put the value of v in mvr =

n 2h 2 or 4 2 mZe2

r

rK

nh 2

n2 z


In the above expression h, , m and e, all are constant. therefore

  h2 K   cons tan t  0.529A   2 2 4 me  

Note : (a) 1A = 10–10m

rn  0.529 

(b) 1 nm = 10–9m

n2 Å Z

(c) 1 pm (picometer) = 10–12 m

2

r

n2 r n Z  1  12  2 Z r2 n 2 Z1

Velocity of the electron of Bohr’s orbt

v

V

Z  2.188  108 cm / sec ond n

Z V Z n  1  1 2 n V2 Z2 n1

Time period =

2 r m3 m3 T n 3 Z2 = 1.51 +10-16 sec 2 T 2  1  13  22 V Z Z T2 n 2 Z1

Energy of Electron in Bohr’s nth orbit En  Er  

Z2 2 2 me4 Z2     13.6 eV n2 h2 n2 2

2

E1 Z1 n  2  22 For two different states E 2 Z2 n1

Kinetic Energy (EK)

Potential Energy (EP)

Z2 K.E. = -Er = 2 × 13.6 eV n

Z2 P.E. = 2En = -27.2× 2 eV n (9)

Reconstruct Your Chemistry With Prince Sir Total energy = Kinetic energy + potential energy E T  E K  E P (i) Total energy = –Kinetic energy (ET – – EK) (ii) (Potential energy = 2 × Total energy) (EP = 2ET).

Quantization of electronic energy Levels


Electronic energy levels of hydrogen atoms E =

hc = h  

Ground state

Ans atom in its lowest energy state or initial is said to be in ground state. This is the most stable of an atom.

Excited State The states of higher energy than the ground state are said to be in excited state. Fcr example, the electron of hydrogen atom in ground state is present in n = 1 orbit. (a) Electron in n = 2 orbit is in first excited state (b) Electron in n = 3 orbit is in second excited state (c) Electron in n = 4 orbit is in third excited state This means that the energy of n + 1 orbit is in first excited state, of n + 2 orbit is in second excited state and of n + 3 orbit is in third excited state, where n = the energy in ground state.

Excitation Potential (a) (b)

The energy required to excite an electron from ground state to any excited state is known as excitation potential. Excitation potential has a positive value. For example, I.E = E.P × charge. First excitation potential of hydrogen atom – E2 – E1 Second excitation potential of hydrogen atom = E3 – E1 Third excitation potential of hydrogen atom – E4 – E1 Energy  ch arg e  Potential (10)

EP =

IE e = = Volt. ch arg e e


Ionisation Energy or Ionisation Potential The energy required to remove an electron from the outermost orbit of a gaseous atom in ground state in called ionisation energy or ionisation potential. Its value is positive. I.E.  E   E1

Separation Energy or Binding Energy The energy required to separate an electron from any given state of an atom is known as separation energy. For example, the first separation energy, i.e. the energy required to remove an electron from the first excited state in hydrogen is + 3.4 eV.

Spectral Evidence for Quantization in Bohr’s Theory (a) (b) (c) (d)

When an electron undergoes transition from lower to higher orbit, it absorbs energy and the spectrum obtained thereby is called absorption spectrum. Whwn an electron undergoes transition from higher to lower orbit, there is emission of energy and the spectrum obtained thereby is called absorption spectrum. A hydrogen atom has only one electron, yet a very large number of lines are visible in its spectrum. The wave number of spectrum can be find out using Rydberg’s equation.

r i S e c n i r P By 

 1 1 1   R  Z2  2  2    n1 n 2 

R = Rydberg constant, n1 = Number of lower energy level

n2 = Number of higher energy level

2 2 me 4 R= Value of R = 109677 em–1  107 m–1. ch 3

SPECTRUM

Emission Spectrum and Absorption Spectrum When a beam of white light passes through a slit or an aperture and then falls on a prism, it gets spilt into many coloured bands. The image of colours so obtained is known as a spectrum. A spectrum is of mainly three types viz. (i) Emission spectrum (ii) Absorption spectrum (11)

Reconstruct Your Chemistry With Prince Sir (i)

Emission Spectrum : When energy is provided to any substance, it starts emitting radiations. These radiations are passed through a spectroscope, they get split up into spectral lines producing emission spectrum. Normally a substance can be excited by any of the following ways. (a) By heating the substance at high temperture (b) By passing electric current through a discharge tube having gaseous substance at very low pressure. (c) By passing electric discharge through a metallic filament. Emission spectra are of the following two types. (i) Continuous spectrum

(ii) Discontinuous spectrum

Continuous Spectrum When sunlight or a glowing heat fluorescent substance like tungsten wire present in an electric bulb, is analysed with the help of a spectroscope, the spectrum obtained on a screen is observed as divided into bands of seven colours, which are in a continuous sequence. Such a spectrum is called a continuous spectrum.

r i S e c n i r P By Continuous Spectrum of white light

Line spectrum or Atomic spectrum When atoms of a substance is excited, it emits radiations are analyzed with the help of a spectroscope, then many fine bright lines of specific colours in a sequence are seen in the spectrum, which is not continuous, i.e. there is dark zone in between any two lines. Such a spectrum is called a line spectrum or atomic spectrum. For example, neon single lamp, sodium vapour, mercury vapour lamp, etc. emit light of different colours anf they give specific line spectra.

Absorption Spectrum When white light emitted by glowing heat fluorescent substance is passed through another substance lime sodium substance. This results in appearance of some black lines in the spectrum. These are present at those places where the line spectrum of the substance i.e. sodium vapour is formed. The spectrum so formed is known as absorption spectrum. (12)


Hydrogen Spectrum Hydrogen atom gives line spectrum. When hydrogen gas is filled at low pressure in a discharge tube and electric discharge is passed through it, a pink coloured is produced in the visible region due to the formation of hydrogen atoms. On studying this light with the help of a spectroscope, series of lines of various wavelengths are obtained in the spectrum. The frequency of spectral lines in the form of wave number can be calculated with the help of the following expression.  higher energy state R hydberg’s constant 2 n lower R = 108677 cm–1 n



 1  1 1 1  1 1   Rz 2  2  2   or  R   2  2     n1 n 2   n1 n 2 

R = 10 m–1 Z Atomic number.

r i S e c n i r P By Series of Lines in Hydrogen Spectrum

Lymen Series : When an electron undergoes transition from a higher energy level (n2), e.g. 2, 3, 4, 5, .....  to ground state or First energy level, the spectrum is said to belong to Lymen series. For this, n1 = 1 and n2 = 2, 3, 4, 5, 6, 7, 8 .......... . Balmer Series : When an electron undergoes transition from a higher energy level (n2), e.g. 2, 3, 4, 5,6, 7, 8 .....  to the second energy level n1 = 2, the spectrum is said to belong to Balmer series. Paschen Series : When an electron falls from a higher energy level to thrid orbit (n = 3). It gives a spectrum that is associated with Paschen series. For this n1 = 3 and n2 = 4, 5, 6, 7, 8 ..... . Brackett Series : When an electron falls from a higher energy level to fourth orbit (n = 4), the spectrum obtained is associated with Brackett Series. For this n1 = 4 and n2 = 5, 6, 7, 8 ..... . Pfund Series : When an electron falls from a higher energy level to fifth orbit (n = 5), the spectrum obtained is associated with Pfund Series. For this n1 = 5 and n2 = 6, 7, 8, 9, 10 ..... . Humphry Series : When an electron falls from a higher energy level to sixth orbit (n = 6), Humphrey seriers of the spectrum obtained. For this n1 = 6 and n2 = 7, 8, 9, 10, 11 ..... .

(13)


Total number of sprectral lines

If an electron jumps from higher energy level n2 to lower energy level n1 than total number of spectral lines emitted

 n 2  n1  n 2  n1  1 2

Drawbacks of Bohr’s Atomic Model (1) (2) (3)

Bohr model cannot explain the elements having more than one electron. Only one-electron species, like hydrogen atom, He+1 ion, Li+2 ion, Be+3 ion, etc. can be explained with the help of Bohr model. Bohr model can explain only circular orbits in the atom and not the elliptical ones. Bohr model cannot explain splitting of spectral lines into finer lines in a magnetic field, which is known as Zeeman effect.


(4)

Bohr model fails to explain the splitting of spectral into finer lines in an electric field, which is known as Stark effect.

(5)

Bohr model fails to explain Hiesenberg uncertainty principle and it cannot be applied for giving any basis to classification of elements and periodicity in their properties.

(6)

Bhor model cannot be used for explaining finer structure of spectrum and calculating intensity of spectral lines.

The Wave Theory of Light Light, X-rays and radiation produced by a radioactive substance are some of the examples of radiation energy. In 1856 Clark Maxwell showed that energy of radiation is of wave nature, i.e. the energy is emitted in the form of a wave. Therefore, he called the emitted energy as electromagnetic wave or electromagnetic radiation. Since energy is a sort of wave, it is explained as were motion. Following are the salient features of this wave motion.

Wave Lenght() The distance between any two successive crests (or troughs) is known as wavelenght. This is expressed as (Lambda). Its common units are as follows. Angstrom (A). Unit  1Å= 10-10 m

1 nm 10–9m

1 pm(pico meter) = 10–12m

1 fm (femto meter) 10–15m (14)


Frequency(v)

The number of vibrations produced in a unit time is called frequency. Here, the time is taken in seconds. The number of wavelengths passing forward in one second from a fixed point is called frequency. Unit :- Hertz (Hz) or sec-1

Relation Between Frequency and Wavelength v=

c 

Wave Number (  ) The reciprocal of wavelength is called wave number. It is represented by  . 1 –1 –1  =  Therefore, the unit of wave number is cm or m

r i S e c n i r P By Planck’s Quantum Theory

If a substance emits or absorbs energy, it does not do so continuously but does but it in the form of discrete series of small packet or bundles, called quanta. This energy could be any of the quantum numbers 1, 2, 3, 4, 5 ........ n but not in the form of fractional quantum number. hc E= = hv 

Therefore

The Duel Nature of Matter (The Wave Nature of Electron) OR Debroglie’s Concept

  

In 1924, a French physicist, Luis De Breoglie suggested that if the nature of light is both that of a particle and of a wave, then this dual behaviour should be true matter also. According to De Broglie, the nature of an electron moving around the nucleus is like a wave the flows in circular orbits around the nucleus. According to De Broglie, the wavelength  of an electron is inversely proportional to its momentum p.

h 1   or= p p K.E. q × 

Formulae : =

Here h = Planck’s constant for one electron.

h h   p m

Where,

h 2m  KE 



h 2qVm

KE = Kineitc energy of the particle, V = Potential difference for a gas molecule K.E =

=

3 mRT T 2

150 Å 

V = Voltage through which electron is accelerated =

(15)

h = 3 2m KT 2

h 3mKT


Quantum Mechanical Theory of Atom The dual nature (particle and wave) of electron led to the use of new system of mechanics called quantum mechanics. This system was first put forward by an Austrian physicist E. Schrodinger and a German physicist W. Heisenberg. The two fundamental principles of quantum nechanics are given below : (i) Heisenberg’s uncertainty principal and (ii) Schrodinger’s wave equation

Heisenberg’s Uncertainty Principle According to this principle, it is impossible to experimentally determine exactly both position and momentum of a minute particle like an electron at the same time. This principal can be depicted mathematically as follows. x × p 

h 4

or x × m × v 

h 4

Here x is uncertainty of position, p is uncertainty of momentum and h is Planck’s constant

r i S e c n i r P By Schrondinger’s Wave Equation

Schrondinger regarded electron as having wave nature and put forward the following complex differential equation.

2 2 2 82 m    2 x 2 y 2 z 2 h

Where

(E – V)  = 0

m = Mass of electron, E = Total energy of electron,  = Wave function

h = Planck constnat, v = Potential energy of electron,

Note :

Quantum Numbers Theset of numbers requried to determine the exact location of electron in an atom are called quantum numbers. Following are the four quantum numbers

Principal Quantum Number (n) (a) (b) (c) (d)

Principal quantum number describes the size and energy of a shell or orbit. An atoms has K, L, M, N, O, P, Q, etc. shell. Each value of ‘n’ represents an energy shell. In any orbit, the number of orbitals is given by n2 and number of electrons is given by 2n2. This is called Bhor-Bury rule.

Azimuthal Quantum Number(  ) (a)

Azimutl al quantum number represents sub-shells in an atom.

(b) (c)

Each value of ‘  ’ represents a subshell. Azimutl al quantum number determines the shape of an orbital.

(d) (e)

The values of ‘  ’ ranges from -  to +  . Values of l 0 1 2 Sub shell s p d

3 f (16)

Reconstruct Your Chemistry With Prince Sir (f)

The number of orbitals in any sub shell = (2  +1) and the number of electrons is determined by the expression 2(2  +1).

(g)

 s Sub-shell  Spherical  p Sub-shell  Dumb-bell  d Sub-shell  Double dumb-bell  =2  f Sub-shell  Complex  =3 The order of enrgy of various sub-shells present in any shell is s < p < d < f < g ......... and so on. The value of orbital angular momentum, i, of an electron can be determined with the help of azimuthal

(h) (i)

 =0  =1

quantum number

i =

l  l  1 

h 2

Here  = Azimuthal quantum number and h = Planck’s constant

Magnetic Quantum Number (m) (a) Magnetic quantum number describes the orientation of orbitals. (b) Each value of m represents an orbital.


(c) The value of m ranges from –  to +  .

(d) Total numberof orbitals present in a subshell = (2  +1) (i) For s sub-shell, l = 0. Thus, m 2 × 0 + 1 = 1 and therfore s sub-shell consists of only one orbital called s orbital. (ii) For p sub-shell, l = 1. Thus, m 2 × 1 + 1 = 3 and therfore p sub-shell consists of three orbitals called px, py and pz orbitals. (iii) For d sub-shell, l = 2. Thus, m 2 × 2 + 1 = 5 and therfore d sub-shell consists of five orbitals called dxy, dyz, dz2, dxz and d x 2  y2 orbitals. (i)

For s sublevel, 1 = 0. Thus, for s orbital, the value of m is 0.

(ii)

For p sub-level, 1 = 1. Thus, the values of m for p orbitals are as follows.

(iii)

For d sub-level, 1 = 2. Thus, the values of m for d orbitals are as follows.

(e) The total number of orbitals present in an energy shell = n2, where n is principal quantum number.

Spin Quantum Number (s) (a) (b)

Spin quantum number gives information about the spin of an electron. The value of s is 1/2 which depicts the direction of spin of the electron.

(c)

If the electron spins in clockwise direction, s is denoted by + electron is denoted by s = –

1 or []. 2 (17)

1 or a sign[]. Anticlockwise spin of the 2

(d) (e)


One orbital can accommodate only two electrons, with opposite spins. The angular momentum of an electron is not only due its motion around the nucleus in an energy level but also due to its spin along its own axis. The angular momentum that arises due to spin of an electron along its axis, is called spin sngular momentum and is depicted by the symbol s. The value of s can be found out with the help of the following expression. s = s(s  1) 

h where ‘s’ is spin quantum number.. 2

In this expression the value of s is always taken as

1 1 and not – . 2 2

WRITINGTHE ELECTRONIC CONFIGURATION Aufbau Principle Aufbau is a German word that means to build up. Therefore, electrons are filled up in increasing order the energies of the orbitals i.e. electrons are first filled in orbitals of lower energy and then in the orbital of higher energy and so on.

r i S e c n i r P By (n +  ) Rule

Accrding to this rule electrons are first filled in those ornitals for which value of ‘n +  ’ is lowest. If two orbitals have same value of  then electrons are first filled in that orbital which has lower value of n. The electrons are filled in a sub-shell according in n + l rule.

Pauli’s Exclusion Principle

(i)

According to Pauli axclusion principle, no two electron can have the same values of all the four quantum numbers. 6s1 and 6s2 n=6 n=6 l=0 l=0 m=0 m=0 s=+

(ii) (iii) (iv)

1 2

s=–

In the above illustrations, the respective values of n ,  and m are same but that of s is different. Pauli exclusion principle can be stated in other words as that “only two electrons can be accommodated in the same orbital only when their spin quantum numbers is different”. If the thrid electron enters in an orbital, the set of four quantum numbers becomes same for any two electron. According to this rule, for any two electrons, a set of maximum three quantum numbers can be same, but the fouth has to be different. For example, two electrons can have same (n, l and m) or (l, m or s) or (n, m or s) Example 1s1 and 1s2 n=1 n=1 l=0 l=0 m=0 m=0 s=+

(v)

1 2

1 2

s=–

1 2

This rule does not apply for hydrogen atom because it contains one electron. (18)


Hund’s Rule of Maximum Multiplicity (a) (b) (c) (d)

Degenerate orbitals The orbitals having same energy are called degenerate orbitals. s sub-shell consists of only one orbital. Thus, it cannot have orbital. According to Hund’s rule, the degenerate orbitals get filled by electrons having parallel spin one by one to give an unpaired state. According to tnis rule, pairing of electrons cannot occur untill all the orbitals have one unpaired electros. For example, C6 can possibly have the following two configurations of 2s2 2p2. (1) (2)

(e)

(f) (g)

The following two conditions have to be fulfilled for Hund’s rule. (1) The orbitals should be degenerate (2) The member of electrons and the degenerate orbitals should be more then one Hund’s rule is not applicable for H, He, Li and Be, because electrons in them go to s sub-shell, which does not have any degenerate orbital. Hund’s rule is not applicable for B5 also, because there is only one electrons in p orbital. Therefour , this rule si applicable from C6 onwards.


Increasing order of energy 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p The maximum number of electrons that can be accommodated in s orbital is 2, that in p orbital is 6, that in d orbital is 10 and that in f orbital is 14.

Exceptions to n + l Rule (a) (b)

There are mainly two exceptions of n + l rule. La57 – 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 5d1 Ac89 – 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6s2, 7s2, 6d1 Due togreater stability of hafl-filled and full-filled orbitals, the configurations d5 ns1 and d10 ns1 are written in place of n4 ns2 and d9 ns2 respectively.

Stability of Half-filled and Fully-Filled orbitals

(a) (b)

(c)

The stability of half-filled orbitals (p3,d5 and f 7) and fully-filled orbitals (p6,d10and f 14) is higher than that in other states. This is due the following reasons. When a sub-shell is half-fully-filled, it means that the distribution of electrons is symmetrical in the orbitals of equal energy. Unsymmetrical distribution of electrons results in lower stability. The electrons present in orbitals of equal energy in an atom can interchange their position, in this process energy is released, resulting stable system. The possibility of interchange of positions is highest in half filled and fully-filled states. This provides greater stability to the system. The exchange energy for helf-filled and fully-filled orbitals is maximum. As the number of electrons increases, electron start pairing resulting in spin coupling. The energy liberated in the process of coupling is called coupling energy. (19)

(d)


The spin of electrons in a fully-filled orbital are opposite to each other or antiparallel. The energy of the system decreses due to neutralization of opposite spins. So fully-filled orbitals are more stable.

S.No. 1. 2. 3.

Orbit If is depicted by n. Maximum number of electrons 2n2 Path of electron around the nucleus is called orbit.

Orbital It is depicted by m Maximum number of electrons 2 with opposite spin. Space around the nucleus where probability of finding an electrons is maximum.

Orbital (a) (b) (c)

The space around the nucleus where probability of finding an electron is maximum, is called an orbital. An electron cloud is negatively charged and the nucleus is positively charged. Therefour, the probability of finding an electron is maximum around the nucleus. The probability of finding an electron is an orbital is 95% to 98%.

r i S e c n i r P By s-Orbital

(a) (b) (c) (d) (e)

Only one s-orbital is possible in an orbit because l = 0 and m = 0 for it. It is spherical in shape and thus the electron density is unform in all directions. The size increases with increase in the value of n. There is vacant space between 1s orbital and 2s orbital, where the probability of finding electron is minimum, it is known as nodal surface. The nodal surface is missing inside 1s orbital because of its proximity with the nucleus. The number of nodel surfaces in an orbit is equal to (n –1)

,p-Orbital (a) (b)

(c) (d)

For p orbitals, l = 1 and m = – 1, 0, + 1. Thus, it can have three configurations, which are distributed in x, y and z axes. Therefore, there are three p-orbitals, which are dumbbell, shaped. Each p-orbital has two lobes and the probability of finding electron inside these two lobes is equal. The plane perpendicualr to the axis two lobes and passing through the point where these two lobes join, is the nodal plane of -orbital, because the probability of finding electron in this plane is negligible or minimum. The value of nodal planes for each of the px, py and pz orbitals is same and these nodal planes are present in xy, yz and xz planes, respectively. The three p-orbitals of a particualr orbit (px, py and pz ) have equal energy and therefour these are called degenerated orbitals. (20)


d-Orbital

(a)


For d orbitals, l = 2 and m = –2, –1, 0, +1,+2. Therefore, there are five orientations and thus five dorbitals. (b) Its shape is like a double dumbbell. (c) The five orientations of d-orbitals are as follows : (i) The double dumbbell of dxy orbital are situated between x and y axes. (ii) The double dumbbell of dyz orbital are situated between y and z axes. (iii) The double dumbbell of dxz orbital are situated between x and z axes. (iv) The double dumbbell of d x 2  y2 orbital are directed on x and y axes. (v)

dz2 orbital is composed of one dumbbell and one ring. The dumbbell is situated on z axis and the ring is present on its middle part.

Shape of d Orbitals (21)


f-orbitals (i) They have complex shapes, (ii) For these, l = 3 and m = –3, –2, –1, 0, +1,+2, +3 (iii) These have seven orientations.

Nodal planes : The surfaces where probability of finding an electron is zero is called nodal surface or nodal plane. There are two type of nodal planes radial nodes oe angualr nodes. Total number of nodes = (n-1) Angualr nodes =  Radial nodes = (n-  -1)

BOOST YOUR BASICS Q.1 If the energy of an elecrton in hydrogen atom is given by expression, –1312/n2 kJ mol–1, then the energy required to excite the elecrton from ground state to second orbit is : (A) 328 kJ/mol (B) 656 kJ/mol (C) 984 kJ/mol (D) 1312 kJ/mol


Q.2 Calculate the de-Broglie wave length of the electron in the ground state of hydrogen atom, given that its knietic energy is 13.6 eV : (1eV = 1.602 × 10–19J) (A) 3.328 × 10–10 m (B) 2.338 × 10–10 m (C) 3.328 × 1010 m (D) 2.338 × 10 m Q.3 Which of the following pair having same number of orbitals : (A) N, O (B) O, F (C) Na, K The correct answer is : (A) a, b, c (B) b, c, d (C) c, d, a

(D) S, Cl (D) a, d, b

Q.4 The speed of a proton is one hundredth of the speed of light in vacuum. What is its de-Broglie wavelength ? Assume that one mole of protons has a mass equal to one gram [h = 6.626 × 10–27 erg sec] (A) 13.31× 10–3m (B) 1.33 × 10–3m (C) 13.13 × 10–2 m (D) 1.31× 10–2 m Q.5 The ratio of ionization energy of H and Be+3 is : (A) 1 :1 (B) 1 : 3

(C) 1 : 9

Q.6 Hydrogen spectrum consists of : (A) An intense line (B) Six series of lines

(C) Three series of lines (D) Four series of lines

Q.7 Which of the following salt has isoelectronic cation and anion : (A) KF (B) NaCl (C) SrCl

(D) 1 : 16

(D) MgF2

Q.8 Which set of quantum numbers is possible for the last electron of Mg– ion : (A) n = 3, l = 2, m = 0, s = +

1 2

(B) n = 2, l = 3, m = 0, s = +

1 2

(C) n = 1, l = 0, m = 0, s = +

1 2

(D) n = 3, l = 0, m = 0, s = +

1 2

(22)


Q.9 If E1,E2 and E3 represent respectively the kinetic energies of an electron, an alpha particle and a proton each having same de Broglie wavelength then : (A) E1 > E3 > E2 (B) E2 > E3 > E1 (C) E1 > E2 > E3 (D) E1 = E2 = E3 Q.10 The value of :[2p(energy) – 1s(energy)] for H-atom would be : (A) 10.2 eV (B) 13.6 eV (C) 3.4 eV

(D) None of these

Q.11 In hydrogen atom, If an electron jumps from n = 6 to n = 2, how many possible spectral lines are obtained: (A) 15 (B) 10 (C) 6 (D) 12 Q.12 In order to have the same wavelength for the electron (mass me) and the neutron (mass mn) their velocities should be in the ratio : (A) mn /me (B) me / mn (C) mn × me (D) one Q.13 A strong argument for particle nature of cathode rays is that they : (A) Produce fluoroscence (B) Travel through vaccumm (C) Get deflected by electric & magnetic field (D) Cast shadows.


Q.14 Ionisation energy of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochomatic radiation of photon energy 12.1 eV. the number of spectral lines emitted by H atom will be: (A) 1 (B) 2 (C) 3 (D) 6 Q.15 The wavelength of Ka line emitted by H like atom is 0.32 Ao . the wavelength of Kb line is emitted by the same element is : (A) 0.18 Ao (B) 0.48 Ao (C) 0.27 Ao (D) 0.38 Ao Q.16 The wavelength of radiation emitted is lo when an electron in H atom jumps from 3rd orbit to 2rd oribt. If in the H atom itself, the electron jumps from fouth orbit to second, wavelength of emitted radiation will be: (A)

25 o 16

(B)

17 o 20

(C)

20 o 27

(D)

16 o 25

Q.17 Hydrogen (H), deutenium (D), singly ionised helium(He+) and dubly ionised lithium (Li2+) all have one electron around the nucles. Consider n = 2 to n = 1 transition. If the wavelengths of emitted radiations are 1, 2, 3 and 4 : (A) 1 = 2 = 43 = 94 (B) 41 =22 = 23 = 4 (C) 1 = 22 = 2 2 o  3 2 o

(D) 1 = 2 = 23 = 34

Q.18 Two paritcles of masses m and 2m have equal knietic energies. Their debroglie wavelengths are in the ratio of : (A) 1 :1

(B) 1 : 2

(C) 1 :

2

Q.19 The Bhor model of atom : (A) Assumes that angular momentoum of electron is quantized (B) Uses Einstien’s photoelectric equation (C) Predicts continuous emission spectra for atoms (D) Predicts the same emission spectra for all type of atoms (23)

(D)

2 :1


Q.20 Which of the following transitions in H atom emit photons of highest frequencuy (A) 1  2 (B) 2  6 (C) 6  2 (D) 2  1

Q.21 Light coming from a discharge tube filled with Hydrogen falls on the cathode of the photoelectric cell. The work fuction of the surface of cathode is 4 eV. Which one of the following values of anode voltage (in V) with respect to cathode will make the photo current zero ? (A) -4 (B) -6 (C) -8 (D) -10 Q.22 When photons of energy hn fall on a photosensitive metallic surface (work function f) elctrons are emitted from the metal surface. It is possible to say that : (A) All ejected elctrons have same KE = hv - hvo (B) The ejected elctrons have a distribution of KE. the most energetic one having KE = hv - hvo (C) The most energetic electrons have KE = hv (D) The KE of ejected elctrons is hvo Q.23 In a photoelectric experiment, photons of energy 5 eV are incident on a metal surface. The liberated photoelectrons are just stopped by an electrode at a potential of -3.5 eV with respect to metal. The work function of the metal sufrace is : (A) 8.5 eV (B) 5 eV (C) 3.5 eV (D) 1.5 eV


Q.24 Light of two different frequencies, whose photons have energies are 1 and 2.5 eV respectively successively illuminate a metal whose work function is 0.5 eV. The ratio of the max. speeds of the emitted elctrons will be : (A) 1 : 2 (B) 1 : 4 (C) 1 : 2 (D) 1 : 1 Q.25 If the electron is to acquire a wavelength of 0.1 nm. what potential difference does it fall througt ? (A) 150.5 eV (B) 110 eV (C) 200 eV (D) 516.6 eV 1 m v 2 bombards a heavy nucleus of charge Ze. The distance of closest approach 2   for the particle will be proportional to :

Q.26 An  particle of enegy

(A) v2

1 (B) m a

(C)

1 4 v

(D)

1 Ze

Q.27 An energy of 24.6 eV is required to remove one of the electrons from a neutral He atom. The energy in eV required remove both the electrons from a neutral from a neutral He atom is (A) 38.2 (B) 49.2 eV (C) 51.8 eV (D) 79 Q.28 Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the elctron. The wavelength of the radiation that will be emitted when this particle jumps from first excited state to ground state is equal to (A)

2 3R

(B)

1 3R

(C)

1 2R

(D) None

Q.29 The longest wavelength in the ultraviolet region of the spectrum is 122 nm. the smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is : (A) 802 nm (B) 823 nm (C) 1882 nm (D) 1648 nm

(24)


Q.30 Work function of a metal X equals ionisation energy of Li2- ion in second excited state. Work function of another metal Y equal ionisation energy of He ion with electron in n = 4. Now a photon of energy x falls on both the metals such that max. K.E. of photoelectrons emitted from metal X is half that of photoelctron emitted from metal Y. Then value of x ? (A) -0.875 eV (B) -1.75 eV (C) 2.72 eV (D) 2.38 eV

Subjective Questions Q.1 Calculate the wavelength emitted during the transition of electron in between two levels of Li2+ ion whose sum is 4 and difference is 2. Q.2 Find the quantum number ‘n’ corresponding to the excited state of He- ion if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4 nm. Q.3 When certain metal was irradiated with light frequency 1.6 × 1016 Hz the photo electrons emitted had twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency 1.0 × 1016 Hz. Calculate threshold frequency (v) for the metal. Q.4 Find the velocity (ms–1) of electron in first Bhor’s orbit of radius a0. Also find the de Broglie’s wavelenght


(in m). Find the orbital angular momentum of 2p orbital of hydrogen atom in units of

h . 2

Q.5 Two hydrogen atoms collide head on and end up with zero kinetic energy. Each atom then emits a photon of wavelength 121.6 nm. Which transition leads to this wavelegth ? How fast were the hydrogen atoms traveling before collision ? Q.6 Calculate the energy in kJ per mole of electronic charge accelerated by a potential of 1 volt. Q.7 (A) (B) (C)

What is highest frequency photon that can be emitted from hydrogen atom? What is the wavelength of this photon? Calculate the longest wavelength transition in the paschen series of He+ Calculate the ratio of the wavelength of first and the ultimate line of Balmer series of Li2+ ?

Q.8 The eyes of a certain member of the reptile family pass a visul signal to the brian when the visual receptors are strudck by photns of wavelength 859 nm. If a total energy of 3.15 × 10–14 J is requires to trip the signal, what is the minimum number of photon that must strike the receptor. Q.9 O2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom 1.967 eV more energetic than normal. The dissociation of O2 into two normal atom of axygen requires 498 kJ mol-1. What is the maximum wavelength effective for photo chemical dissociation of O2? Q.10 Uranium has at.no. 92 and atomic weight 238.029. Calculate the number of protons, neutrons and electrons in uranium atom. Q.11 Find energy of each of the photons which : (A) Correspond to light of frequency 3 × 1015 Hz

(B)

Have wavelength of 0.50 A.

Q.12 H-atom is exposed to electromagnetic radiation of 1028 Ao and gives out induced radiations. Calculate the wavelength of induced radiations. [6563 Ao, 1216 Ao, 1026 Ao] (25)


Q.13 A certain dye absorbs 4530 Ao and fluoroses at 5080 Ao these being the wavelengths of maximum absorption that under given conditions 47% of the absorbed energy is emitted. Calculate the ratio of number of quata emitted to the number absorbed. [0.527] Q.14 Calulate the threshold frequency of metal binding energy is 180.69 kJ mol-1 of electron. Q.15 What is the number of photons of light with a wavelength of 4000 pm that provide: J of energy ?

Q.16 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 um. Calculate the rate of emission of quanta per second. Q.17 A dust particle having mass equial to 10–11 g, diameter 10–4 cm and velocity 10–4 cm sec–1. The error in measurement of velocity is 0.1%. Calculate uncertainty in its position. Q.18 Point out the following : (A) How many energy subshells are possible in n = 3 level. (B) How many orbitals of all kinds are possible in n = 3 level. Q.19 The dissociation energy of H2 is 430.55 kJ/mol. If H2 is exposed to radiant energy of wavelength 253.7 nm. what fraction of radiant energy will be converted into K.E.


Q.20 What values are assigned to quantum numbers n. / m for : (A) 3s

(B) 4Pz

(C) 4d x 2  x 2

(D) 5dz2

Q.21 Given below are the sets of quantum numbers for given orbitals. Name these orbitals. (A) n = 3 (B) n = 5 (C) n = 4 (D) n = 2 (E) l=1 l=2 l=1 l=0 m = –1 m=0 m = 1 m=0 Q.22 What are the numbers of nodes present in : (A) ls (B) 2s (C) 2p

(D)

n=4 l=2 m = 2

3p orbitals

Q.23 An atom of an element has 13 electrons. Its nucleus has 14 neutrons. Find out atomic no. and approcimate atomic weight. Indicate the arrangement of electrons and the electrovalency of the element. Q.24 A neutral atom of an element has 2K.8L.9M and 2N electrons. Find out the following : (A) Atomic no. (B) Total no. of s-electrons (C) Total no. of p-electrons (D) Total no. of d-electrons (E) Valency of element (F) No. of unpaired electrons. Q.25 Calculate total spin, magnetic moment for the atoms having at. no. 7. 24. 34and 36. Q.26 Find out the number of waves made by an electrons in one complete revolution in 3rd bohr orbit. [3] Q.27 Calculate the energy emitted when electrons of 1 g-atom of Hydrogen undergo transition giving the spectral lines of lowest energy in the visible region of its atomic spectra. [182.5 kJ] (26)


Q.28 The H atom in the ground state is excited by means of monochromatic light of wavelength x Ao. The resulting spectrum consist of 15 different lines. Calculate the value of x. [938 Ao]

Q.29 The eyes of a certain member of the raptile family pass a visual singal to the brain when the visual recrptors are struck by photons of wavelength 850 nm. If a total energy of 3.15 × 10–14 J is required to trip the signal. What is the minimum number of photon that must strike the receptor. Q.30 A photon of 300 nm is absorbed by a gas and then re-emits two photons. One re-emitted photon has wavelength 496 nm. Calculate energy of other photon re-emitted out. Q.31 Calculate the wavelength of first line of Lyman series of ten times ionised sodium atom (Z = 11) and compare with the wavelength of first line of Balmer series of H atom. Shows that : (i) The sum of the energy for the transitions n = 3 to n = 2 and n = 2 to n = 1 is equal to the energy of transition for n – 3 to n – 1. (ii) Are wavelength and frequencies of the emitted spectrum are also additive in the same way as their energies are ?

r i S e c n i r P By FACETHE CHALLENGE

Q.1 A photoelectric plate is initially exposed to a spectrum of hydrogen gas excited to second energy level. Later when the same photoelectric plate is exposed to a spectrum of some unknown H like gas, excited to second level it is found that the De-Broglie wavelength of the photoelectrons now ejected has decreased

6.1 times. For this new gas difference of energies of first Lyman line and Blamer series limit is found to be two times the ionisation potential of the H atom. Detect the atom and determine the work function of the photoelectric plate. Q.2 A hydrogen like atom is observed to emit six wavelength originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and – 0.544 eV (including both these levels) (a) Find the atomic number of the atom. (b) Find the quantum number of the levels between which transition is traking place. (c) Calculate the largest wavelength emitted in these radiations. [Given h.c. = 1240 eV – nm] Q.3 Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing  -particles. A maximum energy electron combines with an particle to from He+ ion, emittinf a single photon in this process. He+ ions thus formed are in their fourth excited state. Calculate the energy released during combination of photon with the particle. Q.4 Electrons in a H like atom (Z = 3) make transition from 4th to 5th orbit and from 4th to 3th orbit. The resulting radiations are incdent on a metal plate and eject photoelctrons. The stopping potential for the photoelectrons ejected by shorter wavelength is 3.95 V. Calculate the work function of the metal and the K.E. of the photoelectrons ejected by longer wavelength. Q.5 Find the spectral series to which the spectral line of H belong if its wave number is equal to the difference in wave numbers of wavelengths 500 nm and 400 nm which are the members of Balmer series. Q.6 1.8 g H2 atoms are excited to radiations. The study of spectra indicates that 27% of the atoms are in 3rd energy level and 15% of atoms in 2nd energy level the rest in ground state. JP of H is 13.6 eV.Calculate(a) Number of atoms present in III and II energy level. (b) Total energy evolved when all the atoms return to ground state. (27)


Q.7 A hydrogen like atom is in higher excited state of quantum number ‘n’. This excited atom can make transition to the first excited state by succesively emitting 2 photons of energies 10.2 eV and 17 eV respectively. Alternatively the atom from the same excited state can make transition to the second excited state by successively emitting 2 photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of n and Z. Q.8 Acollection of H-like atoms has some atoms in the lowest energy level ‘A’ and some atoms in a particular upper energy level ‘B’ and there are no atoms in any other energy level. The atoms of gas make transition to a higher energy level by absorbing mono chromatic light of photon of energy 2.55 eV. Subsequent the atoms emit radaition of six different wavelengths. Some of the emitted photons have energy 2.55 eV and some have more and some have less than 2.55 eV. (i) Find the quantum number of intially excited level. (ii) Find the atomic number of gas. (iii) Find the ionisation energy for the gas atoms. (iv) Find the maximum and minimum energies of the emitted photons.

r i S e c n i r P By CHOOSE YOUR FAVOURITES

More then one Correct type Questions : Q.1 Which sets of quantum no. are onsistent with the theory ? (A) n = 2, l = 1, m = 0, s = – 1/2 (B) n = 4, l = 3, m = –2, s = – 1/2 (C) n = 3, l = 2, m = –3, s = + 1/2 (D) n = 4, l = 3, m = –3, s = + 1/2 Q.2 Which statements are correct : (A) The energies of the various sub-level in the same shell of H-atom are in order s > P > d > f. (B) s-orbital is non-directional (C) Electrons in motion behave as if they are waves (D) An orbital can accomodate a maximum of two electrons with parallel spins. Q.3 When  -particles are sent through a thin metal foil most of them go straight through the foil because. (A)  -particles are much heavier than electrons (B)  -particles are positively charged (C) Most part of the atom is empty space (D)  -particles move with high velocity Q.4 Which of the following reflects the wave nature of light ? (A) Photoelectric effect (B) E = mc2 (C) Difeaction

(D) Interference

Q.5 The electron in a hydrogen atom makes a transition n1  n2, where n1 and n2 are the principle quantum numbers of the two states. Assume the Bhor model to be valid the time period of the electron in the inital state is eight times that in the final stte. The possible values of n1 and n2 are : (A) n1 = 4, n2= 2 (B) n1 = 8, n2= 2 (C) n1 = 8, n2= 1 (D) n1 = 6, n2= 3 Q.6 Which of the following product in a hydrogen atom are independent of the principle quantum number n? The symbols have their usual meanings. (A) vn (B) Er (C) En (D) vr Q.7 Which of the following statement/s is/ are wrong ? (A) If the value of l = 0, the electron distribution is spherical (B) The shape of the orbital is given by distribution is spherical (C) Angular momentum of 1s, 2s & 3s electrons are equal (D) In an atom, all electrons travel with the same velocity (28)


Q.8 For the energy levels in an atom, which one of the following statement/s is/are correct ? (A) There are seven principle electron energy levels. (B) The second principle energy level can have four sub-energy levels and contain a maximum of eight electrons. (C)The M energy level can have a maximum of 32 electrons (D) The 4s sub-energy level is at a lower energy than the 3d sub-energy level Q.9 Many elements have non-integral atomic masses because? (A) They have isotopes (B) Their iosotopes have non-integral masses (C) Their iosotopes have different masses (D) The constituents, neutrons, protons and electrons combine to give fractional masses Q.10 Rutherford’s -scattering experiment led to the following conclusions. (A) Atom has largely empty space (B) The centre of the atom has positively charged nucleus (C) The size of the nucleus is very small as compared to the size of the atom (D) The electrons revolve around the nucleus


Q.11 Which statement (s) is (are) correct ? (A) Electrons in motion behave as if they are waves (B) s-orbital is non-directional (C) An orbital can accommodate a maximum of two electrons with parallel spins (D) The energies of the various sub-levels in the same ae in order s > p > d > f Q.12 Which concerning Bhor’s model are true ? (A) Predicts that probability of electron near nucleus is more (B) Angular momentum of electron in H atom =

nh 2

(C) Introdues the idea of stationary states (D) Explains line spectrum of hydrogen Q.13 Heisenberg uncertainty is not valid for. (A) Moving electrons (C) Stationary particles

(B) Motor car (D) All of the above

Q.14 For the energy levals in an atom which one of the following statements is/are correct ? (A) There are seven principle electron energy levels (B) The second principle energy level can have 4 sub-energy levels and contain a maximum of 8 electrons (C) The M energy level can have a maximum of 32 electrons (D) The 4s sub-energy level is at a lower energy than the 3d sub-energy level Q.15 Whcih of the following statements are correct for an electron that has n = 4 and m = –2? (A) The electron may be in a d-orbital (B) The electron is in the fourth principle electronic shell (C) The electron may be in a p-orbital (D) The electron may have the spin quantum number = +1/2 (29)


Q.16 Which of the following is the correct set of four quantum numbers for each of the electrons in the 3d subshell when it is fully occupied ? (A) n = 3,  = 2, m = 0, 1,2, –2 and s = ± 1/2, ±1/2, ± 1/2, ±1/2, and ±1/2 (B) n = 3,  = 1, m = 0, –1,2, –2 and s = ± 1/2, ±1/2, ± 1/2, ±1/2, and ±1/2 (C) n = 4,  = 2, m = 0, 1, –1,2, –2 and s = ± 1/2, ±1/2, ±1/2, ±1/2, and ±1/2, (D) None is correct

GO COMPRHENSIVE Passage-1 Rutherford proposed the atomic after his most striking experiment on -scattering leading to discovery of nucleus. Bhor later on modified the atomic model on the basis of planck’s quantum theory of light and nh . The collection of fine lines in 2 line spectrum led sommerfeld to give the idea of elliptical orbits. He successfully explained the existence of

proposed the concept of stationary circular orbits of quantised energy


h . 2 The emission of a spectral line in atomic spectra was supposed to be due to the jump of electron form one energy level to other. Q.1 The angular momentum of electron in 2p orbital is :

subshells and their number in a shell. The angular momentum of subshells was proposed as l  l  l .

(A)

h 2

(B)

h 2

(C)

h 2

(D)

h 2

Q.2 The volume occupied by the nucleus is about .......... times of volume of atom: (A) 10–15 (B) 1015 (C) 10–12 (D) 10–10 Q.3 Non-directional orbital is : (A) 3s (B) 4f

(C) 4d

(D) 4p

Passage-2 The electrons in a poly-electronic atom are filled one by one in order of increasing energy level. The energy of subshells and orientation of orbitals depends upon the values of three quantum numbers (i.e.,n, l, and m respectively) derived from schrodingr wave equation. The different orbitals of a subshells however possess same energy level and are called degenerate orbitals but their energy level changes in presence of magnetic field and the orbitals are non-degenerate. A spectral line is noticed if an electron jumps from one level to other. The paramagnetic nature of element is due to the presence of unpaired electron. Q.1 Which in each pair is most stable ion: Cu+ or Cu2+ and Fe2+ or Fe3+ (A) Cu+, Fe3+ (B) Cu2+, Fe3+ (C) Cu2+, Fe2+ (D) Cu+, Fe2+ Q.2 The number of unpaired electrons in Cr atom is : (A) 2 (B) 3 (C) 5

(D) 6

Q.3 The total magnetic moment of Ni2+ ion is: (A) 6 BM

(B) 8 BM

(C) 15 BM

Q.4 The number of spherical and angular nodes in 2p orbitals are: (A) 1, 1 (B) 2, 1 (C) 1, 0 (30)

(D) 12 BM

(D) 0, 1


Q.5 The possible number of spectral lines when an electron can jump from 5th shell to 2nd shell is: (A) 4 (B) 2 (C) 3 (D) 6

Q.6 The correct order for energy levels in H-atom is: (A) 3s = 3p = 3d > 2s (B) 3d > 3p > 3s > 2s (C) 3d > 3p = 3s > 2s (D) 3d > 3p > 3s = 2s Q.7 Which elements has 18 electrons in its outer-most shell: (A) Cu (B) Pd (C) Cd (D) Tc (A) Rotation of the electron in clcok wise and anticlockwise direction respectively (B) Rotation of the electron in anticlockwisec and closk wise direction respectively (C) Magnetic moment of the electron pointing up an ddown respectively (D) Two quantum mechanical spin states which have no classical analogues

Assertion & Reason The questions given below consist of two statements as Assertion (A) and Reason (R); while answering these questions, choose any one of them: (A) If (A) and (B) are both correct and (R) is the correct reason for (A). (B) If (A) and (B) are both correct and (R) is not the correct reason for (A). (C) If (A) is correct but (R) is incorrect. (D) If (A) is incorrect but (R) is incorrect. Q.1 A : The first ionization energy of B is less than Be. R : 2p orbitals is lower in energy than 2s.


Q.2 A : Spectral lines would not be seen for 2px – 2pz transition R : 2p orbitals are degernrate orbitals.

Q.3 A : Spectral lines are never obtained when electron de-excites from dxy to dyz because they are degenrate. R : When ligands approach central metal ion, its d-orbitals split into two sets. Q.4 A : Nucliede 30 Al is less stable than 13

40 20

Ca

R : Nuclides having odd number of protons and neutrons are generally unstable. Q.5 A : The 3p orbital has higher energy level than 3s He+ ion. R : The enrgy of an orbital depends upon n and l. Q.6 A : Specific charge of -particles is twice to that of proton. R : Specific charge is given by e/m. Q.7 A : d-orbitals are five fold non-degenerate in presence fo magnetic field. R : In presence of magnetic field, the energy of orbitas becomes aitogether different. Q.8 A : Electronmagnetic radiations will emitted for the transition of 3p to 2s orbital in H-atom. R : Both have same energy level and thus no transition. Q.9 A : 24Cr has more paramagnetic nature than 25Mn R : Cr has more number of unpaired electron than Mn Q.10 A : The possible number of electrons in a sub-shell is (4l + 2). R : The possible number of orientations in a sub-shell is are (2l + 1). (31)


Q.11 A : Aufbau rule is violted in writing electronic configuration of Pd R : Pd shows diamagnetic natture.

Q.12 A : Hydrogen has one electrons in its orbit but it shows several spectral lines. R : There are many excited energy levels available for different H-atom when spectra of H2 gas is studied. Q.13 A : For n = 3.1 may be 0.1 and 2; and m may 0:0. ±1 : and 0. ±1 and ±2. R : For each value of n there are 0 to (n –1) possible values of l: and for each value of l there are 0 to ±1 value of m. Match the following Q.14 Column -I

Column -II h 4

A

Aufbau principle

P.

p.x 

B. C.

Pauli’s exculsion principle Heisenberg’s uncertainty principle

Q. R.

Maximum number of electron is shell is 2n2 (n +  ) rule

Einstein’s equation

S.

hv = hv +

D.


(A) (l,a), (2, b), (3, c), (4, d) (C) (l, a), (2, d), (3, c), (4, b) Q.15 A. B. C. D.

1 mv2 2

(B) (l, a), (2, b), (3, d), (4, c) (D) (l, b), (2, a), (3, d), (4, c)

Column -I Electronic configuration Mn4+ No. of unpaired electrons in Fe2+ No. of electrons in Na (11) having m = 0 Electronic configuration of Sc(21)

Column -II P. [Ar] 4s2 3d1 Q. 7 R. 4 S. [Ar] 4so 3d3

Q.16 ‘l’ and ‘m’ are respectively the azimuthal and magnetic quantum numbers Column -I Column -II A. Number of values of ‘l’ for an energy level P. 0, 1, 2,........... (n –1) B. Actual values of ‘l’ for a particular type of orbital Q. +l to –l through zero C. Number of values of ‘m’ for l = 2 R. 5 D. Actual values of ‘m’ for a particuiar type of orbital S. n (A) (l,a), (2, b), (3, c), (4, d) (C) (l, a), (2, d), (3, c), (4, b) Q.17 A. B. C. D.

(B) (l, a), (2, b), (3, d), (4, c) (D) (l, b), (2, a), (3, d), (4, c)

Column -I Change of photon energy into kinetic energy of electrons Transition between two atomic energy levels Mosley’s law Electron emission from a material

Column -II P. Characteristic X–rays

(32)

Q.

Photoelectric effect

R. S.

Hydrogen spectrum –decay

Q.18 A. B. C. D. Q.19 A. B.

C. D. Q.20 A. B. C. D. E. F.

Column -I Radius of nth orbital Energy of nth shell Angular momentum in lowest orbital Velocity of electron in nth

Reconstruct Your Chemistry With Prince Sir Column -II P. Inversely probability to Z Q. Integral multiple of h/2 R. Probability to n2 S. Inversely proportional to n

Column -I Column -II Orbital angular momentum in the P. Principle quantum number electron in a hydrogen-like atomic orbital A hydrogen-like one-electron wave Q. Azimuthal quantum number function obeying puuli principle integral multiple of h/2 Shape, size and orientation of hydrogen-likeR. Hydrogen spectrum atomic orbitals Probability density of electron at the S. Electron spin quantum number nucleus in hydrogen-like atom

r i S e c n i r P By Column -I Angular momentum Plank’s law Aufbau principle De-Broglie equation

Column -II P. Orientation of an electron in an orbital Q. Photon R. Line spectrum in visible region S. mv r

Balmer series

T.

=

Hund’s rule

U.

Electronic configuration

(33)

h mu


FEEL THE HEAT

Previous year IIT-JEE questions. Q.1 With what velocity should an particle travel towards the nucleus of a Cu atom so as to arrive at a distance 1013 m . [JEE 1997] Q.2

A compound of Vanadium has magnetic moment of 1.73 BM work out electronic configuration of Vanadium Ion in the compound. [JEE 1997]

Q.3

The energy of an electron in the first Bohr orbit of H atom is  13.6 eV . The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is/are : (A)  3.4 eV (B)  4.2 eV (C)  6.8 eV (D) + 6.8 eV [JEE 1998]

Q.4

The number of nodal planes in a px orbital is: (A) one (B) two

(C) three

(D) zero

[JEE 2000]

Q.5

Calculate the energy required to excite one litre of hydrogen gas at 1 atmp and 298K to the first excited state of atomic hydrogen. The energy for the dissociation of H – H is 436 KJ mol–1.

Q.6

The quantum numbers +1/2 and –1/2 for the electron spin represent: (A) rotation of the electron in clockwise and anticlockwise direction respectively. (B) rotation of the electron in anticlockwise and clockwise direction respectively. (C) magnetic moment of the electron pointing up and down respectively. (D) two quantum mechanical spin states which have no classical analogue.


[JEE 2001]

Q.7

Rutherfords experiment , which established the nuclear model of atom, used a beam of :– (A)  - particles, which impinged on a metal foil and get absorbed. (B)  - rays, which impinged on a metal foil and ejected electron. (C) Helium atoms, which impinged on a metal foil and got scattered. (D) Helium nuclie, which impinged on a metal foil and got scattered. [JEE 2002]

Q.8

The magnetic moment of cobalt of the compund Hg[Co(SCN)4] is [Given : Co+2] (A) 3

(B) 8

(C) 15

(D)

24

[JEE 2004]

Q.9

The radius of which of the following orbit is same as that of the first Bohr’s orbit of hydrogen atom? (A) He+ (n = 2) (B) Li2+ (n = 2) (C) Li2+ (n = 3) (D) Be3+ (n = 2) [JEE 2004]

Q.10

Given in hydrogenic atom rn, Vn, E, Kn stand for radius, potential energy, total energy and kinetic energy in nth orbit. Find the value of U,v,x,y. [JEE 2006] (A)

Vn U= K n

(P)

1

(B)

1 x rn  E

(Q)

–2

(R)

–1

(S)

0

(C) (D)

rn  Zy (Z = Atomic number) v = (Orbital angular momentum of electron in its lowest energy )

(34)

Q.11


Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [JEE 2008] (A)

Q.12

Column I Orbital angular momentum of the electron in a hydrogen-like atomic orbital

(P)

Column II Principal quantum number

(B)

A hydrogen-like one-electron wave function obeying Pauli principle

(Q)

Azimuthal quantum number

(C)

Shape, size and orientation of hydrogen like atomic orbitals

(R)

Magnetic quantum number

(D)

Probability density of electron at the nucleus in hydrogen-like atom

(S)

Electron spin quantum number


The work function () of some metals is listed below. The number of metals which will show \ photoelectric effect when light of 300 nm wavelength falls on the metal is [2010] Metal Li Na K Mg Cu Ag Fe Pt W (eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

ANSWERS BOOST YOUR BASICS 1.

C

2.

A

3.

D

4. (35)

B

5.

D

6.

B

Reconstruct Your Chemistry With Prince Sir 7.

D

8.

D

9.

D

10.

A

11.

B

12.

A

13.

C

14.

C

15.

C

16.

C

17.

A

18.

D

19.

A

20.

D

21.

D

22.

B

23.

D

24.

C

25.

A

26.

B

27.

D

28.

A

29.

B

30.

A

2.

(a) 3; (b) 15, 12; (c) 154.06 nm

FACE THE CHALLENGE 1.

Z = 2, = 4.2 eV

4.

= 2.1 eV & K.E. = 0.75 eV 5.

6.

(a) 628.72 × 1021 atoms; (b) [823.5 kJ

8.

Ans. (i) n = 4, (ii) z = 1, (iii) 13.6 eV, (iv) 12.75 eV, 0.66 eV

3.

3.38 eV

Brackett series 7. [n = 6, Z = 3]

CHOOSE YOUR FAVOURITES 1.

A,B,D

7.

A,B

13.

B, C

2.

B,C

3.

A,C

4.

A,B,C

5.

C,D

r i S e c n i r P By 8.

A,C

9.

A,C

10.

A,B,C

11.

A, C

14.

A, D

15.

B, D

16.

C

2.

1. B 2.D 3.B 4.D 5.D 6.A 7.B

6.

A,D

12.

B, C, D

GO COMPREHENSIVE 1.

1. B 2. A 3. A

ASSERTION & REASON 1.

C

2.

A

3.

B

4.

C

5.

A

6.

D

7.

A

8.

D

9.

A

10.

A

11.

A

12.

A

13.

A

14.

B

15.

A-S, B-R, C-Q, D-P

16.

A-S, B-P, C-R, D-Q

17.

A-Q, B-P, S, C-Q, D-QR

18.

A-P,R B-R C-Q D-S

19.

A-Q B-S C-P, Q,R

D-P,Q

20.

A-S B-Q C-Q D-T E-R F-P

FEEL THEHEAT Q.1

6.3 × 106 m/s

Q.2

[Ar] 3d1

Q.3

A

Q.4

A

Q.5

97.819 KJ

Q.6

D

Q.7

D

Q.8

C

Q.9

D

Q.10

(A) Q, (B) P, (C) R, (D) S

Q.11

(A) Q,R (B) P, Q, R,S (C) P, Q, R (D) P, Q

12. 4

(36)

13.



(37)

Atomic Structure

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