IIT Class XI Chem Structure of Atoms

S T U D Y M A T E R I A L PHASE - I STRUCTURE OF ATOM IIT-JEE

CHEMISTRY

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C O N T E N T S

CONTENTS

STRUCTURE OF ATOM 1.

Theory

2.

Solved Problems (i)

Subjective Type Problems

(ii)

Single Choice Problems

(iii) Multiple Choice Problems (iv) Miscellaneous Problems

3.

•

Comprehension Type Problems

•

Matching Type Problems

•

Assertion-Reason Type Problems

Assignments (i)

Subjective Questions

(ii)

Single Choice Questions

(iii) Multiple Choice Questions (iv) Miscellaneous Questions

(v) 4.

•

Comprehension Type Questions

•

Matching Type Questions

•

Assertion-Reason Type Questions

Problems Asked in IIT-JEE

Answers

STRUCTURE OF ATOM IIT-JEE Syllabus Electron, proton and neutron; Constitution of nucleus, Properties of alpha, beta and gamma rays; Rutherford’s scattering experiment; Bohr’s atomic model (mathematical details excluded); Quantum numbers; Pauli’s exclusion principle; Hund’s rule; Aufbau principle; Electronic configuration of elements (upto atomic number 36); Shapes of s, p, and d–orbitals.

CONTENTS Dalton’s Theory of atom Subatomic particles Various atomic models Photoelectric effect Particles & wave nature of electron Heisenberg uncertainty principle Schrodinger wave equation Quantum numbers Rules for filling electrons Shapes and size of orbitals Quantum mechanical treatment of H–atom

INTRODUCTION The classical view of atomic structure was constructed with the body of knowledge accumulated in physics over several centuries. We begin with the focus on atomic nucleus with its protons and neutron. This includes obtaining atomic masses and atomic number. Electrons are at the heart of our modern view of atomic structure. This chapter, firstly provides experimental evidence for the picture of atom. After that it deals with nature of light and finally explains the behaviour of electrons in atoms leading to the modern view of atomic structure.

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Chemistry : Structure of Atom

INSTITUTE OF CORRESPONDENCE COURSES

1. DALTON’S THEORY OF ATOM In 1807, John Dalton developed his atomic theory. According to this theory : • An atom cannot be subdivided. • Atoms are neither created nor destroyed during chemical reactions. • Atoms of the same element are alike; in particular all atoms of an element have the same mass. • Atoms of different elements are not alike; in particular, their masses are different. 1.1 REASON FOR FAILURE OF DALTON’S THEORY In 1833, Michael Faraday showed that there is a relationship between matter and electricity. This was the first major break through to suggest that atom was not a simple indivisible particle of all matter but was made up of small particles. Discovery of electrons, protons and neutrons discarded the indivisible nature of atom proposed by John Dalton.

2. SUBATOMIC PARTICLES The researches done by various eminent scientists and the discovery of radioactivity have established beyond doubt, that atom is not the smallest indivisible particle but had a complex structure of its own and was made up of still smaller particles like electrons, protons, neutrons etc. At present about 35 different subatomic particles are known but the three particles namely electron, proton and neutron are regarded as the fundamental particles. We shall now take up the brief study of these fundamental particles. 2.1 CATHODE RAYS (DISCOVERY OF ELECTRON) Gases are normally poor conductor of electricity at ordinary or high pressure. However when a tube filled with a gas is evacuated to a pressure of 0.01 mm Hg or lower and an electric potential is applied across a pair of electrodes into the tube discharge takes place between the electrodes during which a stream of rays moves from cathode to anode, William Crookes (1879) called these rays as cathode rays. 2.1a PROPERTIES OF CATHODE RAYS 1.

Cathode rays travel in straight lines with a high speed approaching to that of light. Their linear propagation is shown by the fact that they caste shadows of the solid objects placed in their path.

2.

They produce temperature rise in any object they strike.

3.

They pass through thin films of metals but are stopped by thicker foils.

4.

Cathode rays can produce mechanical effect e.g. when allowed to fall on a small paddle wheel they cause rotation of wheel. This experiment shows that cathode rays consist of material particles.


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5.

The cathode rays can be deflected by electric and magnetic field. The direction of deflection is always such that these particles bear a negative charge.

6.

Finally no matter what the nature of the cathode or the gas in the discharge tube, the negatively charged particles are always the same as reflected by the same e/m ratio by J.J. Thomson.

These particles constituting cathode rays have been named electron by J.J. Thomson. The fact that electrons are independent of the nature of source from which they come suggest that they are constituent of all elements. Thomson determined e/m ratio of electrons by applying electric and magnetic field 2.1b METHOD OF EXPERIMENTAL DETERMINATION OF E/M Tube was evacuated through here and then sealed off.

Anode + (metal plugs with slits)

Scale pasted on outside of tube to measure deflection of cathode ray

–

3 Magnetic only 1 Both or neither 2 Electric only Cathode –

Cathode ray

Aluminum plates (5 cm × 2 cm, 1.5 cm apart)

e Determination of m ratio of electron

e of electron = 1.7588 ×1011 C / kg = 1.7588 ×108 C / g m

2.1C DETERMINATION OF CHARGE OF ELECTRON An American physicist R. A. Millikan determined charge on electron by oil drop experiment Charge on electron = 1.6022 × 10 –19 coulombs, from the data obtained from the above two experiments, mass of electron can be calculated. m=

e = 9.1× 10–28 g e m

Eyepiece of ‘‘telescope’’ with three equally spaced crosshairs

Oil was sprayed in fine drops. Air Pinhole with cover X-rays

‘‘Telescope’’ (lowpower magnifier) Brass plates 22 cm in diameter and 1.6 cm apart

Drop under observation

Determination of charge of electron


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2.2 POSITIVE RAYS (ANODE RAYS) OR CANAL RAYS Since electron is an essential constituent of atom and since atom as a whole is electrically neutral. it follows that an equal magnitude of positive charge must be present in an atom. In 1886 Goldstein discovered that not only cathode rays composed of electrons moving towards anode are present but positive rays moving in opposite direction are also present. By perforating the cathode these positive rays can be made to pass through the hole. 2.2a PROPERTIES OF ANODE RAYS (i)

They are deflected in electric and magnetic field just as cathode rays but in the opposite direction showing that they consist of positively charged particles.

(ii) the charge and mass of the particles constituting anode rays vary and depend upon the nature of gas contained in the discharge tube. (iii) Positive rays are not emitted by anode but originate between the electrodes from the ionization of gas atoms. The mass of positively charged particle is virtually the same as that of the atoms from which they are derived, Wein (1898) determined the value of e/m for positive particles and found that it was different for different gases. When hydrogen gas was taken in the discharge tube, the lightest positively charge particle was found. It’s mass was nearly the same as that of H atom (1.673 × 10–24 g) and carried a positive charge exactly equal and opposite to that of an electron. This positively charged particle was named proton. 2.3 DISCOVERY OF NEUTRON The electrically neutral charge particle, neutron was discovered by James Chadwick by bombarding boron or beryllium with α -particles. 9 4

1 Be + 24 He  → 12 6 C + 0n

2.4 CHARACTERISTICS OF THE THREE FUNDAMENTAL PARTICLES ARE Electron Proton Neutron Symbol P n e or e –1 Approximate relative mass

1/1836

1

1

–1

+1

No charge

Mass in kg

9.109 × 10 –31

1.673 × 10 –27

1.675 × 10 –27

Mass in amu

5.485 × 10 –4

1.007

1.008

Actual charge (coulomb)

1.602 × 10 –19

1.602 × 10 –19

0

4.8 × 10 –10

4.8 × 10 –10

0

Approximate relative charge

Actual charge (e.s.u.)




3. VARIOUS ATOMIC MODELS 3.1 THOMSON’S MODEL Based on these discoveries J.J. Thomson proposed the Plum Pudding Model. This model proposed that atoms are blobs of a positively charged jelly like material, with electrons suspended in it like raisins in a pudding.

3.2 RUTHERFORD’S MODEL Rutherford allowed a narrow beam of α–particles to fall on a very thin gold foil (thickness = 4 × 10–4 cm) and

he observed (i) Majority of the α -particles pass straight through the gold strip with little or no deflection. (ii) Some -particles are deflected from their path and diverge. (iii) Very few -particles are deflected backwards through angles greater than 90°. (iv) Some were even scattered in the opposite direction at an angle of 180° [Rutherford was very much surprised by it and remarked that "It was as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back and hit you"]. 3.2a CONCLUSIONS 1. The fact that most of the α particles passed straight through the metal foil indicates the most part of the atom is empty. 2. The fact that few α particles are deflected at large angles indicates the presence of a heavy positively charged body i.e., for such large deflections to occur α -particles must have come closer to or collided with a massive positively charged body, and he named it nucleus. 3. The fact that one in 20,000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.


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3.2b ATOMIC MODEL Rutherford proposed an atomic model as follows. (i) All the protons (+ve charge) and the neutrons (neutral charge) i.e. nearly the total mass of an atom is present in a very small region at the centre of the atom. The atom's central core is called nucleus. (ii) The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10 –13 while the atom has a diameter of the order 10 –8 of cm. So, the size of atom is 105 times more than that of nucleus. (iii) Most of the space outside the nucleus is empty. (iv) The electrons, equal in number to the net nuclear positive charge, revolve around the nucleus with high speed in various circular orbits. (v) The centrifugal force arising due to the high speed of an electron balances the columbic force of attraction of the nucleus and the electron remains stable in its path. Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part. 3.2C DEFECTS OF RUTHERFORD'S ATOMIC MODEL 1. Position of electrons : The exact positions of the electrons from the nucleus are not mentioned. 2. Stability of the atom : Neils Bohr pointed out that Rutherford's atom should be highly unstable. According to the law of electro-dynamics, the electron should therefore, continuously emit radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus. It was calculated that the electron should fall into the nucleus in less than10 –8 sec. But it is known that electrons keep moving outsided the nucleus. To solve this problem Neils Bohr proposed an improved form of Rutherford's atomic model. Before going into the details of Neils Bohr model we would like to introduce you some important atomic terms.

3.3 SOME IMPORTANT CHARACTERISTICS OF A ELECTROMAGNETIC WAVE A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and troughs. Every wave has five important characteristics, namely, wavelength (λ), frequency (ν), velocity (c), wave number ( ν ) and amplitude (a). Magnetic field component Direction of travel

Electro magnetic wave 6 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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Ordinary light rays, X–rays,γ–rays, etc. are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. These radiations have wave characteristics and do not require any medium for their propagation. (i) Wave length (λ) : The distance between two neighbouring troughs or crests is known as wavelength. It is denoted by λ and is expressed in cm, m, nanometers (1nm=10–9m) or Angstrom (1Å=10–10m). (ii) Frequency (ν) : The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. It is denoted by ν(nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps. The frequency of a wave is inversely proportional to its wave length (λ) ν∝

1 c or ν = λ λ

(iii) Velocity : The distance travelled by the wave in one second is called its velocity. It is denoted by c and is expressed in cm sec–1. c = νλ or λ =

c ν

(iv) Wave number ( ν ) : It is defined as number of wavelengths per cm. It is denoted by ν and is expressed in cm–1. ν=

1 ν (or) ν = λ c

(v) Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. It determines the intensity or brightness of the beam of light. If all the components of Electromagnetic Radiation (EMR) are arranged in order of decreasing or increasing wavelengths or frequencies, the pattern obtained is known as Electromagnetic Spectrum. The following table shows all the components of light. S.No. Name 1. Radio wave

Wavelength

Frequency(Hz)

3 × 1014 – 3 × 107

1× 105 – 1×109

2.

Microwave

3 × 107 – 6 × 106

1× 109 – 5 × 1011

Source Alternating current high frequency Klystron tube

3.

Infrared (IR)

5 × 1011 – 3.95 × 1016

Incandescent objects

4.

Visible

6 × 106 – 7600 7600–3800

3.95 × 1016 – 7.9 × 1014

Electric bulbs, sun rays

5.

Ultraviolet(UV)

3800–150

7.9 × 1014 – 2 × 1016

6.

X-Rays

150–0.1

2 × 1016 – 3 × 1019

7.

γ -Rays

0.1–0.01

3 × 1019 – 3 × 1020

8.

Cosmic Rays

0.01–zero

3 × 1020 –Infinity

Sun rays, arc lamps with mercury vapours Cathode rays striking metal plate Secondary effect of radioactive decay Outer space 7

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3.4 ATOMIC SPECTRUM Electromagnetic spectrum consist of radiations of different wavelength and frequencies. An instrument used to separate the radiations of different wavelength is called spectroscope or spectrograph. It consist of a prism or diffraction grating for dispersion of radiations and a telescope to examine the emergent radiations. In spectroscope the telescope is replaced by photographic film. 1. Emission Spectrum: When the radiation emitted from some source e.g. from the sun or by passing electric discharge to high temperature is passed through prism and then received on photographic plate the spectrum obtained is known as emission spectrum. Emission spectra are as two types: (a) Continuous Spectra : When white light from any source such as sun a bulb or hot glowing body is analysed by passing through the prism. It is observed that it splits up into seven different wide band of colours. These colours are so continuous that each merges into the next. Hence spectrum is continuous spectrum. (b) Line Spectra : When some volatile salt like NaCl is placed in Bunsen Burner or an electric discharge is passed through a gas. Light is emitted if this light is resolved and passed in a spectroscope. It is found that no continuous spectrum is obtained but isolated lines are obtained. This spectrum is called line spectrum. Each line corresponds to a wavelength. (2) Absorption Spectra : When white light from any source is first passed through the solution or vapours of a chemical substance and than analysed by spectroscope. It is observed that some dark lines are obtained. These dark lines are supposed to result from the fact that when white light is passed through chemical substance radiations of certain wavelength are absorbed. Dark lines are obtained at same place where coloured lines were obtained in emission spectra. 3.4a HYDROGEN SPECTRUM If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analysed with the help of spectrograph, it is found to consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly seen on the photographic film. Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum consists of six series of lines each series, known after their discoverer as the Balmer, Paschen, Lyman, Brackett, Pfund and Humphrey series. The wavelength of all these series can be expressed by a single formula. 1 1 1 =ν=R 2 – 2 λ  n1 n 1  ν = wave number λ = wave length R = Rydberg constant (109678 cm –1 ) n1 and n 2 have integral values as follows 8 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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Series Lyman Balmer Paschen Brackett Pfund Note :

n1

n2

Main spectral lines

1 2 3 4 5

2, 3, 4, etc 3, 4, 5 etc 4, 5, 6 etc 5, 6, 7 etc 6, 7, 8, etc

Ultra-vio Visible Infra-red Infra-red Infra-red

All lines in the visible region are of Balmer series but reverse is not true, i.e., all Balmer lines will not fall in visible region

The pattern of lines in atomic spectrum is characteristic of hydrogen.

3.5 PLANCK’S QUANTUM THEORY OF RADIATION In order to explain the spectral distribution of energy radiated by a back body. Max Planck (1901) put forward quantum theory of radiation. According to this theory. (i) Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles or packets known as quanta or photon. (ii) The energy of a quantum is directly proportional to the frequency of the radiation E∝v E = hv v = frequency of radiation h = planck’s constant (iii) A body can emit or absorb energy only in whole number multiples of quantum i.e 1hv, 2hv, etc. Energy in fractions of a quantum cannot be lost or absorbed. this is known as quantization of energy. Illustration 1:

Calculate the frequency and wave number of radiation with wavelength 480 nm.

Solution :

Given λ = 480 nm = 480 × 10–9 m [1 nm = 10–9 m] c = 3 × 108 m/s Frequency,

Illustration 2:

Solution :

v=

c 3 × 108 ms −1 = = 6.25 ×1014 s −1 = 6.25 × 1014 Hz. λ 48 × 10−9

Find the quantum numbers of the excited state of electrons in He+ ion which on transition to ground state and first excited state emit two photons of wavelengths. 30.4 nm and 108.5 nm respectively. (RH = 1.09678 × 107 m–1). 1 1 1 = R H Z2  2 − 2  λ  n1 n 2 

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1 1  1 = 1.09678 × 107 × 4  2 − 2  ∴n2 = 2 −9 30.4 × 10  11 n 2 

n1= 2 (first excited state), n2 = ? 1 1 1 = 1.09678 × 107 × 4  2 − 2  n2 = 5 −9 108.5 × 10  2 n2 

Illustration 3:

In the measurement of the quantum efficiency of photosynthesis in green plants, it was found that 8 quanta of red light at 6850 Å were needed to release one molecule of O2. The average energy storage in the photosynthesis process is 112 kcal/mol O2 released. What is the energy conversion efficiency in the photosynthetic process ?

Solution :

Energy possessed by one quanta, E1 = =

hc λ

6.625 × 10−34 × 3 × 108 = 2.9 × 10−19 J / molecule −10 6850 × 10

Average energy storage in photosynthesis = 112 kcal/mol =

112 × 103 × 4.2 = 7.81 × 10−19 J / molecule 6.023 × 10 23

Energy conversion efficiency = =

Average energy storage × 100 Energy of 8 quanta

7.81 × 10−19 × 100 = 33.66% 8 × 2.9 × 10−19

EXERCISE – 1 1.

Calculate the energy associated with photon of light having a wavelength 6000 Å.

2.

How many photons of light having a wavelength 4000 Å are necessary to provide 1.00 J of energy?

3.

How many moles of photon would contain sufficient energy to raise the temperature of 225 g of water 21°C to 96°C? Specific heat of water is 4.18 J g–1 K–1 and frequency of light radiation used is 2.45 × 109 s–1.

3.6 BOHR’S ATOMIC MODEL Bohr developed a model for hydrogen and hydrogen like atoms one-electron species (hydrogenic species). He applied quantum theory in considering the energy of an electron bond to the nucleus.


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3.6a IMPORTANT POSTULATES An atom consists of a dense nucleus situated at the center with the electron revolving around it in circular orbits without emitting any energy. The force of attraction between the nucleus and an electron is equal to the centrifugal force of the moving electron. Of the finite number of circular orbits possible around the nucleus, and electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor h/ 2π . mvr =

nh 2π

where, m = mass of the electron v = velocity of the electron n = orbit number in which electron is present r = radius of the orbit As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence these orbits are called stationary states. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. The greater the distance of the energy level from the nucleus, the more is the energy associated with it. The different energy levels are numbered as 1, 2, 3, 4, (from nucleus onwards) or K, L, M,N etc. Ordinarily an electron continues to move in a particular stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state. If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2, 3, 4, etc) by absorbing one quantum of energy. This new state of electron is called as excited state. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels. Since the excited state is less stable, atom will lose it’s energy and come back to the ground state. Energy absorbed or released in an electron jump, (∆E) is given by ∆E = E 2 – E1 = hv Where E 2 and E1 are the energies of the electron in the first and second energy levels, and v is the frequency of radiation absorbed or emitted. Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appear as kinetic energy of emitted photo electron.

3.6b MERITS OF BOHR’S THEORY (i) The experimental value of radii and energies in hydrogen atom are in good agreement with that calculated on the basis of Bohr’s theory. 11 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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(ii) Bohr’s concept of stationary state of electron explains the emission and absorption spectra of hydrogen like atoms. (iii) The experimental values of the spectral lines of the hydrogen spectrum are in close agreement with the calculated by Bohr’s theory. 3.6C LIMITATIONS OF BOHR’S THEORY (i) It does not explain the spectra of atoms or ions having more than one electron. (ii) Bohr’s atomic model failed to account for the effect of magnetic field (Zeeman effect) or electric field (Stark effect) on the spectra of atoms or ions. It was observed that when the source of a spectrum is placed in a strong magnetic or electric field, each spectral line further splits into a number of lines. This observation could not be explained on the basis of Bohr’s model. (iii) de-Broglie suggested that electrons like light have dual character. It has particle and wave character. Bohr treated the electron only as particle. (iv) Another objection to Bohr’s theory came from Heisenberg’s Uncertainty Principle. According to this principle “it is impossible to determine simultaneously the exact position and momentum of a small moving particle like an electron”. The postulate of Bohr, that electrons revolve in well defined orbits around the nucleus with well defined velocities is thus not attainable. 3.6d BY BOHR’S THEORY (i) Radius and Energy levels of hydrogen atom: Consider an electron of mass `m’ and charge `e’ revolving around a nucleus of charge Ze (where, Z = atomic number and e is the charge of the proton) with a tangential velocity v.r is the radius of the orbit in which electron is revolving. By Coulomb’s Law, the electrostatic force of attraction between the moving electron KZe2 and nucleus is Coulombic force = 2 r 1 K= (where ε0 is permitivity of free space) 4πε0 K = 9 × 109 Nm2 C –2 In C.G.S. units, value of K = 1 dyne cm 2 (esu) –2 mv 2 r Since the electrostatic force balance the centrifugal force, for the stable electron orbit. The centrifugal force acting on the electron is

mv 2 KZe 2 = 2 r r

… (i)

KZe 2 … (ii) mr According to Bohr’s postulate of angular momentum quantization, we have (or) v 2 =


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nh 2π nh v= 2πmr mvr =

n2h2 4π 2 m 2 r 2 Equating (2) and (3) v2 =

… (iii)

KZe 2 n 2h 2 = 2 2 2 mr 4π m r Solving for r we get r =

n2h2 4π 2 mKZe 2

where n = 1, 2, 3, … ∞ Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of n, i.e., farther the level from the nucleus the greater is the radius. The radius of the smallest orbit (n = 1) for hydrogen atom (Z = 1) is r0 . r0 =

n 2h 2 12 × (6.626 × 10 –34 ) 2 = = 5.29 × 10 –11 m 4π2 me2 K 4 × (3.14)2 × 9 × 10 –31 × (1.6 × 10–19 )2 × 9 ×109

r0 = 0.5629Å Radius of nth orbit for an atom with atomic number Z is simply written as n2 n2 = 0.529 × Å z z (ii) Energy level of Hydrogen atom: The total energy, E of the electron is the sum of kinetic energy and potential energy. Kinetic energy of the electron = 1 2 mv 2 rn = r0

Potential energy =

–KZe 2 r

KZe2 r From equation (1) we know that

Total energy = 1 2 mv 2 –

… (iv)

mv 2 KZe 2 = 2 r r KZe2 2r Substituting this in equation (iv) ∴

1

2 2 mv =

Total energy (e) =

KZe2 KZe2 KZe 2 – =– 2r r 2r 13


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1 KE = – PE, KE = –TE 2 Substituting for r, gives us 2π 2 mZe 2e 4 K 2 where n = 1, 2, 3, … n 2h 2 This expression shows that only certain energies are allowed to the electron. Since this energy expression consists so many fundamental constant, we are giving you the following simplified expressions. E=

E = –21.8 × 10 –12 × = – 21.8 × 10 –19 × E n = –13.6

Z2 erg per atom. n2

Z2 Z2 J per atom = – 13. 6 × eV per atom n2 n2

z2 eV per atom n2

(1eV = 3.83 ×10 –23 Kcal) (1eV = 1.602 × 10 –12 erg) (1eV = 1.602 × 10 –19 J) Z2 kcal/mole (1 cal = 4.18 J) n2 The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n = 1, and as the quantum number increases, E become less negative. When n = ∞ , E = 0 which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated. E = –313.6 ×

H  → H + + e – (ionization). (iii) Velocity of electron nh nh ;v= 2π 2πmr By substituting for r we are getting We know that, mvr =

v=

2π KZe 2 nh

Z cm / sec. n Further application of Bohr’s work was made, to other one electron species (Hydrogenic ion) such as He + and Li2 + . In each case of this kind, Bohr’s prediction of the spectrum was correct. Where excepting n and z all are constant, v = 2.18 ×108


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(iv) Explanation for hydrogen spectrum by Bohr’s theory: According to the Bohr’s theory electron neither emits nor absorbs energy as long as it stays in a particular orbit. However, when an atom is subjected to electric discharge or high temperature, and electron in the atom may jump from the normal energy level, i.e., ground state to some higher energy level i.e., excited state. Since the life time of the electron in excited state is short, it returns to the ground state in one or more jumps. During each jumps, energy is emitted in the form of a photon of light of definite wavelength or frequency. The frequency of the photon of light thus emitted depends upon the energy difference of the two energy levels concerned ( n1 , n 2 ) and is given by hv = E 2 – E1 =

–2π 2 mZ2 e 4 K 2 h2

1 1  n2 – n2   2 1 

2π2 mZ2 e 4 K 2  1 1  2 – 2 3 h  n1 n 2  The frequencies of the spectral lines calculated with the help of above equation are found to be in good agreement with the experimental values. Thus, Bohr’s theory elegantly explains the line spectrum of hydrogen and hydrogenic species. Bohr had calculated Rydberg constant from the above equation. v=

C 2π 2 mZ2 e 4 K 2 = λ h3

1 1  2 – 2  n1 n 2 

1 2π2 mZ2 e 4 K 2 =ν= λ h 3c

1 1  2 – 2  n1 n 2 

ν=

where

2 π 2 me 4 K 2 = 1.097 × 10 –7 m –1 or 109678 cm –1 3 hc

i.e. Rydberg constant (R) ∴

ν=

1 1 1 = RZ2  2 – 2  λ  n1 n 2 

ν = wave number. Illustration 4: Solution:

Calculate the ratio of the radius of Li+2 ion in 3rd energy level to that of He+ ion in 2nd energy level. n2 r∝ z 2

r1  n1   z 2  =    r2  n 2   z1  n1 = 3 n2 = 2 z1 = 3

(for Li2+ ) 15


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(for He + )

z2 = 2 2

r1  3   2  3 =  × = r2  2   3  2 Illustration 5:

How far from the nucleus is the electron in a hydrogen atom if it has energy of –0.850 eV.

Solution:

En = –

Z2 × 21.69 × 10 –19 n2

Z2 – 2 × 21.69 ×10 –19 = –0.85 ×1.6 ×10 –19 n Find the wavelengths of the first line of He+ ion spectral series whose 1 1 – = 2.7451× 10 4 cm –1 interval between extreme lines is λ1 λ2

Illustration 6:

Solution:

Extreme lines means first and last. 1 1  1 1 1  1 RZ2 = RZ2  2 – 2  – R 2  2 – – = 2 2 λ λ2  n1 ∞   n1 (n1 + n)  (n1 + 1) 2.7451× 104 =

109677.76 × 2 2 (n1 + 1)2

(n1 + 1) = 4 n1 = 3 Wavelength of first line, 1 1 1 = 109677.76 × 2 2 ×  2 – 2  λ 3 4  λ = 4689 × 10–8 cm = 4689Å Illustration 7 :

The velocity of electron in a certain Bohr's orbit of hydrogen atom bears the ratio of 1 : 275 to the velocity of light (a) What is the quantum number (n) of orbit? (b) Calculate the wave number of radiations emitted when electron jumps from (n+1) state to ground state

Solution :

Velocity of electron = velocity light × Vn =

1 3 × 108 m/sec = = 1.09 × 106 m/sec 275 275

2πZKe 2 2.18 × 106 × Z = m/sec nh n


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For hydrogen ∴

Vn =

2.18 × 106 m/sec n

1.09 × 106 =

2.18 × 106 ⇒n=2 n

When electron jumps from (n+1) i.e., 3 to ground state ν=

1 = 1.09678 × 107 λ

⇒ Illustration 8:

1 1 1 = RH =  2 − 2  λ 1 3  1 1  2 –1 1 − 9  = 9.75 × 10 m  

Two hydrogen atoms collide head on and end up with zero kinetic energy. Each then emits a photon of wave length 121.6 nm which transition leads to this wavelength? How fast were the hydrogen atoms travelling before collision? RH = 1.09678 × 107 m–1 MH = 1.672 × 10–27 Kg

Solution:

As wavelength of emitted radiation is 121.6 nm, which falls in UV region of spectrum Therefore, n1 = 1 For hydrogen atom ∴

1 1 1  = RH  2 − 2  λ 1 n 

1 = 1.09678 × 107 121.6 × 10−9

1 1  1 − n 2   

n=2 Energy released is due to collision and all the kinetic energy released in form of radiant energy 1 hc mv2 = 2 λ

⇒

1 6.625 × 10−34 × 3 × 108 1.67 × 10–27 × V2 = 2 121.6 × 10−9

∴

V = 4.42 × 104 m/sec

EXERCISE –2 1.

Calculate the radii of the first two Bohr orbits of Li2+.

2.

Ionisation energy of H-atom is 13.6 eV. Calculate ionization energy of He+, Li2+ and Be3+ (all isoelectronic of H).

3.

If the electron in a hydrogen atom were in the energy level with n = 3, how much energy would be required to ionize the atom ? 17 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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4. PHOTOELECTRIC EFFECT Emission of electrons from a metal surface when exposed to light radiations of appropriate wavelength is called photoelectric effect. The emitted electrons are called photoelectrons. Work function or threshold energy may be defined as the minimum amount of energy required to eject electrons from a metal surface. According to Einstien, Maximum kinetic energy of the ejected electron = absorbed energy – work function 1 2 2 mv max

1 1  = hv − hv0 = hc  −   λ λ0 

where v0 and λ0 are threshold frequency and threshold wavelength. 4.1 LAWS OF PHOTOELECTRIC EFFECT (i) Rate of emission of photoelectrons from a metal surface is directly proportional to the intensity of incident light. (ii) The maximum kinetic energy of photoelectrons is directly proportional to the frequency of incident radiation; moreover, it is independent of the intensity of light used. (iii) There is no time lag between incidence of light and emission of photoelectrons. (iv) For emission of photoelectrons, the frequency of incident light must be equal to or greater than the threshold frequency. Illustration 9 :

Photo - electrons are liberated by ultraviolet light of wavelength 3000Å from a metallic surface for which the photoelectric threshold is 4000Å. Calculate de-Broglie wavelength of electrons emitted with maximum kinetic energy

Solution:

As,

ν = hν° + KE ⇒

∴

.E. = hc  − o  = λ λ 

1

=

hc hc = + K.E λ λo

1 

hc(λ − λo ) = λ×λ

6.625 × 10−34 × 3 × 108 (4000 × 10−10 − 3000 × 10−10 ) 3000 × 10−10 × 4000 × 10−10

6565 × 10–19 Joule Also

1 mν2 = 1.6565 × 10–19 2

⇒

m2v2 = 2 × 1.65665 × 10–19 × m

⇒

v2 = 2 × 1.6565 × 10–19 × 9.109 × 10–31 ⇒ mν = 5.49 × 10–25 =

h 6.625 × 10−34 = = 1.2 × 10–9 m −25 mv 5.49 × 10


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Illustration 10 : Photo - electrons are liberated by ultraviolet light of wavelength 3000Å from a metallic surface for which the photoelectric threshold is 4000Å. Calculate de-Broglie wavelength of electrons emitted with maximum kinetic energy Solution:

hc hc = + K.E λ λo

As,

hν = hν° + KE ⇒

∴

K.E. = hc  − o  = λ λ 

1

=

1 

hc(λ − λo ) = λ×λ

6.625 × 10−34 × 3 × 108 (4000 × 10−10 − 3000 × 10 −10 ) 3000 × 10−10 × 4000 × 10−10

=1.6565 × 10–19 Joule Also

1 mν2 = 1.6565 × 10–19 2

⇒

m2v2 = 2 × 1.65665 × 10–19 × m

⇒

m2v2 = 2 × 1.6565 × 10–19 × 9.109 × 10–31

⇒

mν = 5.49 × 10–25 λ=

h 6.625 × 10−34 = = 1.2 × 10–9 m mv 5.49 × 10−25

5. PARTICLE AND WAVE NATURE OF ELECTRON In 1924, de Broglie proposed that an electron, like light, behaves both as a material particle and as a wave. This proposal gave birth to a new theory known as wave mechanical theory of matter. According to this theory, the electrons, protons and even atoms, when in motion, possess wave properties. de Broglie derived an expression for calculating the wavelength of the wave associated with the electron. According to Planck’s equation E = hv = h ⋅

c λ

…(i)

The energy of a photon on the basis of Einstein’s mass–energy relationship is E = mc2

…(ii)

where c is the velocity of the electron. Equating both the equations, we get h

c = mc 2 λ

λ=

h h = mc p 19


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Momentum of the moving electron is inversely proportional to its wavelength. Let kinetic energy of the particle of mass ‘m’ is E. E = ½ mv2 2Em = m2v2 2Em = mv = p (momentum) λ=

h h = p 2Em

5.1 DAVISSON AND GERMER MADE THE FOLLOWING MODIFICATION IN DE BROGLIE EQUATION Let a charged particle say an electron be accelerated with a potential of V; then the kinetic energy may be given as : ½mv2 = eV m2v2 = 2eVm mv = 2eVm = p h λ= 2eVm de Broglie waves are not radiated into space, i.e. they are always associated with electron. The wavelength decreases if the value of mass (m) increases, i.e., in the case of heavier particles, the wavelength is too small to be measured. de Broglie equation is applicable in the case of smaller particles like electron and has no significance for larger particles. 5.2 BOHR’S THEORY VERSUS DE BROGLIE EQUATION One of the postulates of Bohr’s theory is that angular momentum of an electron is an h integral multiple of . This postulate can be derived with the help of de Broglie concept 2π of wave nature of electron. Consider an electron moving in a circular orbit around nucleus. The wave train would be associated with the circular orbit as shown in figure.

(a)

(b)

(a) The number of waves around the orbit is a whole number (6) times the wavelength. (b) The number of waves around the orbit is not a whole number multiple of the wavelength. When the wave goes around the orbit many times, it cancles itself out.


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If the two ends of the electron wave meet to give a regular series of crests and troughs, the electron wave is said to be in phase, i.e., the circumference of Bohr’s orbit is equal to whole number multiple of the wavelength of the electron wave. So,

2πr = n λ

or

λ=

2πr n

From de Broglie equation, λ=

h mv

Thus,

h 2πr = mv n

or

mvr = n ⋅

h 2π

[v = velocity of electron and r = radii of the orbit] i.e., Angular momentum = n ⋅

h 2π

This proves that the de Broglie and Bohr–concepts are in perfect agreement with each other. Illustration 11 : What is the mass of a photon of sodium light with a wavelength of 5890 Å. λ=

h mc

or

m=

h λc

So,

m=

6.63 × 10−27 = 3.752 × 10–33 g −8 10 5890 × 10 × 3 × 10

Solution:

Illustration 12 : An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 Å Solution :

If an electron accelerated by potential V, has velocity 'u' and associated de Broglie's wavelength λ. 1/2 mν2 = eV … (i) λ=

h mu

From equation (ii) µ =

… (ii) h mλ


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 h 

∴

2

h2

1/2 m   = e × V ⇒ V = 2mλ 2 c  mλ  =

(6.625 × 10−34 ) 2 = 63.3 volt 2 × 9.108 × 10 × (1.54 × 10−10 )2 × 1.602 × 10−19 −31

Illustration 13 : As the speed of a particle of mass m increases, its de Borglie wavelength λ decreases. What is the rate of change of λ with the kinetic energy, E for an electron with λ = 1 A ? Solution:

According to de Broglie. Matter wavelength, λ =

h h = mv p

We have, kinetic energy E= ∴

λ=

1 p2 h2 = 2 mv2 = 2 2m λ (2m) h 2mE

The rate of change of de Broglie wavelength λ with the kinetic energy E, dλ h d(E) −1/ 2 1 h λ3 m =− 2 = = − × dE dE 2 h 2m 2mE × E (1 × 10−10 ) 3 × 9.1 × 10−31  dλ  = − = 2.07×106 m J -1  dE  (6.62 × 10−34 )2   at λ =14

EXERCISE – 3 1. What accelerating potential must be imparted to a proton beam to give it an effective wavelength of 0.050Å ? 2. A 1.0 g projectile is shot from a gun with velocity of 100 m s–1. What is the de Broglie wavelength ? 3. Two particles A and B are in motion; if wavelength of A is 5 × 10–8 m, calculate wavelength of B so that its momentum is double of A.

6. HEISENBERG UNCERTAINTY PRINCIPLE For heavier objects, the velocity and their position can be measured accurately at any instant of time but same is not the case for subatomic particles. As a consequence of dual nature of matter, Heisenberg in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum of small particles. This principle states that “It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty i.e. if an attempt is made to


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measure any one of these two quantities with higher accuracy the other becomes less accurate” Mathematical expression for the Heisenberg uncertainty principle is simply written as ∆x . ∆p ≥ h / 4π

where ∆p = m∆v Explaination of Heisenberg uncertainty Principle : Suppose we attempt to measure both the position and momentum of an electron, to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting, the position as well as the velocity of the electron are disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of light used the uncertainty in position is ±λ . The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. So if position is fixed velocity will not be fixed. Illustration 14 : Why electron cannot exist inside the nucleus according to Heisenberg’s uncertainty principle? Solution :

Diameter of the atomic nucleus is of the order of 10–15m The maximum uncertainty in the position of electron is 10–15 m. Mass of electron = 9.1 ×10–31 kg. ∆x. ∆p =

h 4π

∆x × (m.∆v) = h/4π ∆v =

h 1 6.63 × 10−34 1 × = × −15 22 4π ∆x.m 10 × 9.1× 10 −31 4× 7

∆v = 5.80 × 1010 ms–1 This value is much higher than the velocity of light and hence not possible. EXERCISE – 4 1. Calculate the wavelength of a soft ball of mass of 100 g traveling at a velocity of 35 ms–1. 2. If uncertainties in the measurement of position and momentum are equal, calculate uncertainty in the measurement of velocity. 3. Calculate the uncertainty in the velocity of wagon of mass 2000 kg whose position is known to an accuracy of ± 10 m.


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7. SCHRODINGER WAVE EQUATION ∇ 2ψ +

8π 2 m (E − U)ψ = 0 h2

This equation implies in essence that a body of mass m, velocity v, potential energy U and total energy E has a wave associated with it of an amplitude given by wave function Ψ. The schrodinger equation has an infinite many solution. For any physical situation Ψ must be finite, single valued and continuous. The particular values of Ψ which yield satisfactory solution of above equation are called wave functions. Wave function for an electron is called an atomic orbital. Each solution is identified by three quantum numbers. A set of three quantum numbers is needed to describe an electron because electrons in atoms are moving in three dimensional space. Schrodinger equation

Solve

Wave function

ψ2

ψ

Probability of finding electron

for electron this wave function is known as atomic orbital

The wave function Ψ has got no meaning but the square of the wave function Ψ2 is a mathematical expression of how the probability of finding an electron in a small volume varies from place to place. Square of wave function is like the square of amplitude of an electromagnetic wave. The intensity of electromagnetic radiation is proportional to square of amplitude.

8. QUANTUM NUMBERS To understand the concept of Quantum Numbers, we must know the meaning of some terms clearly so as to avoid any confusion. Energy Level: The non-radiating energy paths around the nucleus are called as Energy Levels of Shells. These are specified by numbers having values 1, 2, 3, 4, ... or K, L, M, N, ... in order of increasing energies. The energy of a particular energy level is fixed. Sub-Energy Level: The phenomenon of splitting of spectral lines in electric and magnetic fields reveals that there must be extra energy levels within a definite energy level. These were called as Sub-Energy Levels or Sub-Shells. There are four types of sub-shells namely; s, p, d, f. First energy level (K or ) has one sub-shell designated as 1s, the second energy level (L or 2) has two sub-shell as 2s & 2p, the third energy level (M or 3) has three sub shell as 3s, 3p and 3d, and the fourth energy level (N or 4) has four sub-shells as 4s, 4p, 4d and 4f. The energy of sub-shell increases roughly in the order: s < p < d

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To describe or to characterize the electrons around the nucleus in an atom, a set of four numbers is used, called as Quantum Numbers. These are specified such that the states available to the electrons should follows the laws of quantum mechanics or wave mechanics. 8.1 PRINCIPAL QUANTUM NUMBER: (n) This quantum number represents the main energy levels (principal energy levels) designated as n = 1, 2, 3, ... or the corresponding shells are named as K, L, M, N, ... respectively. It gives an idea of position and energy of an electron. The energy level n = 1 corresponds to minimum energy and subsequently n = 2, 3, 4, ..., are arranged in order of increasing energy. Higher is the value of n, greater is its distance from the nucleus, greater is its size and also greater is its energy. It also gives the total electrons that may be accommodated in each shell, the capacity of each shell is given by the formula 2n 2 , where n : principal quantum number. 8.2 AZIMUTHAL QUANTUM NUMBER: (l) This number determines the energy associated with the angular momentum of the electron about the nucleus. It is also called as the angular momentum quantum number. It accounts for the appearance of groups of closely packed spectral lines in electric field. It can assume all integral values from 0 to n–1. The possible values of l are : 0, 1, 2, 3, ..., n–1. Each value of l describes a particular sub-shell in the main energy level and determines the shape of the electron cloud. When n = 1, l = 0, i.e., its energy level contains one sub-shell which is called as a s-subshell. So for l = 0, the corresponding sub-shell is a s-sub-shell. Similarly when l = 1, 2, 3, the sub-shells are called p, d, f sub-shells respectively. As you known for n = 1, l = 0, there is only one sub-shell. It is represented by 1s. Now for n = 2, l can take two values (the total number of values taken by l is equal to the value of n in a particular energy level). The possible values of l are 0, 1. The two sub-shell representing the IInd energy level are 2s, 2p. In the same manner, for n = 3, three sub-shells are designated as 3s, 3p, 3d corresponding to l = 0, 1, 2, and for n = 4, four sub-shells are designated as 4s, 4p, 4d, 4f corresponding to l = 0, 1, 2, 3. h . The orbital Angular momentum of electron = A(A + 1) 2π Note that its value does not depend upon value of n. 8.3 MAGNETIC QUANTUM NUMBER (m) An electron with angular momentum can be thought as an electric current circulating in a loop. A magnetic field due to this current is observed. This induced magnetism is determined by the magnetic quantum number. Under the influence of magnetic field, the electrons in a given sub-energy level prefer to orient themselves in certain specific regions 25 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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in space around the nucleus. The number of possible orientations for a sub-energy level is determined by possible values of m corresponds to the number of orbitals in a given subenergy level). m can have any integral values between –l to +l including 0, i.e., m = –l, 0 +, l, …, 0, 1, 2, 3, 4, . . ., l–1 + l. We can say that a total of (2l + 1) values of m are there for a given value of l– 2, –1, 0, 1, 2, 3. In s sub-shell there is only one orbital [l = 0, ⇒ m = (2l +1) = 1]. In p sub-shell there are three orbitals corresponding to three values of m : –1, 0 +1. [l = 1 ⇒ m = (2l +1) = 3]. These three orbitals are represented as p x , p y , p z along X, Y, Z axes perpendicular to each other. In d sub-shell, there are five orbitals corresponding to –2, –1, 0 +1, +2, [l = 2 ⇒ m = ( 2 × 2 + 1) = 5] . These five orbitals are represented as d xy , d yz , d zx , d x 2 – y2 ,dz 2 . In f sub-shell there are seven orbitals corresponding to –3, –2, –1, 0, +1, +2, +3 [l = 3 ⇒ m = (2 × 3 + 1) = 7] . 8.4 SPIN QUANTUM NUMBER (s) When an electron rotates around a nucleus it also spins about its axis. If the spin is clockwise, its spin quantum number is +1/2 and is represented as ↑ . If the spin is anticlockwise, its value is –1/2 and is represented as ↓ . If the value of s is +1/2, then by convention, we take that electron as the first electron in that orbital and if the value of s is – 1/2, it is taken as second electron. Illustration 15: Solution:

Write down the values of quantum numbers of all the electrons present in the outermost orbit of argon (At. No. 18) 1s 2 , 2s 2 2p 6 ,3s 2 3p 2x 3p 2y 3p z2 The electronic configuration of argon is Values of quantum numbers are: n

Illustration 16:

1

m

s

3s 3p 2x

3

0

0

+1/2, –1/2

3

1

±1

+1/2, –1/2

3p 2y

3

1

±1

+1/2, –1/2

3p 2z

3

1

0

+1/2, –1/2

2

(a) An electron is in 5f-orbital. What possible values of quantum numbers n, l, m and s can it have? (b) What designation is given to an orbital having (i) n = 2, l =1, and (ii) n = 3, l = 0?

Solution:

(a) For an electron in 5f-orbital, quantum number are: n = 5, l = 3; m = –3, –2, –1, 0, +1, +2, +3 and s = either +

1 1 or – 2 2

(b) (i) 2p, (ii)3s 26 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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EXERCISE – 5 1.

2.

3.

4.

Give the possible values for the missing quantum numbers : (a) n = 3, l = 0, ml = ? (b) n = 3, l = ?, ml = –1 (c) n = ?, l = 1, m1 = + 1 (d) n = ?, l = 2, ml = ? Write appropriate values of n, l and ml quantum numbers for each of the following orbital designations : (a) 4s (b) 3p (c) 5f (d) 6g (e) 3d (f) 7s Also arrange them in increasing energy. Which of the following sets of quantum numbers is/are not allowable ? Why not ? (a) n = 2, l = 1, ml = 0 (b) n = 2, l = 2, ml = –1 (c) n = 3, l = 0, ml = 0 (d) n = 3, l = 1, ml = 1 (e) n = 2, l = 0, ml = –1 (f) n = 2, l = 2, ml = –1 What type of electron orbital is designated for : (a) n = 2, l = 1, ml = –1 (b) n = 2, l = 2, ml = –1 (c) n = 5, l = 0, ml = 0

9. RULES FOR FILLING ELECTRONS 9.1 AUFBAU PRINCIPLE The aufbau principle states that in the ground state of an atom, the orbital with a lower energy is filled up first before the filling for the orbital with a higher energy commences. In other words : The electrons enter the various orbitals in the order of increasing energy. 1s 2s

2p

3s

3p

3d

4s

4p

4d

5s

5p

5d

6s

6p

4f

7s

This order of orbitals is only a guidance and exceptions are known in few cases of atoms. 27 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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9.2 PAULI'S EXCLUSION PRINCIPLE This an important generalisation given by Wolfgang Pauli, in 1925, which determines the maximum number of electrons that an energy level can accommodate. Pauli's exclusion principle states that It is impossible for any two electrons in the same atom to have all the four quantum numbers identical. Thus, in the same atom, any two electrons may have three quantum numbers identical but the fourth must be different. This principle is very useful in determining the maximum number of electrons that can occur in any shell/sub shell. For K-shell, for instance, since n=1, A can have only one value (viz., 0) and m can also have only one value (viz., 0). Hence s can be either +1/2 or -1/2. Thus, there are two combinations of the quantum number, as shown below: m = 0 s = +1/2 n =1 A = o n =1 A = o m = 0 s = -1/2 This shows that in the K-shell, there is only one subshell A = 0 and there are only two electrons of opposite spins. For L-shell, since n = 2, A can have two values ( 0 and 1,) m can have one value of A = 0 and three values of A =1 (-1,0 and +1) and s can have two values (=+1/2 and-1/2) for each value of m. These possibilities would give rise to eight combinations of the four quantum numbers, keeping in view the exclusion principle . The maximum number of electrons in a shell is given by 2n2. 9.3 HUND'S RULE OF MAXIMUM MULTIPLICITY This rule states that Electron pairing in orbitals of same energy (degenerate orbitals) will not take place unless all the available orbitals of a given subshell contain one electron each with parallel spin. For example, we know that there are three p orbitals (px, py and pz) of the p-subshell in a principle energy level. According to Hund's rule, each of the three p orbitals must get one electron each of parallel spin before any one of them gets the second electron of opposite spin as shown below

9.4 RULE OF EXTRA STABILITY OF COMPLETELY HALF FILLED OR COMPLETELY FILLED ORBITAL When the energy of a wave function is evaluated by using the appropriate Schrodinger equation, two types of energies, viz., Colulomb and exchange energies, are involved. The coulomb energy is a measure of the mean repulsion of two electrons occupying different orbitals, the exchange energy is the lowering of energy for a set of two electrons of parallel spin which arises as a result of in distinguish ability of the electron coordinates. For a system containing more than two electrons of parallel spin, the lowering of energy due to exchange of electrons is given by ∆E = N × K


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where N is the total number of possible exchanges of coordinates between sets of two electrons of parallel spin and K is the average exchange energy per set of electrons of parallel spin. For a system containing n electrons of parallel spin. For a system containing n electrons of parallel spin, N is given by N=

n! 2(n − 2)!

The exchange energies for the two possible configurations of Cr are as follows: 3d4, 4s2 n = 5 ∆E = 10K 1 n = 6 ∆E = 15K 3d5, 4s that is, the exchange energy stabilization of 3d5, 4s1 configuration is larger than 3d4, 4s2 configuration. Besides this, since the two electrons with parallel spins remain farther apart than two electrons with opposed spins, the Columbic repulsion energy is also smaller in case of two parallel spins. Thus the configuration 3d54s1, which contains more number of parallel spins, is more stable than 3d54s2. The stability of the 3d104s1 configuration for the copper atom may be explained in the same way. Thus, it may be concluded that the unusual stability of electronic configurations involving half-filled and completely filled orbitals is due to the larger exchange energy and lesser columbic repulsion. The simplified explanation of the above illustration is given below : Exchange Energy : This stabilizing effect arises whenever two or more electrons with the same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half filled or completely filled (Figure given below). As a result the exchange energy is maximum and so is the stability. 1

3

2

4 exchange by electron 1



4


Fig. Possible exchanges for a d5 configuration. 9.5 ELECTRONIC CONFIGURATIONS OF ELEMENTS Based on the rules, we can easily determine the electronic configurations of most elements. We just need to know the atomic number of an element, the order in which orbitals are to be


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filled and the maximum number of electrons in a shell, sub-shell or orbital. The configuration so obtained can be represented in two ways. As an illustration, let us consider fluorine (Z = 9): F (Z = 9) 1s2 2s2 2p 2x 2p 2y 2p 1z or ↑ ↓ ↑ ↓ ↑ ↑ ↓ ↑ ↓ 1s

2s

2px

2py

2pz

Importance of knowing the exact electronic configuration of an element lies in the fact that the chemical properties of an element are dependent on the behaviour and relative arrangement of its electrons. Electronic configurations of heavier elements (beyond Z = 56) deviate a little from the order mentioned previously. These are brief listed below a) Lanthanides La (Z = 57) : [Xe] 6s25d1 (not 4f1) Ce (Z = 58) : [Xe] 6s25d14f1 Pr (Z = 59) : [Xe] 6s25d14f2 b) Actinides Ac (Z = 89) : [Rn] 7s26d1 (not 5f1) Th (Z = 90) : [Rn] 7s26d15f1 Pa (Z = 91) : [Rn] 7s26d15f2 c) Beyond Z = 103 Z = 104 : [Rn] 5f146d27s2 Z = 105 : [Rn] 5f146d37s2 Z = 106 : [Rn] 5f146d47s2 Z = 112 : [Rn] 5f146d107s2 EXERCISE – 6 1. State the basic ideas that are violated by each of the following electronic configuration and replace each by the correct configuration : (a) B5 – 1s2 2s2 (b) Na11 – 1s2 2p6 2p6 2d1 (c) K19 – [Ar] 3d1 (d) Ti22 – [Ar] 4s2 4p2 (e) Hg80 – [Xe] 4f10 5d10 6s2 6p2 2. Which of the following will be coloured ion ? (a) Fe2+ (b) Cu+ (c) Sc3+ (d) Mn2+ Which has the maximum magnetic moment ? 3. Magnetic moment of X3+ ion of 3d series is 35 B.M. What is atomic number of X3+ ?

10. SHAPES AND SIZES OF ORBITALS By shape of an orbital we mean the shape of the region in space in which there is probability of finding an electron. It is basically determined by the azimuthal quantum number ( A ), while the orientation of an orbital depends on the magnetic quantum number (m). The size of an orbital increases with the value of the principal quantum number (n). Shapes of orbitals in the various sub-shells are as discussed below: The place where probability of finding an electron is 0 (Ψ = 0) is known as node and a plane passing through node is known as nodal plane. 30 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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Total number of nodal planes = A There are two types of nodes: (i) Radial node: These are the points at some distance from the nucleus where there is zero probability of finding the electrons. S-orbital being spherically symmetric and hence non-directional, can’t have angular node. (i.e. points where 4πr 2 Ψ 2 = 0 ) (ii) Angular nodes: These are directional in nature so these are associated with p and d orbitals. For a particular quantum number n Total nodes = n – 1 Radial nodes = n – A – 1 Angular nodes = A 10.1 s – ORBITALS s - orbitals ( A = 0) : The orbitals are spherical and symmetrical about the nucleus. There is vacant space between two successive s-orbitals, known as ‘radial node’. Due to nucleus symmetricity, these are non-directional.

y

2s

1s x

nucleus

radial node

Z

10.2 p – ORBITALS p - orbitals ( A = 1): These have a dumb - bell shape, having three possible mutually perpendicular orientations. As they are not symmetrical, they have a directional character. The two lobes of p-orbitals are separated by a nodal plane, where the probability of finding electron is zero. z

yz plane y

y

y

xy plane

xz plane x

x

px orbital

z

z

py orbital

x

pz orbital

The three 2p orbitals and Nodal Planes

10.3 d – ORBITALS d - orbitals ( A = 2): They have relatively complex geometry. Out of the five d-orbitals, the first two d z2 & d x2 − y2 lie along the axes, while the other three (dxy, dzx, dyz) project in

(

)

between the axis.


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z

z

z

y

y

y

x

x

x dxy orbital z

dyz orbital

dxz orbital z y

y

x

x dx2 – y 2 orbital

dz2 orbital The five d orbitals

10.4 NODES AND NODAL PLANES FOR d-ORBITALS z

z Nodal cone y

y

x

x

Two nodal surface forming two nodal cones about z axis for d Z2 orbital.

z

z

xz - nodal plane

y

xy - nodal plane

z

y

x

Two nodal planes xy and xz for d yz orbital


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z

yz - nodal plane

z

y

y xy - nodal plane

x

x

Two nodal planes yz and xy for dxz orbital

z

yz - nodal plane z

y

y xz - nodal plane x

x

Two nodal planes xz and yz for dxy orbital z

z

y

y

x

x

Nodal planes for d x2 − y2 at angle 45° to x and y axis

Illustration 17 : Name the orbitals corresponding to given set of quantum numbers

Solution:

(a) n = 3, A = 2, m = ± 2

(b) n = 4, A = 0, m = 0

(c) n = 3, A = 1, m ± 1

(d) n = 2, A = 1, m = 1

(a) 3d x 2 − y2

(b) 4s

(c) 3p x or 3p y

(d) 2px or 2py


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Illustration 18 : (a) Find the orbital angular momentum of an electron in the following orbital (i) 3p

(ii) 3d

(iii) 3s

(b) Arrange the electrons represented by the following sets of quantum number in decreasing order of energy (i) n = 4, l = 0, m = 0, ms = ±1/2 (ii) n = 3, l = 2, m = 0, ms = ±1/2 (iii) n = 3, l = 1, m = 1,ms = –1/2 (iv) n = 3, l = 0, m = 0,ms = –1/2 Solution :

(a) µ =

A(A + 1)

h = 2π

A(A + 1) =

(i)

For 3p, A = 1 µA = 1(1 + 1) = 2 =

(ii)

For 3d , A = 2 µA =

(iii)

For 3s, A = 0, µl = 0

2(2 + 1)

h = 6 = 2π

(b) For the same value of n, higher the value of (n + A), higher is the energy. (c) For the same value of n + A, higher the value of n, higher will be the energy. For (i) ( n + A) = 4 for 4s orbital For (ii), (n+A) = 4, for 3d orbital For (iii) ( n + A) = 4 for For (iv), (n+A) = 3

3p orbital

3s orbital

Decreasing order of energy = (ii) > (i) > (iii) > (iv)

11. QUANTUM MECHANICAL TREATMENT OF HYDROGEN ATOM Hydrogen atom is the simplest system consisting of one electron and one proton. Schrodinger equation is written in terms of Cartesian coordinates (x, y, z). If electron moves at a distance from the stationary nucleus and atom is spherical in shape. If the position of electron is expressed in polar coordinates (r, θ, φ), the schrodinger equation can be solved easily. The Cartesian coordinates can be converted to polar coordinates by following relation. x = r sin θ cos φ y = r sin θ sin φ 34 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

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z = r cos θ and also, x2 + y2 + z2 = r2 When Schrodinger equation is solved in polar coordinates, for hydrogen atom. It gives the possible energy states and corresponding wave function [Ψ (r, θ, φ)], called atomic orbitals. The schrodinger equation cannot be exactly solved for a multi electron atom. Atomic Orbitals and their Pictorial Representation : An atomic orbital is an electron wave function, Ψ (r, θ, φ) obtained from the solution of the schrodinger equation. Wave function Ψ is a mathematical function of the three coordinates of electron (r, θ, φ). This wave functionΨ can be written into three separate parts each of which is a function of only one coordinate. Ψ = f(r, θ, φ) Ψ (r, θ, φ) = R(r), α(θ), ω(φ) R is function of r, α is function of θ and ω is function of φ. Where, R is radial function and α and ω are angular function. The radial function depends on quantum numbers n and A. The angular function depends on A and m. Significance of Wave Function Ψ : The orbital wave function Ψ has no physical significance. It is the square of the absolute value of wave function |Ψ|2, which has a physical significance. It measure the electron probability density at a point in an atom. If the variations of Ψ or |Ψ|2 with r, θ, φ are to be observed graphically. It would need a four dimensional graph in space but on the plane of paper we can draw only two dimensional graph to overcome this difficult. We draw separate diagrams for (i) Variation of radial function (ii) Variation of Angular function R is known as radial wave function R2 is known as radial probability density and 4π2r2R2 is known as radial probability. 11.1 FUNCTION VARIATION OF RADIAL FUNCTION (i) Plot of Radial Wave Function (R) vs r: In all cases R approaches zero as r approaches infinity. There is a node (where the probability of finding electron is zero) in 2s orbital radial function. At the node the value of radial function changes from positive to negative for 2s orbital. In general ns orbitals have (n – 1) nodes.

Node

The plots of the radial wave function R Vs. distance r


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(ii) Radial Probability Density (R2)vs r: The square of radial wave function R2 for an orbital gives the radial density. The radial density gives the probability density of finding the electron at a point along a particular line. Node

2

The plots of the radial probability density R Vs. distance r

(iii) Radial Probability Functions (4π2r2R2) vs r: Radial probability function gives the probability of finding the electron at a distance r from the nucleus regardless of direction.

2

2

The plots of the radial density function 4 π r R Vs. distance r


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SECTION - I SUBJECTIVE TYPE PROBLEMS Problem 1 : Solution :

Estimate the wavelength of second line in Balmer series. T147he transition responsible for second Balmer line is 4 → 2.  1 1 ∆E = 13.6 Z  n 2 − n 2  1 ( 2) 2

Problem 2 :

Solution :

  = 2.55 eV  

hc 6.63 × 10−19 × 3 ×108 = 2.55 × 1.6 × 10−19 ∆E

Now

λ =

⇒

λ = 4.862 × 10–7 m 4862 Å

A spectral line in the spectrum of H-atom has a wave number of 82200 cm–1 . What transition is responsible for this radiation? (Rydberg constant RH=1.096 × 107 m-1). λ = 1 / ν = 1/8220000 ⇒ it lies in UV region or in Lyman series Hence, n1 = 1 Using the relation for wave umber for H-atom:  1 1  − 2 2  n1 n 2 

v = 1/λ = RH 

 1 1  82200 = 1.096 × 105  2 − 2   n1 n 2  ⇒ ⇒ Problem 3 :

Solution :

1–

1 n 22

=

3 4

n2 = 2

Calculate the wavelength of light radiation that would be emitted, when an electron in the fourth Bohr’s orbit of He+ ion falls to the second Bohr’s orbit. To what transition does this light radiation correspond in the H-atom ? Let us calculate ∆E first.  1

Substituting Now

1 

∆E = 21.7 × 10–19 Z2  n 2 − n 2  2   1 n1 = 2 and n2 = 4, Z = 2 we get ; ∆E = 1.632 × 10–8 J λ =

hc = 1.218 × 10–7 m = 1218 Å ∆E

The value of λ = 1218 Å implies that, in H-atom this transition would lie in Lyman Series. Hence our aim is now to find the transition : n2 → 1. (1st orbit) 

1 

∆E = 21.7 × 10–18 = 21.7 × 10-19 1 − n 2  2   ⇒ n2 = 2 Hence the corresponding transition in H-atom is 2 → 1.

⇒

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Problem 4 :

Find the wavelength of radiation required to excite the electron in ground level of Li++ (Z = 3) to third energy level. Also find the ionisation energy of Li 2+ . R = 1.097 × 107 m-1  1 1  ν = RZ2  2 − 2  n n 2   1

Solution : Putting the values : We get

n 1 = 1, n2 = 3, Z = 3 ν = 8.77 × 107 m–1

⇒

λ =

1 = 113.9 Å ν

Ionisation energy is the energy required to remove the electron from ground state to infinity i.e. corresponding transition responsible is 1 → ∞ i.e.

Problem 5 :

Solution :

1

Calculate the uncertainty in position assuming uncertainty in momentum within 0.1 % for : (a) a tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s. (b) a electron moving in an atom with a velocity of 2 × 106 m/s. Using Uncertainty Principle, ∆x ∆p =

p = mv = 0.2 × 10 = 2.0 kgm/s ∆p = 0.1 % of p = 2 × 10–3

⇒

∆x =

⇒

h 6.63 × 10 −34 = 4π∆p 4 × 3.14 × 2 ×10−3

= 2.135 × 10–32 m. p = mv = 9.1 × 10–31 × 2 10–6 = 1.82 × 10–24 Kgm/s ∆p = 0.1 % of p = 1.82 × 10–27 Kgm/s ∆x =

Solution :

h 4π

(a)

(b) For an electron,

Problem 6 :

1 

∆E = 13.6 × 32  2 − 2  1 ∞  ∆E = 122.4 eV = 1.95 × 10–7 J = ionisation energy

h 6.63 × 10 −34 = 4π∆p 2 × 3.14 ×1.82 ×10 −21

∆x = 2.89 × 10–8 m

Hydrogen when subjected to photo-dissociation, yields one normal atom and one atom possessing 1.97 eV more energy than normal atom. The bond dissociation energy of hydrogen molecule into normal atoms is 103 kcals mol–1 . Compute the wave length of effective photon for photo dissociation of hydrogen molecule in the given case. H2 → H + H* where H is normal H-atom and H* is excited H-atom. So the energy required to dissociate H2 in this matter will be greater than the usual bond energy of H2 molecule. E(absorbed) = dissociation energy of H2 + extra energy of excited atom energy required to dissociate in normal manner = 103 × 103 cal (given)


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=

103 × 103 × ( 4.18 ) 6 × 10 23

= 7.17 × 10−19 J / atom

extra energy possessed by excited atom is 1.97 eV = 1.97 × 1.6 × 10–19 J = 3.152 × 10–19 –19 E (abosorbed)= 7.17 × 10 + 3.152 × 10–19 = 1.032 × 10–18 J Now calculate the wavelength of photon corresponding to this energy. λ = Problem 7 : Solution :

hc 6.63 × 10−34 × 34 × 108 = = 1.926 × 10–7 m = 1926 Å E 1.032 ×10 J

Find out the number of waves made by a Bohr electron in one complete revolution in its 3rd orbit. For an electron to be in a particular energy level of radius r, with de Bronglie wavelength λ: 2πr = nλ (where n = number of waves in one revolution) λ =

h 2πr mv ⇒n= mv h

for third orbit ; r = 0.53 × 10–10 (3)2 m ; substituting the value of r, v, m and h, we get, n =3

v = 2.165 × 106 (1/3)

Problem 8 :

With what velocity should an α particle travel towwards the nucleus of a copper atom so as to arrive at a distance 10–13 metre from the nucleus of the copper atom?

Solution :

To arrive at a distance ‘d’ from the nuceleus, kinetic energy of α - particle equals electrostatic potential energy. KE = PE K ( q α )(q cu ) 1 m α Vα2 = 2 d

qα qα mα K ⇒

= 2 × 1.6 × 10–19 C d = 10–23 m = 29 × 1.6 × 10–19 C = 4 × 1.67 × 10–27 kg = 1/4πε0 = 9 × 109 N-m/C2

Vα =

2 k q α q cu mα d

Substitute the given values, V α = 6.325 × 106 m/s. Problem 9 :

Find the energy released (in ergs), when 2.0 gm atom of Hydrogen undergo transition giving spectral line of lowest energy in visible region of its atomic spectrum.

Solution :

∆E = 2.178 × 10–18 (Z2)  2 − 2  J / atom  n1 n 2 

1

1

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For lowest energy transition : n2 = 3 1

1

⇒ ∆E = 2.178 × 10–18 (12)  2 − 2  J / atom 2 3  For 2.0 gm atom,  5   

∆E = (2 × 6.023 × 1023 ) 2.178 ×10–18  36  Joules. = 3.63 × 105 J = 3.63 × 1012 ergs Problem 10 : Solution :

Find the ratio of frequencies of violet light (λ = 4.10 × 10–5) to that of red light (λ = 6.56 × 10 –5 cm). Also determine the ratio of energies carried by them. C = νλ ; c : speed of light, ν : frequency & λ : wavelength ν1 ν2

⇒

ν1 ν2

λ2

= λ 1 =

1 : red & 2 : violet

6.56 × 10−5 = 1.6 4.10 × 10−5

Now the energy associated with electromagnetic radiation is given by E = hν. ⇒

E1 E2

ν

λ

1 2 = ν = λ = 1.6 2 1

Hence the ration of energies is same as that of frequencies.


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SECTION - II SINGLE CHOICE PROBLEMS Problem 1 :

A photon of wavelength 300 nm is absorbed by a gas and then reemitted as two photons. One photon is red with wavelength of 760 nm. The wave number of the second photon will be (a) 2.02 × 106 m–1 (b) 3.02 × 106 m–1 6 –1 (c) 1.02 × 10 m (d) 2.2 × 106 m–1 E =

Solution :

hc λ

...(i)

hc

E1 = λ 1

...(ii)

hc

E2 = λ 2

...(iii)  1 1  +   λ1 λ 2 

E = E1 + E2 = hc  1 λ

1 -9

(300 ×10 ) 1 λ2

1

1

= λ +λ 1 2 =

1 -9

(760 ×10 )  1

1 

+

1 λ2 9

=  ×10  300 760  = (0.00333 – 0.00131) × 109 = 2.02 × 106 m–1

(a) Problem 2 :

A certain laser transition emits 6.37 × 1015 quanta per second per square meter. The power output in joules per square meter per second would be (l = 632.8 nm). (a) 5 × 10–5 J m–2 sec–1 (b) 2 × 10–3 J m–2 sec–1 (c) 9.2 × 10–3 J m–2 sec–1 (d) none of the above

Solution :

Energy falling per square meter per sec. No. of quanta falling per square meter per second 6.37 ×1015 × 6.626 ×10-34 × 3×108 = = Energy of one qunta 632.8 ×10-19

= 2 × 10–3 J m–2 sec–1 (b) Problem 3 :

The uncertainity in the location of circulating electron is equal to its de Broglie wavelength. The minimum percent error in its measurement of velocity under this circumstance will be approximately (a) 4 (b) 8 (c) 18 (d) 22

Solution :

Let the electron be moving with momentum, p, its wavelength will be equal to h/p. 41

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∆x =

h p

From Heisenberg’s uncertainity principle, ∆x × ∆p = ∆p =

h 4π

h×p 4π × h

∆p 1 p = 4π

Minimum percent error in measuring its velocity would be 100 ×

∆p ∆v 100 100 = p ×100 = = = 7.96 » 8 v 4π 4 × 3.14

(b) Problem 4 :

A mono electronic species in energy level with energy X was provided with excess of energy so that it jumps to higher energy level with energy Y. If it can emit 6 wavelenths originated from all possible transition between these group levels, then which of the following relation is correct? (a)

X 3 = 1+ Y n

(b)

X n = Y 6

(c)

X = (n -1)2 Y

(d)

X 3 = 1+ Y n

(where n is the principal quantum number of energy level X). Solution :

If n is the principal quantum number of energy level corresponding to energy X, the principal quantum number of energy level Y is (n + 3) as it emit 6 wavelengths.

X (n + 3) 2 = Y n2 X 3 = 1+ Y n

(a) Problem 5 :

If we plot a graph of stopping potential of photoelectrons against the frequency of radiation that has emitted it, then its sope would be equal to (a) 6.625 × 10–34 (b) 3.656 × 1012 (c) 2.414 × 1014 (d) 5.796 × 1018

Solution :

From Fintein’s equation for photoelectric effect, we have hv = hv0 + 1/2 mv2max; where v0 is threshold frequency If Vstop be the stopping potential of photoelectrons then its maximum possible kinetic energy must be equal to the opposing work i.e. eVstop. hv = hv0+ eVstop


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e

v = v0 +   Vstop h So the given graph will be a straight line with slope equal to e h

=

1.6 ×10−19 6.626 ×10−34

= 2.414×1014 (c) Problem 6 :

The radius of hydrogen atom in its ground state is 5.3 × 10–11 m. After collision with an electron it is found to have a radius of 21.2 × 10–11m. The principal quantum number of the final state of the atom is (a) 2 (b) 3 (c) 4 (d) 5

Solution :

Radius of H– atoms in its ground state, r1 = 5.3 × 10–11 meters Radius of Bohr atom in its nth orbit, rn = 5.3 × 10–11

n2 meters z

For H-atom, z = 1 and rn = 2.2 × 10–11 meters 1/2

 21.2 ×10−11   n =  −11   5.3×10 

=2

(a) Problem 7 :

Suppose 10–17 J of energy is needed by theinterior of human eye to see an object. How many photons of green light (l = 550 nm) are needed to generate this minimum amount of energy? (a) 14 (b) 28 (c) 39 (d) 42

Solution :

Let the number of photons required be n. nhc = 10–17 λ

n=

10−17 × 550 ×10 −9 10−17 × λ = = 27.6 » 28 photons hc 6.626 × 10−34 × 3 × 108

(b) Problem 8 :

Which element has a hydrogen like spectrum whose lines have wavelength one fourth of atomic hydrogen? (a) He+ (b) Li2+ 3+ (c) Be (d) B4+


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Solution :

 1

1 λH

= R ×12 

1 λX

= RZ2 

2

 n1

−

1   n 22 

...(i)

 1 1   −  n2 n2   1 2

...(ii) λx

1

Dividing equation (i) by equation (ii), we get λ = 2 H Z 1 1 = 2 4 Z

Z = 2 (He+) (a) Problem 9 :

Solution :

Ionisation potential of hydrogen is 13.6 eV. Hydrogen atom in the ground state are excited by monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen according to Bohr’s theory (a) One (b) Two (c) Three (d) Four The electron in H atom is excited to III shell after absorbing 12.1 eV. The possible transition state = (3-1) = 3. (c)

Problems 10 : Which of the following statements is not correct? (a) Special stability of half-filled and fully-filled atomic configurations amongst s- and p-block elements is reflected in ionization potential trends along a period (b) Special stability of half-filled and fully-filled atomic configurations amongst s- and p-block elements is reflected in electron affinity trends along a period (c) Aufbau order is not obeyed in cases where energy difference between ns and (n – 1)d subshell is large. (d) Special stability of half-filled subshell is attributed to higher exchange energy of stabilization Solution : Aufbau order is obeyed in cases where energy difference between ns and (n–1)d subshell is large.


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SECTION - III MULTIPLE CHOICE PROBLEMS Problem 1:

Solution :

Problem 2 :

Solution :

Problem 3:

In hydrogen like sample two different types of photons A and B are produced by electronic transition. Photon B has it’s wavelength in the infrared region if photon A has more energy than B, then the photon A may belong to the region (a) ultraviolet (b) visible (c) infrared (d) none (a, b, c) Since B is in infrared region and A has more energy than B hence it will have lesser wave length i.e. ultra violet, visible or infrared region. A hydrogen like atom in ground state absorbs ‘n’ photons having the same energy and it emits exactly ‘n’ photons when electronic transition takes place. Then the energy of the absorbed photon may be (a) 91.8 eV (b) 40.8 eV (c) 48.4 eV (d) 54.4 eV (a, b) Since it absorbs ‘n’ photons and it also emits exactly n-photons therefore transition must have taken place from 1 to 2. ∴ Energy of photon = 10.2 Z2 where Z = 1, 2, 3, 4 Identify the correct statement(s) (a) |Ψ| is the probability of finding the electron in an orbital (b) p-orbital is directional in nature (c)

d x 2 − y2 has dumb bell shape along x-and y-axis

(d)

d z 2 has dumb bell shape along x-axis and y-axis.

Solution :

(b, c) It is obvious from the fact.

Problem 4 :

If uncertainty in momentum is twice the uncertainty in position of an electron then uncertainty



h 

in velocity is :  ! = 2π  

Solution :

(a)

1 ! 2m

(b)

! 4πm

(c)

1 ! 4m

(d)

1 ! m

(d)

∆P.∆x ≥ ∵

h 4π

2∆x = – ∆P (given)


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∴

∆P 2 h ≥ 2 4π

∴

m 2 ( ∆V) 2 ≥

h 2π

{∵ ∆P = m∆V}

∆P 2 h ≥ 2 4π

Problem 5 :

∴

∆V ≥

h 2m 2 π

∴

∆V ≥

1 h m 2π

or

∆V ≥

1 ! m

In a hydrogen like sample electron is in 2nd excited state, the binding energy of 4th state of this sample is 13.6 eV, then (a) A 25 eV photon can set free the electron from the second excited state of this sample (b) 3 different types of photon will be observed if electrons make transition up to ground state from the second excited state (c) If 23 eV photon is used then K.E. of the ejected electron is 1eV

Solution :

(d) 2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atoms (a, b) B.E. of

4th

state = 13.6

z2 n2

⇒ 13.6

z2 = 13.6 42

⇒ z=4

sample is Be3+ ∴ energy of elecron in 3rd state state = 1.5 × 42 = 24 eV therefore 25 eV photon will cause ionisation. Problem 6 :

Solution :

Problem 7 :

Solution :

The orbitals which have same number of nodes are (a) 2s, 2p (b) 3p, 3d (c) 2s, 3p (d) 3s, 4d (a, b) 2s has one spherical node and 2p has one planar node, 3p has one planar and one spherical node whereas 3d has two planar nodes but no spherical node. The probability of finding the electron in the px orbital is (a) Zero at the nucleus (b) Maximum on two opposite sides of the nucleus along X-axis (c) Zero on Z-axis (d) Same on all sides around the nucleus. (a, b, c) Fact.


NARAYANA



Problem 8 :

Solution : Problem 9 :

The possible correct sets of quantum numbers for the unpaired electron of oxygen atom in the orbitals of p-subshel are (a) n = 2, l = 1, m = 0 (b) n = 2, l = 1, m = –1 (c) n = 2, l = 1, m = + 1 (d) n = 2, l = 1, m = – 2 (a, b, c) Option (d) is not possible because if l = 1, m has three values i.e. –1, 0, + 1. The diagram illustrates a possible electronic configuration of which of the following species

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↓ ↓

Solution :

Problem 10 :

Solution :

(a) Cl+ ion (c) Ar+2 ion (a, b) Total number of electrons = 16 Cl+ ion and S atom have 16 electrons.

(b) S atom (d) P–2 ion

Which of the following statements are wrong? (a) Heisenberg’s uncertainity principle is applicable only to microscopic particles (b) de Broglie concept of dual nature is applicable to all material particles (c) The electronic configuration of palladium (Z = 46) is [Kr] 4d85s2 (d) dxy, dyz and dxz orbitals have identical shapes. (a, c) Heisenberg’s uncertainity principle is applicable to microscopic as well as macroscopic particles but has significance only for microscopic particles. Hence, (a) is wrong. Electronic configuration of Pd is [Kr] 4d105s0. Hence, (c) is wrong.


NARAYANA



MISCELLANEOUS PROBLEMS SECTION - IV COMPREHENSION TYPE PROBLEMS Write up – I In the Rutherford’s experiment, α-particles were bombarded towards the copper atoms so as to arrive at a distance of 10–13 metre from the nucleus of copper and then getting either deflected or traversing back. The α-particles did not move further closer. Problem 1 :

Solution :

The velocity of the α-particles must be (a) 8.32 × 108 cm/sec

(b) 6.32 × 108 cm/sec

(c) 6.32 × 108 m/sec

(d) 6.32 × 108 km/sec

(b) 2 At the distance of closest approach v =

2Ze 2e 1 × 4πε 0 × r0 m

Hence, substituting the value of m = 4, z = 29, e = 1.66 × 10–19, r0 = 10–13 mts. v = 6.32 × 108 cm/sec Hence, choice (b) is correct while (a), (c), (d) are incorrect as from above calculation. Problem 2 :

Solution :

Which of the following metals can be used instead of gold in α-scattering experiment (a) Pt

(b) Na

(c) K

(d) Cs

(a) It is because ‘Pt’ is highly malleable and ductile and least ractive. (b), (c) and (d) are soft metals and highly reactive, therefore, cannot be used.

Problem 3 :

From the Rutherford’s α-particle scattering, it can be concluded that (a)

N α sin 4

(c)

Nα

θ 2

1 sin θ / 2 4

(b) N α

1 sin 4 θ

(d) N = sin

θ 2

Where N = Number of α-particles scattered by an angle θ. Solution :

It is observed that N = no. of particles of scattering ∝

1 sin (θ / 2) 4

Hence, choice (c) is correct while choice (a), (b) and (d) are incorrect. 48 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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Write up – II The branch of science which deals with dual behaviour of mater is said to be quantum mechanics. The fundamental equation of quantum mechanics is Schrodinger wave equation. The important features of quantum mechanics are (i) The energy of electron is quantized as a result of wave like properties of electrons. (ii) The position and momentum of an electron cannot be determined simultaneously. The path of electron cannot be determined. We can only talk of probability of finding electron. (iii) An atomic orbitals is represented by wave function ‘Ψ’ for an electron. ‘R’ is radial wave function whose value varies with distance from the nucleus Ψ 2 determines total probability (angular and radial) whereas R2 is called raidal probability. All information about electron in an atom is stored in orbital wave function Ψ. ‘R’ is radial wave function. (iv) R can be +ve or –ve but R2 is always +ve. Similarly, Ψ can be +ve or – ve but Ψ2 is always +ve Ψ2 can be equal to zero. Problem 4 :

Solution : Problem 5 :

Solution :

Problem 6 :

The regions or space where Ψ2 = 0 are called (a) nodes (b) antinodes (c) orbitals (d) energy levels (a). Nodes are region where there is no probability of finding elctron, i.e., Ψ 2 = 0. Number of nodal planes in f-orbitals are (a) 0 (b) 1 (c) 2 (d) 3 (d) There are three nodal planes in f-orbitals No. of nodal planes = l = 3 Which of the following probability distribution curves represents 2s orbital for H-atom

(a)

(b) R2

2

R

r

(c)

r

(d) R2

2

R

r Solution :

r

(b) It is because 2s orbital has one node where R2 = 0. 49


NARAYANA



MATCHING TYPE PROBLEM 7.

Sol.

Column (I)

Column (II)

(a)

Binding energy of He+ atom in an excited state

(p)

infrared region

(b)

7 → 3 transition in hydrogen atom

(q)

3.4 eV

(c)

5 → 1 transition in hydrogen atom

(r)

13.6 eV

(d)

Series limit of Balmer series in hydrogen atom

(s)

10 spectral lines observed

(a)

B.E. of He+ atom =

13.6 × 22 n2

n = 1, 2, 3 ..... Hence it can be 13.6 eV, 3.4 eV both (b)

In 7 → 3 transition ∆n = 7 – 3 = 4 ∴ Maximum no. of spectral line observed =

4(4 + 1) = 10 2

It is line of Paschen series Hence infrared region (c)

5 → 1 : 10 lines

(d)

Series limit of Balmer series is the last line i.e. 3.4 eV energy.

(a) — (q, r)

(b) — (p, s)

(c) — (s)

(d) — (q)

ASSERTION-REASON TYPE PROBLEMS The question given below consist of an ASSERTION and the REASON. Use the following key for the appropriate answers (a) If both Assertion and Reason are correct and Reason is the correct explanation for Assertion (b) If both Assertion and Reason are correct and Reason is not the correct explanation for Assertion (c) If Assertion is correct but Reason is not correct. (d) If Assertion is incorrect but Reason is correct. 8. Assertion : In third energy level there is no f-subshell. Reason : For n = 3, the possible values of l are 0,1 and 2. Solution : Both assertion and reason are true and reason is a correct explanation for assertion Hence Ans. (a) 9. Assertion : Heisenberg’s uncertainty principle holds good for all objects but it is of significance only for microscopic particles. Reason : Heisenberg’s uncertainity principle is due to the limitation of the measuring instruments to measure accurately the position and velocity of microscopic particles. Solution : Heisenberg uncertainity principle is not due to limitation of the measuring instrument. Ans. (c) 50 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320



10.Assertion : Electronic configurations of Fe3+ (containing 23 electrons) is not same as that of Vanadium (Z = 23), i.e. isoelectronic species do not have the same electronic configuration. Reason :

Here, Fe3+ follows Aufbau principle where as Vanadium does not.

Solution :

Assertion is correct but reason is false. Electronic configuration of Fe3+ : [Ar]3d5 Electronic configuration of V : [Ar]3d34s2 Ans. (c)


NARAYANA



ASSIGNMENTS SECTION - I SUBJECTIVE QUESTIONS

LEVEL - I 1.

Magnetic moment of X3+ ion of 3d series is

35 BM. What is the atomic number of X3+ ?

2.

The electromagnetic radiation of wave length 3186 Å is just sufficient to ionise the outermost electron of a gaseous Cs atom. Calculate the ionization energy of Cs in kJ/mole.

3.

Calculate the energy emitted when electrons 1 g atom of hydrogen undergo transition giving spectral lines of lowest energy in the visible region of its atomic spectrum.

4.

Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.

5.

Calculate the de Broglie wavelength of the electron in the ground state of hydrogen atom, given that its kinetic energy is 13.6 eV. (1eV = 1.602 × 10–19 J).

6.

With what velocity must an electron travel so that its momentum is equal to that of a photon of wavelength 560 nm?

7.

A laser emits monochromatic radiation of wavelength 663 nm. If it emits 1015 quanta per second per square metre, calculate the power output of the laser in joule per square metre per second.

8.

If astronomical observations, signals observed from the distant starts are generally weak. If the photon detector receives a total of 3.15 × 10–8 J from the radiations of 600 nm, calculate the number of photons received by the detector.

9.

The energy of an α-particle is 6.8 × 10–18 J. What will be the wavelength associated with it?

10.

The ionization energy of hydrogen in excited state is + 0.85 eV. What will be the energy of the photon emited when it returns to the ground state?

LEVEL - II 1.

Electromagnetic radiation of wavelength 242 MM is just sufficient to ionise the sodium atom. Calculate the ionisation energy in kJ mol–1.

2.

Find the number of quanta of radiations of frequency 4.67 × 1013 s–1 that must be absorbed in order to melt 5g of ice. The energy required to melt 1 g of ice is 333 J.

3.

Calculate the wavelength and energy of radiation emitted for the electronic transition from infinite to stationary state of hydrogen atom.

4.

If the energy difference between two electronic states is 46.12 kcal mol–1, what will be the frequency of the light emitted when the electrons drop from higher to lower states?

5.

Calculate the ratio of the velocity of light and the velocity of electron in the first orbit of a hydrogen atom.

6.

The wavelength of a certain line in Balmer series is observed to be 4341 Å. To what value of ‘n’ does this correspond? 52 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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7.

The uncertainty in position and velocity of a particle are 10–10 m and 5.27 × 10–24 m s–1 respectively. Calculate the mass of the particle.

8.

Calculate the momentum of a particle which has a de Broglie wavelength of 2.5 × 10–10 m.

9.

Find the number of waves made by a Bohr electron in one complete revolution in the 3rd orbit.

10.

The ionisation energy of hydrogen atom is 13.6 eV. What will be the ionisation energy of He+.

LEVEL - III 1.

Helium can be excited to 1s1 2p1 configuration by supplying 33.12 × 10−19 J of energy. The lowest excited singlet state 1s1 2s1 lies 5000 cm-1 below 1s1 2p1 state. What should be the average He–X bond energy so that HeX2 could form non-endothermically ? Assume the compound would be formed from lowest excited singlet state of He+. Heat of atomization of X = 200 kJ mol–1 .

2.

A uranyl actinometer containing oxalic acid is irradiated with light of wavelength 4350 Å for 15 minutes. At the end of this time, it was found that oxalic acid equivalent to 12 ml of 0.001 M KMnO4 solution has been decomposed by light. At this wavelength, the quantum efficiency of actinometer is 0.58. Find the power of radiation source in 10–4 watt assuming that one photon decomposes only one molecule of oxalic acid.

3.

The reaction between H2 and Br2 to form HBr in the presence of light is initiated by the photodecompostion of Br2 into free bromine atoms (free radicals) by absorption of light. The bond dissociation energy of Br2 is 192 kJ/mole. What is the longest wavelength of the photon that would inititate the reaction?

4.

A hydrogen like atom (described by the Bohr’s model) 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 values). (a)

Find the atomic number of the atom.

(b)

Calculate the smallest wavelength emitted in these transitions. (use hc = 1240 eV-nm, E1H = – 13.6 eV)

5.

Electrons in hydrogen like atom (Z = 3) make transition from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and ejects photoelectrons. The stopping potential for the photo-electrons ejected by the shorter wavelength is 3.95 eV. Calculate the work function of the metal and stopping potential for the photo-electrons jected by the longer wavelength. (RH = 1.094 × 107 ms–1).

6.

A particle of charge equal to that of an electron and mass 208 times the mass of electron (i.e., mu-meson) moves in a circular orbit around a nucleus of charge + 3e. (Take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system. (RH = 1.097 × 107 m–1). (i)

Derive an expression for the radius of Bohr orbit.

(ii)

Find the value of n for which the radius of the orbital is approximately the same as that of the first Bohr orbit for H-atom.

(iii)

Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit to the first orbit.


NARAYANA



7.

Estimate the difference in energy between 1st and 2nd Bohr’s orbit for a H atom. At what minimum atomic number a transition from n = 2 to n = 1 energy level would result in the emission of X-rays with l = 3.0 × 10–8 m? Which hydrogen atom like species does this atomic number corresponds to?

8.

A series of lines in the spectrum of atomic H lies at wavelengths 656.46 , 480.27 , 434.17 , 410.29 nm. What is the wavelength of next line in this series?

9.

When a beam of 10.6 eV photons of intensity 2.0 W/m 2 falls on a platinum surface of area 1 × 10 -4 m2 and work function 5.6 eV , 0.53 % of the incident photons eject photo electrons. Find the number of photo electrons emitted per second and their mimimum energies in (eV). Take 1 eV = 1.6 × 10 -19 J .

10.

Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites . It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 Å . A similiar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d . Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described can be formed .


NARAYANA



SECTION - II SINGLE CHOICE QUESTIONS 1.

2.

3.

If the speed of electron in the Bohr’s first orbit of hydrogen atom be x, then speed of the electron in the 3rd orbit is (a) x/9

(b) x/3

(c) 3x

(d) 9x

Which one of the following set of quantum numbers is not possible for 4p electron? (a) n = 4, l = 1, m = – 1, ms = + 1/2

(b) n = 4, l = 1, m = 0, ms = + 1/2

(c) n = 4, l = 1, m = 2, ms = + 1/2

(d) n = 4, l = 1, m = – 1, ms = – 1/2

The orbital angular momentum of an electron in 2s orbitals is

1 h (a) + ⋅ 2 2π (c) 4.

5.

6.

7.

8.

9.

(b) zero

h 2π

(d)

2⋅

h 2π

The uncertainity in the momentum of an electron is 1.0 × 10–5 kg ms–1. The uncertainity in its position will be (h = 6.62 × 10–34 kg m2 s–1) (a) 1.05 × 10–28 m

(b) 1.05 × 10–26 m

(c) 5.27 × 10–30 m

(d) 5.25 × 10–28 m

Naturally occuring boron is 20% 5B10 and 80% 5B11. The atomic weight of boron is (a) 10.50

(b) 11.0

(c) 10.80

(d) 10.20

A particle A moving with a certain velocity has a de Broglie wavelength of 1 Å. If particle B has mass 25% of that A and velocity 75% of that of A, the de Broglie wavelength of B will be approximately (a) 1 Å

(b) 5.3 Å

(c) 3 Å

(d) 0.2 Å

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 in the He+ spectrum. (a) n = 4 to n = 1

(b) n = 3 to n = 2

(c) n = 3 to n = 1

(d) n = 2 to n = 1

Which of the following is violation of Pauli’s exclusion principle? (a)

↑↓

↑↓

(c)

↑↓

↑

↓

↑

(b)

↑↑

(d)

↑

↑ ↑

↑ ↑

↑ ↑

The probability of finding an electron residing in a px orbital is zero (a) in the yz plane

(b) in the xy plane

(c) in the y-direction

(d) in he z-direction 55


NARAYANA



10.

11.

12.

13.

14.

15.

Which of the following has more unpaired d-electrons? (b) Fe2+ (a) Zn+ 3+ (c) Ni (d) Cu+ 4000 Å proton is used to break the iodine molecule, then the % of energy converted to the KE of iodine atoms if bond dissociation energy of I2 molecule is 246.5 kJ/mole (a) 8% (b) 12% (c) 17% (d) 25% What will be the energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving the spectral line of lowest energy in the visible region of its atomic spectrum. (b) 11.91 × 103 J (a) 18.25 × 104 J (d) 63.12 × 104 J (c) 52.18 × 105 J 2 Probability of finding the electron Ψ of ‘s’ orbital doesn’t depend upon : (a) Distance from nucleus (r) (b) Energy of ‘s’ orbital (c) Principal quantum number (d) Azimuthal quantum number Consider the following statements. The d-orbitals have 1. four lobes and two nodes 2. four lobes and one node 3. same sign in the opposite lobes 4. opposite sign in the opposite lobes Which of the above statements are correct? (a) 1 alone (b) 1 and 3 (c) 2 and 3 (d) 1 and 4 The maximum number of electrons in a subshell having the same value of spin quantum number is given by (a) l + 2 (b) 2l + 1 (c) l (l + 1)

16.

(d)

l (l + 1)

Potential energy of Li2+ electron is −3e 2 (a) 4π∈0 r

3e2 (b) 4π∈0 r

−2e 2 −e 2 (d) 4π∈0 r 4π∈0 r 2 The energy of an electron in the second Bohr orbit of H-atom is –E. The energy of the electron in the Bohr’s first orbit is (a) – E/4 (b) – 4E (c) 4E (d) – 2 E The wave number of the shortest wavelength of absorption spectrum of H-atom is (Rydberg constant = 109700 cm–1) (a) 109700 cm–1 (b) 3/4 × 109700 cm–1 (c) 1/2 × 109700 cm–1 (d) 9 × 10 × 109700 cm–1 In the Bohr’s orbit, what is the ratio of total kinetic energy and total energy of the electron? (a) – 1 (b) – 2 (c) + 1 (d) + 2 In two H atoms A and B the electrons move around the nucleus in circular orbits of radius r and 4r respectively. The ratio of the times taken by them to complete one revolution is (a) 1 : 4 (b) 1 : 2 (c) 1 : 8 (d) 2 : 1

(c)

17.

18.

19.

20.


NARAYANA



SECTION - III MULTIPLE CHOICE QUESTIONS 1.

2.

Which of the following quantum numbers is/are not allowed (a) n = 3, l = 2, m = 0

(b) n = 2, l = 2, m = –1

(c) n = 3, l = 0, m = 1

(d) n = 5, l = 2, m = –1

For radial probability distribution curves, which of the following is/are correct. (a) The number of maxima in 2s orbital are two (b) The number of spherical or radical nodes is equal to n – l – 1 (c) The number of angular nodes are ‘l’ (d) 3dz2 has two angular nodes.

3

4.

5.

‘g’ orbital is possible if (a) n = 5, l = 4

(b) It will have 18 electrons

(c) It will have 9 types of orbitals

(d) It will have 22 electrons

Which of the following transitions are allowed in the normal electronic emission spectrum of an atom? (a) 2s → 1s

(b) 2p → 1s

(c) 3d → 2p

(d) 5p → 3s

Choose the correct relations on the basis of Bohr’s theory. (a) Velocity of electron ∝

1 n

(b) Frequency of revolution ∝

(d) Force on electron ∝

(c) Radius of orbits ∝ n 2 Z 6.

1 n4

The magnitude of the spin angular momentum of an electron is given by (a) S = s(s + 1)

(c) S = 7.

1 n3

h 2π

(b) S = s

h 2π

1 h (d) S = ± × 2 2π

3 h × 2 2π

The ground-state electronic configuration of the nitrogen atom can be represented as (a)

↑↓

↑↓

↑

↑

↑

(b)

↑↓

↑↓

↑

↓

↑

(c)

↑↓

↑↓

↑

↓

↓

(d)

↑↓

↑↓

↓

↓

↓ 57


NARAYANA



8.

Which of the following statements relating to the spectrum of hydrogen atom are false? (a) The lines can be defined by quantum number (b) The line of longest wavelength in the Balmer series corresponds to the transition between n = 3 and n = 2 levels (c) The spectral lines are closer together at long wavelengths (d) A continuum occurs at n = ∞

9.

Which of the following statements are correct? (a) Angular momentum of 3s electron is equal to that of 2s electron (b) Energy levels of H-atom are not equally spaced (c) Energy of electron is zero only when it is stationary (d) Humphrey series lies in the far infrared region

10.

Which of the following statements are not correct? (a) An orbital containing an electron having quantum number n = 2, l = 0, s = + 1/2 is spherical (b) All photons have same energy (c) the frequency of X-rays is less than radiowaves (d) As intensity of light increases, the frequency increases.


NARAYANA



MISCELLANEOUS QUESTIONS SECTION - IV COMPREHENSION TYPE QUESTIONS Write up - I The wave numbers of the spectral lines of Lyman Series may be calculated using the equation given below as derived by Niels Bohr for an atom or ion free from electron-electron repulsion.

1   ν = R H ⋅ Z2 1 − 2  where n = 2, 3, 4 5….¥  n  where 2π2 m e k 2 e 4 = Rydberg constant Ch 3 me is the mass of electron but for actual calculation we will have to consider the nuclear motion and then me will have to be replaced by reduced mass (m) where RH =

µ=

m N ⋅ me m N + m e , mN = mass of nucleus

All the other symbols have their usual significances. 1.

2.

3.

If proton in the nucleus of hydrogen atom is replaced by positron (+1e0) having the charge of proton but mass that of electron, the wave number of the lowest energy transition in the Lyman series of the above designated H-atom considering the nuclear motion will be equal to (a)

3 RH 4

(b)

3 RH 8

(c)

RH 2

(d)

3 RH 2

What will be the ratio of the wave number of the 1st line to that of the 5th line of Lyman series of He+ ion? (a)

25 32

(b)

27 35

(c)

100 32

(d)

108 35

The 3 rd IP of Be (Z = 4) per atom/ion (R H = 109677 cm –1 and h = 6.625 × 10 –34 J.s, C = 3 × 108 ms–1 and 1.6 × 10–19 J = 1 eV) is equal to (a) 217.6 eV (b) 163.2 eV (c) 123.0 eV (d) Can’t be calculated using Bohr’s expression 59 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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Write up - II

ˆ for every physically observable property like momentum In the quantum mechanics there is operator (A) (linear or angular), position, K.E., total energy etc., and whenever the function (ψ) is operated upon by the operator of any property , we get back ψ multiplied by a constant (a) which is the value of the property under consideration. The function ψ obeying the condition is called eigenfunction of the operator ˆ and the constant “a” is called eigen-value. Mathematically A ˆ ψ = a ⋅ψ A This equation is called eigen-value equation. The Schrodinger wave equation is the energy eigen value equation. The operator for total energy is called Hamiltonion operator (H) defined as H=−

∂2 ∂2  h 2  ∂2 + +  +V 8π2 m  ∂x 2 ∂y 2 ∂z 2 

Where V = PE, m = mass of microparticle. Solving Schrodinger wave equation for H-atom after putting proper value of V, we get wave functions (ψ) and corresponding energy values. Each wave function denotes an orbital. ψ2 measures the probability of finding the electron and ψ2dx gives the probability of finding the electron in the region x and x + dx. 4πr2ψ2(r)dr gives the probability of finding the electron in spherical shell of thickness dr at a distance r from the nucleus. A plot of radial distribution function [4πr2ψ2(r)] vs. distance from the nucleus (r) gives the radial probability distribution curve of electron. For 1s and 2s orbitals each containing only one electron, these curves are as given below

4π r 2ψ 2 (r )

1s

4π r 2ψ 2 (r ) 2s r

r

The wave functions of 1s and 2s orbitals of hydrogenic atom or ion are given below: 1/ 2

ψ1s =

Z 2   a0 

e − Zr / a 0

1/ 2

 Z   Zr  − Zr / 2a 0 ψ2s =    2 − e a0   2a 0   In 2s-orbital there occurs a point at the radius at which curve touches X-axis. Here the probability of finding the electron is zero. This is called radial node. On the basis of the above write-up, answer the following questions : 4.

An electron moves along x-axis and restricted to move only between x = 0 and x = L. Hence L

L

2 (a) ∫ ψ dx = 1

(b)

∫ ψ dx > 1

(d)

0 L

(c)

2

0

∫ψ

2

dx = 0

0 L

∫ ψ dx < 1 2

0


NARAYANA



5.

The value of r at which radial node of 2s-electron of H-atom will appear is: (a) r = a0 (b) r = 2a0 (c) r =

6.

a0 2

(d) r =

a0 3

The value of r at which the maxima of the probability distribution curve of 1s orbital electron of H-atom will occur is (a) r = a0 (b) r = 2a0 (c) r =

a0 3

(d) r =

a0 2

MULTIPLE MATCHING TYPE QUESTIONS Match the following 7.

Column – I

Column – II

(a) Angular momentum

(p) increases by increasing shell numbers

(b) Kinetic energy

(q) decreases by decreasing atomic numbers

(c) Potential energy

(r) increases by decreasing atomic numbers

(d) Velocity

(s) decreases by increasing shell numbers

8.

Column – I (a) Visible, infrared, ultraviolet –

Column – II (p) increases

wavelength of light (b) Lyman, Balmer, Paschen –

(q) decreases

number of lines in respective spectrums (c) As we move outwards from an atom

(r) first increases then decreases

distance between two consecutive shells (d) As we more outwards from an atom

(s) Remains same

energy differnce between two consecutive shells

ASSERTION-REASON TYPE QUESTIONS The question given below consist of an ASSERTION and the REASON. Use the following key for the appropriate answers (a)

If both Assertion and Reason are correct and Reason is the correct explanation for Assertion

(b)

If both Assertion and Reason are correct and Reason is not the correct explanation for Assertion

(c)

If Assertion is correct but Reason is not correct.

(d)

If Assertion is incorrect but Reason is correct. 61


NARAYANA



Assertion : In Rutherford’s gold foil experiment, very few α-particles are deflected back.

9.

10.

11.

12.

13. 14.

15.

Reason :

Nucleus present inside the atom is heavy.

Assertion :

Limiting line in the Balmer series has a wavelength of 364.4 mm.

Reason :

Limiting line is obtained for a jump of electron from n = ∞.

Assertion :

There are two spherical nodes in 3s-orbital.

Reason :

There is no planar node in 3s-orbital

Assertion :

An orbital cannot have more than two electrons.

Reason :

The two electrons in an orbital create opposite magnetic field.

Assertion :

The configuration of B-atom cannot be 1s22s2.

Reason :

Hund’s rule demands that the configuration should display maximum multiplicity.

Assertion :

The plots of radial probability density and radial probability function versus distance r from the nucleus for any particular orbital are identical

Reason :

Radial probability density (R2) is along the radial distance where as radial probability function (4πr2R2) is the probability in a shell of thickness dr.

Assertion :

The 19th elecron in potassium atom enters into 4s-orbital and not the 3d-orbital.

Reason :

(n + l) rule is followed for determining the orbital of the lowest energy state.


NARAYANA



SECTION - V (PROBLEMS ASKED IN IIT-JEE) A.

Only one option is correct (Objective Questions)

1.

The energy of an electron in the first Bohr orbital of H atom is – 13.6 eV. The possible energy values(s) of the excited state(s) for electrons in Bohr orbitals of hydrogen is (are) :

2.

(a) –3.4 eV

(b) – 4.2 eV

(c) – 6.8 eV

(d) + 6.8 eV

(1998)

The electrons, identified by quantum numbers n and l, (i) n = 4, l = 1, (ii n = 4, l = 0, (iii) n = 3, l = 2 and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as : (1999)

3.

4.

5.

6.

(a) (iv) < (ii) < (iii) < (i)

(b) (ii) < (iv) < (i) < (iii)

(c) (i) < (iii) < (ii) < (iv)

(d) (iii) < (i) < (iv) < (ii)

The electronic configuration of an element is 1s2, 2s2, 2p6, 3s2, 3p6, 3d5 4s1. This represents its : (a) excited state

(b) ground state

(c) cationic form

(d) anionic form

(2000)

The number of nodal planes in 2 px orbital is :

(2001)

(a) one

(b) two

(c) three

(d) zero

The wavelength associated with a golf ball weighing 200g and moving at a speed of 5 m/h is of the order: (2001) (a) 10–10 m

(b) 10–20 m

(c) 10–30 m

(d) 10–40 m

1 1 The quantum numbers + and − for the electron spin represents : 2 2

(2001)

(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 7.

8.

If the nitrogen atom had electronic configuration 1s7, it would have energy lower than that of the normal ground state of configuration 1s2 2s2 2p3, because the electrons would be closer to the nucleus, yet 1s7 is not observed because it violates : (2002) (a) Heisenberg uncertainty principle

(b) Hund’s rule

(c) Pauli’s exclusion principle

(d) Bohr’s postulate of stationary orbitals

Rutherford’s experiment, which established the nuclear model of the atom, used a beam of : (a) β–particles, which impinged on a metal foil and got absorbed

(2002)

(b) γ–ray, which impinged on a metal foil and ejected electrons (c) helium atoms, which impinged on a metal foil and got scattered (d) helium nuclei, which impinged on a metal foil and got scattered 63 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

NARAYANA



9.

Which hydrogen like species will have same radius as that of Bohr orbit of hydrogen atom? (a) n = 2, Li2+

(b) n = 2, Be3+

(c) n = 2, He+

(d) n = 3, Li2+

10.

The number of radial nodes of 3s and 2p orbitals are respectively : (a) 2, 0 (b) 0, 2 (c) 1, 2 (d) 2, 11

B.

More than one options are correct (Objective Questions)

1.

An isotone of

2.

(b)

77 33 As

(c)

77 34 Se

(d)

78 34 Se

Many elements have non–integral atomic masses because :

When alpha particles are sent through a thin metal foil, most of them go straight through the foil because : (1984) alpha particles are much heavier than electrons alpha particles are positively charged most part of the atom is empty space alpha particle move with high velocity

The sum of the number of neutrons and proton in the isotope of hydrogen is :

The atomic nucleus contains :

(1988) (b) neutrons (d) photons

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) : (1998) (a) – 3.4 eV (c) – 6.8 eV

7.

(1986)

(b) 5 (d) 3

(a) protons (c) electrons 6.

(1984)

they have isotopes their isotopes have non–integral masses their isotopes have different masses the constituents, neutrons, protons and electrons, combine to given fractional masses

(a) 6 (c) 4 5.

(2005)

(1984)

77 32 Ge

(a) (b) (c) (d) 4.

is :

(a)

(a) (b) (c) (d) 3.

76 32 Ge

(2004)

(b) – 4.2 eV (d) + 6.8 eV

Which of the following statement (s) is (are) correct ? [Ar]3d5

(1998)

4s1,

(a) The electronic configuration of Cr is (Atomic number of Cr = 24). (b) The magnetic quantum number may have a negative value (c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of Ag = 47) (d) The oxidation state of nitrogen in NH3 is – 3.


NARAYANA



8.

Decreases in atomic number is observed during : (a) alpha emission (c) positron emission

9.

(1998) (b) beta emission (d) electron capture

Ground state electronic configuration of nitrogen atom can be represented by : (a)

↑↓ ↑↓ ↑ ↑ ↑

(b)

↑↓ ↑↓ ↑ ↓ ↑

(c)

↑↓ ↑↓ ↑ ↓ ↓

(d)

↑↓ ↑↓ ↑ ↑ ↑

(1999)

SUBJECTIVE QUESTIONS 1.

Estimate the difference in energy between 1st & 2nd Bohr’s orbit for a H atom . At what minimum atomic number a transition from n = 2 to n = 1 energy level would result in the emission of x - rays with l = 3.0 ´ 10 -8 m ? Which hydrogen atom like species does this atomic number corresponds to? [JEE 1993]

2.

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum ? [JEE 1993] Find out the number of waves made by a Bohr electron in one complete revolution in its3rd

3.

orbit.

[JEE 1994]

4.

Iodine molecule dissociates into atoms after absorbing light of 4500 Å . If one quantum of radiation is absorbed by each molecule , calculate the kinetic energy of iodine atoms . (Bond energy of [JEE 1995] I2 = 240 k J mol -1).

5.

A bulb emits light of l = 4500 Å . The bulb is rated as 150 watt and 8 % of the energy is emitted as light. How many photons are emitted by the bulb per second. [JEE 1995]

6.

Consider the hydrogen atom to be a proton embedded in a cavity of radius a0 (Bohr’s radius) whose charge is neutralized by the addition of an electron to the cavity in vacuum infinitely slowly . (a) Estimate the average of total energy of an electron in its ground state in a hydrogen atom as the work done in the above neutralization process . Also if the magnitude of the average kinetic energy is half the magnitude of the average potential energy , find the average potential energy. (b) Also derive the wavelength of the electron when it is a0 from the proton . How the wavelength of an electron in the ground . State Bohr’s orbit . [JEE 1996]

7.

An electron beam can undergo diffraction by crystals . Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 Å. [JEE 1997]

8.

A compound of vanadium has a magnetic moment of 1.73 BM . Work out the electronic configuration of the vanadium ion in the compound . [JEE 1997]

9.

Calculate the energy requird to excite one litre of hydrogen gas at 1 atm and 398 k to the first excited state of atomic hydrogen . The energy for the dissociation of H - H bond is 436 k J mol -1 Also calculate the minimum frequency of photon to break this bond. [JEE 2000]


NARAYANA



ANSWERS EXERCISE - 1 –12

1.

3.312 × 10

erg

3.

7.22 × 104 mol

2.

2.01 × 1018

EXERCISE - 2 1.

r1 = 0.1763 Å; r2 = 0.7053 Å

2.

(IE) He+ = 54.4 eV (IE) ; (IE) Li2+ = 122.4 eV ; (IE) Be3+ = 217.6 eV

3.

(IE) H in 3rd orbit = 1.51 eV

EXERCISE - 3 1.

32.77 eV

3.

2.5 × 10–8 m

2.

6.63 × 10–33 m

2.

∆V =

EXERCISE - 4 1.

1.893 × 10-34 m

3.

∆V = 2.64 × 10–39 ms–1

1 h 2m π

EXERCISE - 5 1.

(a) ml = 0

(b) l = 1

(c) n = 2

(d) n = 3, ml = – 2 to + 2 (any value)

2.

3p < 4s < 3d < 7s < 5f < 6g

3.

(b), (e) and (f) are not allowed.

4.

(a) 2p

(b) 4d(c) 5s

EXERCISE - 6 1.

2.

3.

2

2

1

(a)

1s 2s 2p : maximum capacity of 2s orbital is 2.

(b)

1s2 2s2 2p6 3s1 : 2d orbital does not exist.

(c)

[Ar] 4s1 : 3d is filled only after 3s has been completely filled.

(d)

[Ar] 4s2 : 3d2 : 3d is filled before 4p.

(e)

[Xe] 4f14 5d10 6s2 : maximum capacity of 4f suborbit is 14.

(a)

Fe2+ = [Ar] 3d6

(b)

Cu+ = [Ar] 3d10 no unpaired electron in 3d

(c)

Sc3+ = [Ar] no unpaired electron.

(d)

Mn2+ = [Ar] 3d5

Z = 26


NARAYANA



SECTION - I (Subjective Questions) LEVEL - I 1.

26

2.

375.25 kJ / mol

3.

182.5 kJ

4.

27419.25 cm–1

5.

0.3328 nm

6.

1300 ms–1

7.

3 × 10–4 Jm–2 s–1

8.

9.51

10.

12.75 eV

9.

2.31 ×

10–16

m

LEVEL - II 1.

493.6 kJ mol–1

2.

5.38 × 1022

3.

λ = 9.11 × 10–8 m ; E = 2.17 × 10–18 J

4.

4.84 × 1014 cycle sec–1

5.

c/v = 137

6.

n=5

7.

0.099 kg

8.

2.64 × 10–24 kg m sec–1

9.

3

10.

54.5 eV

1.

1167.5 kJ mol–1

2.

158

3.

6235 Å

4.

Li2+, 4060 nm

5.

w = 3.193 × 10–19 J, Vs = 0.756 V

6.

(i)

(ii)

n = 25

LEVEL - III

r = n2 × 10–14 m

(iii) 5.48 × 10–11 m 7.

10.2 eV, 2, He+

8.

397.2 nm

9.

6.25 × 1011

10.

d = 0.5 Å, E = 150.8eV.

SECTION - II (Single Choice Questions) 1. 3. 5. 7. 9. 11. 13. 15. 17. 19.

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

2. 4. 6. 8. 10. 12. 14. 16. 18. 20.

(c) (c) (b) (b) (b) (a) (b) (a) (a) (c)


NARAYANA



SECTION - III (Multiple Choice Questions) 1. 3. 5. 7. 9.

(b, c) (a, b, c) (b, c, d) (a, c) (b, c, d)

2. 4. 6. 8. 10.

(a, b, c) (a, b, d) (a, b, d) (a, d) (b, c, d)

SECTION - IV (Comprehension Type Questions) 1. 3. 5.

(b) (a) (b)

2. 4. 6.

(b) (a) (a)

(Multiple Matching type questions) 7. 8.

(a) (c) (a) (c)

(p) (p, q) (r) (p)

(b) (d) (b) (d)

(q, s) (q, s) (p) (q)

(Assertion Reason Type Questions) 9. 11. 13. 15.

(b) (b) (c) (a)

10. 12. 14.

(a) (b) (d)

SECTION - V (Problems asked in IIT-JEE) OBJECTIVE QUESTIONS A.

Only one 1. 3. 5. 7. 9.

B.

option is correct (a) (b) (c) (c) (b)

2. 4. 6. 8. 10.

(a) (a) (d) (d) (a)

2. 4. 6. 8.

(a, c) (d) (a) (a)

2. 4. 7. 9.

n1 = 1 , n2 = 2 0.216 ´ 10 -19 J 63.3 volt 98.19 k J , 10.93 ´ 1014 Hz

More than one option are correct 1. 3. 5. 7. 9.

(b, d) (a, c) (a, b) (a, b, c) (a, d)

SUBJECTIVE QUESTIONS 1. 3. 5. 8.

10.2 eV , z = 2 , He+ 3 27.2 × 1018 1 s2 , 2 s2 , 2 p6 , 3 s2 3 p6 3 d1


IIT Class XI Chem Structure of Atoms

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