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Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

New Delhi – 110 018, Ph. : 9312629035, 8527112111

CA – 1

ATOMIC STRUCTURE

Sy l labus :

Discovery of sub-atomic particles (electron, protonand neutron); Thomson and Rutherford atomicmodels and their limitations; Nature of electromag-netic radiation, photoelectric effect; Spectrum ofhydrogen atom, Bohr of hydrogen atom - its postu-lates, derivation of the relations for energy of theelectron and radii of the different orbits, limitationsof Bohr’s model; Dual nature of matter, de-Broglie’srelationship, Heisenberg uncertainty principle.Elementary ideas of quantum mechanics, quantummechanical model of atom, its important features, and 2, concept of atomic orbitals as oneelectron was functions; Variation of and 2 withr for 1s and 2s orbitals; various quantum numbers(pricipal, angular momentum and magneticquantum numbers) and their significance; shapesof s, p and d-orbitals, electron spin and spinquantum number; Rules for filling electrons inorbitals - aufbau principle, Pauli’s exclusionprinciple and Hund’s rule, electronic configurationof elements, extra stability of half-filled andcompletely filled orbitals.


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CA – 2

CONCEPTS

C1 Atomic Hypothesis (Dalton)

It assumed that –

(i) an atom cannot be subdivided.

(ii) atoms are neither created not destroyed during chemical reactions.

(iii) atoms of the same element are alike; in particular all atoms of an element have the same mass.

(iv) atoms of different elements are not alike; in particular, their masses are different.

C2 Properties of Electron, Proton and Neutron

Symbol Proton (p) Neutron (n) Electron (e)

Mass 1.67252 × 10–27 kg 1.67482 × 10–27 kg 9.1091 × 10–31 kg

Charge 1.60210 × 10–19 C 0 1.60210 × 10–19 C

Mass relative to 1836 1839 1the electron

Charge relative +1 0 –1to the proton

Discovery Goldstein Chadwick Thomson

C3 Some Sub-nuclear Particles

Particle mass Charge

Antiproton Same as that of proton Negative

µ-meson (muon) 210 times that of an electron Positive and negative

-meson 276 times that of an electron Positive and negative

(pion) 265 times that of an electron Zero

Neutrino Very much less than that of an Zeroelectron

Positron Same as that of an electron Positive

C4 Rutherford’s Nuclear Model of Atom

(i) Rutherford’s scattering experiments disproved the Thomson model and led to the nuclear modelof the atom in which positive charge is spread over a sphere of radius 10–15m, the so-callednucleus, and that the outer most electron clouds are about 10–10 m from the centre of the nucleus.

(ii) The positive charge of a nucleus is due to the positively charged particles called protons. Butmass of the nucleus is due to the protons and neutrons.

(iii) The total number of protons and neutrons in a nucleus determine the nuclear mass.

C5 Nature of Light and Electromagnetic Waves

Wave theory consides light to be a form of wave motion of wavelength , related to frequency v and

velocity of light c be equation

c

v .

Light waves are also considered electromagnetic in nature (i.e, they are oscillations of electric andmagnetic fields in space). Various types of electromagnetic radiations having various wavelengths(or frequencies) are known and they constitute the so-called electromagnetic spectrum.

By quantum theory put forward by planck E = nh =

nhc.

where n is number of photons.

Practice Problems :

1. Number of photons of light of wavelength 4000 Å required to provide 1.00 J of energy is

(a) 2.01 × 1018 (b) 12.01 × 1031 (c) 1.35 × 1017 (d) none is correct


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CA – 32. One quantum is absorbed per molecule of gaseous iodine for converting into iodine atoms. If light

absorbed has wavelength of 5000 Å, the energy required in kJmol–1 is

(a) 2.38 × 10–5 (b) 2.38 × 105 (c) 1.38 × 102 (d) none

3. A near ultraviolet photon of 300 nm is absorbed by a gas and then re-emitted as two photons. Onephoton is red with wavelength 760 nm. The wavelength of second photon is

(a) 300 nm (b) 460 nm (c) 760 nm (d) 495.65 nm

[Answers : (1) a (2) b (3) d]

C6A Bohr Theory of H-atom

This theory is used to calculate the radius (r) and energy (E) of a permissible orbit for one-electron specieslike H, He+, Li2+ etc.

C6B Postulates of Borh Theory

(i) The electrons continue revolving in their respective orbits without losing energy. Thus each orbitis associated with a definite energy hence it is also called fixed energy levels.

(ii) Angular momentum (mvr) of an electron in a given orbit is quantised.

2

nhmvrn .....

[m is the mass of electron, v is the velocity of electron, r is radius of orbit in which it is revolving,n is the number of orbit]

(iii) Energy is emitted or absorbed by an atom only when an electron moves from one level to another. Thus

hc)EE(E

12 nn where2nE = energy of the n

2 level,

1nE = energy of the n1 level

Thus wave number v

ishc

E1v

C6C Results of Bohr Theory

1. Radius of nth orbit

Z

n53.0r

2

n Å where Z = atomic number

2. Velocity of the electron in the nth orbit

137

c

n

Zvn where c = 3 × 108 m/s

3. Energy of the electron in the nth orbit

2

2

nn

Z6.13E ev, )J(

n

Z)1018.2(E

2

218

n

ET = K + U, K = -E

T =

2

U , U = 2E = –2K

4. Wavelength of photon emitted for a transition from n2 to n

1

22

21

2H

n

1

n

1ZR

1

where R = 1.096 × 107 m–1 (Rydberg’s constant)


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CA – 45. Ionisation Energy

Energy required to remove the electron from the outermost orbit of the atom in gaseous phase iscalled ionisation energy (I.E.).

E = 0, thus the difference between the ground state of an atom and the excited state that

correspond to n2 = is called ionisation energy.

atom/eVn

Z6.13)IE(

2

2

z

6. Hydrogen Spectrum

Series n1

n2

Region of Spectrum

Lyman 1 2, 3, ... Ultraviolet

Balmer 2 3, 4, ... Visible

Paschen 3 4, 5, ... Infrared

Brackett 4 5, 6, ... Infrared

Pfund 5 6, 7, ... Infrared

Practice Problems :

1. The ratio of the energy of the electron in ground state of hydrogen to that of the electron in firstexcited state of Be3+ is :

(a) 1 : 4 (b) 1 : 8 (c) 1 : 16 (d) 16 : 1

2. If the shortest wavelength of H atom in Lyman series is x. then longest wavelength in Balmer seriesof He+ is :

(a)5

x9(b)

5

x36(c)

4

x(d)

4

x5

3. The wavelength of the first line in Lyman series of hydrogen and that of the first line in Balmer seriesof lithium are in the ratio

(a) 9 : 4 (b) 3 : 5 (c) 5 : 3 (d) 3 : 2

[Answers : (1) a (2) a (3) a]

C7A Particle and Wave Nature (Dual nature of Electron)

de Broglie based on Millikan’s oil drop experiment (which showed particle nature) and diffraction study(which showed wave nature) suggested the dual nature of electron, both as a material particle and as awave. According to de Broglie’s equation.

)KE(m2

h

mv

h

p

h , also

)eV(m2

h

e

where p (= mv = )KE(m2 ) is called momentum of the particle of mass m moving with velocity v..

The circumference of nth orbit is equal to n times of wavelength of electrons.

2rn = n

Practice Problems :

1. How fast is an electron moving if it have a wavelength equal to the distance it travels in onesecond ?

(a)m

h(b)

h

m(c)

p

h(d)

)KE(2

h


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CA – 52. A hydrogen molecule at 2000C is moving with a speed of 2.4 × 105 cm per sec. The de-Broglie’s wave

length is of the order of

(a) 10,000 Å (b) 1. Å (c) 5.000 Å (d) 5Å

3. The accelerating potential is needed to produce an electron beam with an effective wavelength of0.090 Å is

(a) (7.33 × 10–23)/meq

e(b) (7.33 × 10–23)2/m

eq

e

(c) (7.33 × 10)2/meq

e(d) (9.33 × 10–23)2/m

eq

e

[Answers : (1) a (2) b (3) b]

C7B Photoelectric Effect : When a beam of light of suitable wavelength or frequency is allowed to fall on thesurface of metal, the electrons are emitted from the surface of metal. This phenomena of emittion ofelectrons is known as photoelectric effect. It was suggested that some portion of the photon energy is usedup in removing the electrons from the surface by overcoming the attractive force of nucleus (known asthreshold energy or work function of the metal) and rest portion is utilised in imparting velocity of these

electrons (into K.E. of the photoelectrons). 20 mv

2

1Wh ...[m, v is the mass and velocity of electron]

W0 = h

0. [where

0 is the threshold frequency]

Practice Problems :

1. When a certain metal was irradiated with a light of frequency 3.2 × 1016 Hz, the photoelectronsemitted had twice the kinetic energy as did photoelectrons emitted when the same metal wasirradiated with light frequency 2.0 × 1016 Hz. Hence threshold frequency is

(a) 1.6 × 1016 Hz (b) 0.8 × 1015 Hz (c) 8 × 1015 Hz (d) 8 × 1016 Hz

2. The threshold frequency for photo electric emission of electrons from platinum is 1.3 × 1015 sec–1.What is the minimum energy that photons of a particulars radiation must possess to produce thephoto electric effect with platinum metal ?

(a) energy must be equal to 6.63 × 10–34 × 1.3 × 1015 J

(b) energy must be greater than 6.63 × 1034 × 1.3 × 1015 J

(c) energy corresponding to 400 nm

(d) none of these

[Answers : (1) c (2) b]

C8 Heisenberg’s Uncertainity Principle

It is not possible to determine precisely both the position and the momentum (or velocity) of a small movingparticle (e.g. electron, proton etc.)

m4

hv.x

4

hp.x

where x, p and v are the uncertainties with regard to position, momentum and velocity respectively.

In terms of uncertainty in energy, E and uncertanty in time t, this principle is written as,

4

ht.E .

Practice Problems :

1. The uncertainly in the momentum of an electron is 1 × 10–6 kg m sec–1. The uncertainly in positionwould be

(a) 1.05 × 10–23 m (b) 2.1 × 10–25 m (c) 1.05 × 10–27 m (d) 1.05 × 10–25 m


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CA – 62. If a 1.0 gm body is travelling along x-axis at 100 m sec–1 within 1 m sec–1. The theoretical uncertainity

in its position is

(a) 103/4h (b) 103h/4 (c) h/4 (d) none

[Answers : (1) d (2) b]

C9A Schrodinger wave mechanical model :

As it was given by the Heisenberg that it is impossible to determine the position and velocity ofe– simultaneously. To overcome this a new model of atom was introduced which was based on dual behaviourof matter. It was introduced by Erwin Schrodinger.

He introduced a new concept for determining the position of e– i.e., orbital.

C9B Orbital : It is the region in space where there is high probability of finding the electron.

Difference between orbit and orbital

Orbit Orbital

1. It is circular or eliptical path traced by an electron 1. It is the region in space where there iswhile revolving round the nucleus of atom giving high probability of finding the electron.fixed value of the distance of e– from the nucleus.

2. It violates the Heisenberg principle 2. It does not violate the Heisenbergprinciple

3. It is not in accordance to the dual character 3. It is in accordance of dual character ofof matter matter.

C9C Quantum numbers : Each orbitals is designated by three quantum number n, l and m.

The principle quantum number (n) :

1. It determines the size and to a large extent the energy of the orbital.

2. The larger the value of n, the larger the energy of the orbital.

3. Principle quantum number also identifies the shell

i.e.,NMLK

4,3,2,1n

shell.

4. There are n2 orbitals in a shell.

5. All the orbitals of a given volume of n constitute a single shell of atom.

6. Each shell consists of one or more subshells or sublevels.

7. The number of subshells in a principal shell is equal to the value of n.

Azinuthal or Subsidiary quantum number : (l)

Each subshell in a shell is designated by l. l can have n values ranging from o to (n – 1) e.g. when n = 1,l = 0, n = 2, l = 0, 1 etc. i.e. for n = 1, l = 0 it means there is one subshell for n = 2, l = 0, 1. It means thereare two subshell.

Subshells corresponding to different values of l are represented by following symbols

l = 0, 1, 2, 3, 4, 5

notations s, p, d, f, g, h

Each subshell consists of one or more orbitals.

The number of orbitals in a subshell is given by (2l + 1).

e.g. In any l = 0 subshell, there are 2(0) + 1 = 1 orbitals.

For l = 1 subshell there are 2(1) + 1 = 3 orbitals.

In any l = 2 subshell, there are 2(2) + 1 = 5 orbitals.

In other words,

subshell notation = s, p, d, f, g

value of l = 0, 1, 2, 3, 4

Number of orbitals = 1, 3, 5, 7, 9


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CA – 7

The orbital angular momentum =

2

h)l(ll

The quantum number l also gives the shape of the orbital in the subshell.

Magnetic quantum number (m) :

It gives information about the orientation of the orbital.

For any subshell (defined by l values), (2l + 1) values of m are possible and these values are given as

m = –l, –(l – 1), ...... O, ....... + (l – 1), +l

e.g. for l = 2(d), m can have total = 2(2) + 1 = 5 values.

These values are –2, –1, 0, +1, +2 [Five orbitals]

For l = 1(p) m = 2(1) + 1 = 3 values.

These values are –1, 0, +1 [Three orbitals]

i.e., there are three preffered orientation of p in space.

Thus each orbital is defined by set of values of n, l and m

e.g. If 4s is given then n = 4, l = 0

If 5p is given then n = 5, l = 1

Spin quantum number : Electrons spin about the axes.

Some spinning in one direction and some in other direction.

Two orientation of electron which is possible is one is in clockwise direction and other one is in anticlockwisedirection.

They are represented by two arrows (spin up) and (spin down)

The two spins have either +½ value or –½ value.

Practice Problems :

1. Correct set of quantum numbers for the unpaired electron in chlorine atom is

(a) n = 2, l = 1, m = 0 (b) n = 2, l = 1, m = 1

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

2. Arrange the electrons represented by the following sets of quantum numbers in the decreasing orderof energy for a multi-electron atom

(i) n = 4, l = 0, m = 0, s = +½ (ii) n = 3, l = 1, m = 1, s = –½

(iii) n = 3, l = 2, m = 0, s = +½ (iv) n = 3, l = 0, m = 0, s = – ½

(a) (iii) > (ii) > (i) > (iv) (b) (iii) > (iv) > (i) > (ii)

(c) (i) > (ii) > (iii) > (iv) (d) (iii) > (i) > (ii) > (iv)

3. For a d-electron the orbital angular momentum is

(a) 6 (b) 2 (c) (d) 2

4. Among the following the possible orbitals are 1p, 2s, 2p and 3f.

(a) 1p, 2s (b) 1p, 3f (c) 3f, 2p (d) 2s, 2p

5. The number of electrons in an atom having the following quantum numbers are respectively

(1) n = 4, ms = –½ (2) n = 3, l = 0

(a) 16, 9 (b) 32, 18 (c) 16, 2 (d) 9, 16

6. The lowest value of n that allows g orbitals to exist is

(a) 2 (b) 3 (c) 4 (d) 5

7. An electron is in one of the 3d orbital. The possible values of n, l and ml for this electron are

respectively

(a) 3, 1, 1 (b) 3, 2, –2 (c) 3, 0, 2 (d) 3, 1, –1

[Answers : (1) c (2) d (3) a (4) d (5) d (6) d (7) b]


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CA – 8C9D Electronic configuration of atoms : The distribution of electrons into orbitals of atom is called its

electronic configuration.

The filling of electron into different orbitals takes place according to following three rules :

1. Aufbau Principle : In the ground state of the atoms, the orbitals are filled in order of their increasingenergies.

In other words,electrons first occupy the lowest energy orbital available to then and then enther to higherenergy orbital.

The order of increase of energy of orbitals can be calculated by (n + l) rule.

Lower the value of (n + l) for an orbital, the lower is its energy.

If two orbitals have the same (n + l) value, the orbital with lower value of n has the lower energy.

2. Pauli Exclusion Principle : The number of electrons to be filled in various orbitals is restricted by thisprinciple.

No two electrons in an atom can have the same set of four quantum numbers.

e.g. If an electron in an atom has particular set of quantum number say n = 1, l = 0, m = 0, s = +½ then noother electron in an atom can have this set of quantum number.

3. Hund’s Rule of maximum multiplicity : This rule deals with the filling of electrons into orbitalsbelonging to same subshell.

Orbital of same subshells like px, py, pz are of equal energy and these are called as degenrate orbitals.

Similarly d has five orbitals of same energy i.e., dxy

, dyz

, dzx

, 222 zyxd,d

, f has 7 degenrate orbitals etc.

Hund’s Rule States that : pairing of electrons in the orbitals belonging to the same subshell (p, d or f) doesnot takes place until each orbital belonging to that subshell has got one electron each i.e. is singly occupied.

e.g. zyx ppp

as there are 3-orbitals the pairing of electron starts in p with the entry of 4th electron.

4. Magnetic Moment µ = )2n(n , n number of unpaired electron.

Practice Problems :

1. A compound of vanadium has magnetic moment of 1.73 B.M. The electronic configuration ofvanadium in the compound is

(a) [Ar]3d1 (b) [Ar]3d3 (c) [Ar]3d2 (d) [Ar]3d0

2. The ground state electronic configuration of Nitrogen (z = 7) atom. Can be represented as

(a) (b)

(c) (d)

3. If Hund’s rule is not followed, magnetic moment of Fe2+, Mn+ and Cr all having 24 electrons will bein order :

(a) Fe2+ < Mn+ < Cr (b) Fe2+ = Cr < Mn+

(c) Fe2+ = Mn+ < Cr (d) Mn2+ = Cr < Fe2+

4. Magnetic moment of Xn+ (Z = 26) is 24 B.M. Hence number of unpaired electrons and value of n

respectively are :

(a) 4, 2 (b) 2, 4 (c) 3, 1 (d) 0, 2

[Answers : (1) a (2) a (3) b (4) a]


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CA – 9C10 Isotopes

Atoms of the same element having same atomic numbers but different mass numbers are called isotopese.g.,

92U238

92U235,

84PO213,

84PO216.

Isobars : Atoms of different elements having same mass numbers but different atomic numbers are calledisobars, e.g., Po216 and

85At216,

88Ra228,

89Ac228 and

90Th228.

Isosters : Molecules having same number of atoms and same number of electrons are called Isosters.

For e.g., CO and N2, each has two atoms and the total number of electrons are 14.

Isodiaphers : Atoms having the same difference of neutrons and protons or same isotopic numbers. It isnoticed that nucleide and its decay product after emission are isodiaphers. e.g.,

92U235 and

90Th231,

29Cu65

and 24

Cr55.

Isotones : Nucleides having same no. of neutrons are known as isotones 2N14 &

8O15,

54Xe136 &

56Ba139,

50Ce140

59Pr141.


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CA – 10

E X E R C I S E

1. Which electronic level would allow the hydrogenatom to absorb a photon but not to emit a photon

(a) 3s (b) 2p

(c) 1s (d) 3d

2. In Aufbau principle is not followed in filling ofsub-shells then block of the element will be changein

(a) Na (z = 11) (b) Ni (z = 28)

(c) Sc (z = 21) (d) Fe (z = 26)

3. Maximum number of electrons present in anorbital having n + l = 4

(a) 8 (b) 4

(c) 16 (d) 2

4. Which of the following statement is correct ?

(a) The electronic configuration of Cr(z = 24) is [Ar] 3d5 4s

(b) The magnetic quantum number mayhave a fractional.

(c) In silver atom (z = 47), 23 electrons havea spin of one type and 24 of oppositetype

(d) The oxidation state of Nitrogen in NH3

is –3

5. Following ions will be coloured if Aufbau rule isnot followed :

(a) Cu2+ (b) Fe2+

(c) Sc3+ (d) a, b true

6. Magnitude of the charge on the helium ion [He2+]is :

(a) 4.8 × 10–10 esu (b) 2.4 × 10–10 esu

(c) 9.6 × 10–10 esu (d) 1.6 × 10–19 eV

7. A photon was absorbed by a hydrogen atom in itsground state and the electron was promoted to thefifth orbit. When the excited atom returned to itsground state, visible and other quanta wasemitted. Other quanta are :

(a) 2 1 (b) 5 2

(c) 3 1 (d) 4 1

8. Which of the following orbitals has/have zeroprobability of finding the electron in xy plane ?

(a) pz

(b) dyz

(c) dzx

(d) px

9. The radial distribution curve of 2s sublevelconsists of x nodes, x is :

(a) 1 (b) 3

(c) 2 (d) 0

10. If Aufbau rule is not used, 19th electron isSc (Z = 21) will have :

(a) n = 3, l = 0 (b) n = 3, l = 1

(c) n = 3, l = 2 (d) n = 4, l = 0

11. If each orbital can hold a maximum of 3 electrons,the number of elements in 4th period of periodictable (long form) is :

(a) 48 (b) 54

(c) 27 (d) 36

12. Which orbital gives an electron the greatestprobability of being found close to thenucleus ?

(a) 3p (b) 3d

(c) 3s (d) equal

13. Which describes orbital :

(a) (b) 2

(c) |2| (d) none

14. Suppose 10–17 J of energy is needed by the humaneye to see an object. How many photons of greenlight ( = 550 nm) are needed to genrate thisminimum amount of energy ?

(a) 14 (b) 28

(c) 39 (d) 42

15. The energy of an electron in the first Bohr orbit ofH-atom is –13.6 eV. The possible energy values ofthe excited state for electrons in Bohr orbit ofhydrogen is :

(a) –3.4 eV (b) –4.2 eV

(c) – 6.8 eV (d) 6.8 eV

16.h

is the angular momentum of the electron in the

_______________orbit of He+.

(a) 1 (b) 2

(c) 3 (d) 4

17. If nuclear radius is 10–13 cm and atomic radius is10–8 cm, the ratio of atomic volume to nuclearvolume would be

(a) 10+12 (b) 10+15

(c) 10+8 (d) 10+20

18. Calculate the work function for Na if thresholdfrequency is 4.39 × 1014 sec–1

(a) hv0

(b) h × c/0

(c) 6.63 × 10–34 × 4.39 × 1014

(d) all of these


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CA – 1119. The ionization energies of H, He+ and Li2+ are in

the ratio of

(a) 1 : 4 : 9 (b) 1 : 2 : 3

(c) 1 : 1/4 : 1/9 (d) 1 : 1 : 1

20. The Rydberg relation between all hydrogen atomsspectral lines is

]n/1n/1[R1 2

221

The shortest wave length of a photon that can beemitted when an electron jumps from the n = 4 stateis

(a) 1/R (b) 16/15 R

(c) 36/5R (d) 144/7R

21. Consider the four orbitals in a calcium atom :2p, 3p, 3d and 4s. These orbitals are arranged inorder of increasing energy is

(a) 2p < 3p < 3d < 4s

(b) 2p < 3p < 4s < 3d

(c) 2p < 4s < 3p < 3d

(d) 4s < 2p < 3p < 3d

22. The quantum numbers +½ and –½ for the electronspin represent.

(a) Rotation of the electron in clockwise andanti clockwise direction respectively.

(b) Rotation of electron in anticlockwise andclockwise direction respectively.

(c) Magnetic moment of the electronpointing up and down respectively

(d) Two quantum mechanical spin stateswhich have no classical analogue.

23. The mass of an electron is me. If its K.E. is E, then

its wavelength is

(a) h/2meE (b) h/2E

(c) h/2me

(d) h/meE

24. Then the wavelength of an electron moving with avelocity of 2.05 × 107 m sec–1 is

(a) 3.53 × 10–10 m (b) 3.53 Å

(c) 3.53 × 10–11 m (d) None

25. Two particles A and B are in motion ; if wavelengthof A is 5 × 10–8 m, the wavelength of B as itsmomentum is double of A is

(a) 2.5 × 10–8 (b) 10–7

(c) 0.5 × 10–8 (d) 5 × 10–10

ANSWERS (EXERCISE)

11. c

12. c

13. b

14. b

15. a

16. b

17. b

18. d

19. a

20. b

21. b

22. d

23. a

24. c

25. a

1. c

2. a

3. a

4. c

5. d

6. a

7. a

8. a

9. a

10. c


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CA – 12

AIEEE ANALYSIS [2004/2005/2006]

AIEEE ANALYSIS [2003]

3. The orbital angular momentum for an electron re-

volving in an orbit is given by

2

h.)1(ll . This

momentum for an s-electron will be given by

(a)2

h.2 (b)

2

h.

2

1

(c) zero (d)2

h

4. The number of d-electrons retained in Fe2+

(At. no. of Fe = 26) ion is

(a) 6 (b) 3

(c) 4 (d) 5

5. In Bohr series of lines of hydrogen spectrum, thethird line from the red end corresponds to whichone the following inter-orbit jumps of the electronfor Bohr orbits in an atom of hydrogen ?

(a) 3 2 (b) 5 2

(c) 4 1 (d) 2 5

8. Which of the following sets of quantum numbers iscorrect for an electron in 4f orbital ?

(a) n = 4, l = 3, m = +4, s = +1/2

(b) n = 4, l = 4, m = –4, s = –1/2

(c) n = 4, l = 3, m = +1, s = +1/2

(d) n = 3, l = 2, m = –2, s = +1/2

[2004]

9. Consider the groun state of Cr atom (Z = 24). Thenumbers of electrons with the azimuthal quantumnumbers l = 1 and 2 are, respectively

(a) 12 and 4 (b) 12 and 5

(c) 16 and 4 (d) 16 and 5

[2004]

10. Which one of the following ions has the highestvalue of ionic radius ?

(a) LI+ (b) B3+

(c) O2– (d) F–

[2004]

11. The wavelength of the radiation emitted, when in ahydrogen atom electron falls from infinity tostationary state 1, would be (Rydberg constant =1.097 × 107m–1)

(a) 91 nm (b) 192 nm

(c) 406 nm (d) 9.1 × 10–8 nm

[2004]

6. The de-Broglie wavelengths of the tennis ball ofmass 60 g moving with a velocity of 10 metres persecond is approximately

(Planck’s constant. (h = 6.63 × 10–34)

(a) 10–33 metre (b) 10–31 metre

(c) 10–16 metre (d) 10–25 metre

7. Which one of the following groupings represents acollection of isoelectronic species ?

(At. numbers Cs-55, Br-35)

(a) Na+, Ca2+, Mg2+

(b) N3—, F—, Na+

(c) Be, Al3+, Cl—

(d) Ca2+, Cs+, Br


1. In a hydrogen atom, if energy of an electron inground state is 13.6 eV, then that in the 2nd excitedstate is

(a) 13.6 eV (b) 6.04 eV

(c) 1.51 eV (d) 3.4 eV

2. Uncertainty in position of a minute particle of mass25 g in space is 10–5m. What is the uncertainty in itsvelocity (in ms–1) ?

(a) 0.5 × 10–23 (b) 0.5 × 10–34

(c) 2.1 × 10–34 (d) 2.1 × 10–28


New Delhi – 110 018, Ph. : 9312629035, 8527112111

CA – 1312. Which one of the following sets of ions represents

the collection of isoelectronic species ?

(a) K+, Ca2, Sc3+, Cl–

(b) Na+, Ca2+, Sc3, F–

(c) K+, Cl–, Mg2+, Sc3+

(d) Na+, Mg2+, Al3+, Cl–

(Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12,Al = 13, K = 19, Ca = 20, Sc = 21)

[2004]

13. Hydrogen bomb is based on the principal of

(a) nuclear fission

(b) natural radioactivity

(c) nuclear fusion

(d) artificial radioactivity

[2005]

14. In a multi-electron atom, which of the followingorbitals described by the three quantum memberswill have the same energy in the absence ofmagnetic and electric fields ?

(i) n = 1, l = 0, m = 0

(ii) n = 2, l = 0, m = 0

(iii) n = 2, l = 2, m = 1

(iv) n = 3, l = 2, m = 0

(v) n = 3, l = 2, m = 0

(a) (i) and (ii) (b) (ii) and (iii)

(c) (iii) and (iv) (d) (iv) and (v)

[2005]

15. Of the following sets which one does NOT containisoelectronic species ?

(a) 4

24

34 ClO,SO,PO

(b) 2

22 C,N,CN

(c) 3

23

23 NO,CO,SO

(d) 3

23

33 NO,CO,BO

[2005]

16. According to Bohr’s theory, the angularmomentum of an electron in 5th orbit is

(a) 2.5 h/ (b) 25 h/

(c) 1.0 h/ (d) 10 h/

[2006]

17. Uncertainty in the position of an electron(mass = 9.1 × 10–31 kg) moving with a velocity300 m s–1, accurate upto 0.001%, will be

(a) 3.84 × 10–2 m (b) 19.2 × 10–2 m

(c) 5.76 × 10–2 m (d) 1.92 × 10–2 m

(h = 6.63 × 10–34 J s)

[2006]


18. Which of the following sets of quantum numbersrepresents the highest energy of an atom ?

(a) n = 4, l = 0, m = 0, s = ± ½

(b) n = 3, l = 0, m = 0, s = ± ½

(c) n = 3, l = 1, m = 1, s = ± ½

(d) n = 3, l = 2, m = 1, s = ± ½

ANSWERS AIEEE ANALYSIS

1. c 2. d 3. c 4. c 5. b 6. a 7. b

8. c 9. b 10. c 11. a 12. a 13. c 14. b

15. a 16. a 17. d 18. d


New Delhi – 110 018, Ph. : 9312629035, 8527112111

CA – 14

1. Which of the following have higher mass

(a) Electron (b) Proton

(c) Neutron (d) All equal

2. Angular momentum of an electron in a given orbitis given by

(a)2

nh(b)

nh

(c)4

nh(d)

2

nh

3. Radius of nth orbit is given by

(a) ÅZ

n53.0

2

2

(b) ÅZ

n53.0

(c) ÅZ

n53.0

2

(d) ÅZ

n53.0

2

4. The uncertainity in momentum of electron, whenuncertainity in position is 5.28 × 10–12, is

(a) 10–14 (b) 10–34

(c) 10–20 (d) 10–23

5. When K.E. of electron gets 4 times then wavelengthreduce to

(a) times4

1(b) times

2

1

(c) times3

1(d) no effect

6. Ratio of charge/mass is zero for

(a) proton (b) electron

(c) neutron (d) positron

7. Wavelength of spectra line emitter when theelectron in n = 2 in H atoms de-excites to groundstate is

(a)R3

4(b)

R4

3

(c)R4

1(d)

R3

1

8. “No two electrons in an atom can have the same setof four quantum numbers” is a

(a) Aufbau principle

(b) Pauli Exclusion Principle

(c) Hund’s Rule

(d) None of these

TEST YOURSELF

ANSWERS

1. c

2. d

3. c

4. d

5. b

6. c

7. a

8. b

9. c

10. d

9. CO and N2 are

(a) Isodiaphers (b) Isotones

(c) Isosters (d) isobars

10. Which of the following incorrect about theorbital ?

(a) It is the region in the space where highprobability of finding the electron

(b) It is in a accordance with the dualcharacter of matter.

(c) It violate the Heisemberg principle.

(d) all incorrect

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