Atomic Structure IIT

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Prepared by V. V. Aditya vardhan var dhan [email protected]

ATOMIC STRUCTURE IIT-JEE Fundamental Particles

1) The The mas masss of an elec electron tron is 1) 0.000549 amu 2) 1.00727amu Note : Mass of an electron

3) 1.00867 amu

4) None

= 9.1x10-31 Kg = 5.49 x 10-4 amu

2) The The ch charge arge on an elec electro tronn is is -31 1) 9.1 x 10 coulombs 2) 1.602 x 10-19 coulombs 3) 4.8 x 10-10 coulombs 4) None 3) Th Thee mas masss of a proton proton is equa equall to -24 1) 1.6726 x 10 grams 2) 1.00727 amu -27 3)1.6726 x 10 Kg 4) All 4) Th Thee mas masss of of a neu neutr tron on is -27 1) 1.6749 x 10 Kg 2) 1.00867 amu 3) Both 1 & 2 4) None 5) Th Thee ch charge rge on a neu neutro tronn is is 1) + 1.6 x 10 -19 coulombs 2) -1.6 x 10-19 coulombs 3) 4.8 x 10 -10 e.s.u 4) None 6) 1 amu is equal ual to 1) 1.6726 x 10-27 Kg 2) 1.6605 x 10-27 Kg 3) 1.6749 x 10-27 Kg 4) None 7) Th Thee sy symbol mbol used used to repr repres esen entt a β - particle is 1) +1e0 2) -1e0 3) +1p1 4) 2He4

N o m A c H i . l D a R m A V g A @ Y i T d I D m a A . e V c h i d a

Note : β - particle is nothing but an electron

8) Th Thee ratio ratio of mass mass of of proton proton to that that of an elec electron tron is is 1 1) 2) 1836 3) 1839 1836 9) Th Thee ratio ratio of of mas masss of neu neutron tron to that of an elec electron tron is is 1 1) 1839 2) 3) 2 1836 10) The ratio of specif specific ic charge charge of electron electron to that of a proton is 1 1) 1836 2) 3) 1839 1836 Charge

4) 1

4) 1836

4) None

e

Note : Specific charge = mass = m

11) 11) The equation equation relatin relatingg the characteris characteristic tic frequen frequency cy ( ) of X-rays X- rays emitted emitted by anti-cathode anti-cathode and atomic at omic number (Z) of the t he metal used as anti-cathode, ant i-cathode, given by Moseley Moseley is 1)   a( z  b ) 2)   a( z  b ) 3)   z 2 4)   a  ( zb ) 12) Charge Charge on electron was was experim experimental entally ly determi determined ned by by 1) Millikan 2) Goldstein 3) Chadwick 4) J.J.Thomson 13) Choose Choose the incorre incorrect ct stateme statement nt 1) Nucleus occupies very small small fraction of atom at om and most of the atom at om is empty, empty, hence most of the alpha particles passed through the t he gold foil undeflected undeflected during d uring  - particle part icle scattering experiment. experiment. 2) Entire Ent ire mass of the atom at om is concentrated in the nucleus and the mass of electrons is negligible. negligible. 3) Nucleus carries positive charge due to presence pr esence of protons prot ons in it. 4) Most of o f the atom is not empty and hence the  -particles are deflected deflected back. 14) The e/m value of cathode rays rays in in a discharge discharge tube tube is is

Prepared by V. Aditya vardhan [email protected]

15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26)

1) always same irrespective of the gas taken 2) different for different gases 3) same only when noble gases are taken 4) None Neutron was discovered by 1) Thomson 2) Chadwick 3) Goldstein 4) Rutherford The highest value of e/m is observed for anode rays when the discharge tube is filled with 1) N2 2) H2 3) O2 4) He Which of the following pair have identical values of e/m 1) A proton and a neutron 2) A proton and a deuterium nucleus 3) A deuterium nucleus and an  -particle 4) An electron and  -rays Choose the correct increasing order of e/m values for e,p,n and  - particle 1) p < n <  < e 2) e < p <  < n 3) n < p < e <  4) n <  < p < e The e/m value of electron is 1) 9.1 x 10-31 Kg 2) 1.602 x 10-19C 3) 1.759 x 1011C.Kg -1 4) Zero The relation between 6C13and 6C14 is 1) isobars 2) isotopes 3) isotones 4) None 14 The isobar of 6C is 1) 8O16 2) 6C13 3) 7N14 4) 2Be4 Isotones contain same number of 1) electrons 2) protons 3) positrons 4) Neutrons The triad of nuclei that is isotonic is 1) 6C14, 7N15, 9F17 2) 6C12, 7N14, 9F19 3) 6C14, 7N14, 9F17 4) 6C14, 7N14, 9F19 The radius of atomic nucleus is of the order of 1) 10-10 cm 2) 10-13cm 3) 10-15cm 4) 10-8cm The natural abundance of 17Cl35and 17Cl37 is approximately 75% and 25% respectively. The molar atomic mass of chlorine will be 1) 36 g 2) 35.5g 3) 36.5g 4) 37g Which of the following reaction led to the discovery of neutrons. 1) 6 C16 1 p1 7 N 14  0 n1 2) 4 Be9  2 He4 6 C12  0 n1 3)

5


11 2 11 1 B 1 D 6 C  0 n

4)

4

8 4 11 1 Be  2 He 6 C  0 n

Quantum Theory

1) Quantum theory had suggested 1) Emission of energy is continuouss 2) Energy is emitted discontinuously in the form of quanta 3) Energy of the radiation is quantized. 4) Both 2 & 3 2) The wavelength of a radiation emitted by a sodium lamp is 600 nm. The frequency is 1) 5 Hz 2) 5 x 1014 Hz 3) 500 Hz 4) 5 x 1015 Hz 3) If the light radiation from neon atom has a wavelength of 300 nm, then the energy of the photon being emitted is 1) 6.626 x 10-19 J 2) 1.1 x 10-10 J 3) 2 x 10-19 J 4) 3.3 x 10-12 J 4) Violet light is able to eject electrons from the surface of potassium metal, whereas red light cannot.It is because, 1) the intensity of violet light is greater than that of red light. 2) the frequency of red light is lower than that of violet light and its energy is not sufficient to knock out the electrons. 3) the wavelength of violet light is greater than that of red light 4) None 5) Electrons with a kinetic energy of 6.023 x 104 J/mol are evolved from the surface of a


metal, when it is exposed to a radiation of wavelength of 300nm. The minimum amount of energy required to remove one electron from the metal atom is 1) 5.626 x 10-19J 2) 3 x 10-19J 3) 6.02 x 10-19 4) 6.62 x 10-34J Formula :

E = W + KE E = Energy of radiation W = Work function KE = Kinetic energy.

6) Electrons with kinetic energy of 3.313 x 10-24 J are evolved when a light radiation of 5 x109 s-1 of frequency falls on a metal surface. The threshold frequency of the metal is 1) 5 Hz 2) 4.5 x 109 Hz 3) 500 Hz 4) 3 x 1024 Hz Formula :

K.E = h (  -

 0)

7) When the frequency of light incident on a metalic plate is doubled, the kinetic energy of emitted photoelectrons will be 1) Doubled 2) Halved 3) More than doubled 4) Unchanged. Hint :


h  = W + KE1 2h  = W + KE2  2W + 2KE1 = W + KE2 KE2 = 2KE1 + W

8) Photo electrons are ejected from metals A & B when a light beam of wavelength  1 is used to irradiate them. But electrons are ejected only from metal ‘B’, when another radiation of wavelength  2 is used. Then the false statement among the following is 1) Electrons need more energy to escape from metal ‘A’ 2) 2   1 3) Threshold frequency of ‘A’ is higher than that of metal ‘B’ 4) When irradiated by light beam of wavelength  1 , the electrons ejected out from metal ‘A’ have more kinetic energy. 9) Photo-electrons are evolved when a metal is exposed to violet light. But no electrons are evolved when yellow light is used. If the metal is exposed to red light, the photo-electrons are 1) evolved when the intensity (of red light) is increased 2) not evolved 3) evolved even at low intensities 4) evolved only when a thin sheet of metal is used 10) The value of Planck’s constant is 1) 6.625 x 10-34 cal.sec 2) 1.584 x 10-34 cal.sec 3) 6.625 x 10 -27J.sec 4) 6.625 x 10-34 erg.sec Note: 1 calorie = 4.184 Joule

11) The number of photoelectrons emitted during photoelectric effect is proportional to 1) Intensity of incident beam 2) Work function 3) Frequency of incident beam 4) Velocity of incident beam Bohr’s Atomic Model

1) The angular momentum of an electron revolving in L - shell of hydrogen atom is h h 2h 3h 1) 2) 3) 4) 2   2 2) The angular momentum of electron in Bohr’s first orbit of hydrogen atom is ‘x’ what will be its angular momentum in the Bohr’s first orbit of He+ ? 1) 2x 2) 4x 3) x 4) x2 3) The force of attraction between nucleus and an electron is given by 1)

e r

2)

e 2 r

3)

 Ze 2 r

4)

 Ze2 r 2


4) Bohr’s equation to calculate the radius of an orbit is 1)

n 2 h2

n2h2

2) 2 2 2 2 2 me 2 4 me Z 5) The radius of Bohr’s orbit is proportional to 1) n

2)

n2 Z

2 2

3)

n h

2 2

.4 0

4)

4 2 me2 Z

3) Z 2

4)

n h

4 2 me 2 Z  4 0  n Z

6) The radius of Bohr’s second orbit in hydrogen atom is 1) 0.539 A0 2) 2.116 A0 3) 0.2116 A0 4) None 2+ 7) The radius of Bohr’s first orbit in Li is 1) 0.0587 pm 2) 17.63 pm 3) 176.3 pm 4) 0.529 pm + 3+ 8) The ratio of radius of first orbit of He to that of Be is 1) 2 : 1 2) 1 : 2 3) 1 : 4 4) 4 : 1 9) The radius of first orbit in H-atom is equal to the radius of 1) 1st orbit in He+ 2) 2nd orbit in Li2+ 3) 2nd orbit in Be3+ 4) 2nd orbit in He+ 10) The ratio of radii of first three orbits in hydrogen atom is 1) 1 : 4 : 9 2) 1 : 2 : 3 3) 1 : 2 : 4 4) 9 : 4 : 1 11) The distance between successive orbits while going away from nucleus 1) decreases 2) increases 3) first decreases and then increases 4) unchanged 12) Bohr’s expression to calculate the energy of electron in a given orbit is


2 2 me 4 z 2 1)  2 2 2 n h  4 0 

3) 

2) 

4 2 me4 z 2  4 0  n2 h2

2

4) 

2 2 me 4 z  4 0  nh

4 2 me4 z nh 2

13) The potential and kinetic energies of Bohr’s orbits are expressed as 1) 

2 Ze

1 & 2

r2 2 Ze

2 Ze

r 2

2) 

2

1 & 2

Ze

r

Ze

r

2 1 Ze & & 3)  4)  2 r r r r 14) The total energy and potential energy of electron in a given orbit are expressed as

1 1)  2

15) 16) 17) 18)

2 Ze

2

2 Ze

r

1 & 2

2 Ze

2 Ze

r

2) 

2 Ze

r

1 & 2

2 Ze

r

2 2 2 1 Ze Ze 1 Ze2 Ze & & 3)  4)  2 r r 2 r r The energy of electron in the 1st orbit of hydrogen atom is 1) -13.6 eV 2) -2.18 x 10-18J 3) -5.21 x 10-19 Calories 4) All The energy of one mole of electrons present in Bohr’s first orbit of hydrogen atoms is 1) - 313.6 Kcal 2) -1312 kJ 3) Both 1 & 2 4) None The energy of electron in second shell of hydrogen atom is 1) +3.4 eV 2) -3.4 eV 3) -13.6 eV 4) + 13.6 eV + The energy of electron in the ground state of He ion is 1) -13.6 eV 2) + 54.4 eV 3) - 27.2 eV 4) - 54.4 eV


19) The kinetic and potential energies of electron in the ground state of hydrogen atom are respectively. 1) +13.6 eV & -13.6 eV 2) + 13.6 eV & - 27.2 eV 3) -136 eV & + 13.6 eV 4) 27.2 eV & - 13.6 eV 20) The energy of an electron in the M-shell of hydrogen atom is x eV. The energy of electron in its L-shell is 1) 

 x

eV

 x

3) 

eV

 x

eV

 x

eV 9 4 4 9 21) The ratio of energies of electrons in first three Bohr’s orbits in hydrogen atom is 1 1 1 1 1 1 1 1 1) 1: : 2) 1: : 3) : :1 4) : :1 4 9 2 3 3 2 9 4 22) The energy difference between two successive orbits while going away from the nucleus 1) decreases 2) increases 3) remains same 4) first decreases and then increases

2)

4)


23) The ratio of energies of electrons in the ground states of H, He+ and Li2+ is 1 1 1 1 1) 9 : 4 : 1 2) : :1 3) 1: : 4) 1 : 4 : 9 9 4 4 9 24) The energy of an electron in the first orbit of hydrogen atom is -y Joule. The kinetic energy of electron in the second orbit will be y y y y 1) - Joule 2) Joule 3)  Joule 4) Joule 4 2 2 4 25) The potential energy of electron present in the ground state of Li2+ ion is 3e 3e2 3e2 3e2 1) + 4 r 2)  4 r 3)  4 r 4)  4 r 2 0 0 0 0 26) The energy of electron in the first orbit of hydrogen atom is equal to the energy of electron in the 1) 2nd orbit of He+ 2) 3rd orbit of Li2+ 3) 4th orbit of Be3+ 4) All -18 27) The energy of electron in the ground state of hydrogen atom is -2.18 x 10 J. The potential energy of electron in the first excited state of He+ ion is 1) - 2.18 x 10-18 J 2) + 2.18 x 1018 J 3) -4.36 x 10-18J 4) - 1.09 x 10-18J 28) The velocity of an electron in Bohr’s orbit is expressed as

1)

2 Ze2

2) 

nh

2 Z 2e

3)

nh

4 2 me2

4)

nh

2 Ze n2h2

29) Choose the incorrect statement 1) The energy of electron increases with ‘n’ value of orbit 2) The kinetic energy of electron decreases with ‘n’ value of orbit 3) The potential energy of electron increases with ‘n’ value orbit 4) The velocity of electron increases with ‘n’ value of orbit 30) The ratio of velocities of electrons in first three orbits of hydrogen atom will be 1 1 1 1 1) 1 : 4 : 9 2) 1: : 3) 1: : 4) 1 : 2 : 3 2 3 4 9 31) The velocity of electron in first orbit of hydrogen atom is 1) 2.188 x 108 cm.sec-1 2) 13.6 cm sec-1 3) 2.188 x 10 8m.sec-1 4) 0.529 x 10 -18 cm sec-1 32) If the velocity of an electron is x cm.sec-1 in the ground state of hydrogen atom, the velocity in the second orbit is 1)

x

4

-1

cm.sec

2) x cm.sec 2

-1

3)

x

2

-1

cm.sec

4)

x

2

4

cm.sec-1


33) Rydberg’s constant can be expressed as 1)

2 2 me4 z 2

2)

ch3

2 2 me4 n2 h2

3)

2 2 me4 ch3

4)

4 2 me4 ch3

34) The value of Rydberg’s constant is 1) 1.096 x 10 5 cm-1 2) 109677 cm 3) 109.677 m-1 4) 1.096 x 10 3 m-1 35) An electron jumps from 4th orbit to 2nd orbit. The spectral line corresponding to this transition will be observed in 1) Lyman series 2) Paschen series 3) Pfund series 4) Balmer series 36) The wave number of limiting line in Lyman series of hydrogen atomic spectrum is 1) 109677 cm-1 2) 27892 cm-1 3) 13142 cm-1 4) 42314 cm-1 37) The ionisation energy of hydrogen atom is 1) 13.6 eV 2) 3.4 eV 3) 1.51 eV 4) 0.85 eV 38) The spectral line corresponding to longer wavelength in hydrogen atomic spectrum is 1) Limiting line in Pfund series 2) First line in Lyman series 3) Limiting line in Lyman series 4) First line in Pfund series 39) A spectral line is observed in the visible region of emission spectrum of hydrogen. The probable transition corresponding to this line 1) n = 2  n = 5 2) n = 2  n = 3 3) n = 4  n = 2 4) n = 5  n = 3 40) The ratio of energies of first three lines in Lyman series is 1 1 1 3 8 15 4 9 16 1) 1 : 2 : 3 2) : : 3) : : 4) : : 2 3 4 4 9 16 3 8 15 41) The wave number of a spectral line for a given tranistion is x cm-1 for He+, then its value for Be3+ for same transition is


1) x cm-1

2) 4 x cm-1

3)

x

4

cm-1

4) 16 x cm-1

42) The energy of electron in the infinite orbit is 1) +13.6 eV 2) 1 eV 3)  4) zero 43) A hydrogen atom is supplied with 10.2 eV of energy in its ground state. The highest energy level into which the electron can be excited is 1) 3rd 2) 4th 3) 2nd 4) 10th 44) The energy required to remove an electron from a hydrogen atom in its 1st excited state is 1) 13.6 eV 2) 3.4 eV 3)1.51 eV 4)0.85 eV 45) The wavelength corresponding to  -line in Balmer series is 1) 6564 A0 2) 3282 A0 3) 15232 A0 4) None 46) The number of spectral lines formed when an electron undergoes all possible transitions between 6th orbit and 1st orbit is 1) 6 2) 15 3) 30 4) 21 Use formula :





n n 1

2

; where n = n 2  n1

47) An electron in the ground state of hydrogen atom absorbed 12.75 eV of energy and get excited. The number of spectral lines formed due to all possible transitions when electron jumps back from highest possible energy level, during this excitation, to L-Shell of the atom is 1) 15 2) 3 3) 1 4) 10 h 48) The difference in angular momentum of two bohr’s orbits is 3 . The number of all possible elec2π tronic transitions between these two orbits is 1) 3 2) 12 3) 6 4) 1


49) The ratio between Kinetic energy and total energy of electron in atoms according to Bohr’s model is 1) 1 : 1 2) 1 : -1 3) 1 : 2 4) -2 : 1 50) The wave number of limiting line of Paschen series of hydrogen atomic spectrum is equal to R R R 1) RH 2) H 3) H 4) H 25 9 16 51) The lines in Lyman series of hydrogen atomic spectrum are formed in 1) Visible region 2) Far IR region 3) UV region 4) Near IR region 52) Bohr’s theory could not explain 1) Zeeman effect 2) Fine atomic spectrum 3) Stark effect 4) All

N o m A c H i . l D a R m A V g A @ Y i T d I D m a A . e V c h i d a de Broglie’s wave concept

1) The wave length of an electron can be expressed as

2)

3) 4) 5) 6) 7)

h h h h2 1) 2) 3) 4) 2π mv nπ nπ The electron revolving around the nucleus behaves as a stationary wave only when the circumference of its orbit is equal to 1) Fractional multiple of its wavelength. 2) Integral multiple of its wavelength. 3) Velocity of electron 4) Radius of orbit Quantization of angular momentum of an electron in a stationary orbit can be justified by using. 1) Bohr’s theory 2) Heisenberg’s theory 3) de Broglie’s theory 4) Quantum theory The velocity of an electron is 6.626 x 104 m.s-1. The wavelength of electron is 1) 110 A0 2) 232 A0 3) 91 A0 4) 323 A0 The ratio of wavelength of proton to that of electron which are moving with same velocity is 1) 1 : 9 2) 1836 : 1 3) 1 : 1836 4) 1 : 2 The number of waves produced by electron wave in Bohr’s 3rd orbit is 1) 1 2) 2 3) 3 4) 9 The wavelength of electron wave in the first orbit of hydrogen atom is 1) 3.33 A0 2) 0.529 A0 3) 2.21 A0 4) 1.06 A0 Note : Use the formula - n =2 r

where r = 0.529 x n 2 A0

8) The wavelength of electron revolving in circular orbit in Bohr’s atom can be expressed as nh 2

n 2h 2

nh 4) None 2π m 2πmZe2 4π2 me 2 9) The wavelength of electron wave in the 2nd orbit of hydrogen atom is ‘ x’ cm. The wavelength of electron in the third orbit will be 9 x 2 x 4 x 3 x cm cm cm 1) cm 2) 3) 4) 4 3 9 2 10) The relation between wavelength of electron and its kinetic energy is

1)

2)

3)

h2 h h nh 1)  = 2)  = 3)  = 4)  = KE KE 2.KE.m 2.KE.m 11) The kinetic energy of electron in an orbit of hydrogen atom is 2.18 x 10-11erg. The wavelength of electron wave will be. 1) 0.529 A0 2) 2.21 A0 3) 3.33 A0 4) None 12) The ratio of wavelengths of electron waves of first three orbits in hydrogen atom is 1 1 1) 1: : 2) 1 : 2 : 3 3) 3 : 2 : 1 4) 1 : 4 : 9 2 3 13) The ratio of radii of two successive orbits is 4 : 9. The ratio of the wavelengths of electron waves in these orbits is 1) 9 : 4 2) 3 : 2 3) 16 : 81 4) 2 : 3


14) The ratio of wavelengths of electron waves in two orbits is 3 : 5. The ratio of kinetic energies of electrons will be 1) 25 : 9 2) 5 : 3 3) 9 : 25 4) 3 : 5 Quantum Numbers

1) Match the following A) Principal quantum number (n) B) Azimuthal quantum number (l) C) Magnetic quantum number (m) D) Spin quantum number (s)


Choose the correct matching A B C

2) 3) 4)

5) 6)

1) Neils Bohr 2) Uhlenbeck & Goudsmit 3) Lande 4) Sommerfeld D

1) 1 2 4 3 2) 1 4 3 2 3) 1 3 2 4 4) 3 1 4 2 In order to explain fine hydrogen atomic spectrum, Sommerfeld proposed. 1) Principal quantum number 2) Spin quantum number 3) Azimuthal quantum number 4) Magnetic quantum number Magnetic quantum number (m) is proposed by Lande, in order to explain 1) Low resolution hydrogen atomic spectrum 2) Fine hydrogen atomic spectrum 3) Zeeman effect 4) Spin - Spin coupling The wrong statement about principal quantum number ‘n’ is 1) It is used to identify the main energy level or shell 2) It is used to determine the size and energy of orbit 3) It is used to calculate the number of orbitals in the nth shell and is given by n2. 4) It is used to determine the orientation of orbital The quantum number which is used to determine the shape of orbital is 1) Principal quantum number (n) 2) Azimuthal quantum number (l) 4) Spin quantum number (m s) 3) Magnetic quantum number (m) The orbital angular momentum of electron is precisely given by 1)

nh

2)

l  l  1

h

3)

s  s  1

h

2 2 2 7) The number of subshells in a given main shell ‘n’ is equal to

4)

n  n  2

h

2

1) n - 1 2) n2 3) n  n  1 4) n 8) The possible subsidiary quantum number values of sublevels in the 3rd shell are 1) l = 1, 2 & 3 2) l = 1 & 2 3) l = 0, l & 2 4) l = 1, 2, 3 & 4 9) Match the following Azimuthal quantum number A) l = 0 B) l = 1 C) l = 2 D) l = 3 The correct matching is A B C D

1) 3 2) 3 3) 3 4) 2

2 1 1 3

1 2 4 1

4 4 2 4

Shape of orbital

1) Dumb-bell 2) Diffused 3) Spherical 4) Double dumb-bell


10) The n & l values of 4f orbital are 1) 4 & 3 2) 4 & 4 3) 4 & 1 4) 3 & 3 11) The number of orbitals in a given shell ‘n’ is equal to 1) 2n 2) 2n2 3) n2 4) 2n - 1 12) The number of orbitals in given sub shell ‘l’ is equal to 1) l + 1 2) l + 2 3) 2l + 2 4) 2l + 1 13) The number of orbitals in l = 2 subshell will be 1) 2 2) 3 3) 5 4) 1 14) The significance of magnetic quantum number (m) is 1) It denotes the size of orbital 2) It represents the shape of orbital 3) It indicate the orientation of orbital 4) It represents angular momentum of orbital 15) The possible values of magnetic quantum numbers of orbitals in a subshell l= 2 are 1) -2, -1, +1, +2 2) -2, -1, 0, +1, +2 3) -1, 0, +1 4) 0, +1, +2, +3 16) The maximum number of electrons that can be accommodated in sub shell l = 2 1) 2 2) 5 3) 4 4) 10


Note : The no.of electrons in subshell ‘ l’ = 4 l + 2

17) The maximum number of electrons that can be accommodated in an orbital is 1) 2 2) 1 3) 4 4) 1 / 2 18) The values of spin quantum number can be 1 1 1 1 1)  2)  3)  or  4) 0 or 1 2 2 2 2 19) Spin quantum number was proposed by 1) Goudsmidt & Uhlen beck 2) Lande 3) Neils Bohr 4) de Broglie 20) The number of orbitals present in the shell with n = 4 is 1) 4 2) 8 3) 16 4) 32 Note : The number of orbitals in given shell, n is given by n 2

21) The maximum number of electrons that can be present in a main energy level n = 3 is 1) 3 2) 18 3) 6 4) 9 22) The quantum number that is not possible for electron in 4d orbital is 1) l = 2 2) s = + 1 / 2 3) m = 0 4) m = + 3 23) 2px and 2py orbitals differ in their 1) energy 2) shape 3) orientation 4) ‘l’ value 24) The notation of orbital with n = 3 and l = 1 is 1) 3s 2) 3p 3) 3d 4) 2p 25) The incorrect set of quantum numbers for an electron will be 1) n = 3, l = 1, m = 0, s = + 1 / 2 2) n = 2, l = 0, m = 0, s = - 1 / 2 3) n = 4, l = 3, m = -3, s = - 1 / 2 4) n = 3, l = 2, m = +3, s = + 1 / 2 26) The number of ‘p’ orbitals present in a shell with n = 4 is 1) 4 2) 12 3) 10 4) 3 Heisenberg’s uncertainty principle

1) The mathematical expression for Heisenberg’s uncertainty principle is 1)  x. v 

h n

2)  x.  p 

h2



3)

.

 x  p



h n

4)  x.

m v 

h n

2) Uncertainty in position of a 0.25g particle is 10-5m. Uncertainty in velocity will be 1) 1.2 x 1034 m.sec-1 2) 1.6 x 10-20 m.sec-1 3) 1.7 x 10-19 m.sec-1 4) 2.1 x 10-26 m.sec-1 3) Uncertainty in the position of an electron moving with a velocity 300 m.s-1 accurate upto 0.001% will be 1) 19.3 x 10 -2 m 2) 5.76 x 10-2 m 3) 1.93 x 10-2 m 4) 3.84 x 10-2 m


Schrodinger wave equation


1) The Schrodinger wave equation which describes the three dimensional electron wave with potential energy ‘V’ is given by  2  2  2 8 2 m 1) 2  2  2   E  V   0 x y z h  2  2  2 8 m 2) 2  2  2  2  E  V   0 x y z h

 2  2  2 8 2 m 3) 2  2  2  2  E  V   0 x y z h  2  2  2 8 m  E  V   0 4) 2  2  2  h x y z

2) Choose the incorrect statement 1)  is called orbital wave function. It represents the amplitude of electron wave, but has no physical significance. 2)  2 is called probability density of electron and always has positive values. 3) Schrodinger’s wave equation is time dependent. 4) None 3) Which of the following is not a boundary condition required to get valid Eigen functions for  ? 1)  must be finite and continous 2)  must be single valued at a given point 3) The probability of finding the electron over the space from +  to -  must be equal to zero 4) All ˆ in the schrodinger equation, Hˆ  E  ,is known as 4) H 1) Kinetic energy operator 2) Heisenberg operator 3) Hamiltonian operator 4) Potential energy operator  Note : The total energy operator also called as Hamiltonian operator is the sum of Kinetic energy operator (T) and potential  energy operator ( V ) .

i.e., Hˆ = Tˆ + Vˆ .

5) The orbital wave function   written in terms of polar co-ordinates : r, θ and  can be expressed as   r, θ,    R  r    θ    

Choose the incorrect statement related to above expression. 1) R (r) is called radial wave function which depends on quantum numbers ‘n’ & ‘l’ 2)   θ  &    are the angular wave functions which depend upon quantum numbers ‘l’ and ‘m’ 3) R (r) gives dependence of orbital upon radial distance of electron from the nucleus. 4)   θ  and    are the angular wave functions and depend only on principal quantum number ‘n’ Note : The orbital wave functions ( ) can be expressed in terms of spherical polar co-ordinates (r, θ &  ). The relation between polar co-ordinates and cartesian co-ordinates can be given as x = r sin θ cos  y = r sin θ sin  z = r cos θ



6) The energy of quantized energy states in hydrogen atom can be given by 1) E n 

  e e4

8 02 h 2 n 2

Note :

2) E n 

  e e2

4 0 h 2 n 2

3) E n 

  e e4

2 0 h 2 n 2

4) E n 

  e e 4

8 0 hn

1)  e is called reduced mass of electron

1

 e



1

m e



1

mn

Where

me = mass of electron mn = mass of nucleus 2) For hydrogen atom electron energy depends only on ‘n’ and independent of ‘ l’ and ‘n’.

7) The probability of finding an electron at a distance ‘r’ from the nucleus regardless of direction iscalled 1) radial wave function 2) radial probability density 3) radial probability function 4) All Note : 4 r

R = radial wave function R2 = radial probability density 2

2

drR  radial

probability function

8) The correct graph of radial wave function (R) of 2s orbital plotted against radial distance ‘r’ is

Note :

Option 1) is the graph for 1s orbital


Option 2) is the graph for 3s orbital Option 4) is the graph for 2p orbital

9) The maximum electron density can be found at the nucleus in case of orbital 1) s 2) p 3) d 4) none Note: Consider above graphs of ‘s’ orbitals. The ‘R’ value is maximum at r=0. Also consider the ‘R 2’ vs ‘r’ graphs in the following question.

10) The correct graph of radial probability density (R2) of 2p orbital plotted against radial distance ‘r’ is

Note :

Option 1) is the graph for 1s orbital Option 2) is the graph for 2s orbital Option 4) is the graph for 2p orbital

11) The correct graph of radial probability function (4 r 2 R 2 ) of 2s orbital plotted against radial distance ‘r’ is

Note : 2s orbital has one radial node


12) The space around the nucleus where the  2 > 90% is called 1) atomic orbital 2) nodal region 3) nodal plane 4) All 13) The point at a radial distance from the nucleus where the probability of finding the electron is zero is called 1) Atomic orbital 2) Nodal point 3) Molecular orbital 4) None 14) The radius of maximum probability for 1s orbital of hydrogen atom is 1) 52.9 pm 2) 105.8 pm 3) 264.5 pm 4) 2.116 pm 15) The radial distance from the nucleus at which the nodal region of 2s orbital of hydrogen atom is 1) 0.529 Å 2) 1.050 Å 3) 2.645 Å 4) 2.116 Å 16) In 2s orbital of hydrogen atom the radial probability function has its maximum value at the radial distance, 1) 2.645 Å 2) 2.116 Å 3) 0.529 Å 4) 1.058 Å 17) In 2p orbital of hydrogen, the radial probability function has it’s peak vlaue at the radial distance. 1) 2.645 Å 2) 2.116 Å 3) 0.529 Å 4)1.058 Å 18) Assertion : The energy of 2s orbital will be less than that of 2p orbital even though the radius of maximum probability for 2s orbital is at 264.5 pm and for 2p orbital is at 211.6 pm. Reason : There is a small additional maxima at 52.9 pm in case of 2s orbital. Hence the electron in this orbital penetrates little closer to the nucleus and binds strongly to it than that of 2p orbital. Choose the correct answer 1) Both ‘A’ & ‘R’ are correct ; and ‘R’ is the correct explanation of ‘A’ 2) Both ‘A’ & ‘R’ are correct ; but ‘R’ is not the correct explanation of ‘A’. 3) ‘A’ is correct but ‘R’ is incorrect 4) ‘A’ is incorrect but ‘R’ is correct. 19) The number of radial nodes present in a given orbital is equal to 1) l 2) n - l - 1 3) n - 1 4) n - l - 2


Note : Radial node, also called as nodal region or spherical node, does not pass through the nucleus

20) The number of angular nodes present in a given orbital is equal to 1) l 2) n - l - 1 3) n - l 4) n - l - 2 Note : Angular node, also called as nodal plane, passes through the nucleus.

21) The number of radial nodes in 1s orbital is 1) 1 2) 2 3) 3

4) 0

22) The number of radial nodes in 2s orbital is 1) 2 2) 3 3) 1

4) 0


23) The number of angular nodes in 2s orbital is 1) 1 2) 2 3) 3

4) 0

Note: Angular nodes are absent in ‘s’ orbitals.

24) The number of angular nodes in 2p orbital is 1) 1 2) 2 3) 3

4) 0

Note: ‘p’ orbitals have only one angular node perpenducular to the axis of orientation of the orbital. For example, p x orbital has one angular node along yz plane.


25) The number of radial nodes in 4p orbital is 1) 1 2) 2

26)

3) 3

4) 4

The number of radial nodes and angular nodes respecti vely present in 4d xy orbital will be

1) 0 & 1

2) 1 & 2

3) 2 & 2

4) 1 & 1

Note: 4d orbital has one radial node. And usually ‘d’ orbitals have two angular nodes. In case of 4d xy orbital the angular nodes are present along yz and xz planes.

27) The total number of radial and angular nodes present in a given orbital is equal to 1) n 2) n - l 3) n - l - 1 4) n - 1


28) Which of the following radial distribution graph corresponds to orbital with n = 3, l = 2 ?


29) The orbital with two nodal planes but with no nodal regions is 1) 4d 2) 4s 3) 3d 4) 4f 30) The number of nodal planes in 4f orbital is 1) 1 2) 2 3) 3 4) zero 31) The  2 value is zero along xy and yz planes in a d-orbital. The notation of that d-orbital will be 1) dxy 2) dxz 3) dyz 4) 1 or 3 Electronic configuration

1) The orbital with least energy before filling up of electrons is 1) 3d 2) 4s 3) 4p 4) 5s Note : Orbital with least (n + l) value has least energy and is filled first

2) The orbital which is filled first will be 1) 4f 2) 5d 3) 6p

4) 7s

Note : If the orbitals possess same (n + 1) value, the orbital with least ‘n’ value has least energys

3) Which of the following electronic configuration violates Hund’s rule 1) 1s 2 2s 2 2 p1x p1y p1z 2) 1s 2 2 s 2 2 p 2x p1y p oz 3) 1s 2 2 s 2 2 px2 p 2y p1z 4) 1s 2 2 s 2 2 p1x p1y pz0 4) No two electrons in an atom can have same set of quantum numbers. This statement is known as 1) Hund’s rule 2) Heisenberg’s uncertainty principle 3) Pauli’s exclusion principle 4) None 5) The correct ground state electronic configuration of chromium is 1) [Ar] 3d4 4s2 2) [Ar] 3d5 4s2 3) [Ar] 3d5 4s0 4) [Ar] 3d5 4s1 6) The set of quantum numbers of valence shell electron in potassium atom is 1) n = 3, l = +1, m = +1, s = + 1 / 2 2) n = 4, l = 0, m = 0, s = + 1 / 2 3) n = 3, l = 0, m = 0, s = + 1 / 2 4) n = 3, l = l, m = 0, s = + 1 / 2 2 3 1 7) [Ne] 3s 3p 3d is the electronic configuration of 1) ‘P’ in ground state 2) ‘S’ in 1st existed state 3) ‘P’ in 1st excited state 4) ‘S’ in 2nd existed state 8) The number of electrons in the valence shell of calcium is 1) 2 2) 3 3) 1 4) 6


9) In a multi electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields ? a) n = 1, l = 1, m = 0; b) n = 2, l = 0, m = 0; c) n = 2, l = 1, m = 1; d) n = 3, l = 2, m = +1; e) n = 3, l = 2, m = 0 The correct combinations are 1) a & b 2) b & c 3) a & d 4) d & e 10) Which orbital is nearest to the nucleus after filling with electrons ? 1) 5d 2) 6s 3) 6p 4) 4f 2+ 11) The number of electrons in 3d - orbital in Zn ion is 1) 8 2) 10 3) 9 4) 0 12) The number of electrons with l = l in the ground state of sulfur atom is 1) 6 2) 10 3) 8 4) 4 13) The number of unpaired electrons in 3d orbital of copper atom is 1) 5 2)10 3) 1 4) None Note: Copper has anomalous electronic configuration i.e., [Ar] 3d104s1 14) The magnetic moment of Mn2+ ion is 1) 3.87 BM 2) 5.9 BM 3) 7.93 BM 4) 6.92 BM


Formula: magnetic moment( )= n(n+2) BM where n=no. of unpaired electrons

15) The correct increasing order of energy of 4f, 5p, 6s & 5d orbitals is 1) 4f < 5p < 5d < 6s 2) 4f < 6s < 5d < 5p 3) 5p < 6s < 4f < 5d 4) 5p < 5d < 6s < 4f 16) The number of exchanges possible between parallel electrons in ‘d’ orbitals of chromium is 1) 6 2) 10 3) 5 4) 4 Note: In [Ar] 3d54s1 configuration, there are five unpaired ‘d’ electrons. The number of exchanges are calculated by following formula. These exchanges between degenerate d-electrons lower the energy of the atom and thus by confering more stability to the atom in this anomalous configuration. If the configuration is [Ar] 3d 44s2 as predicted, then the exchanges possible are only 6. Hence this configuration is less stable. Formula : 5C  2

17)

5  5  1 2!

= 10

  n!  this is the modification of equation !  !  r  n r   

The maximum number of electrons possible in an atom with n = 4 and m s = - 1/ 2 is

1) 32 2) 8 3) 4 4) 16 18) The electronic configuration of hydride ion is 1) 1s1 2) 1s0 3) 1s2 4) 2s1 19) Assertion[A] : [Ar] 3d54s1 configuration of chromium is more stable than [Ar] 3d44s2 configuration Reason[R] : The exchange energy will be maximum in case of atoms with half filled subshells 1) Both ‘A’ & ‘R’ are correct ; and ‘R’ is the correct explanation of ‘A’ 2) Both ‘A’ & ‘R’ are correct ; but ‘R’ is not the correct explanation of ‘A’. 3) ‘A’ is correct but ‘R’ is incorrect 4) ‘A’ is incorrect but ‘R’ is correct. 20) Which of the following pairs has two unpaired electrons 1) Mg, Si 2) S, Mg 3) S, Si 4) S, Fe Note: Key to the questions and updates, if any, can be downloaded from http://groups.google.com/group/adichemadi

Atomic Structure IIT

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