Q. 1 Which is not basic postulate of Dalton s atomic theory? Option 1 Atoms are neither created nor destroyed in a chemical reaction Option In a given compound, the relative number and kinds of atoms are constant. Option 3 Atoms of all elements are alike, including their masses. Option 4 Each element is composed of extremely small particles called atoms. Dalton souid atoms of same element will be alike but atoms of different elements will be different in all aspect including their masses. Q. The number of electrons in a neutral atom of an element is equal to its: Option 1 Atomic weight Option Atomic number Option 3 Equivalent weight Option 4 Electron affinity Correct Answer For a neutral atom, number of electrons = number of protons = atomic number Q.3 The e/m for positive rays in comparison to cathode rays is: Option 1 Very low Option High Option 3 Same Option 4 none Cathode rays contains electrons and positive rays contain positive charged gaseous atom mass of electron is negligible in comparison to positive charge atom e ratio for cathode rays is very low. m Q.4 Which has highest e/m ratio? Option 1 He + Option H + Option 3 He + Option 4 H Correct Answer e/m ratio for + + 1 + +1 He = = H = =1 4 1 + +1 1 0 He = = H= =0 4 4 1 Q.5 Cathode rays have: Option 1 Mass only Option Charge only Option 3 Neither mass nor charge Option 4 Mass and charge both Both mass and charge according to the Properties of cathode rays.
Q.6 Mass of neutron is times the mass of electron. Option 1 1840 Option 1480 Option 3 000 Option 4 None of these Mass neutron = 1840 M electron Q.7 Positive rays or canal rays are: Option 1 Electromagnetic waves Option a steam of positively charged gaseous ions Option 3 A stream of electrons Option 4 neutrons Correct Answer Positive rays are a steam of positively charged gaseous ions. Q.8 The triad of nuclei that is isotonic is: Option 1 14 15 17 6 C, 7 N, 9 F Option 1 14 19 6 C, 7 N, 9 F Option 3 14 14 17 6 C, 7 N, 9 F Option 4 14 14 19 6 C, 7 N, 9 F For Isotonic species, they should have equal number of neutrons. C = 14-6 = 8 N = 15-7 = 8 F = 17 9 = 8 Number of neutrons = Mass number atomic number Q.9 The ion that s isoelectronic with CO is : Option 1 CN - Option O + Option 3 O Option 4 N + For Isoelectronic species, they should have equal number of electrons. CO = 6 + 8 = 14 CN = 6 + 7 + 1 = 14 due to negative charge Q. An isotone of 3 Ge 76 is (i) 3 Ge 77 (ii) 33 As 77 (iii) 34 Se 77 (iv) 34 Se 78 Option 1 Only (i) and (ii) Option Only (ii) and (iii) Option 3 Only (ii) and (iv) Option 4 (ii), (iii), and (iv) For Isotonic species, Number of neutrons should be same 76 3Ge=76-3=44 77 33 As=77-33=44
76 78 3Ge=77-3=45 34 Se=78-34=44 34Se 77 = 77 34 = 43 Q.11 Which of the following atoms and ions are isoelectronic i.e. have the same number of electron with the neon atom Option 1 F - Option Oxygen atom Option 3 Mg Option 4 N - For isoelectronic species number of electrons should be equal Ne = F - = a + 1 = 0 = 8 Mg = 1 N - = 7 + 1 = 8 Q.1 When a gold sheet is bombarded by a beam of α particles, only a few of them get deflected whereas most go straight, undeflected. This is because Option 1 The force of attraction exerted on the α particles by the oppositely charged electrons is not sufficient. Option A nucleus has a much smaller volume than that of an atom. Option 3 The force of repulsion acting on the fast moving α particles is very small. Option 4 The neutrons in the nucleus do not have any effect on the α particles Correct Answer Only a few gut defeated whereas most of them go straight, undefeated because most of the space in the atom is empty and nucleus has very small volume because of which diffraction occurs. Q.13 Discovery of the nucleus of an atom was due to the experiment carried out by Option 1 Bohr Option Mosley Option 3 Rutherford Option 4 Thomson Nucleus was discovered during the a-rays scattering experiment performed by Rutherford Q.14 Rutherford s scattering experiment is related to the size of the Option 1 Nucleus Option Atom Option 3 Electron
Option 4 Neutron Nucleus Q.15 8 The velocity of light is 3.0 ms. Which value is closest to the wavelength in 15-1 nanometers of a quantum of light with frequency of 8 s Option 1 7 3 Option -5 Option 3-18 5 Option 4 1 3.7 8 c=3 m/sec λ. υ =c 3 mlsec λ= 8 15 sec -1 3-7 λ = m 8 λ =0.37 λ =3.7 nm 8-7 9 Q.16 14 The frequency of a wave of light is 1 s. The wave number associated with this light is Option 1-7 5 m Option -8-1 4 cm Option 3-7 -1 m Option 4 4-1 4 cm λ. υ =c 8 3 1-6 λ = = 14 1 4 1 λ =wavenumber= =4 m λ 6 4 cm -1 0 4-1 4 cm Q.17 Rank the following types of radiations from the highest energy to the lowest. Ultraviolet/visible/X-ray/microwave/infrared Option 1 X-ray, ultraviolet, microwave, infrared, visible Option Ultraviolet, X-ray, visible, infrared, microwave Option 3 infrared, microwave, ultraviolet, visible, X-ray Option 4 X-ray, ultraviolet, visible, infrared, microwave Highest Energy lowest Energy X-ray, ultraviolet, visible, infrared, microwave 6-1
Q.18 The frequency of a green light is 6 14 Hz. Its wavelength is : Option 1 500 nm Option 5 mm Option 3 50,000 nm Option 4 None of these υ. λ =c 8 3 λ= 6 14 1-6 λ = m -6 9 λ =0.5 nm λ =500nm Q.19 Which wave property is directly proportional to energy of electromagnetic radiation: Option 1 Velocity Option Frequency Option 3 Wave number Option 4 All of these hc 1 E=h λ = =hc. =hc υ λ λ Q.0 The number of photons of light of ν =.5 m necessary to provide 1 J of energy are Option 1 18 Option 17 Option 3 0 Option 4 19 E=nhυ 1J=nhc.υ 1 J n=.5 m -5 J.m 1 n= 5-19 n=0. 19 n= 8 6-1 -5 hc= J.m Q.1 The number of photons emitted in hours by a 60 W sodium lamp Option 1 6.50 Option 6.40 Option 3 8.40 Option 4 3.40 ( λ of photon = 6000A )
Energy Power= time -5 hc= J.m E 60= 60 60sec E = 160000 Joules nhc E= λ 160000 6000-5 - =n=6.40 Q. Minimum number of photons of light of wavelength 4000A, which provide 1J energy: Option 1 18 Option 9 Option 3 0 Option 4 nhc E= λ -5 n 1J= 4000 - -7 4 n= -5 n= 18 Q.3 The energy Ecorresponding to intense yellow line of sodium of λ 589 nm is: Option 1. ev Option 43.37 ev Option 3 47.1 ev Option 4.11 kcal hc E= λ 140 cv.nm E= 589nm E=.ev Q.4 The relation between energy of a radiation and its frequency was given by: Option 1 De Broglie Option Einstein Option 3 Planck Option 4 Bohr Planck 4
Q.5 Which is not characteristics of Planck s quantum theory of radiation? Option 1 Radiation is associated with energy. Option Energy is not absorbed or emitted in whole number or multiples of quantum. Option 3 The magnitude of energy associated with a quantum is proportional to the frequency. Option 4 Radiation energy is neither emitted nor absorbed continuously but in small packets called quanta. Correct Answer Plank did not say that energy is absorbed or emitted in whole number or multiple of quantum Q.6 Which of the following is not a characteristics of Planck s quantum theory of radiation? Option 1 Energy is not absorbed or emitted in whole number or multiples of quantum. Option Radiation is associated with energy. Option 3 Radiation is associated with energy emitted or absorbed continuously but in the form of small packets called quanta. Option 4 The magnitude of energy associated with quantum is proportional to frequency. Energy is not absorbed or emitted in whole number or multiples of quantum. Q.7 Einstein s theory of photoelectric effect is based on: Option 1 Newtons corpuscular theory of light Option Huygen s wave theory of light Option 3 Maxwell s electromagnetic theory of light Option 4 Planck s quantum theory of light Planck s quantum theory of light Q.8 In photoelectric effect the number of photo-electrons emitted is proportional to : Option 1 Intensity of incident beam Option Frequency of incident beam Option 3 Velocity of incident beam Option 4 Work function of photo cathode According to photoelectric effect frequency decides whether there will be current or not but the amount of current (no. of photon ejected out) will be decided by the intensity of light Q.9 Increase in the frequency of the incident radiations increase the : Option 1 Rate of emission of photo-electrons Option Work function Option 3 Kinetic energy of photo-electrons Option 4 Threshold frequency h υ -h υ 0 = K.E So if we increase υ then K.E. will increases as υ 0 or hυ 0 -w 0 is constant for a given metal Q.30 Threshold wavelength depends upon: Option 1 Frequency of incident radiation
Option Velocity of electrons Option 3 Work function Option 4 None of the above hc Work function = w 0 = (Threshold wavelength) λ 0 it only depends on work function. Q.31 Photoelectric effect shows Option 1 Particle-like behavior of light Option Wave-like behavior of light Option 3 Both wave-like and particle behaviour of light Option 4 Neither wave-like nor particle-like behaviour of light Particle-like nature of light. (Theroy Part) Q.3 Ultraviolet light of 6. ev falls on aluminium surface (work function = 4. ev). The kinetic energy (in joule) of the fastest electron emitted is approximately : Option 1-1 3 Option -19 3 Option 3-17 3 Option 4-15 3 Correct Answer h υ -h υ 0 = K.E max 6. - 4. = ke k.e = ev Q.33 Option 1 Option Option 3 Option 4-19 = 1.6 Joules -19 =3 The threshold wavelength for photoelectric effect on sodium is 5000A. Its work function is : -19 4 J 1 J -19 J 3 - J hc w 0 = λ 0-5 J.m w 0 = 5000 - m = 5-5 7-18 =4-19 =4 Joules Q.34 The kinetic energy of the photoelectrons does not depend upon Option 1 Intensity of incident radiation
Option Frequency of incident radiation Option 3 Wavelength of incident radiation Option 4 Wave number of incident radiation Intensity of incident radiation Q.35 If the threshold frequency of a metal for photoelectric effect is ν 0, then which of the following will not happen? Option 1 If the frequency of the incident radiation is ν 0, the kinetic energy of the electrons ejected is zero. Option If the frequency of the incident radiation is ν, the kinetic energy of the electrons ejected will be h ν h ν 0 Option 3 If the frequency is kept same at ν but intensity is increased, the number of electrons ejected will increase. Option 4 If the frequency of incident radiation is further increased, the number of photoelectrons ejected will increase. h υ -h υ 0 = K.E a) if υ - υ 0,KE=0 b) If υ - υ thenke=h υ -h υ 0 c) no. of electrons ejected intensity of light but Independent of frequency Q.36 The line spectrum of two elements is not identical because Option 1 They do not have same number of neutrons Option They have dissimilar mass number Option 3 They have different energy level schemes Option 4 They have different number of valence electrons Elements can have same no. of neutrons Isotones Elements can have same mass no. Isobar Elements can have same no of valence e Elements in a group But Elements cannot have same Energy level and because of that line spectrum of two elements is different Q.37 The energy of electron in 3 rd orbit of hydrogen atom is Option 1-1311.8 kj mol -1 Option -8.0 kj mol -1 Option 3-145.7 kj mol -1 Option 4-37.9 kj mol -1 ( -131 z ) E= KJ/mol ( n) ( ) ( 3) -131 1 E = 3 E 3 = -145.7 kj/mol Q.38 The ionization energy of H atom is 13.6 ev. The ionization energy of Li + ion will be
Option 1 54.4 ev Option 40.8 ev Option 3 7. ev Option 4 1.4 ev E + = E ( Z ) IonisationLi ionisation H =13.6 ( 3) =1.4ev Q.39 The ratio of the difference in energy between the first and the second Bohr orbit to that between second and third Bohr orbit is Option 1 1 Option 1 3 Option 3 7 5 Option 4 4 9 1 1-13.6( z ) - E -E = 1 1 E-E3 1 1-13.6( z ) - 3 3 = 4 5 36 3 36 = 4 5 7 = 5 Q.39 Energy of electron of hydrogen atom in second Bohr orbit is Option 1-19 -5.44 J Option Option 3 Option 4-19 -5.44 kj -19-5.44 cal -19-5.44 ev z E=-13.6 n E=-13.6 ( 1) ( ) -13.6 E= =-3.4ev 4
-19 E=-3.4 1.6 =-5.44-19 Joules Q.41 The energy of second Bohr orbit in the hydrogen atom is -3.4 ev. The energy of fourth orbit of He + ion would be Option 1-3.4 ev Option -0.85 ev Option 3-13.64 ev Option 4 +3.4 ev -RHz E = n H H ( ) ( ) -R 1-3.4= R H = 13.6 z E =-RH He + n =-13.6 = -3.4ev ( ) ( 4) Q.4 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 to hydrogen is (are) Option 1-3.4 ev Option -4. ev Option 3-6.8 ev Option 4 +6.8 ev -13.6z E= n For Q.43 The ionization energy of hydrogen atom is 13.6 ev. The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is Option 1-18 1.69 J Option Option 3 Option 4-3 1.69 J 3 1.69 J 5 1.69 J First excited state means n =.
Ionisation Energy = -E st 1 state E i = -13.6ev ( ) ( ) -13.6 1 E=E -E = 1. 1.6-18 1.6 J -19 ( ) - -13.6 =.ev Q.44 In a Bohr s model of atom when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed (1erg = + J) Option 1-11.15 erg Option Option 3 Option 4-0.1911 erg -1.389 erg - 0.39 erg Correct Answer E=E3-E1 1 1 =-13.6z - 3 1 1-9 =-13.6 ( 1) 9 8 = 13.6ev 9-19 = 1.08 1.6 Joules -19 7 = 19.34 erg =19.34-1 = 0.1934 - erg Q.45 An electron in H-atom is moving with a kinetic energy of 5.45 J. What will be energy level for this electron? Option 1 1 Option Option 3 3 Option 4 None of these Correct Answer T.E. = -K.E -19 T.E.=-5.45 J -19 ( ) -5.45-13.6 1 = ev= -19 1.6 n 13.6 1.6 =n = 5.45 n =3.99 4 n = Q.46 The energy required to dislodge electron from excited isolated H-atom, IE 1 = 13.6 ev is Option 1 = 13.6 ev Option > 13.6 ev
Option 3 Option 4 < 13.6and > 3.4 ev 3.4 ev IE 1 = +13.6eV IE = +3.4ev 1 st excited state IE 3 = +1.51ev nd excited state 3.4eV Q.47 The radius of first Bohr s orbit for hydrogen is 0.53 A. The radius of third Bohr s orbit would be Option 1 0.79 A Option 1.59 A Option 3 3.18 A Option 4 4.77 A n 0 r=0.59 A z ( ) ( 1) 0.59 3 r = 3 =0.53 9 =4.77A Q.48 The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530A. The radius for the first excited state (n = ) orbit is Option 1 0.13 A Option 1.06 A Option 3 4.77 A Option 4.1 A First excited state n = ( ) 0.53 r= =.1A Z Q.49 According to Bohr model, angular momentum of an electron in the 3rd orbit is : Option 1 3h π Option 1.5h π Option 3 3π h
Option 4 9h π Correct Answer nh mvr= π For 3 rd orbit 3h 1.5h mvr= = π π (According to Bohr s Model) π Q.50 Electronic energy is a negative energy because Option 1 Electron carries negative charge Option Energy is zero near the nucleus and decrease as the distance form the nucleus increases. Option 3 Energy is zero at an infinite distance from the nucleus and decreases as the electron comes closer to the nucleus. Option 4 There are interelectronic repulsions. Energy is zero at n=, it decreases as we go close to the nucleus It is negative Q.51 Ratio of frequency of revolution of electron in the second excited state of He + and second state of hydrogen is Option 1 3 7 Option 7 3 Option 3 1 54 Option 4 7 velocity.18z z f= = Distance n 0.59 n z f 3 n ( ) rd + ( ) ( ) ( ) ( ) 3 f3 ofhe 3 = = nd 3 f ofh 3 1 7 Q.5 The line spectrum observed when electron when electron jumps from higher level to M level is known as Option 1 Balmer series Option Lyman series Option 3 Paschen series Option 4 Brackett series M level means n = 3 It comes under Paschen series
Q.53 How many spectral lines are produced in the spectrum of hydrogen atom from 5 th energy level? Option 1 5 Option Option 3 15 Option 4 4 Correct Answer Number of spectral lines n( n-1) = where, n = higher Energy level n = 5 5( 5-1) = 0 = = Q.54 What transition in He + ion shall have the same wave number as the first line in Balmar series of H atom? Option 1 7 5 Option 5 3 Option 3 6 4 Option 4 4 First line of Balmer series means 3 1 1 1 wave number = =Rz - λ 3 5 5R υ=rz = ( z=1) 36 36 a) 7 5 for He + 1 1 υ=rz - 5 49 X b) 5 3 For He + 1 1 υ=r( ) - 9 5 X c) 6 4 For He + 1 1 υ=r( ) - 16 36 =4R 5 36 16 5R = 36 Q.55 An electron jumps from 6 th energy level to 3 rd energy level in H-atom, how many lines belong to visible region? Option 1 1 Option Option 3 3 Option 4 zero
6 3 In visible region (lines which comes to n = ) No lines will come in visible region Q.56 The wavenumber for the shortest wavelength transition in the Balmer series of atomic hydrogen is Option 1 740 cm -1 Option 840 cm -1 Option 3 940 cm -1 Option 4 1186 cm -1 Shortest wavelength for balmer series 1 1 1 = υ =Rz - λ 9670 1 = 4 = 7417.5cm -1 Q.57 The difference in wavelength of second and third lines of Balmer series in the atomic spectrum is Option 1 131 A Option 54 A Option 3 34 A Option 4 6 A Correct Answer second line of balmer series 4 1 1 1 3R =Rz - = λ 4 16 16 1 16 = λ 3R 3 rd line of Balmer series 5 1 1 1 1R =Rz - = λ 4 5 0 1 0 = 3 1R 16 0 1 λ - λ 3 = - 3 1 R
1 =911.5A R 11-0 1 = 911.5 = 911.5 1 1 =50.8A Q.58 The third line in Balmer series corresponds to an electronic transition between which Bohr s orbits in hydrogen atom Option 1 5 3 Option 5 Option 3 4 3 Option 4 4 Correct Answer 3 rd line in Balmer series is5 1 1 1 =Rz - λ 4 5 1 1R = λ 0 Q.59 When electrons in N shell of excited hydrogen atom return to ground state, the number of possible lines spectrum is : Option 1 6 Option 4 Option 3 Option 4 3 N shell means n = 4 no. of spectral line 4( 4-1) = = 6 n( n-1) = Q.60 The wave number of the first line of Balmer series of H atom is 1500 cm -1. What is the wave number of the first line of Balmer series of Li + ion? Option 1 1500 cm -1 Option 6080 cm -1
Option 3 76000 cm -1 Option 4 136800 cm -1 First line of balmer 3 1 1 1 =Rz - λ 4 9 1 5R = =1500 λ 36 1500 R= 36 5 (z = 1 for hydrogen) (i) First line of Balmer for Li + 1 1 1 =RZ - λ 4 9 1 =R( 3) 5 λ 36 1500 36 5 = =9 (after putting volume of R from (i)) 5 36 = 136800cm -1 Q.61 In hydrogen spectrum, the series of lines appearing in ultra violet region of electromagnetic spectrum are called Option 1 Balmer lines Option Lyman lines Option 3 Pfund lines Option 4 Brackett lines Correct Answer Lyman lines (Theory) Q.6 The wave number of the first line of Balmer series of hydrogen is 1500 cm -1. The wave number of the first Balmer line of Li + ion is Option 1 1500 cm -1 Option 60800 cm -1 Option 3 76000 cm -1 Option 4 136800 cm -1 First line of Balmer for H 3 1 1 1 5R =R( 1) - = =1500 λ 4 9 36 First line of Balmer for Li + 3 1 1 1 5R =R( 3) - = 9=1500 9 λ 4 9 36 =136800cm -1 Q.63 A certain transition in H spectrum from an excited state to the ground state in one or more steps gives rise to a total of lines. How many of these belong to the UV spectrum? Option 1 3 Option 4 Option 3 6
Option 4 5 Correct Answer Number of spectral lines n( n-1) = n -n = n - n 0 = 0 n 5n + 4n - 0 = 0 (n- 5) (n + 4) = 0 n = 5 as n = -4 is not possible For uv spectrum line has to come form any level to n = 1. Q.64 The de-broglie wavelength associated with a material particle is Option 1 Directly proportional to its energy Option Directly proportional to momentum Option 3 Inversely proportional to its energy Option 4 Inversely proportional to momentum h h λ db = = mv p p momentum of particle λ db de-broglie wavelength Q.65 The de Broglie wavelength of a tennis ball of mass 66 g moving with the velocity of metres per second is approximately Option 1-35 metres Option -33 metres Option 3-31 metres Option 4-36 metres Correct Answer h λ = mv -34 6.6 = -3 66-34 = -33 = m Q.66 The wavelength of a cricket ball weighing 0 g and travelling with a velocity of 50 m/s is Option 1-8 1.3 m Option -37 1.3 m
Option 3 Option 4-34 1.3 m -30 1.3 m h λ = mv -34 6.6 = 50-3 0 6.6 = 5-34 -34 =1.3 m Q.67-3 An electron has kinetic energy.8 Jde-Broglie wavelength will be nearly -31 ( =9.1 kg) m e Option 1-4 9.4 m Option -7 9.4 m Option 3-8 9.4 m Option 4-9.4 m h h λ db = = mv mke = = 6.6-34 -31-3 9.1.8 6.6-34 50.96-34 6.6 = 50.9 =0.94-8 -7-54 -7 =9.4 m Q.68 A cricket ball of 0.5 kg is moving with a velocity of 0 ms -1. The wavelength associated with its motion is Option 1 1 cm 0 Option -34 66 m Option 3-35 1.3 m Option 4-8 6.6 m h λ db = mv -34 = 6.6 0.5 0 66 = 5-36 =13. -36
-35 =1.3 m Q.69 An electron with velocity v is found to have a certain value of de Broglie wavelength. The velocity that the neutron should possess to have the same de Broglie wavelength is Option 1 V Option v 1840 Option 3 1840 v Option 4 1840 v Correct Answer λ = λ electron h m v e e h = m v n n me v n = v m n neutron 1 m = m 1840 v v n = 1840 e n Q.70 If uncertainty in the position of an electron is zero, the uncertainty in its momentum would be Option 1 Zero Option h 4 π Option 3 h < 4π Option 4 Infinite h x. p 4π x = 0 h p 4 π Q.71 For an electron, if the uncertainty in velocity is ν, the uncertainty in its position ( x ) is given by : Option 1 h π m ν Option π hm ν Option 3 h 4π π m ν Option 4 πm h ν
h x. p 4π h x. mv 4 π P=mv h x. v 4 π m h x 4 π m v Q.7 A ball of mass 00 g is moving with a velocity of m sec -1. If the error in measurement of velocity is 0.1%, the uncertainty in its position is: Option 1 Option Option 3 Option 4-31 3.3 m -7 3.3 m -5 5.3 m -3.64 m v = 0.1% 0.1 = =0.01 0 h x 4 π m v -34 6.6 4 3.14 00-3 0.01-34 3 = 0.67-3 =.6 m Q.73 The Heisenbergs uncertainty principle states that. Option 1 No two electrons in the same atom can have the same set of four quantum numbers Option Two atoms of the same element must have the same number of protons Option 3 It is impossible to determine accurately both the position and momentum of an electron simultaneously Option 4 Electrons of atoms in their ground states enter energetically equivalent sets of orbitals singly before they pair up in any orbital of the set It is impossible to determine accurately both the position and momentum of an electron simultaneously By theory Q.74 If the uncertainty in the position of an electron is zero, the uncertainty in its momentum be Option 1 Zero Option h π Option 3 h 4π Option 4 Infinity
h x. p 4π x=0 p Q.75 If uncertainty in the measurement of position and momentum of an electron are equal then uncertainty in the measurement of its velocity is approximately: Option 1 1-1 8 m s Option Option 3 Option 4 1-1 6 m s 1-1 4 m s 1-1 m s h x p 4 π x= p h ( p) 4 π h ( m v) 4 π h v 4 π m v= 6.6-34 -31 ( ) 4 3.14 9.1 6.6 = 40 6.6 = 40-34 6 14 =0.079 14 =8 1 Q.76 In the Schrodinger s wave equation Ψ represents Option 1 Orbit Option Wave function Option 3 Wave Option 4 Radial probability Correct Answer Ψ represents wave function Q.77 For each value of l, the number of m s values are Option 1 l Option nl Option 3 l +1 Option 4 n -l
l +1 (Theory) Q.78 A subshell with n = 6, l = can accommodate a maximum of Option 1 electrons Option 1 electrons Option 3 36 electrons Option 4 7 electrons n = 6, l = means 6d will have 5 orbitals. max electrons can be aceomalid as each orbital can have maximum of electrons. Q.79 Which of the following sets of quantum number is correct for an electron in 4f orbital? Option 1 +1 n=3, l =, m=-, s= Option -1 n=4, l =4, m=-4, s= Option 3 +1 n=4, l =3, m=+1, s= Option 4 +1 n=4, l =3, m=+4, s= +1 n=4, l =3, m=+1, s= 4f orbital means n=4, l =3, m 1 =-3,-,-1,0,1,,3 1 ms = ± +1 n=4, l =3, m=+1, s= Q.80 For a d-electron, the orbital angular momentum is Option 1 h 6 π Option h π Option 3 h π Option 4 Zero orbital angular momentum = l ( l l+1 ) h π for d electron l =
( ) h = +1 π h = 6 π Q.81 The correct designation of an electron with n = 4, l = 3, m =, and s = 1/ is : Option 1 3d Option 4f Option 3 5p Option 4 6s Correct Answer n = 4, l = 3, m =, and s = 1/ means electron is present in 4f Q.8 A 3d-electron having s = +1/ can have a magnetic quantum no : Option 1 + Option +3 Option 3-3 Option 4 +4 For 3d n = 3, l = m can be -, -1, 0, 1, Q.83 The electrons identify by n and I (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l = (iv) n = 3, l = 1 Can be placed in order of increasing energy, from lowest to highest Option 1 (iv) < (ii) < (iii) < (i) Option (ii) < (iv) < (i) < (iii) Option 3 (i) < (iii) < (ii) < (iv) Option 4 (iii) < (i) < (iv) < (ii) n = 4 l = 1 4p n + l = 5 n = 4 l = 0 4s n + l = 45 n = 3 l = 3d n + l = 5 n = 3 l = 1 3p n + l = 4 n + l rule state the shell will less value of n + l will have lesser energy and in case if n + l is same for two orbitals we have to give preference to n, 3p < 4s < 3d < 4p Q.84 According to (n + l) rule after completing np level the electron enters to : Option 1 (n - l) d Option (n + l) s Option 3 Nd Option 4 (n + l) p Correct Answer Configuration order is 1s s p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p we can see that after p There is 3s after 3p there is 4s and so on after np level electron will enter (n + 1)S.
Q.85 The correct ground state electronic configuration of chromium atom (Z = 4) is Option 1 [Ar] 3d 5 4s 1 Option [Ar] 3d 4 4s Option 3 [Ar] 3d 6 4s 0 Option 4 [Ar] 4s 1 4p 5 z = 4 chromium is exceptional electronic configuration n[ar] 3d 5 4s 1 Q.86 In manganese atom, Mn (Z = 5), the total number of orbitals populated by one or more electrons (in ground state) is Option 1 15 Option 14 Option 3 1 Option 4 Q.87 The correct set of quantum numbers for the unpaired electron of chlorine atom is n l m Option 1 1 0 Option 1 1 Option 3 3 1 1 Option 4 3 0 0 Cl 1s s p 3s 3p Q.88 The maximum number of 4d-electrons having spin quantum number 1 s=+ are Option 1 Option 7 Option 3 1 Option 4 5 4d will have 5 orbitals
5 electrons will have rust 5 will have 1 - spin 1 + spin Q.89 Which of the following has maximum number of unpaired electrons? Option 1 Mg + Option Ti 3+ Option 3 V 3+ Option 4 Fe 3+ Mg + 1s s p 6 0 Ti +3 1s s p 6 3s 3p 6 3d 1 1 V +3 1s s p 6 3s 3p 6 3d Fe +3 1s s p 6 3s 3p 6 3d 5 5 Q.90 Azimuthal quantum number for the last electron in Na atom is Option 1 1 Option 0 Option 3 Option 4 3 Correct Answer Na 1s s p 6 3s 1 l = 0 for 3s Q.91 Presence of three unpaired electrons in phosphorus atom can be explained by Option 1 Pauli s rule Option Uncertainty principle Option 3 Aufbau s rule Option 4 Hund s rule P= 1s s p 6 3s 3p 3 Hund s rule (Theory) Q.9 Consider the following ion 1. Ni +. Co + 3. Cr + 4. Fe 3+ (Atomic number : Cr = 4, Fe = 6, Co = 7 and Ni = 8) The correct sequence of increasing number of unpaired electrons in these ions is Option 1 1,, 3, 4 Option 4,, 3, 1 Option 3 1, 3,, 4 Option 4 3, 4,, 1
1,, 3, 4 Q.93 The quantum numbers for the outermost electron of an element are given below 1 n=, l=0, m=0, ms =+ The atom is Option 1 hydrogen Option lithium Option 3 beryllium Option 4 boron Correct Answer 1 n = l = 0 m = 0 ms = + s H 1s 1 Li 1s s 1 I 3 1s s p 1 Q.95 Which electronic configuration does not follow the Pauli s exclusion principle? Option 1 1s, s p 4 Option 1s, s p 4, 3s Option 3 1s, p 4 Option 4 1s, s p 6, 3s Pauli exclusion principle says an orbital can have max. of electrons 3s 3 is not possible 1s, s p 4, 3s Q.96 Magnetic quantum number for the last electron in sodium is : Option 1 3 Option 1 Option 3 Option 4 zero Na 1s s p 6 3s 1 n = 3 l = 0 m = 0 Q.97 1 1 1 Nitrogen has the electronic configuration 1 s,s px py pz and not 1 0 1 s,s px py pz. It was proposed by :
Option 1 Aufbau principle Option Pauli s exclusion principle Option 3 Hund s rule Option 4 Uncertainty principle Hund s rule Q.98 Which of the following has maximum number of unpaired electron (atomic number of Fe 6) Option 1 Fe Option Fe(II) Option 3 Fe(III) Option 4 Fe(IV) Fe(III) Q.99 A compound of vanadium has magnetic moment of 1.73 B.M. The electron configuration of the vanadium ion in the compound is Option 1 [Ar] 4s 0 3d 1 Option [Ar] 4s 1 3d 0 Option 3 [Ar] 4s 3d 0 Option 4 [Ar] 4s 0 3d 3 Magnetic moment µ b = n( n+) 1.73= n( n+ ) 3 = n + n n + n - 3 = 0 n + 3n - n - 3 = 0 (n + 3) (n - 1) = 0 n = 1 one unpaired electron should be there V = [Ar]4s 3d 3 For one unpaired e will have to remove 4 electrons V +4 = [Ar]3d 1 Q.1 How many spherical nodes are present in 4s orbital in a hydrogen atom? Option 1 0 Option Option 3 3 Option 4 4 spherical nodes = n - l 1 For 4s = 4-0 - 1 = 3
Q. The number of nodes possible in radial wave function of 3d orbital is Option 1 1 Option Option 3 0 Option 4 3 Radial nodes = n - l - 1 For 3d = 3 - - 1 = 0 Q.3 The d-orbital with the orientation X and Y axes is called : Option 1 d z Option Option 3 d zy d yz Option 4 d x - y d x - y Q.4 Which d-orbital does not have four lobes? Option 1 d x - y Option d xy Option 3 d z Option 4 d xz d z Q.5 3p y orbital has nodal plane Option 1 XY Option YZ Option 3 ZX Option 4 Any
Q.6 The number of angular nodes in a 3s atomic orbital is Option 1 0 Option 1 Option 3 Option 4 3 Angular nodes = l For 3s = 0 Q.7 The number of radial nodes in a 3s atomic orbital is Option 1 0 Option 1 Option 3 Option 4 3 radial nodes = n-l-l For 3s = 3-0-1 = Q.8 The quantum number not obtained from Schrodinger equation is Option 1 n Option l Option 3 m Option 4 s s Theory (n,l,m can be derived from schrodinger equation)