1. Rutherford’s experiment, which established the nuclear model of the atom, used a beam
of
(A)
particles, which impinged on a metal foil and got absorbed
(B)
rays, which impinged on a metal foil and ejected electrons
(C) helium atoms, which impinged on a metal foil and got scattered
(D) helium nucleus, which impinged on a metal foil and got scattered
Ans: (D) alpha particles are helium atoms stripped off their two electrons.
2.
Column A
i.
ii.
iii.
iv.
Column B
Isotopes
Isotones
Isobars
Isodiaphers
a. 40S, 40Cl, 40Ar, 40K
,
36
,
S, 37Cl, 38Ar, 39K
,
,
(A) i – a, ii – b, iii – c, iv – d
(B) i – d, ii – c, iii – b, iv – a
(C) i – d, ii – c, iii – a, iv – b
(D) i – d, ii – a, iii – c, iv – b
Ans: (C)
Isotopes: Same number of protons (p or atomic number) but different mass number ( ,
Isotones: Same number of neutrons(n) but different atomic number (Z) (36S, 37Cl, 38Ar, 39K)
IsobArs: Same mass number (A) (40S, 40Cl, 40Ar, 40K)
IsodiAPhers: Difference of [No. of neutrons – P (No. of protons)] is the same (
,
,
,
3. If the mass attributed to a neutron were halved and that attributed to the electron were
doubled, the atomic mass of
would
(A) remain approximately the same
(B) be approximately doubled
(C) be approximately halved
(D) be reduced approximately by 25%
Ans: (D)
Atomic mass is approximately the mass of the neutrons + protons (as mass of electrons is
negligible compared to them).
So, Suppose the mass of a proton = mass of a neutron = 1 gm
Then, the mass of Carbon is 12 gm (6 gm from 6 protons and 6 gm from 6 neutrons)
Now, mass of a neutron is halved, therefore, the contribution from neutrons = 3 gm
)
)
The new mass of Carbon = 6 (from protons) + 3 (from neutrons) = 9 gm
Hence, the mass has reduced by approximately 25% (from 12 gm to 9 gm).
4. Evaluate the following ratios for the energy of the electron in a particular orbit:
[Kinetic : Potential ] and [Total : Kinetic]
(A) [1 : -2] and [-1 : 1]
(B) [1 : 2] and [1 : 1]
(C) [1 : 1] and [1 : 2]
(D) [1 : -1] and [1 : -2]
Ans: (A)
Total energy of an electron = Kinetic Energy (K.E) + Potential Energy (P.E)
= m
For an electron (charge ‘e’) moving in any orbit of radius ‘r’ around a nucleus of charge ‘Ze’, the
electrostatic force of attraction causing the centripetal acceleration is given by the equation:
(
Multiplying both sides by
(
=>

)
, we get:
)
P.E =
5. If the I.E. of He+ is 54.4eV then
(A) I.E.of H is 13.6eV and that of Li+2 is 122.4eV
(B) I.E.of H is 13.6eV and that of Li+2 cannot be determined
(C) I.E.of H is 13.6eV and that of Li+2 is 27.2eV
(D) All of the above are wrong
Ans: (A)
I.E. is the energy required to take an electron from the 1st shell of He+ to shell, n = infinity
We know that the energy required to make electronic transitions is proportional to Z2 (Atomic
Number)
For I.E (the same electronic transition, 1st shell to shell n = infinity), I.E for H (Z = 1) will be (I.E
of He+/22) = 13.6 eV
For I.E of Li+2 (Z = 3), it will be (I.E of He+/22)*32 = 122.4 eV
6. It is possible for the radius of which of the following to be equal to the 1st shell of Hydrogen:
(A) He+
(B) Li+2
(C) Be+3
(D) Dueterium
Ans: (C)
Radius of the nth shell of a H-like species of Atomic No. Z is given by 0.53
According to the question, 0.53
= 0.53
Or, n =
n can only have positive integral values, therefore, only Be+3 (Z = 4) will satisfy the equation.
7. In a Hydrogen like atom, once the electrons move around the nucleus in circular orbits of
radius R & and once in orbit of radius 4R. The ratio of the time taken by it to complete one
revolution is :
(A) 1 : 4
Ans: (C)
(B) 4 : 1
(C)1 : 8
(D) 8 : 7
Time taken to complete one revolution is = (
(
)
(
)
(as r is proportional to
Velocity of an electron is proportional to (1/n)
Therefore, ratio of the time-taken woud be:
(
)=( )
= 1:8
8. The orbital angular momentum of an electron in 3p sub-shell is:
(A)
(B) 0
(C)
(D)
Ans: (A)
Orbital angular momentum of an electron = √
( )
9. The shortest wavelength (in angstrom) of the Lymann series for H-atom is:
(A) 100 ̇
(B) 1000 ̇
(C) 911 ̇
(D) 1213 ̇
Ans: (C)
Shortest wavelength means longest jump. For Lymann series longest jump is from
For Short-cut, take ( )
(
)
10. The minimum and maximum kinetic energy of an electron ejected from a metal surface
when radiation of 200 nm falls on it. Work function of the metal is 5 eV. (1eV=1.6 x 1019J)
(A) 0 and 1.21 eV
(B) 1.21 eV and 5 eV
(C) 0 and 5 eV
(D) 0 and 0
Ans: (A)
Short Cut:
( )
(
)
(where
Energy of the incoming radiation) here is, E = (1242/200) = 6.21 eV
Min. Kinetic Energy is zero (when the electron that absorbs the energy loses so much
energy that it is unable to overcome the work function also)
Max Kinetic Energy =
= 6.21- 5 = 1.21 eV
11. A compound contains vanadium (Z = 23) as
B.M. Value of n is:
(A) 3
(B) 1
and has a magnetic moment of 1.73
(C) 2
Ans: (D)
Magnetic moment = √
Bohr Magneton
(where
1.73 B.M. =
Now,
means that the number of unpaired electrons = 1)
(D) 4
So, we need to take 4 electrons out V till it configuration reaches V+4
when it will
have once unpaired electron (Remember that when we take electrons out, they are taken
out firstly from 4s then from 3d).
12. The two electrons present in an orbital are distinguished by
(A) principal quantum number
(C) magnetic quantum number
(B) azimuthal quantum number
(D) spin quantum number
Ans: D
13.
An
electron is moving with a kinetic
de Broglie wavelength for this electron?
–3
(A) 5.28 10 m
–10
(C) 2 10 m
energy
of
4.55
10
–25
J.
What
will
be
–7
(B) 7.28 10 m
–5
(D) 3 10 m
Ans: (B)
Modifying de-Broiglie’s equation, we get
Where K is the Kinetic energy.
rd
14. Number of waves made by an electron in one complete revolution in 3 Bohr orbit is
(A) 2
(C) 4
(B) 3
(D) 1
Ans: 3
( ) with de-Broiglie’s equation,
Combine the equations:
We get,
Which means, that the number of waves (LHS) is equal to the circumference of that given orbit and is
equal to the shell number.
15. The quantum numbers of four electrons (el to e4) are given below
n
l
m
s
el
3
0
0
+1/2
e3
3
2
2
-1/2
n
l
m
s
e2
4
0
0
1/2
e4
3
1
-1
½
the correct order of decreasing energy of these electrons is:
(A)
(B)
(C)
(D) none
Ans: C
Use n+l rule. When it is equal, then the one with a lower value of n has lower energy.
16. If the uncertainty in velocity & position is same, then the uncertainty in momentum will
be
(A) √
(B)
√
(C) √
(D)
√
Ans:
Given,
Hence,
Or,
Therefore, m
=√
17. The correct order of wavelength of hydrogen
, Deuterium
, Tritium (
)(moving with
same kinetic energy is
(A)
(B)
(C)
(D)
Ans: (A)
Modifying de-Broiglie’s equation, we get
Where K is the Kinetic energy.
K is the same for all, therefore, it boils down to their respective masses. Hydrogen has the least mass,
therefore its wavelength will be the highest.
18. The wavelength of the emitted radiation when an electron jumps from shell n = 4 to n = 2 will be the
same as for which transition of an electron in
(A) n = 2 to n = 1
(B) n = 4 to n = 2
(C) n = infinity to n=1
Ans: (A)
(
or, (
(
)
)
(
)
(
)
Comparing we get,
None of the options given here match this. This question remains cancelled.
)
(D) Not possible
19. ‘M’ g of an element gave ‘N’ g of oxide Equivalent weight of the element is
(a)
(b)
(c)
(d)
Ans: (D)
Applying the law of equivalence,
No. of equivalents of the metal = No. of equivalents of oxygen
Or, (wt/Eq. wt) of the metal = (wt/Eq. wt) of oxygen
i.e. ( )
wt of Oxygen
20. In the disproportionation reaction,
(A)
(B)
, the equivalent mass of
(C)
will be:
(D)
In a disproportionation reaction, two reactions are taking place in parallel. The n-factor is given by is
given by,
)
Therefore,
Therefore, Equivalent mass of
21. Hydrogen peroxide in its reaction with
respectively, is acting as a:
(A) reducing agent, reducing agent
(B) oxidising agent, reducing agent
(C) reducing agent, oxidising agent
(D) oxidizing agent, reducing agent
Ans: (C)
exists but
22.
does not exist because
(A) nitrogen has no vacant d – orbitals
(B)
is unstable
(C) nitrogen atom is much small
(D) nitrogen is highly inert
Ans: (A)
23. The state of hybridization of Xe in
(a)
(b)
is
(c)
(d)
Ans: (D)
XeF4 has 2 lone pairs and 4 bond pairs, therefore 6 orbitals will be involved, therefore,
24. The correct order of dipole moment is
(a)
(b)
(c)
(d)
Ans: (A)
25. . The hybridization of atomic orbitals of nitrogen in
(a)
respectively
(b)
(c)
respectively
(d)
respectively
Ans: (B)
26. Amongst
(a)
and
and
(b)
and
the compounds with greatest and least ionic character are
(c)
and
(d)
and
Ans: (B)
27. Which one of the following is the correct order of interactions?
(a) Covalent < hydrogen bonding < van der Waals < dipole- dipole
(b) van der Waals < hydrogen bonding < dipole-dipole < covalent
(c) van der Waals < dipole- dipole < hydrogen bonding < covalent
(d) dipole- dipole < van der Waals < hydrogen bonding< covalent
Ans: (C)
Covalent bonds are the strongest and van der Waals forces are the weakest.
The reason hydrogen bonds are stronger than dipole-dipole is because they involve large differences in
electronegativity (H being not very electronegative, bonded with extremely electronegative O, F or N),
thus creating a large electric dipole.
Just for reference, a hydrogen bond is about 1/10th of the strength of an ionic bond, but about 10 times
stronger than a dipole-dipole force.
28. Which of the following species is diamagnetic?
(a)
(b)
(c)
(d)
Ans: (D)
29. Incorrect order about bond angle is
(A)
(B)
(C)
(D)
Ans: (C)
Bond angle of OF2 is less than H2O, Bond angle of
30. Which
of the following is true?
(A) Bond order bond-length bond energy
(B) Bond order (1/bond-length
bond energy
(C) Bond order bond-length (1/bond energy)
(D) Bond order (1/bond-length
(1/bond energy)
Ans: (B)
Bond order indicates the number of bonds between two atoms in a molecule. Greater the number
of bonds, smaller will be the bond length and greater will be the bond energy