JEE Main Chemistry Atomic Structure Online Test
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JEE Main Chemistry Atomic Structure Online Test. JEE Main Online Test for Chemistry Atomic Structure. JEE Main Full Online Quiz for Chemistry Atomic Structure. JEE Main Free Mock Test Paper 2019. JEE Main 2019 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Chemistry Atomic Structure. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n Take JEE Main Chemistry Atomic Structure…
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Question 1 of 20
1. Question
A continuum in spectrum is observed due to the following.
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Question 2 of 20
2. Question
The correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving with same kinetic energy is
Correct
Incorrect
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Question 3 of 20
3. Question
Two atoms are said to be isobars if
Correct
Isobars have the same mass number (i.e., sum of protons and neutrons) but different atomic number (i.e, number of protons).
Incorrect
Isobars have the same mass number (i.e., sum of protons and neutrons) but different atomic number (i.e, number of protons).
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Question 4 of 20
4. Question
Selected the correct alternate.
Correct
Incorrect
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Question 5 of 20
5. Question
The following transition occurs when Lithium atoms are sprayed into hot flame. The various steps are numbered for identification.
Which of these transition result in emission of light?
Correct
Energy of 3d is more than 3p therefore it will emit energy, similarly from 4s → 3p transition.
Incorrect
Energy of 3d is more than 3p therefore it will emit energy, similarly from 4s → 3p transition.
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Question 6 of 20
6. Question
Which has [Ar] 3d54s1 configuration each having 24 e?
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Question 7 of 20
7. Question
The quantum numbers of six electrons are given below. Arrange them in the order of increasing energies.
I. n = 4, l = 2, ml = – 2, ms = –1/2
II. n = 3, l = 2, ml = 1, ms = +1/2
III. n = 4, l = 1, ml = 0, ms = +1/2
IV. n = 3, l = 2, ml = – 2, ms = –1/2
V. n = 3, l = 1, ml = – 1, ms = +1/2
VI. n = 4, l = 1, ml = 0, ms = +1/2
Correct
Subshell notation n + l
1. n = 4, l = 2, ml = – 2, ms = – 1/2 4d 4 + 2 = 6
2. n = 3, l = 2, ml = 1, ms = + 1/2 3d 3 + 2 = 5
3. n = 4, l = 1, ml = 0, ms = + 1/2 4p 4 + 1 = 5
4. n = 3, l = 2, ml = – 2, ms = – 1/2 3d 3 + 2 = 5
5. n = 3, l = 1, ml = – 1, ms = + 1/2 3p 3 + 1 = 4
6. n = 4, l = 1, ml = 0, ms = + 1/2 4p 4 + 1 = 5
5
3p
(Arrangement of orbitals in order of their increasing energies)
Incorrect
Subshell notation n + l
1. n = 4, l = 2, ml = – 2, ms = – 1/2 4d 4 + 2 = 6
2. n = 3, l = 2, ml = 1, ms = + 1/2 3d 3 + 2 = 5
3. n = 4, l = 1, ml = 0, ms = + 1/2 4p 4 + 1 = 5
4. n = 3, l = 2, ml = – 2, ms = – 1/2 3d 3 + 2 = 5
5. n = 3, l = 1, ml = – 1, ms = + 1/2 3p 3 + 1 = 4
6. n = 4, l = 1, ml = 0, ms = + 1/2 4p 4 + 1 = 5
5
3p
(Arrangement of orbitals in order of their increasing energies)
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Question 8 of 20
8. Question
if each orbital can hold a maximum of 3 electrons, the number of elements in 4th period of Periodic Table (long form) is
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Question 9 of 20
9. Question
The ionization energy of gaseous Na atoms is 495.80 kJ mol–1. The lowest possible frequency of light that can ionize a Na atom is
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Incorrect
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Question 10 of 20
10. Question
A mixture contains atoms of fluorine and chlorine. The removal of an electron from each atom of sample absorbs 284 kJ while the addition of an electron to each atom of mixture release 68.89 kJ. Determine the percentage composition of the F in the mixture, given that the ionization energies of F and Cl are 27.91 × 10–22 and 20.77 × 10–22 kJ, respectively, and electron affinities are 5.53 × 10–22 and 5.78 × 10–22 kJ respectively.
Correct
Let atoms of chlorine and fluorine in the mixture be x and y respectively
∴ Energy absorbed 284 kJ
= (x) × 20.77 × 10–22 + (y) × 27.91 × 10–22 … (i)
Energy released = 68.8 kJ = (x) × 5.78 × 10–22 + (y) × 5.53 × 10–22 …(ii)
On solving equation (i) and (ii) we can get the value of x and y respectively.
% of Cl = 62.19
% of F = 37.81
Incorrect
Let atoms of chlorine and fluorine in the mixture be x and y respectively
∴ Energy absorbed 284 kJ
= (x) × 20.77 × 10–22 + (y) × 27.91 × 10–22 … (i)
Energy released = 68.8 kJ = (x) × 5.78 × 10–22 + (y) × 5.53 × 10–22 …(ii)
On solving equation (i) and (ii) we can get the value of x and y respectively.
% of Cl = 62.19
% of F = 37.81
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Question 11 of 20
11. Question
Select the correct matching for 3d orbital.
Correct
Angular nodes = I
Radial nodes = (n – I – 1)
Incorrect
Angular nodes = I
Radial nodes = (n – I – 1)
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Question 12 of 20
12. Question
The uncertainty in position and velocity of the particle are 0.2 nm and 10.54 × 10–27 ms–1 respectively then the mass of the particle is: (h = 6.625 × 10–34 Js)
Correct
∆x = 0.2 × 10–9m, ∆V = 10.54 × 10–27 ms–1.
∴ ∆x × m∆V = h
∴ 0.2 × 10–9 × m × 10.54 × 10–27 = 0.527 × 10–34
∴ m = 24.86 gm.
Incorrect
∆x = 0.2 × 10–9m, ∆V = 10.54 × 10–27 ms–1.
∴ ∆x × m∆V = h
∴ 0.2 × 10–9 × m × 10.54 × 10–27 = 0.527 × 10–34
∴ m = 24.86 gm.
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Question 13 of 20
13. Question
Which of the following conclusions could not be derive from Rutherford’s a–particle scattering experiment?
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Incorrect
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Question 14 of 20
14. Question
If n = 6, the correct sequence of filling of electrons will be
Correct
6s, 4f, 5d, 6p. It is the correct sequence of filling of electrons.
Incorrect
6s, 4f, 5d, 6p. It is the correct sequence of filling of electrons.
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Question 15 of 20
15. Question
If the nitrogen atom had electronic configuration 1s7, it would have energy lower that of normal ground state configuration 1s2 2s2 2p3, because the electrons would be closer to the nucleus. Yet 1s7 is not observed because it violates:
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Question 16 of 20
16. Question
Given this set of quantum numbers for a multi–electron atom 2, 0, 0, 1/2 and 2, 0, 0,–1/2 . What is the next higher allowed set of n and l quantum numbers for this atom in its ground state?
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Question 17 of 20
17. Question
In the following electronic configuration, some rules have been violated
I : Hund II : Pauli’s exclusion III : Aufbau
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Question 18 of 20
18. Question
The atomic weight of an element is double of its atomic number. If there are four electrons in 2p orbital, the element is isotonic with
Correct
Since, there are four electrons in 2p orbital, the outer configuration of the element is 2s2, 2p4. The complete configuration of the element is 1s2, 2s2, 2p4.
The element contains total 8 electrons, so its atomic number is 8.
Then, the number of neutrons = 16 – 8 = 8
It is isotonic with 7N15 (as it also contains 8 neutrons).
Incorrect
Since, there are four electrons in 2p orbital, the outer configuration of the element is 2s2, 2p4. The complete configuration of the element is 1s2, 2s2, 2p4.
The element contains total 8 electrons, so its atomic number is 8.
Then, the number of neutrons = 16 – 8 = 8
It is isotonic with 7N15 (as it also contains 8 neutrons).
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Question 19 of 20
19. Question
S-orbital is spherically symmetrical hence
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Question 20 of 20
20. Question
The sub shell that arises after f sub shell is called g sub shell. What is the total number of orbitals in the shell in which the g sub shell first occur?
Correct
For f sub shell l = 3; ∴ g sub shell l = 4
For principal shell, l = 4, n = 5
Total no. of orbital in shell = n2 = 52 = 25
Incorrect
For f sub shell l = 3; ∴ g sub shell l = 4
For principal shell, l = 4, n = 5
Total no. of orbital in shell = n2 = 52 = 25
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