Hi Buddy,
Surf here Structure of Atom Class 11 Question Bank and download the pdf also,
Happy Preparation..
Q-1. Packet of energy is called
(a) Electron (b) Photon (c) Position (d) Proton
Q-2. Orbital which is not possible
(a) 2p (b) 3d (c) 3s (d) 3f
Q-3. the magnetic quantum number of an atom is related to the
(a) size of the orbital (b) spin angular momentum (c) orbital angular momentum (d) orientation of the orbital in space
Q-4. The principal quantum number of an atom is related to the
(a) size of the orbital (b) spin angular momentum (c) orbital angular momentum (d) orientation of the orbital in Spence
Q-5. The designation of an orbital with in = 4 and 1 = 3
(a) 4s (b) 4p (c) 4d (d) 4f
Q-6. What transition in the hydrogen spectrum would have the same wavelength
as the Balmer transition n = 4 to n = 2 in the He+ spectrum?
(a) n = 4 to n =1 (b) n = 3 to n = 2
(c) n = 3 to n = 1 (d) n = 2 to n =1
Q-7. The wave number of first line of Balmer series of hydrogen in 15200 cm-1.
The wave number of the first Balmer line of Li2+ ion is
(a) 15200 cm-1 (b) 60800 cm-1 (c) 76000 cm-1 (d) 136,800 cm-1
Q-8. An electron is moving in Bohr’s orbit. Its de Broglie wavelength is λ. What
is the circumference of the forth orbit?
(a) 2/λ (b) 2λ (c) 3λ (d) 3/λ
Q-Which of the following statements in relation to the hydrogen atom is correct?
(a) 3s-orbital is lower in energy than 3p-orbital
(b) 3p-orbital is lower in energy than 3-d-orbital
(c) 3s and 3p orbitals all have the same energy.
(d) 3s, 3p and 3d orbitals all have the same energy.
Q-10. For principle quantum number, n = 4, the total number of orbitals having
1 = 3 is
(a) 3 (b) 7 (c) 5 (d) 9
Q-11. The number of d-electrons retained in Fe2+ (At. no. of Fe = 26) ion is
(a) 3 (b) 4 (c) 5 (d) 6
Q-12. Pauli exclusion principle helps to calculate the maximum number of electrons
that can be accommodated in any
(a) orbital (b) subsell (c) shell (d) All of these
Q-13. An electron in a H-like atom is in an excited state. It has a total energy of –3.4 ev,
calculate the de-Broglie’s wavelength?
(a) 66.5Å (b) 6.66Å (c) 60.6Å (d) 6.06Å
Q-14. When the frequency of light incident on a metallic plate is doubled, the KE of the emitted
photoelectron will be;
(a) doubled
(b) Halved
(c) Increased but more than doubled of the previous KE
(d) Remains unchanged
Q-15. The ratio of the difference in energy of electron between the first and second Bohr’s orbit
to that between second and third Bohr’s orbit is;
(a) 13 (b) 275 (c) 94 (d) 49
Q-16. The approximate quantum no. of a circular orbit of diameter 20.6 nm of the hydrogen
atom according to Bohr’s theory is
(a) 10 (b) 14 (c) 12 (d) 16
Click Here To download the complete/ chapter wise e-book of NCERT Chemistry
Ans. 1. (b), 2. (d), 3. (d), 4. (a), 5. (d), 6. (d), 7. (d), 8. (c), 9. (d), 10. (b), 11. (d), 12. (a)13. (b) 14. (c) 15. (b) 16. (b)
Structure of Atom Class 11 Question Bank: Assertion & Reasoning Questions
Directions: (Questions 1 to 4)
A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion.
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion.
C. If Assertion is true statement but Reason is false.
D. If both Assertion and Reason are false statements.
1. Assertion : Number of orbitals in 3rd shell is 9.
Reason : Number of orbitals for a particular value of n = n2.
2. Assertion : Two nodal planes are present in 3dxy.
Reason : Number of nodal planes = l
3. Assertion : The energy of an electron is largely determined by its principal
quantum number.
Reason : The principal quantum number is a measure of the most probable
distance of finding the electrons around the nucleus.
4. Assertion : An orbital cannot have more than two electrons, moreover, if
an orbital has two electrons they must have opposite spins.
Reason : No two electrons in an atom can have same set of all the four
quantum numbers.
Ans. 1. A 2. A 3. A 4.
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