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Q. No.The quantum number of four electrons are given below :
| I. | n = 4, l=2 , ml = -2, ms = \(-\frac{1}{2}\) |
| II. | n=3, l=2, ml = 1, ms = \(+\frac{1}{2}\) |
| III. | n=4 , l=1, ml = 0, ms = \(+\frac{1}{2}\) |
| IV. | n=3, l=1, ml=1, ms=\(-\frac{1}{2}\) |
The correct order of their increasing energies is:
1. I < III < II < IV
2. IV < II < III < I
3. I < II < III < IV
4. IV < III < II < I
Subtopic: Shell & Subshell |
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Level 1: 80%+
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The electrons identified by quantum numbers n and l can be placed in order of increasing energy as:
a. n = 4, = 1
b. n = 4, = 0
c. n = 3, = 2
d. n = 3, = 1
| 1. | (d) < (b) < (c) < (a) | 2. | (b) < (d) < (a) < (c) |
| 3. | (a) < (c) < (b) < (d) | 4. | (c) < (d) < (b) < (a) |
Subtopic: Shell & Subshell |
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Level 1: 80%+
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Total number of orbitals that is/are considered as axial orbital(s) is:
\(p_x, p_y, p_z, d_{xy}, d_{yz}, d_{zx}, d_{x^2−y^2}, d_{z^2}~\)
1. 8
2. 3
3. 2
4. 5
\(p_x, p_y, p_z, d_{xy}, d_{yz}, d_{zx}, d_{x^2−y^2}, d_{z^2}~\)
1. 8
2. 3
3. 2
4. 5
Subtopic: Shell & Subshell |
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Level 2: 60%+
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Consider the following combination of n, l, and m values.
(i) n=3; l=0; m=0
(ii) n=4; l=0; m=0
(iii) n=3; l=1; m=0
(iv) n=3; l=2; m=0
The correct order of energy of the corresponding orbitals for multielectron species:
1. (ii) > (i)> (iv)> (iii)
2. (iv) > (ii) > (iii) > (i)
3. (i) > (iii) > (iv) > (ii)
4. (iv) > (iii) > (i) > (ii)
(i) n=3; l=0; m=0
(ii) n=4; l=0; m=0
(iii) n=3; l=1; m=0
(iv) n=3; l=2; m=0
The correct order of energy of the corresponding orbitals for multielectron species:
1. (ii) > (i)> (iv)> (iii)
2. (iv) > (ii) > (iii) > (i)
3. (i) > (iii) > (iv) > (ii)
4. (iv) > (iii) > (i) > (ii)
Subtopic: Shell & Subshell |
87%
Level 1: 80%+
JEE
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