Question: 80: Match List-I with List-II :
| List – I (Quantum Number) | List – II (Orbital) | ||
|---|---|---|---|
| (a) | \mathrm{n}=2, \ell=1 | (i) | 2 s |
| (b) | \mathrm{n}=3, \ell=2 | (ii) | 3 s |
| (c) | \mathrm{n}=3, \ell=0 | (iii) | 2 p |
| (d) | \mathrm{n}=2, \ell=0 | (iv) | 3 d |
Choose the correct answer from the options given below :
(1) (a) – (iii), (b) – (iv), (c) – (i), (d) – (ii)
(2) (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
(3) (a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
(4) (a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
Answer: Option (4)
Explanation:
The azimuthal quantum number \ell determines the type of orbital.
For \ell=0, the orbital is an s-orbital; for \ell=1, it is a p-orbital;
and for \ell=2, it is a d-orbital.
For (a), \mathrm{n}=2, \ell=1 corresponds to a p-orbital in the second shell,
which is the 2p orbital. Hence, (a)-(iii).
For (b), \mathrm{n}=3, \ell=2 corresponds to a d-orbital in the third shell,
which is the 3d orbital. Thus, (b)-(iv).
For (c), \mathrm{n}=3, \ell=0 corresponds to an s-orbital in the third shell,
which is the 3s orbital. Therefore, (c)-(ii).
For (d), \mathrm{n}=2, \ell=0 corresponds to an s-orbital in the second shell,
which is the 2s orbital. Hence, (d)-(i).
Thus, the correct matching is (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i),
which corresponds to option (4).