Match List I with List II :

| | List I (n) | List I (l) | | List II (Orbital) |
|---|---|---|---|---|
| A. | 2 | 1 | I. | 3d |
| B. | 4 | 0 | II. | 2p |
| C. | 5 | 3 | III. | 4s |
| D. | 3 | 2 | IV. | 5f |

Choose the correct answer from the options given below.

Question

Match List I with List II :

| | List I (n) | List I (l) | | List II (Orbital) |
|---|---|---|---|---|
| A. | 2 | 1 | I. | 3d |
| B. | 4 | 0 | II. | 2p |
| C. | 5 | 3 | III. | 4s |
| D. | 3 | 2 | IV. | 5f |

Choose the correct answer from the options given below.

Accepted Answer

C. 1. The **principal quantum number ($n$)** gives the shell number and appears as the first digit of the orbital name, while the **azimuthal quantum number ($l$)** decides the subshell letter using the code:
$$
l=0 	o s,\quad l=1 	o p,\quad l=2 	o d,\quad l=3 	o f
$$

2. For entry A, $n=2$ and $l=1$, so combining these values gives the orbital
$$
2p
$$
This matches II.

3. For entry B, $n=4$ and $l=0$, so the orbital is
$$
4s
$$
This matches III.

4. For entry C, $n=5$ and $l=3$, so the orbital is
$$
5f
$$
This matches IV.

5. For entry D, $n=3$ and $l=2$, so the orbital is
$$
3d
$$
This matches I.

Hence, the correct matching is A-II, B-III, C-IV, D-I, so the answer is C.

Match List I with List II : List I (n)List I (l)List II (Orbital)A.21I.3dB.40II.2pC.53III.4sD.32IV.5f Choose the correct answer from the options given below.

Match List I with List II :

List I (n) List I (l) List II (Orbital)
A. 2 1 I. 3d
B. 4 0 II. 2p
C. 5 3 III. 4s
D. 3 2 IV. 5f

Choose the correct answer from the options given below.