The electrons identified by quantum number n and l are
(I) n = 1, l = 2 (II) n = 4, l = 0
(III) n = 3, l = 2 (IV) n = 3, l=1
can be placed in order of increasing energy as

Options

A. `IV lt II lt III lt I`
B. `II lt IV lt I lt III`
C. `I lt III lt II lt IV`
D. `III lt I lt IV lt II`

Answer

To solve the problem of ordering the electrons identified by quantum numbers \( n \) and \( l \) in terms of increasing energy, we will use the \( n + l \) rule. According to this rule, the energy of an electron in an atom is determined by the sum of its principal quantum number \( n \) and its azimuthal quantum number \( l \). The higher the \( n + l \) value, the higher the energy of the electron.

### Step-by-Step Solution:

1. **Identify \( n \) and \( l \) values:**
- (I) \( n = 1, l = 2 \)
- (II) \( n = 4, l = 0 \)
- (III) \( n = 3, l = 2 \)
- (IV) \( n = 3, l = 1 \)

2. **Calculate \( n + l \) for each case:**
- (I) \( n + l = 1 + 2 = 3 \)
- (II) \( n + l = 4 + 0 = 4 \)
- (III) \( n + l = 3 + 2 = 5 \)
- (IV) \( n + l = 3 + 1 = 4 \)

3. **List the \( n + l \) values:**
- (I) \( n + l = 3 \)
- (II) \( n + l = 4 \)
- (III) \( n + l = 5 \)
- (IV) \( n + l = 4 \)

4. **Order the \( n + l \) values:**
- The smallest \( n + l \) is from (I) which is 3.
- The next is (II) and (IV) which both have \( n + l = 4 \). Since \( n \) is larger in (II) than in (IV), (II) will have higher energy than (IV).
- The largest \( n + l \) is from (III) which is 5.

5. **Final order of increasing energy:**
- (I) \( n = 1, l = 2 \) (Energy 3)
- (IV) \( n = 3, l = 1 \) (Energy 4)
- (II) \( n = 4, l = 0 \) (Energy 4)
- (III) \( n = 3, l = 2 \) (Energy 5)

Thus, the order of increasing energy is:
1. (I) \( n = 1, l = 2 \)
2. (IV) \( n = 3, l = 1 \)
3. (II) \( n = 4, l = 0 \)
4. (III) \( n = 3, l = 2 \)