The elctron identified by quantum numbers n and l
(i) `n=4,l=1` (ii) `n=4,l=0`
(iii) `n=3,l=0` (iv) `n=3,l=1`
can be placed in order of increasing energy

Options

A. `(iii) lt (iv) lt (i) lt (ii)`
B. `(iii) lt (iv) lt (ii) lt (i)`
C. `(i) lt (iii) lt (ii) lt (iv)`
D. `(iii) lt (i) lt (iv) lt (ii)`

Answer

To determine the order of increasing energy for the electrons identified by the given quantum numbers, we will follow these steps:

### Step 1: Identify the quantum numbers
The quantum numbers provided are:
1. \( n = 4, l = 1 \) (4p orbital)
2. \( n = 4, l = 0 \) (4s orbital)
3. \( n = 3, l = 0 \) (3s orbital)
4. \( n = 3, l = 1 \) (3p orbital)

### Step 2: Calculate the \( n + l \) values
Next, we calculate the \( n + l \) values for each orbital:
1. For \( n = 4, l = 1 \):
\[ n + l = 4 + 1 = 5 \]
2. For \( n = 4, l = 0 \):
\[ n + l = 4 + 0 = 4 \]
3. For \( n = 3, l = 0 \):
\[ n + l = 3 + 0 = 3 \]
4. For \( n = 3, l = 1 \):
\[ n + l = 3 + 1 = 4 \]

### Step 3: Compare \( n + l \) values
Now, we compare the \( n + l \) values:
- \( 3s \) has \( n + l = 3 \)
- \( 3p \) has \( n + l = 4 \)
- \( 4s \) has \( n + l = 4 \)
- \( 4p \) has \( n + l = 5 \)

### Step 4: Determine the order of increasing energy
The order of increasing energy is determined by the \( n + l \) values:
- The lowest \( n + l \) value is \( 3 \) (3s).
- The next is \( 4 \) (3p and 4s). Since both have the same \( n + l \) value, we compare their \( n \) values. The 3p orbital (n=3) has a lower energy than the 4s orbital (n=4).
- The highest \( n + l \) value is \( 5 \) (4p).

Thus, the increasing order of energy is:
1. \( 3s \)
2. \( 3p \)
3. \( 4s \)
4. \( 4p \)

### Final Order
The final order of increasing energy is:
\[ 3s < 3p < 4s < 4p \]