An electron present in which of the following orbitals has the minimum value for `(n + l + m + s)`? Consider the minimum possible value for m and s (where ever applicable) .

Options

A. `3p`
B. `5p`
C. `4d`
D. `5s`

Answer

To solve the problem, we need to calculate the value of \( n + l + m + s \) for each of the given orbitals and find the minimum value. Let's break it down step by step.

### Step 1: Identify the values of \( n \), \( l \), \( m \), and \( s \) for each orbital.

1. **For 3p orbital:**
- \( n = 3 \) (principal quantum number)
- \( l = 1 \) (for p orbitals, \( l = 1 \))
- \( m \) can take values from \( -l \) to \( +l \): \( m = -1, 0, +1 \). The minimum value is \( -1 \).
- \( s \) (spin quantum number) can be \( +\frac{1}{2} \) or \( -\frac{1}{2} \). We take the minimum value, which is \( -\frac{1}{2} \).

Now, calculate \( n + l + m + s \):
\[ n + l + m + s = 3 + 1 - 1 - \frac{1}{2} = 3 + 1 - 1 - 0.5 = 2.5 \]

2. **For 5p orbital:**
- \( n = 5 \)
- \( l = 1 \)
- \( m = -1 \) (minimum value)
- \( s = -\frac{1}{2} \)

Now, calculate \( n + l + m + s \):
\[ n + l + m + s = 5 + 1 - 1 - \frac{1}{2} = 5 + 1 - 1 - 0.5 = 4.5 \]

3. **For 4d orbital:**
- \( n = 4 \)
- \( l = 2 \) (for d orbitals, \( l = 2 \))
- \( m = -2 \) (minimum value)
- \( s = -\frac{1}{2} \)

Now, calculate \( n + l + m + s \):
\[ n + l + m + s = 4 + 2 - 2 - \frac{1}{2} = 4 + 2 - 2 - 0.5 = 3.5 \]

4. **For 5s orbital:**
- \( n = 5 \)
- \( l = 0 \) (for s orbitals, \( l = 0 \))
- \( m = 0 \) (minimum value)
- \( s = -\frac{1}{2} \)

Now, calculate \( n + l + m + s \):
\[ n + l + m + s = 5 + 0 + 0 - \frac{1}{2} = 5 + 0 + 0 - 0.5 = 4.5 \]

### Step 2: Compare the calculated values
- For **3p**: \( 2.5 \)
- For **5p**: \( 4.5 \)
- For **4d**: \( 3.5 \)
- For **5s**: \( 4.5 \)

### Conclusion
The minimum value of \( n + l + m + s \) is \( 2.5 \) for the **3p orbital**.

### Final Answer
The electron present in the **3p orbital** has the minimum value for \( n + l + m + s \).

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