Question

Correct statements for an element with atomic number 9 are 
A. There can be 5 electrons for which $ m_s = +\frac{1}{2} $ and 4 electrons for which $ m_s = -\frac{1}{2} $ 
B. There is only one electron in $ p_z $ orbital. 
C. The last electron goes to orbital with $ n = 2 $ and $ l = 1 $. 
D. The sum of angular nodes of all the atomic orbitals is 1. 
Choose the correct answer from the options given below:

• A. A and B Only
• B. A, C and D Only
• C. C and D Only
• D. A and C Only

Hint:

- For p-block elements: - \( n \) = principal quantum number - \( l \) = 1 for p-orbitals - Angular nodes = \( l \) - \( m_s \) distribution follows Hund's rule - Fluorine has 1 unpaired electron in \( 2p \) subshell

Answer 1

The Correct Option is D

Element Analysis:
- Atomic number 9 is Fluorine (F)
- Electronic configuration: 1s² 2s² 2p⁵
- Orbital diagram: 1s² ↑↓ 2s² ↑↓ 2p⁵ ↑↓ ↑↓ ↑

Statement Evaluation:
- A: TRUE - Total electrons = 9 - 5 with m_s = +1/2 (all unpaired + one paired)
- 4 with m_s = -1/2 (remaining paired electrons)
- B: FALSE - p_z orbital contains 1 electron (↑), but same applies to p_x and p_y
- Not a unique characteristic
- C: TRUE - Last electron enters 2p orbital (n=2, l=1)
- D: FALSE - Angular nodes = l
- Sum: 1s (0) + 2s (0) + 2p (1 each × 3) = 3 ≠ 1