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

To determine the correct statements for an element with atomic number 9, we begin by identifying this element and analyzing its electronic configuration.

1. Identify the element with atomic number 9:
   • The element with atomic number 9 is Fluorine (F).
2. Determine the electronic configuration of Fluorine:
   • The electronic configuration is \(1s^2 2s^2 2p^5\).
   • This means there are 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, and 5 electrons in the 2p orbitals.
3. Explanation for statement A:
   • There can be 5 electrons for which \( m_s = +\frac{1}{2} \) and 4 electrons for which \( m_s = -\frac{1}{2} \) in Fluorine.
   • This is because the s and p orbitals can each hold two electrons with spins \(\pm \frac{1}{2}\).
   • In the electronic configuration \(1s^2 2s^2 2p^5\), there are precisely 5 electrons with spin \(\frac{1}{2}\) and 4 with spin \(-\frac{1}{2}\).
   • Thus, statement A is correct.
4. Explanation for statement B:
   • The 2p orbital configuration of Fluorine is \(2p^5\), indicating that there are a total of 5 electrons in the p orbitals. The \(p_x\), \(p_y\), and \(p_z\) orbitals will accommodate these electrons.
   • Since the p orbitals fill up in such a way to minimize energy (with one electron in each orbital before pairing occurs), there will actually be 2 electrons in one of the \(p\) orbitals and 1 each in the other two.
   • So, statement B is incorrect.
5. Explanation for statement C:
   • The last electron in the electronic configuration \(1s^2 2s^2 2p^5\) indeed goes into the 2p orbital, where \(n = 2\) and \(l = 1\).
   • Thus, statement C is correct.
6. Explanation for statement D:
   • The term "angular nodes" refers to the nodal planes in an atomic orbital. The number of angular nodes is equal to the azimuthal quantum number \(l\).
   • For the given configuration, for each of the 1s, 2s, and 2p orbitals, they have the following nodes:
     • 1s orbital: 0 angular node (s-orbitals have 0 angular nodes).
     • 2s orbital: 0 angular node (s-orbitals have 0 angular nodes).
     • 2p orbitals: 1 angular node per orbital (p-orbitals have 1 angular node).
   • Thus, the sum of angular nodes for all orbitals is not 1; statement D is incorrect.

Based on the above analysis, the correct statements are A and C. Hence, the correct option is A and C Only.

Answer 2

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