Question

Consider the last electron of element having atomic number 9 and choose the correct option.

• A. Sum total nodes = 1
• B. \( n = 2; l = 0 \)
• C. Last electron enters 2s subshell
• D. There are 5 \( e^- \) with \( l = 0 \)

Hint:

The total number of nodes in an orbital is equal to \( n - 1 \), where \( n \) is the principal quantum number. For \( 2p \), \( n = 2 \), so the total nodes are 1.

Answer 1

The Correct Option is A

The electron configuration for the element with atomic number 9 is \(1s^2 2s^2 2p^5\). The final electron occupies the \(2p\) orbital.
Step 1: Quantum numbers for the last electron:
Based on the electron configuration, the last electron is in the \(2p\) orbital. Therefore:

• The principal quantum number is \( n = 2 \).
• The azimuthal quantum number is \( l = 1 \) (for a p orbital).
• The magnetic quantum number \( m_l \) can be -1, 0, or +1 (for a p orbital).

Step 2: Calculate the total number of nodes:
The total number of nodes \( N \) is determined by the formula:\[N = n - 1\]For \( n = 2 \), the total number of nodes is:\[N = 2 - 1 = 1\]Conclusion:
There is 1 total node for the last electron, which corresponds to option (1).