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

For the element with atomic number 9, we have the following electron configuration: \[ 1s^2 2s^2 2p^5 \] The last electron will be in the \( 2p \) orbital.
Step 1: Quantum numbers for the last electron:
The electron configuration shows that the last electron is in the \( 2p \) orbital, so:
The principal quantum number \( n = 2 \),
The azimuthal quantum number \( l = 1 \) (since it is a p orbital),
The magnetic quantum number \( m_l \) can range from -1 to +1 (for a p orbital).
Step 2: Calculate the total number of nodes:
The total number of nodes \( N \) is given by: \[ N = n - 1 \] For \( n = 2 \), the total number of nodes is: \[ N = 2 - 1 = 1 \] Conclusion:
The total number of nodes for the last electron is 1, corresponding to option (1).