Question

Identify a correct set of quantum numbers that represents the highest energy of an electron in an orbital from following.

• A. \(n=3,l=0,m_{l}=0,s=+1/2\)
• B. \(n=3,l=1,m_{l}=+1,s=+1/2\)
• C. \(n=3,l=2,m_{l}=+1,s=+1/2\)
• D. \(n=4,l=0,m_{l}=0,s=+1/2\)

Hint:

Compare the n + l values of the four sets; the largest n + l is the highest in energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In a multi-electron atom, orbital energy increases with \(n+l\). If \(n+l\) is equal, the orbital with higher \(n\) has higher energy.

Step 2: Calculation
A: 3s, \(n+l=3\).
B: 3p, \(n+l=4\).
C: 3d, \(n+l=5\).
D: 4s, \(n+l=4\).

Step 3: Conclusion
3d has the greatest \(n+l\) value, so it has the highest energy. This is option C.

Final Answer:
The set with \(n=3, l=2\) (3d orbital) has the highest energy. \[ \boxed{\text{(C) }n=3,\ l=2,\ m_{l}=+1,\ s=+1/2} \]