Question

All the four quantum numbers of last electron of sodium atom are
1. n = 3, l = 0, m = 0, S = \(+\frac{1}{2}\) 
2. n = 3, l = 1, m = +1, s = \(+\frac{1}{2}\)
3. n = 3, l = 2, m = +1, s = \(+\frac{1}{2}\)
4. n = 2, l = 0, m = 0, s = \(+\frac{1}{2}\)

Answer 1

Correct Answer - Option 1 : n = 3, l = 0, m = 0, S = \(+\frac{1}{2}\) 

Concept:
There are 4 quantum numbers to consider : 

• The principal quantum number ( n ) defines the shell number. 
• The angular momentum quantum number ( l ) defines the size and energy of the orbital. 
• The magnetic quantum number ( m_l ) defines the information about the number of orbitals in a sub-shell and their orientation.
• The spin quantum number ( m_s ) defines the direction of spin of an electron within an orbital. 

Explanation:

Electronic configuration of sodium \({}_{11}Na\) : 1s² 2s² 2p⁶ 3s¹

• Last electron enters into 3s orbital. 
• n=3 that is in the third shell. 
• l= 0 that is s-orbital.
• m=0 orientation of s-orbital.
• s=\(+\frac{1}{2}\) spin quantum number. 

All the four quantum numbers of last electron of sodium atom are n = 3, l = 0, m = 0, S = \(+\frac{1}{2}\)