Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
The maximum number of electrons that can have principal quantum number, n=3, and spin quantum number, m s =-1 / 2, is
Q. The maximum number of electrons that can have principal quantum number,
n
=
3
, and spin quantum number,
m
s
=
−
1/2
, is
1916
199
JEE Advanced
JEE Advanced 2011
Report Error
Answer:
9
Solution:
For principal quantum number
(
n
=
3
)
Number of orbitals
=
n
2
=
9
So, number of electrons with
m
s
=
−
2
1
will be 9