The maximum number of electrons that can have principle quantum number, n=3 and spin quantum number, ms=-(1/2), is

Question

The maximum number of electrons that can have principle quantum number, n=3 and spin quantum number, ms=-(1/2), is (A) 3 (B) 5 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

When n=3, l=0,1,2 i.e., there are 3 s, 3 p and 3 d-orbital's. If all these orbitals are completely occupied as <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/adeb004d720b742db16d856905b70341-.png" /> Total 18 electrons, 9 electrons with s=+(1/2) and 9 with s=-(1/2)

Q. The maximum number of electrons that can have principle quantum number, $n = 3$ and spin quantum number, $m_{s} = - \frac{1}{2}$ , is

3

5

7

9

When $n = 3, l = 0, 1, 2$ i.e., there are $3 s, 3 p$ and $3 d$ -orbital's. If all these orbitals are completely occupied as

Total 18 electrons, 9 electrons with $s = + \frac{1}{2}$ and 9 with
$s = - \frac{1}{2}$

Q. The maximum number of electrons that can have principle quantum number, n=3 and spin quantum number, ms​=−21​, is

3

5

7

9

When n=3,l=0,1,2 i.e., there are 3s,3p and 3d-orbital's. If all these orbitals are completely occupied as Total 18 electrons, 9 electrons with s=+21​ and 9 with s=−21​

Q. The maximum number of electrons that can have principle quantum number, $n = 3$ and spin quantum number, $m_{s} = - \frac{1}{2}$ , is

When $n = 3, l = 0, 1, 2$ i.e., there are $3 s, 3 p$ and $3 d$ -orbital's. If all these orbitals are completely occupied as

Total 18 electrons, 9 electrons with $s = + \frac{1}{2}$ and 9 with
$s = - \frac{1}{2}$