The value of magnetic quantum number of the last electron of " Na " is

Question

The value of magnetic quantum number of the last electron of " Na " is (A)  3  (B)  2  (C)  1  (D)  0

Tardigrade Admin · Accepted Answer

Electronic configuration of Na is 1s22s22p63s1 . Here the valence electron is in 3 s orbital. The l value of this orbital is zero So, the magnetic quantum number of this orbital is zero

Q. The value of magnetic quantum number of the last electron of " $N a$ " is

$3$

$2$

$1$

$0$

Electronic configuration of $N a$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ . Here the valence electron is in $3 s$ orbital.
The $l$ value of this orbital is zero So, the magnetic quantum number of this orbital is zero

Q. The value of magnetic quantum number of the last electron of " Na " is

3

2

1

0

Electronic configuration of Na is 1s22s22p63s1 . Here the valence electron is in 3s orbital. The l value of this orbital is zero So, the magnetic quantum number of this orbital is zero

Q. The value of magnetic quantum number of the last electron of " $N a$ " is

$3$

$2$

$1$

$0$

Electronic configuration of $N a$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ . Here the valence electron is in $3 s$ orbital.
The $l$ value of this orbital is zero So, the magnetic quantum number of this orbital is zero