Q.
The electrons, identified by quantum numbers $n$ and $l$
1. $n=4, l=1$
2. $n=4, l=0$
3. $n=3, l=2$
4. $n=3, l=1$
Can be placed in the order of increasing energy, from the lowest to highest, as
Structure of Atom
Solution:
iv $<$ ii $<$ iii $<$ i
According to Aufbau's principle, filling of electrons in various subshells of an atom takes place in the increasing order of energy, starting with the lower most
The following order is observed:
$1 s \, 2 s \, 2 p\, 3 s\, 3 p \, 4 s\, 3 d\, 4 p \, 5 s \, 4 d \, 5 p\, 6 s \, 4 f \, 5 d \, 6 p \,7 s .$.
According to Bohr-Bury rule, $(n+l)$ rule, the subshell with the lower value of $(n+l)$ is filled first. If the values for $(n+l)$ are equal, the one with the smaller value of $n$ is filled first
n
l
(n+1)
i
4
1
5
ii
4
0
4
iii
3
2
5
iv
3
1
4
| n | l | (n+1) | |
|---|---|---|---|
| i | 4 | 1 | 5 |
| ii | 4 | 0 | 4 |
| iii | 3 | 2 | 5 |
| iv | 3 | 1 | 4 |