The electronic configuration for the following atoms are given below :
a. `1s^(2) 2s^(2) 2p^(5)` , b. `1s^(2) 2s^(2) 2p^(4)` , c. `1s^(2) 2s^(2) 2p^(6) 3s^(2)`
d. `1s^(2)2s^(2) 2p^(6)` , e. `1s^(2)2s^(2)2p^(6) 3s^(1)`
i. From the above configuration, arrange them in decreasing `IE`.
ii. Which of the electronic configuration given above wil have the lowest `IE` ?
iii. Which of the electronic configuration given above will be for noble gases ?

Explanation

i. Arrange the electronic configuration of all the atoms in decreasing order of atomic number, in such a manner that atoms containing the same outer energy shell are grouped together, e.g.
`underset((c ))(1s^(2)2s^(2)2p^(6)3s^(2))(Z=12) gt underset((e))(1s^(2) 2s^(2) 2p^(6)3s^(1))(Z = 11).... "M-shell" gt underset((d))(1s^(2) 2s^(2) 2p^(6))(Z=10) gt underset((a))(1s^(2) 2s^(2) 2p^(5)) (Z=9) gt underset((b))(1s^(2) 2s^(2) 2p^(4)) (Z=8)...L "shell"`
ii. Since the `M`-shell is far away from the nucleus than `L`-shell, so lesser amount of energy is required to remove an electron from `M`-shell than from `L`-shell.
Thus, the `IE` of atoms `(c)` and `(e)` should be lower than that of atoms `(a), (b)` and `(d)`.
Moreover, in case of atom `(c)` the electron is to be removed from the more stable completely filled `3s`-orbitals, whereas in case of atom `(e)` is not so.
Therefore, the `IE` of atom `(c)` should be higher than that of atom `(e) [i.e. (c) gt (e)]`.
iii. The nuclear charge on atoms `(d), (a)` and (b) is `+10, +9` and `+8` repectively. Since the `IE` increases with the increase in nuclear charge so the decreasing order of `IEs` are:
`(d) gt (a) gt (b)`
iv. Therefore, decreasing order of `IEs` of all the atoms is `(d) gt (a) gt (b) gt (c) gt (e)`
v. From the above discussion it is clear that atom `(e)` has the lowest `IE`.
vi. The electronic configuration of atom `(d)` represents the noble gas `1s^(2) 2s^(2) 2p^(6)` since the configuration of noble gases is `ns^(2)np^(6)`.