The electronic configuration for the following neutral atoms are given for use in questions
(a)` 1s^(2) , 2s^(2), 2p^(6) , 3s^(2) , ( b) 1s^(2) , 2s^(2) , 2p^(6) ,3s^(1) , ( c) 1s^(2) ,2s^(2), 2p^(4) , ( d) 1s^(2) , 2s^(2) 2p^(5) , ( e) 1s^(2) , 2s^(2) ,2p^(6)`
(i) List the above configuration in order of increasing ionization enthalpy.
(ii) Which of the electronic configuration given above would you expect to have the lowest ionization enthalpy.
(iii) Which of the electronic configuration given above would you expect for the noble gas ?

Explanation

(i) Arrange the electronic configuration of all the atoms in order of increasing atomic number in such a way that atoms containing the same outer energy shell are grouped together. Thus we have
`underset(c) (1s^(2) 2s^(2) 2p^(4))" " underset(d)(1s^(2) 2s^(2) 2p^(5)) " " underset(e)(1s^(2) 2s^(2) 2p^(6))`...... L- shell
`underset(b)(1s^(2) 2s^(2) 2p^(6) 3s^(1))" "underset(a)(1s^(2)2s^(2) 2p^(6) 3s^(2))" "`......M- shell
Since the M-shell is more distant from the nucleus than L - shell amount of energy is required to remove an electron from M- shell from the L-shell. In other words ionization enthalpy (IE) of atom (a) and (b) should be lower than that of atoms (c) , (d) and (e),
Further in case of atom (a) the electron is to be removed from the more stable completely filled 3s- orbital whereas in case of atom (b) it is not so. Therefore the IE of atom (b) should be lower than the of atom (a)
The nuclear charge on atoms (c), (d) and (e) is + 8, +9 and + 10 respectively . Since, the IE increase with increase in nuclear charge so the IEs of atoms (c), (d) and (e) follow the sequence `: c lt d lt e`
From the above discussion we conclude that the ionization enthalpies of the five atoms increase in the following order :
`b lt a lt c lt d lt e, i.e., 1s^(2) 2s^(2) 2p^(6) 3s^(1) lt 1s^(2) 2s^(2) 2p^(6) 3s^(2) lt 1s^(2) 2s^(2) 2p^(4) lt 1s^(2) 2s^(2) 2p^(5) lt 1s^(2) 2s^(2) 2p^(6)`
(ii) Evidently atom (b) with electronic configuration `1s^(2) 2s^(2) 2p^(6) 3s^(1)` has the lowest ionization enthalpy.
(iii) Since the outer electronic configuration of noble gases is `ns^(2) np^(6)` therefore the electronic configuration, `1s^(2) 2s^(2) 2p^(6)` [atom (e)] represents an noble gas.