Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.
| Column I (Electronic configuration) |
Column II (Electron gain enthalpy/kJ mol$$^{-1}$$ |
|
|---|---|---|
| A. | $$ 1 s^2 2 s^2 2 p^6 $$ |
$$-$$53 |
| B. | $$ 1 s^2 2 s^2 2 p^6 3 s^1 $$ |
$$-$$328 |
| C. | $$ 1 s^2 2 s^2 2 p^5 $$ |
$$-$$141 |
| D. | $$ 1 s^2 2 s^2 2 p^4 $$ |
$$+$$48 |
A. $\rightarrow$ (4)
B. $\rightarrow$ (1)
C. $\rightarrow(2)$
D. $\rightarrow(3)$
A. This electronic configuration corresponds to the noble gas i.e., neon. Since, noble gases have $+\Delta_{\text {eg }} H$ values, therefore, electronic configuration (A) corresponds to the $\Delta_{\text {eg }} H=+48 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
B. This electronic configuration corresponds to the alkali metal i.e., potassium. Alkali metals have small negative $\Delta_{\mathrm{eg}} H$ values, hence, electronic configuration (B) corresponds to $\Delta_{\mathrm{eg}} H=-53 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
C. This electronic configuration corresponds to the halogen i.e., fluorine. Since, halogens have high negative $\Delta_{\text {eg }} H$ values, therefore, electronic configuration ( $C$ ) corresponds to $\Delta_{\mathrm{eg}} H=328 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
D. This electronic configuration corresponds to the chalcogen i.e., oxygen. Since, chalcogens have $\Delta_{\text {eg }} H$ values less negative than those of halogens, therefore, electronic configuration (D) corresponds to $\Delta_{\mathrm{eg}} H=-141 \mathrm{kJmol}^{-1}$.
Assertion (A) Generally, ionisation enthalpy increases from left to right in a period.
Reason (R) When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
Assertion (A) Boron has a smaller first ionisation enthalpy than beryllium.
Reason (R) The penetration of $2 s$ electron to the nucleus is more than the $2 p$ electron hence $2 p$ electron is more shielded by the inner core of electrons than the $2 s$ electrons.
Assertion (A) Electron gain enthalpy becomes less negative as we go down a group.
Reason (R) Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.
Electron gain enthalpy of an element is equal to the energy released when an electron is added to valence shell of an isolated gaseous atom.
$$A(g)+e^{-} \longrightarrow A^{-}(g) ; \Delta_{\text {eg }} H=\text { negative }$$
Factors affecting electron gain enthalpy
(i) Effective nuclear charge Electron gain enthalpy increases with increase in effective nuclear charge because attraction of nucleus towards incoming electron increases.
(ii) Size of an atom Electron gain enthalpy decreases with increase in the size of valence shell.
(iii) Type of subshell More closer is the subshell to the nucleus, easier is the addition of electron in that subshell.
Electron gain enthalpy (in decreasing order) for addition of electron in different subshell ( $n$-same) is $s>p>d>f$
(iv) Nature of configuration Half-filled and completely-filled subshell have stable configuration, so addition of electron in them is not energetically favourable.
Variation in the periodic table As a general rule, electron gain enthalpy becomes more and more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom. Electron gain enthalpy becomes less negative as we go down a group because the size of the atom increases and the added electron would be farther from the nucleus. Electron gain enthalpy of O or F is less than that of the succeeding element ( S or Cl ) because the added electron goes to the smaller $n=2$ level and suffers repulsion from other electrons present in this level. For the $n=3$ level ( S or Cl ), the added electron occupies a larger region of space and suffers much less repulsion from electrons present in this level.