The radius of $\mathrm{Na}^{+}$cation is less than that of Na atom. Give reason.
When an atom loses an electron to form cation, its radius decreases. In a cation, per electron nuclear forces increases due to decrease in number of electrons. As a result of this, effective nuclear charge increases and the radius of cation decreases. e.g., ionic radius of $\mathrm{Na}^{+}$ is smaller than the radius of its parent atom Na .
\begin{equation} \begin{array}{ll} &\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+1 \mathrm{e}^{-} \\ \text {Electrons } & 11 \qquad \quad10 \\ \text { Nuclear charge } & 11 \qquad \quad 11 \\ \text { lonic size } & 186 \mathrm{pm} \quad 95 \mathrm{pm} \end{array}\end{equation}
Among alkali metals which element do you expect to be least electronegative and why?
On moving down the group, electronegativity decreases because atomic size increases. Fr has the largest size, therefore it is least electronegative.
Match the correct atomic radius with the element.
| Element | Atomic radius (pm) |
|---|---|
| Be | 74 |
| C | 88 |
| O | 111 |
| B | 77 |
| N | 66 |
All the given elements are of same period and along a period, atomic radii decreases because effective nuclear charge increases. Thus, the order of atomic radii is $\mathrm{O}<\mathrm{N}<\mathrm{C}<\mathrm{B}<\mathrm{Be}$ or, $\mathrm{Be}=11 \mathrm{pm}, \mathrm{O}=66 \mathrm{pm}, \mathrm{C}=77 \mathrm{pm}, \mathrm{B}=88 \mathrm{pm}, \mathrm{N}=74 \mathrm{pm}$.
Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.
| Elements | $$\Delta H_1$$ | $$\Delta H_2$$ | $$\Delta_{eg} H$$ | ||
|---|---|---|---|---|---|
| (i) | Most reactive non-metal | A. | 419 | 3051 | $$-$$48 |
| (ii) | Most reactive metal | B. | 1681 | 3374 | $$-$$328 |
| (iii) | Least reactive element | C. | 738 | 1451 | $$-$$40 |
| (iv) | Metal forming binary halide | D. | 2372 | 5251 | $$+$$48 |
(i) Most reactive non-metal has high $\Delta_i H_1$ and $\Delta_i H_2$ and most negative $\Delta_{\text {eg }} H$. Therefore, the element is $B$.
(ii) Most reactive metal has low $\Delta_i H_1$ and high $\Delta_i H_2$ (because the second electron has to be lost from noble gas configuration) and has small negative $\Delta_{\text {eg }} H$. Therefore, the element is $A$.
(iii) Noble gases are the least reactive elements. They have very high $\Delta_i H_1$ and $\Delta_i H_2$ and have positive $\Delta_{\mathrm{eg}} H$ values. Thus, the element is $D$.
(iv) Metal forming binary halides are alkaline earth metals. They have $\Delta_i H_1$ and $\Delta_i H_2$ values little higher than those of most reactive metals (such as $A$ ) and have comparatively slightly less negative $\Delta_{\mathrm{eg}} H$ values. Thus, the element is $C$.
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 $\Delta_{\mathrm{eg}} H=-141 \mathrm{kJmol}^{-1}$.