Given below are some elements of the modern periodic table: $4Be, 9Fe, 14Si, 19K, 20Ca$. (i) Select the element that has one electron in the outermost shell and write its electronic configuration. (ii) Select two elements that belong to the same group. Give reason for your answer. (iii) Select two elements that belong to the same period. Which one of the two has bigger atomic size?

Based on the given elements $_4\text{Be}$, $_9\text{F}$ (printed as $_9\text{Fe}$), $_{14}\text{Si}$, $_{19}\text{K}$, and $_{20}\text{Ca}$:

(i) Element with one valence electron:

• Element: Potassium ($_{19}\text{K}$)
• Electronic Configuration: $2, 8, 8, 1$ (or $\text{K}=2, \text{L}=8, \text{M}=8, \text{N}=1$)

(ii) Two elements belonging to the same group:

• Elements: Beryllium ($_4\text{Be}$) and Calcium ($_{20}\text{Ca}$)
• Reason: Both elements have $2$ valence electrons in their outermost shell (electronic configurations: $\text{Be} = 2, 2$ and $\text{Ca} = 2, 8, 8, 2$). Elements with the same number of valence electrons belong to the same group (Group 2).

(iii) Two elements belonging to the same period and their atomic size comparison:

• Pair 1 (Period 2): Beryllium ($_4\text{Be}$) and Fluorine ($_9\text{F}$, given as $_9\text{Fe}$)
• Bigger Atomic Size: Beryllium ($_4\text{Be}$) is larger than Fluorine ($_9\text{F}$).
• Reason: Across a period from left to right, atomic size decreases due to an increase in nuclear charge which pulls the electrons closer to the nucleus.
• Pair 2 (Period 4): Potassium ($_{19}\text{K}$) and Calcium ($_{20}\text{Ca}$)
• Bigger Atomic Size: Potassium ($_{19}\text{K}$) is larger than Calcium ($_{20}\text{Ca}$).
• Reason: Potassium lies to the left of Calcium in Period 4. Atomic radius decreases from left to right across a period.