Question

Choose the correct sequence by arranging the given ions in the order of increasing size :
A. $Cl^-$
B. $Mg^{2+}$
C. $Br^-$
D. $S^{2-}$
E. $Na^+$
Choose the correct answer from the options given below :

• A. A < D < C < B < E
• B. B < E < C < A < D
• C. B < E < A < D < C
• D. E < B < C < A < D

Hint:

When comparing atomic or ionic sizes, rule #1 is the number of shells (period number on the periodic table). Rule #2 is for isoelectronic series: the more highly positive the charge (or the less negative), the smaller the ion, due to a stronger pull by the protons.

Answer 1

The Correct Option is C

Step 1: Concept:
The objective is to arrange five different monoatomic ions in order of increasing ionic radius. This requires comparing isoelectronic series and analyzing the effect of principal quantum numbers (electron shells) and effective nuclear charge.

Step 2: Key Formula or Approach:
1. Ions with a higher principal quantum number ($n$) generally have a larger size because their valence electrons occupy shells further from the nucleus.
2. For isoelectronic species (ions with the exact same number of electrons), the size decreases as the nuclear charge (number of protons, $Z$) increases. More protons pull the same number of electrons more tightly inward.

Step 3: Step-by-step Explanation:

• Let's break the given ions into groups based on their electron configurations:
- Group 1 ($Ne$ core, 10 electrons): $Na^+$ ($Z=11$) and $Mg^{2+}$ ($Z=12$). Since they are isoelectronic, the one with more protons is smaller. Thus, $Mg^{2+} < Na^+$.
- Group 2 ($Ar$ core, 18 electrons): $Cl^-$ ($Z=17$) and $S^{2-}$ ($Z=16$). Isoelectronic again, more protons means smaller size. Thus, $Cl^- < S^{2-}$.
- Group 3 ($Kr$ core, 36 electrons): $Br^-$ ($Z=35$).

• Now, we compare the groups. The size increases significantly as we add entirely new electron shells.
- $Ne$ core ions ($n=2$ valence shell) are strictly smaller than $Ar$ core ions ($n=3$ valence shell).
- $Ar$ core ions are strictly smaller than $Kr$ core ions ($n=4$ valence shell).

• Therefore, combining our observations, the complete size order from smallest to largest is:
$Mg^{2+}$ (10 e-, $Z=12$) $<$ $Na^+$ (10 e-, $Z=11$) $<$ $Cl^-$ (18 e-, $Z=17$) $<$ $S^{2-}$ (18 e-, $Z=16$) $<$ $Br^-$ (36 e-, $Z=35$).

• Translating this to the given letters: B ($Mg^{2+}$) < E ($Na^+$) < A ($Cl^-$) < D ($S^{2-}$) < C ($Br^-$).

Step 4: Final Answer:
The correct sequence is B < E < A < D < C, which matches option (C).