Question:

Match the following

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When matching atomic properties like atomic radius, ionization energy, or electronegativity, always recall their general trends across periods and down groups in the periodic table. Atomic radius decreases across a period (due to increasing effective nuclear charge) and increases down a group (due to increasing number of electron shells).
Updated On: Jun 3, 2025
  • A-III, B-I, C-IV, D-II
  • A-III, B-IV, C-II, D-I
  • A-I, B-III, C-II, D-IV
  • A-I, B-IV, C-III, D-II
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The Correct Option is A

Solution and Explanation

Step 1: Understand the Trend of Atomic Radius in the Periodic Table
Atomic radius generally follows these trends: \begin{itemize} \item Decreases across a period from left to right: As you move across a period, the effective nuclear charge increases, pulling the electrons closer to the nucleus, thus decreasing the atomic radius. \item Increases down a group: As you move down a group, new electron shells are added, increasing the distance of the outermost electrons from the nucleus, thus increasing the atomic radius despite an increase in nuclear charge. \end{itemize} Step 2: Identify the Elements and Their Positions
The given elements are: \begin{itemize} \item A) Al (Aluminum): Period 3, Group 13. \item B) F (Fluorine): Period 2, Group 17. \item C) N (Nitrogen): Period 2, Group 15. \item D) Si (Silicon): Period 3, Group 14. \end{itemize} Ordering them by expected atomic radius (smallest to largest): % Option (A) Fluorine (F) - Period 2, Group 17 (smallest due to highest effective nuclear charge in Period 2 among these and only 2 shells). % Option (B) Nitrogen (N) - Period 2, Group 15 (larger than F as it's to the left of F in Period 2). % Option (C) Silicon (Si) - Period 3, Group 14 (larger than Period 2 elements due to more shells, but smaller than Al as it's to the right of Al in Period 3). % Option (D) Aluminum (Al) - Period 3, Group 13 (largest among these as it's furthest left in Period 3 among these, and has 3 shells). So, the expected order of increasing atomic radii is F< N< Si< Al. Step 3: Match Elements with Given Atomic Radii
The given atomic radii in pm are: I) 64, II) 117, III) 143, IV) 74. Ordering these radii from smallest to largest: 64 pm, 74 pm, 117 pm, 143 pm. Now, let's match based on our expected order: \begin{itemize} \item B) F (Fluorine): Expected to be the smallest. Matches with I) 64 pm. \item C) N (Nitrogen): Expected to be the second smallest. Matches with IV) 74 pm. \item D) Si (Silicon): Expected to be the third smallest (or second largest). Matches with II) 117 pm. \item A) Al (Aluminum): Expected to be the largest. Matches with III) 143 pm. \end{itemize} Therefore, the matches are:
A) Al - III) 143 pm
B) F - I) 64 pm
C) N - IV) 74 pm
D) Si - II) 117 pm
Step 4: Compare with Options
Let's check Option (1): A-III, B-I, C-IV, D-II.
This perfectly matches our derived mapping.
Step 5: Analyze the Options
\begin{itemize} \item Option (1): A-III, B-I, C-IV, D-II. This is the correct match. \item Option (2): A-III, B-IV, C-II, D-I. Incorrect. \item Option (3): A-I, B-III, C-II, D-IV. Incorrect. \item Option (4): A-I, B-IV, C-III, D-II. Incorrect. \end{itemize}
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