Question

Match List – I with List – II:

Choose the correct answer from the options given below.

• A. a – iv, b – i, c – ii, d – iii
• B. a – i, b - ii, c - iii, d – iv
• C. a – iv, b – ii, c – iii, d – i
• D. a – iii, b – iv, c – i, d – ii

Hint:

To quickly find the period and group:

• \textbf{Period:} The highest principal quantum number (n) in the electron configuration.
• \textbf{Group:} For s-block, it's the number of ns electrons. For p-block, it's 10 + ns + np electrons. For d-block, it's ns + (n-1)d electrons.

Knowing the atomic numbers of noble gases helps in writing configurations quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The position of an element in the periodic table (Group and Period) is determined by its electronic configuration, specifically its valence shell and the number of valence electrons.
Step 2: Key Formula or Approach:
Determine the group and period by writing the electronic configuration based on the atomic number, usually by finding the preceding noble gas.
Step 3: Detailed Explanation:
(a) Radium (Ra - 88): Preceding noble gas is Radon (86). Configuration: $[Rn] 7s^{2}$. Highest principal quantum number is 7 (Period 7). It has 2 valence electrons in the s-orbital (Group 2, Alkaline earth metal). (a $\rightarrow$ iv)
(b) Gallium (Ga - 31): Preceding noble gas is Argon (18). Configuration: $[Ar] 3d^{10} 4s^{2} 4p^{1}$. Highest $n=4$ (Period 4). It is a p-block element, Group = $12 + \text{p-electrons} = 12 + 1 = 13$. (b $\rightarrow$ i)
(c) Tungsten (W - 74): Preceding noble gas is Xenon (54). Configuration: $[Xe] 4f^{14} 5d^{4} 6s^{2}$. Highest $n=6$ (Period 6). It is a d-block element, Group = $\text{s-electrons} + \text{d-electrons} = 2 + 4 = 6$. (c $\rightarrow$ ii)
(d) Palladium (Pd - 46): Preceding noble gas is Krypton (36). Configuration: Exceptionally $[Kr] 4d^{10} 5s^{0}$. It belongs to Period
5. Group = $10 + 0 = 10$. (d $\rightarrow$ iii)
Step 4: Final Answer:
Matching gives: a-iv, b-i, c-ii, d-iii.