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: Determine the position of each element.
(a) Ra (Z = 88):
The noble gas preceding Radium is Radon (Rn, Z=86).
Electronic configuration: [Rn] 7s\(^2\).
The principal quantum number of the valence shell is n=7, so it is in the 7\(^{th}\) period.
It is an s-block element with 2 valence electrons, so it is in Group 2.
This matches with (iv). So, a \(\rightarrow\) iv.
(b) Ga (Z = 31):
The noble gas preceding Gallium is Argon (Ar, Z=18).
Electronic configuration: [Ar] 3d\(^{10}\) 4s\(^2\) 4p\(^1\).
The principal quantum number of the valence shell is n=4, so it is in the 4\(^{th}\) period.
It is a p-block element. For p-block, Group number = 10 + (valence s-electrons + valence p-electrons) = 10 + 2 + 1 = 13.
This matches with (i). So, b \(\rightarrow\) i.
(c) W (Z = 74):
The noble gas preceding Tungsten is Xenon (Xe, Z=54).
Electronic configuration: [Xe] 4f\(^{14}\) 5d\(^4\) 6s\(^2\).
The principal quantum number of the valence shell is n=6, so it is in the 6\(^{th}\) period.
It is a d-block element. Group number = number of ns + (n-1)d electrons = 2 + 4 = 6.
This matches with (ii). So, c \(\rightarrow\) ii.
(d) Pd (Z = 46):
The noble gas preceding Palladium is Krypton (Kr, Z=36).
Electronic configuration is exceptional: [Kr] 4d\(^{10}\) 5s\(^0\).
The highest principal quantum number is n=5, so it is in the 5\(^{th}\) period.
It is a d-block element. Group number = number of ns + (n-1)d electrons = 0 + 10 = 10.
This matches with (iii). So, d \(\rightarrow\) iii.
Step 2: Compile the matches.
The correct matches are: a \(\rightarrow\) iv, b \(\rightarrow\) i, c \(\rightarrow\) ii, d \(\rightarrow\) iii.
Step 3: Final Answer:
This combination corresponds to option (A).