Question

What is the position of an element in long form of periodic table if the expected electronic configuration is [Kr]4d$^{10}$5s$^2$?

• A. Group - 6, Period - 5
• B. Group - 9, Period - 5
• C. Group - 5, Period - 6
• D. Group - 11, Period - 5

Hint:

For d-block elements, the group number can be identified based on the number of electrons in the d and s orbitals. Group 11 corresponds to d$^{10}$5s$^2$ configuration.

Answer 1

The Correct Option is D

Step 1: Understanding the electronic configuration.
The given electronic configuration is [Kr]4d$^{10}$5s$^2$, which means the element is in the 4d block and belongs to the 5th period. The 4d$^{10}$ configuration indicates that the element has filled 4d orbitals with 10 electrons, and the 5s$^2$ configuration indicates two electrons in the 5s orbital.
Step 2: Identifying the group.
In the long-form periodic table, elements with the electronic configuration ending in 4d$^{10}$5s$^2$ belong to Group 11. This is because the d-block elements in this configuration are in the transition metal series and belong to the coinage metals group.
Step 3: Conclusion.
The correct position of the element is in Group 11, Period 5, so the correct answer is (D).