Question

An element 'E' has exceptional valence electron configuration as given 4d^10 . what is the period of 'E' in the modern periodic table ? ​

Answer 1

Explanation:

Answer:→ 5

. Explanation:

Expected electron configuration should be [Kr]4d^10 5s^0 . The principal quantum number (n) in valence shell is thus equal to 5. So the period of the element to which it belongs is 5 .

Hence, it is solved

Answer 2

Answer

5 period

Given

Exceptional valence electronic configuration is 4d^10

so electronic configuration is

1s^2,2s^2,3p^6,3s^3,3p^6,4s^2,3d^10,4P^6,5S^2,4f^14,4d^10

since

S orbital containing 2 electron in its outer cell are in 5 th subcell so the period is 5