Question

What is the position of the element in the periodic table satisfying the electronic configuration $(n-1)d^{1}n{s}^2$ for $n=4$?

• A. $3^{rd}$ period and $3^{rd}$ group
• B. $4^{th}$ period and $4^{th}$ group
• C. $3^{rd}$ period and $2^{nd}$ group
• D. $4^{th}$ period and $3^{rd}$ group

Answer 1

Answer: (D)

$4^{th}$ period and $3^{rd}$ group

From the given date as $n=4$ the electronic configuration will be $3d^{1}4s^2$ i.e. it will belong to period 4 as 'd' orbital is incompletely filled the element belongs to transition series as 1 electron is in 'd' orbital it will belong to the $3^{rd}$ period.