The atom with the electron configuration 1s$^{2}$2s$^{2}$2p63s$^{2}$3p$^{6}$3d$^{10}$4s$^{2}$4p$^{4}$ is in

Step 1: Analyze the question We need to find the period and block of an element from its electron configuration.

Step 2: Concept

• Period = the highest principal quantum number (n) occupied. Here the highest is n = 4, so the element is in period 4.
• Block = the type of subshell that holds the last electron added. The last electrons go into 4p, so it is in the p-block.

Step 3: Solution Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴

• Total electrons = 2 + 2 + 6 + 2 + 6 + 10 + 2 + 4 = 34, so the element is selenium (Se).
• The outermost shell is n = 4, so it is in period 4.
• The valence electrons are 4s² 4p⁴, which is 6 valence electrons, so it is in Group 16 (VIA). The last subshell filling is 4p, so it is in the p-block.
• The 3d subshell is full and lies inside the 4th shell, so it does not make the element a d-block element. d-block elements have their last electron in a d subshell (such as 3d¹ to 3d⁹).

Step 4: Answer Period 4, p-block, which is option A.

Tip: Look at the last subshell in the filling order, and use the highest shell number for the period. Do not be misled by 3d¹⁰ appearing in the middle. If the configuration ends in p, it is a p-block element.