Solved: Consider three elements with the following

Solution 117CQStep 1 of 6The given electronic configurations are X = [Ar] 4s2Y = [Ne] 3s23p4Z = [Ar] 4s23d104p4First, let’s find the position X, Y and Z elements in the periodic table.Position of X element:The electronic configuration of the X is [Ar] 4s2The outer shell electronic configuration is s-subshell.Therefore X is a s -block element.The electrons in s-subshell is 2. Therefore, it is a 2A group element,The Principal quantum number is 4 (n=4) .Therefore, it is a 4th period element.Position of Y element:The electronic configuration of the Y is [Ne] 3s23p4.The outer shell electronic configuration is p-subshell. Therefore Y is a P -block element.Total valence electrons are 2+4 = 6 Therefore it is a 6A group element.The Principal quantum number is 3 (n = 3) .Therefore, it is a 3rd period element.Position of Z element:The electronic configuration of the Z is [Ar] 4s23d104p4.The outer shell electronic configuration is p-subshell. Therefore Z is a P -block element.Total valence electrons are 2+4 = 6 Therefore it is a 6A group element.(Here, we are not considering the fully filled orbital and only consider the same energy levels)The Principal quantum number is 4 (n = 4) .Therefore, it is a 4th period element.Let’s write the position of X,Y and Z elements in the periodic table:ElementElectronic configuration Period Group X [Ar] 4s2 4 2A Y [Ne] 3s23p4 3 6A Z [Ar] 4s23d104p4 4 6A______________________________________________________________________________