The element X in group 17 has 3 shells. If so,
i. Write the subshell electronic configuration of the element.
ii. Write the period number.
iii. What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?
i. As given in the question, the element X has 3 shells, which are K, L, M. The subshells are 1s, 2s, 2p, 3s, 3p, 3d that mean it belongs to the period 3 and group 17.
Group number: 17
Electrons in last shell: 7
Shell electronic configuration: 2,8,7
Sub-shell electronic configuration 1s22s2 2p63s23p5
ii. As the element X has 3 subshells so it belongs to the 3rd period of the periodic table.
iii. Chemical formula when element X reacts with element Y= YX3
Element X has 3 shells means it is the element of 3 period. So, element X is Chlorine which is the element of 3 period and 17 group.
And element Y of the third period has 3 valence electrons because there is one electron in a 3p subshell and 2 electrons are in 3s subshell because s subshell filled first than p subshell. Hence we have concluded that element Y is boron with electronic configuration of 1s22s22p1. We have concluded that the boron has 3 electrons on its outermost shell. So its reacts with chlorine to form BCl3.