This question reads an atomic diagram of a Group I atom and links its electron arrangement to its position in the Periodic Table and to ion formation.
(a) The diagram shows three rings of electrons around the nucleus, so the atom has 3 occupied electron shells. [1]
(b) Counting the electrons shell by shell from the inside out gives 2 in the first, 8 in the second and 1 in the third, so the electronic configuration is 2,8,1. (This matches an atom with 11 electrons, balancing the 11 protons.) [1]
(c) The nucleon (mass) number is the total of protons and neutrons:
\[ \text{nucleon number} = 11 + 12 = 23 \]
[1]
(d) An atom with 11 protons has proton (atomic) number 11, which the Periodic Table identifies as sodium. [1]
(e) The single outer-shell electron places it in Group I, and the three occupied shells place it in Period 3. One mark each. [2]
(f) To reach a full outer shell, this atom loses (gives away) its one outer electron, forming a positive ion with the formula Na+. Losing one negative electron leaves the ion with one more proton than electrons, hence the single positive charge. One mark for the electron loss, one for the formula. [2]
Exam takeaway: the number of occupied shells gives the period and the number of outer electrons gives the group, so a 2,8,1 atom must be in Group I, Period 3, which is sodium.