Answer
a) Beryllium (Be)
b) chlorine ion ($Cl^{-}$)
c) Rubidium ion ($Rb^{+}$)
Work Step by Step
a) If we look at the periodic table, it is seen that group 2 elements usually form the +2 cation. Now, period 2 element which is at group 2 is Beryllium (Be).
b) group 7A elements are: F, Cl, Br, I etc. Among them, Cl has 17 electrons. Now, $Cl^{-}$ ion should have one extra electron i.e. 18 electrons.
c) Find group 1A elements (Li, Na, K, Rb, Cs etc.) and their atomic number. Among them, Rb has the atomic number of 37. Now, no. of electrons for neutral Rb atom = 37. So, $Rb^{+}$ ion will have one electron less than in the neutral atom i.e. 36 electrons.
