Answer
a) Ra - Radium - Loses electrons - Forms cation - $Ra^{2+}$
b) In - Indium - Loses electrons - Forms cation - $In^{3+}$
c) P - Phosphorus - Gains electrons - Forms anion - $P^{3-}$
d) Te - Tellurium - Gains electrons - Forms anion - $Te^{2-}$
e) Br - Bromine - Gains electrons - Forms anion - $Br^{-}$
f) Rb - Rubidium - Loses electrons - Forms cation - $Rb^{+}$
Work Step by Step
All metals lose their electrons to form cations (positively-charged ions) :
(a) Radium (Ra) - Has 2 valence electrons, and thus loses 2, forming $Ra^{2+}$ . Also , Radium lies in group 2, where all elements have valency = 2+
(b) Indium (In) - Has 3 valence electrons, and thus loses 3, forming
$In^{3+}$ . Also , Indium lies in group 13, where all elements have valency = 3+
f) Rubidium (Rb) - Lies in group 1, where all elements have valency = +1.
Non Metals gain electrons to complete their octet to form anions (negatively-charged ions):
(c) Phosphorus (P) - Has 5 valence electrons, hence gains 3 , forming
$P^{3-}$ . Also , Phosphorus lies in group 15, where all elements have valency = 3-
d) Tellurium (Te) - Has 6 valence electrons, hence gains 2 , forming
$Te^{2-}$ . Also , Tellurium lies in group 16, where all elements have valency = 2-
e) Bromine (Br) - Has 7 valence electrons, hence gains 1 , forming
$Br^{-}$ . Also , Indium lies in group 17 (Halogens) , where all elements have valency = 1-
