Exercise 2 solution

(a) Fe2(SO4)3: two iron ions at 3+ and three sulfate ions at 2- (Ionic compound article).
    2 x (+3) = +6 ; 3 x (-2) = -6 ; total = +6 - 6 = 0. Zero net charge, as required.

(b) X is 2+, Y is 1- (charges given for this exercise, not real elements).
    One X gives +2, so two Y are needed to give -2.
    Formula: X Y2 (that is, one X and two Y).
    Check: (+2) + 2 x (-1) = 0.

(c) X is 3+, Y is 2-.
    The smallest common total is 6: two X give +6 and three Y give -6.
    Formula: X2 Y3.
    Check: 2 x (+3) + 3 x (-2) = 6 - 6 = 0.
    Swap method: the 3 and the 2 swap to become subscripts, giving X2 Y3 (the same answer;
    the method is my own summary, not from a page - check).

In each case the test is the one the Ionic bonding article states: strict ratios are
kept so that the compound is charge neutral.