What is the order of decreasing ionic radius?A.     S2− > Cl− > Al3+ > Mg2+B.     Cl− > S2− > Al3+ > Mg2+C.     S2− > Cl− > Mg2+ > Al3+D.     Mg2+ > Al3+ > Cl− > S2−

The order of decreasing ionic radius is determined by the number of electrons and the effective nuclear charge.

1. S2-: This ion has the most electrons (18) and a relatively low effective nuclear charge due to its position in the third period of the periodic table. This results in a larger ionic radius.

2. Cl-: This ion has fewer electrons (18) than S2-, but it is in the same period. However, it has a higher effective nuclear charge, which pulls the electrons closer to the nucleus, resulting in a smaller ionic radius than S2-.

3. Al3+: This ion has fewer electrons (10) and is in the third period. It has a higher effective nuclear charge than both S2- and Cl-, which pulls the electrons even closer to the nucleus, resulting in a smaller ionic radius.

4. Mg2+: This ion has the same number of electrons (10) as Al3+, but it is in the second period. It has a higher effective nuclear charge than Al3+, which pulls the electrons even closer to the nucleus, resulting in the smallest ionic radius.

So, the correct order of decreasing ionic radius is S2- > Cl- > Al3+ > Mg2+, which corresponds to option A.