Answer:
a.
ZnS(s) + O2(g) → ZnO(s) + SO2(g)
b.
HCl (aq) + Ba(OH)2(aq) → BaCl2(aq) + H2O
First let us determine the electronic configuration of
Bromine (Br). This is written as:
Br = [Ar] 3d10 4s2 4p5
Then we must recall that the greatest effective nuclear
charge (also referred to as shielding) greatly increases as distance of the
orbital to the nucleus also increases. So therefore the electron in the
farthest shell will experience the greatest nuclear charge hence the answer is:
<span>4p orbital</span>
