Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is

KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
Currently in this equation, you have 2 hydrogen atoms and 2 oxygen atoms on the left, and then 2 hydrogen atoms and 1 oxygen atom on the right. To balance, you would need to even out the oxygens, so we can first place a 2 in front of H2O to get:
H2 + O2 -> 2H2O
Now, however, you can see that we have too many hydrogen atoms on the right, so to get the final answer, we add a 2 in front of hydrogen on the left:
2H2 + O2 -> 2H2O
I hope this helps!
It would be:
B. HCl
A metal will react with an acid to form a salt (MgCl2) and H2.