CHEM HELP!!!?

A)
HCl (aq) + LiOH (aq) → LiCl (aq) + H₂O (l)
O.N. (oxidation number) of H on the LHS = O.N. of H on the RHS = +1
O.N. of Cl on the LHS = O.N. of Cl on the RHS = -1
O.N. of Li on the LHS = O.N. of Li on the RHS = +1
O.N. of O on the LHS = O.N. of O on the RHS = -2
As oxidation numbers of all elements involved are unchanged in the reaction, the reaction is NOT oxidation-reduction.

B)
NaI(aq) + AgNO₃(aq) → AgI(s) + NaNO₃(aq)
O.N. of Na on the LHS = O.N. of Na on the RHS = +1
O.N. of I on the LHS = O.N. of I on the RHS = -1
O.N. of Ag on the LHS = O.N. of Ag on the RHS = +1
O.N. of N on the LHS = O.N. of N on the RHS = +5
O.N. of O on the LHS = O.N. of ) on the RHS = -2
As oxidation numbers of all elements involved are unchanged in the reaction, the reaction is NOT oxidation-reduction.

C)
Pb(C₂H₃O₂)₂ (aq) + 2NaCl (aq) → PbCl₂(s) + 2NaC₂H₃O₂(aq)
O.N. of Pb on the LHS = O.N. of Pb on the RHS = +2
O.N. of C on the LHS = O.N. of C on the RHS, as both are in C₂H₃O₂⁻
O.N. of H on the LHS = O.N. of H on the RHS, as both are in C₂H₃O₂⁻
O.N. of O on the LHS = O.N. of O on the RHS, as both are in C₂H₃O₂⁻
O.N. of Na on the LHS = O.N. of Na on the RHS = +1
O.N. of Cl on the LHS = O.N. of Cl on the RHS = -1
As oxidation numbers of all elements involved are unchanged in the reaction, the reaction is NOT oxidation-reduction.

D)
Mg (s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Mg is oxidized because the O.N. of Mg increases from 0 (in Mg) to +2 (in MgCl₂).
H is reduced because the O.N. of H decreases from +1 (in HCl) to 0 (in H₂).
Hence, it is an oxidation-reduction reaction.

The answer: D) Mg (s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

D) Mg (s) + 2HCl (aq) = MgCl2 (aq) +H2 (g)

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