1. We check each option for both:
   • correct balancing, and
   • chemically correct product(s).

2. Option A

$[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \xrightarrow{\text{excess NaOH}} [Mg(EDTA)]^{2+} + 6H_2O$

• Charge check:
  • LHS charge $= +2 + (-4) = -2$
  • RHS charge $= -2 + 0 = -2$ So charge is balanced.
• Atom count is also fine.

But chemically, in excess NaOH, $Mg^{2+}$ generally precipitates as $Mg(OH)_2$ because magnesium does not form a very stable soluble hydroxo complex in excess alkali like some amphoteric metals do.

• Also, the statement as written ignores the strong tendency of $Mg^{2+}$ to give hydroxide in basic medium.
• Hence this is not the correct represented reaction under the stated condition.

So, A is not correct.

3. Option B

$CuSO_4 + KCN \to K_2[Cu(CN)_4] + K_2SO_4$

Check stoichiometry first:

• RHS has 4 cyanide ligands, so at least $4KCN$ would be required.
• Potassium count on RHS is $4$ total: $2$ in $K_2[Cu(CN)_4]$ and $2$ in $K_2SO_4$. So LHS would need $4KCN$.

Balanced form would have to look like:

$CuSO_4 + 4KCN \to K_2[Cu(CN)_4] + K_2SO_4$

But even this is chemically not correct in simple form because $Cu^{2+}$ with cyanide undergoes redox, giving initially $CuCN$ and $(CN)_2$, and with excess cyanide forms cuprous complex such as $[Cu(CN)_3]^{2-}$ or related species, not straightforwardly $[Cu(CN)_4]^{2-}$ from $Cu^{2+}$.

So, B is not correct.

4. Option C

$Li_2O + 2KCl \to 2LiCl + K_2O$

This is formally balanced in atoms.

But chemically, such displacement does not occur this way in ordinary conditions. Alkali metal salts/oxides do not undergo this kind of exchange because $K_2O$ and $LiCl$ are not formed by a driving force from $Li_2O$ and $KCl$ in this manner.

So, C is not correct.

5. Option D

$[CoCl(NH_3)_5]^+ + 5H^+ \to Co^{2+} + 5NH_4^+ + Cl^-$

This represents acid decomposition of the ammine complex.

Let us verify balancing.

(i) Charge balance

• LHS charge: $+1 + 5(+1) = +6$
• RHS charge: $+2 + 5(+1) + (-1) = +6$

Charge balanced.

(ii) Atom balance

• Co: 1 on both sides
• Cl: 1 on both sides
• N: 5 on both sides
• H:
  • LHS: $5NH_3$ contributes $15H$, plus $5H^+$ contributes $5H$, total $20H$
  • RHS: $5NH_4^+$ contributes $20H$

All balanced.

(iii) Chemical correctness

In acidic medium, coordinated $NH_3$ ligands are protonated to $NH_4^+$, causing breakdown of the ammine complex and release of chloride, giving cobalt(II) ion.

So, D is correct.

6. Final conclusion

The only balanced equation representing the correct product(s) is:

$\boxed{D}$