1. Find $X$ for $\mathrm{KMnO_4}$ in acidic medium

In acidic medium, permanganate ion $\mathrm{MnO_4^-}$ is reduced to $\mathrm{Mn^{2+}}$.

• Oxidation state of Mn in $\mathrm{KMnO_4}$: $+7$
• Oxidation state of Mn in product $\mathrm{Mn^{2+}}$: $+2$

So, the difference is $X = 7 - 2 = 5$

2. Find $Y$ from the acetate ion test with neutral ferric chloride

Acetate ion gives a brown-red precipitate with neutral ferric chloride due to formation of ferric acetate.

Ferric ion is $\mathrm{Fe^{3+}}$.

Electronic configuration of Fe: $\mathrm{Fe}: [Ar]3d^6 4s^2$

For $\mathrm{Fe^{3+}}$:

• remove two $4s$ electrons and one $3d$ electron $\mathrm{Fe^{3+}}: [Ar]3d^5$

Hence, the number of $d$ electrons is $Y = 5$

3. Compute $X+Y$

$X+Y = 5+5 = 10$

Therefore, the required integer is $\boxed{10}$