1. We need to find in how many given compounds sulphur has oxidation state $+4$.

Given compounds: $\mathrm{SO}_3,\ \mathrm{H}_2\mathrm{SO}_3,\ \mathrm{SOCl}_2,\ \mathrm{SF}_4,\ \mathrm{BaSO}_4,\ \mathrm{H}_2\mathrm{S}_2\mathrm{O}_7$

2. Find oxidation state of sulphur in each compound.

(i) $\mathrm{SO}_3$

Let oxidation state of S be $x$. Oxygen is $-2$. $x+3(-2)=0$ $x-6=0 \Rightarrow x=+6$ So, sulphur is not in $+4$ state.

(ii) $\mathrm{H}_2\mathrm{SO}_3$

Let oxidation state of S be $x$. Hydrogen is $+1$, oxygen is $-2$. $2(+1)+x+3(-2)=0$ $2+x-6=0$ $x=+4$ So, this compound has S in $+4$ state.

(iii) $\mathrm{SOCl}_2$

Let oxidation state of S be $x$. Oxygen is $-2$, each chlorine is $-1$. $x+(-2)+2(-1)=0$ $x-2-2=0$ $x=+4$ So, this compound has S in $+4$ state.

(iv) $\mathrm{SF}_4$

Let oxidation state of S be $x$. Fluorine is $-1$. $x+4(-1)=0$ $x-4=0$ $x=+4$ So, this compound has S in $+4$ state.

(v) $\mathrm{BaSO}_4$

This is ionic: $\mathrm{Ba}^{2+}$ and $\mathrm{SO}_4^{2-}$. In $\mathrm{SO}_4^{2-}$, let oxidation state of S be $x$. $x+4(-2)=-2$ $x-8=-2$ $x=+6$ So, sulphur is not in $+4$ state.

(vi) $\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7$

Let oxidation state of each sulphur be $x$. Hydrogen is $+1$, oxygen is $-2$. $2(+1)+2x+7(-2)=0$ $2+2x-14=0$ $2x=12 \Rightarrow x=+6$ So, sulphur is not in $+4$ state.

3. Compounds with sulphur in $+4$ oxidation state are:

• $\mathrm{H}_2\mathrm{SO}_3$
• $\mathrm{SOCl}_2$
• $\mathrm{SF}_4$

Therefore, the number of such compounds is: $\boxed{3}$

4. Comparison with stored correct answer: Stored correct answer = $3$

Our derived answer also = $3$. So the stored answer is correct.