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(B) To find the oxidation state of Sulphur $(S)$ in each compound:$1$. $SO_3$: $x + 3(-2) = 0 \implies x = +6$$2$. $H_2SO_3$: $2(+1) + x + 3(-2) = 0 \

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$3$

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(B) To find the oxidation state of Sulphur $(S)$ in each compound:$1$. $SO_3$: $x + 3(-2) = 0 \implies x = +6$$2$. $H_2SO_3$: $2(+1) + x + 3(-2) = 0 \implies x = +4$$3$. $SOCl_2$: $x + (-2) + 2(-1) = 0 \implies x = +4$$4$. $SF_4$: $x + 4(-1) = 0 \implies x = +4$$5$. $BaSO_4$: $(+2) + x + 4(-2) = 0 \implies x = +6$$6$. $H_2S_2O_7$: $2(+1) + 2x + 7(-2) = 0 \implies 2x = +12 \implies x = +6$The compounds with $+4$ oxidation state are $H_2SO_3, SOCl_2$,and $SF_4$.Therefore,the total number of such compounds is $3$.

From the given list,the number of compounds with $+4$ oxidation state of Sulphur is $ . . . . . . $.
$SO_3, H_2SO_3, SOCl_2, SF_4, BaSO_4, H_2S_2O_7$

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From the given list,the number of compounds with $+4$ oxidation state of Sulphur is $ . . . . . . $.$SO_3, H_2SO_3, SOCl_2, SF_4, BaSO_4, H_2S_2O_7$