Number of moles of $K_{2}C{r}_2{O}_7$ in acidic medium required to oxidise one mole of $C{u}_{3}PtoCuS{O}_4$ and $H_{3}P{O}_4$ is:

The correct option is D $\frac{11}{6}$
Chemical reaction involved:

$11K_{2}C{r}_2{O}_7+6C{u}_{3}P+62H_{2}S{O}_4\to$
$11C{r}_2(S{O}_4{)}_3+18CuS{O}_4+6H_{3}P{O}_4+$
$11K_{2}S{O}_4+53H_{2}O$

Given moles of $C{u}_{3}P=1$

Salts react in such a way that two or more atoms in the salt undergo a change in oxidation state as a result of either oxidation or reduction. In such a case, we can calculate the n - factor of the salt as the total increase or decrease in oxidation state per mole of the salt.

The oxidation state of $Cu$ changes from +1 to +2 and the oxidation state of $P$ changes from -3 to +5 (both $Cu$ and $P$ are oxidised). n-factor of $C{u}_{3}P=3|2-1|+1|5-(-3)|$
$=11$

For the given reaction,
Gram equivalents of $K_{2}C{r}_2{O}_7=$ Gram equivalent of $C{u}_{3}P$

Number of moles $\times$ n-factor = Number of moles $\times$ n-factor.
$n\times 6=1\times 11$

$n=\frac{11}{6}$

Thus ,$\frac{11}{6}$ moles of $K_{2}C{r}_2{O}_7$ are required.
So, option (A) is the correct answer.

Additional Information:

Reaction Involved:

$C{r}_2{O}_7^{-2}+C{u}_{3}P\to C{r}^{3+}+P{O}_4^{2-}+C{u}^{2+}$

$O.SCr=+6Cu=+1Cr=+3P=+5Cu=+2$
$O=-2$ $P=-3$ $O=-2$

n-factor based on $C{r}_2{O}_7^{2-}=(\Delta O.S)\times atoms$ of $Cr$ in $C{r}_2{O}_7^{2-}=(6-3)\times 2=6$

n-factor based on $Cu=(\Delta O.S)\times atoms$ of $Cu$ in $C{u}_{3}P=(2-1)\times 3=3$

n-factor based on $P=(\Delta O.S)\times atoms$ of $P$ in $C{u}_{3}P=[5-(-3)\times 1]=8$

Thus, n-factor of $C{u}_{3}P=11$