Explanation

(a)
Let oxidation state of Cr in $K_{2}{Cr}_{2}O_{7} = x$
$( + 1 \times 2) + 2x + ( - 2 \times 7) = 0$
or $+ 2 + 2x - 14 = 0$
$\therefore$ $x = + 6$
Let oxidation state of Cr in $K_{2}{CrO}_{4} = x$
$+ 1 \times 2 + x + ( - 2 \times 4) = 0$
2+$x - 8 = 0$
$x = 6$
$\therefore$ Change in oxidation state of Cr is zero when it changes from
$\ \ \ \ \ \ \ K_{2}{Cr}_{2}O_{7}\ to\ K_{2}{CrO}_{4.}$