Correct option is (D) 0.75
\(M_xY_2O_4\)
\(M^{+2} = \frac X3, M^{+3} = \frac{2X}3\)
So, total of O.N. of all atoms
\(\frac{2X}3 + 3\left(\frac{2X}3\right) + 2(+3) + 4(-2) = 0\)
\(\frac{2X}3 + 2X + 6 - 8 = 0\)
\(\frac{8X}3 = 2\)
\(X = \frac 68 = \frac34 = 0.75\)