Fe3+=x Fe2+=0.93−x
Using charge balance Fe0.9S1.0= Charge on neutral compound =0 3×x2×(0.93−x)+2×1=0
Charge on Fe0.9 (Fe2+&Fe3+) ⇒3x+2(0.9−x)=2×2 3x+1.8−2x=2 x=2−1.8 x=0.2
Thus the no. of Fe+2 ion =0.93−x=0.93−0.2=0.73
Ratio Fe2+Fe3+=0.730.2=0.28