Solution:
SOLUTION
In [Cr(H2O)]Cl3, let oxidation number of Cr be x, Here the molecule separates as its ions as, [Cr(H2O)]3+ and 3Cl−. As H2O is a neutral molecule its oxidation number is zero. Hence, x=+3
In [Cr(C6H6)2] let oxidation number of Cr be y, Benzene is a neutral molecule and hence on observing the cross multiplied factor we get the oxidation number of Cr as 0. Hence,y=0
In K2[Cr(CN)2(O)2(NH3)],let oxidation number of Cr be z, Here the ions are- 2K+ and [Cr(CN)2(O)2(NH3)]2−. As NH3 is neutral molecule and O−and CN− is 2×(−1)+4×(−1)=−6 and hence oxidation number of Cr is, z-6=-0?z=+6