Question

How would you account for d following:

1. The oxidising power of oxoanions are in the order VO2 + < Cr2O7 2- < MnO4 -.

2. Cr2+ is a stronger reducing agent than Fe2+.

Answer 1

1. The oxidising power of oxo-anions are in the order of :

$V{O}^{2+} < Cr_{2}{O_{7}}^{2-} < Mn{O_{4}}^{-}$

Reason:

For any sequential oxo-anions from similar period of the Periodic Table , higher the positive oxidation state of central metal , more is the oxidising power is that oxo-anion.

• Vanadium (V) has +4 oxidation state

• Chromium (Cr) has +6 oxidation state

• Manganese (Mn) has +7 oxidation state.

Since oxidation state of manganese is highest between these three central metals , Permanganate ion has highest oxidising power

2. Cr2+ is a strong reducing agent than Fe2+

Reason:

Reduction process refers to gain in electrons. Oxidation means loss of electrons.

Chromium in +2 oxidation state shows higher reducing power as compared to Ferrous ion because :

• Chromium in +2 state undergoes oxidation to form $C{r}^{+3}$ which would result in formation of half filled $t_{2}g$ and eg levels in d orbital splitting

Answer 2

in vo2+

oxidation state of v is +5

in cr2o7 2-

oxidation state of cr is +6

in mbo4-

oxidation state of mn is +7