$$\eqalign{ & F{e^{2 + }} + {C_2}O_4^{2 - } \to F{e^{3 + }} + 2C{O_2} + 3{e^ - } \cr & MnO_4^ - + 5{e^ - } \to M{n^{2 + }} \cr} $$
$$1\,mole$$  of $$KMn{O_4}$$  accepts 5 electrons
$$1\,mole$$  of ferrous oxalate loses 3 electrons
$$5{e^ - } \equiv 1\,mole\,\,{\text{of}}\,\,KMn{O_4}$$
$$3{e^ - }$$  will be equivalent to $$\frac{3}{5}\,mole$$  of $$KMn{O_4}$$

Little bit acid is required for reaction
$$\therefore \,\,M{n^{2 + }}$$   acts as autocatalyst.

Due to lanthanoid contraction atomic radii of $$Zr$$  and $$Hf$$  is almost similar.

The configuration of Lanthanides show that the additional electron enters the $$4f$$ subshell. The shielding of one $$4f$$ electron by another is very little or imperfect. The imperfect shielding of $$f$$ electrons is due to the shape of $$f$$ orbitals which is very much diffused. Thus as the atomic number increases, the nuclear charge increases by unity at each step. While no comparable increase in the mutual shielding effect of $$4f$$ occurs. This causes a contraction in the size of the $$4f$$ subshell. as a result atomic and ionic radii decreases gradually from $$La\,to\,Lu.$$

$$Z{n^{2 + }} \to 3{d^{10}}$$   has no unpaired electrons for excitation.

In (i), $$C{u^ + }$$  is oxidised as well as reduced. In (ii), $$MnO_4^{2 - }$$  ions are oxidised as well as reduced.

Reducing agent

$$Gd\left( {64} \right) = \left[ {Xe} \right]4{f^7}5{d^1}6{s^2}$$
∴ No. of unpaired electrons $$=8$$

$$\because $$  Basicity of hydroxides decreases on moving left to right in a period.

$$F{e^{3 + }} - 3{d^5}$$     No. of unpaired electrons = 5
$$C{r^{3 + }} - 3{d^3}$$     No. of unpaired electrons = 3
$$N{i^{2 + }} - 3{d^8}$$     No. of unpaired electrons = 2
$$C{u^{2 + }} - 3{d^9}$$     No. of unpaired electrons = 1