(a) (i) Transition metals form large number of complexes due to following reasons:
a. Small size of metal ion
b. High charge density
c. Availability of empty $\mathit{d}$-orbitals
(ii) In the lowest oxides of transition metals, the metal atom has low oxidation state. This means some of the valence electrons of metal atom are not involved in the bonding. So, these electrons are not available for donation. Hence, they are basic in nature.
However, in the higher oxides of transition metals, the metal atom has high oxidation state. This means all of the valence electrons of metal atom are involved in the bonding.
So, these electrons are not available for donation. Also, these metal ions have high effective nuclear charge. As a result, they can accept electrons. Hence, they are acidic in nature.
(iii) ${\mathrm{Mn}}^{2+}$ exists in half-filled ${\mathit{d}}^5$ state which is very stable while ${\mathrm{Mn}}^{3+}$ is ${\mathit{d}}^4$ which is not so stable. Conversion from ${\mathit{d}}^4$ to ${\mathit{d}}^5$ will be quick and have negative $∆\mathit{G}$ value. Hence, because the stability factor of the ${\mathrm{E}}^{\circ }$ value is high for this process.
While ${\mathrm{Cr}}^{3+}$ is ${\mathit{d}}^3$ is half-filled $\left({\mathit{t}}_{2\mathrm{g}}^3\right)$ is stable in nature and ${\mathrm{Cr}}^{2+}$ is ${\mathrm{d}}^4$ has one extra electron which it would like to donate to attain the stable half-filled $\left({\mathit{t}}_2^3\right)$ configuration. Hence, for the process ${\mathrm{Cr}}^{3+}$ to ${\mathrm{Cr}}^{2+}$, the value of ${\mathrm{E}}^{\circ }$ is less.
(b) Similarities between lanthanoid and actinoid:
(i) They have prominent oxidation states of +3 .
(ii) They are electropositive.
(iii) They show magnetic properties
Differences between lanthanoid and actinoid :
(i) Lanthanoids are involved in filling of 4f-orbitals, actinoids are involved in filling of 5 f-orbitals.
(ii) Binding energy of $4\mathrm{f}$-orbitals or lanthanoids is less than $5\mathrm{f}$-orbitals or actinoids.
OR
(a) (i) Variable oxidation states shown by transition element can differ by one unit while oxidation state shown by nontransition elements differ by two units (sometimes due to inert pair effect).
For example, oxidation state of
$\mathrm{S}=-2,-4,-6$
Oxidation state of $\mathrm{Co}=+2,+3,+4$
(ii) ${\mathrm{Cu}}^{+}$ is unstable in aqueous solution while ${\mathrm{Cu}}^{2+}$ is more stable. Stability in aqueous medium depends upon the hydration energy, although it will require energy to remove one electron from ${\mathrm{Cu}}^{+}$to convert it into ${\mathrm{Cu}}^{2+}$ but that is compensated by high hydration energy of ${\mathrm{Cu}}^{2+}$. Hence ${\mathrm{Cu}}^{2+}$ is more stable than that of ${\mathrm{Cu}}^{+}$.
(iii) The change in colour is due to the formation of chromate ion.
${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{Cr}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}$
(b) The reasons for this are as follows :
(i) Actinoids display a large number of oxidation states while lanthanoids primarily show only three oxidation states.
(ii) Actinoids are radioactive in nature.