Account for the following. 

(i) ${\mathrm{Mn}}^{2+}$ is more stable than ${\mathrm{Fe}}^{2+}$ towards oxidation to $+3$ states.

(ii) The enthalpy of atomisation is lowest for Zn in 3d-series of the transition elements. 

(iii) Identify the metal in ${\mathrm{MO}}_3\mathrm{F}$ and justify your answer.

(iv) The ${\mathrm{E}}^0$ value for the ${\mathrm{Mn}}^{3+}/{\mathrm{Mn}}^{2+}$ couple is much more positive than for ${\mathrm{Cr}}^{3+}/{\mathrm{Cr}}^{2+}$ couple.

(v) Transition metals form a large number of complexes.

(i) Electronic configuration of ${\mathrm{Mn}}^{2+}={\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^5$

Electronic configuration of ${\mathrm{Fe}}^{2+}={\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^6$

${\mathrm{Mn}}^{2+}$ having half-filled d-orbitals will be more stable than ${\mathrm{Fe}}^{2+}$ because it has half filled d-orbitals.

(ii) Zinc has completely filled d-orbitals, it does not form metallic bonds. Thus, it requires least enthalpy to get atomised. 

(iii) ${\mathrm{MO}}_3\mathrm{F} \mathrm{is} {\mathrm{MnO}}_3\mathrm{F}. \mathrm{In} {\mathrm{MO}}_3\mathrm{F},$

Let the oxidation state of M is x.
$\begin{array}{l}\mathrm{x}+3\times (-2)+(-1)=0\\ \mathrm{x}=+7\end{array}$
i.e. M is in oxidation state of +7.
Hence, the given compound is MnO3F.
(iv) ${\mathrm{E}}_{{\mathrm{Cr}}^{3+}}^{\circ }/{\mathrm{Cr}}^{2+}$is negative $(-0.4\mathrm{V})$ It shows the stability of ${\mathrm{Cr}}^{3+}$ ions i.e. ${\mathrm{Cr}}^{3+}$ in
solution cannot be reduced to ${\mathrm{Cr}}^{2+}$ ions. ${\mathrm{Mn}}^{3+}$has high positive value of E° due to
extra stability of half-filled electronic configuration. Thus, Cr³⁺ is the most stable,
Mn³⁺ is least stable.

(v) Transition elements form a large number of complexes due to the comparatively smaller sizes of the metal ions, their high conic charges and the availability of d-orbitals for bond formation.