A solution of $\mathrm{KMnO}_4$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Oxidising behaviour of $\mathrm{KMnO}_4$ depends on pH of the solution.
In acidic medium $(\mathrm{pH}<7)$
$$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \underset{\text { Colourless }}{\mathrm{Mn}^{2+}}+4 \mathrm{H}_2 \mathrm{O}$$
In alkaline medium $(\mathrm{pH}>7)$
$$\mathrm{MnO}_4^{-}+\mathrm{e}^{-} \longrightarrow \underset{\text { (Green) }}{\mathrm{MnO}^{2-}_4}$$
In neutral medium $(\mathrm{pH}=7)$
$$\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \longrightarrow \underset{\text { (Brown ppt) }}{\mathrm{MnO}_2}+4 \mathrm{OH}^{-}$$
The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain, why?
Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.
$\mathrm{E}^{\mathrm{s}}$ of Cu is +0.34 V while that of Zn is $-$0.76 V . Explain.
$E^{\mathrm{s}}$ value of Cu is positive because of the fact that sum of sublimation enthalpy and ionisation enthalpy to convert $\mathrm{Cu}(\mathrm{s})$ to $\mathrm{Cu}^{2+}(\mathrm{aq})$ is so high that it is not compensate by its hydration enthalpy. $E^{\circ}$ value for Zn is negative because of the fact that after removal of electrons from 4 s orbital, stable $3 d^{10}$ configuration is obtained.
The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
As the oxidation state increases, size of the ion of transition element decreases. As per Fajan's rule, as the size of metal ion decreases, covalent character of the bond formed increases.
Therefore, the halide of transition elements become more covalent with increasing oxidation state of the metal.
While filling up of electrons in the atomic orbitals, the 4 s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?
During filling up of electrons follow $(n+l)$ rule. Here $4 s$ has lower energy than $3 d$ orbital. After the orbitals are filled 4 s goes beyond $3 d$, i.e., $4 s$ is farther from nucleus than $3 d$. So, electron from $4 s$ is removed earlier than from $3 d$.