Although Zr belongs to 4 d and Hf belongs to 5 d transition series but it is quite difficult to separate them, Why?
Separation of Zr and Hf are quite difficult because of lanthanoid contraction. Due to lanthanoid contraction, they have almost same size ( $\mathrm{Zr}=160 \mathrm{pm}$ and $\mathrm{Hf}=159 \mathrm{~pm}$ ) and thus, similar chemical properties. That's why it is very difficult to separate them by chemical methods.
Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
It is due to the fact that after losing one more electron Ce acquires stable $4 f^{\circ}$ electronic configuration. So, Ce shows +4 oxidation state also alongwith +3 oxidation state.
Explain why does colour of $\mathrm{KMnO}_4$ disappear when oxalic acid is added to its solution in acidic medium?
When oxalic acid is added to acidic solution of $\mathrm{KMnO}_4$, its colour disappear due to reduction of $\mathrm{MnO}_4^{-}$ion to $\mathrm{Mn}^{2+}$. Chemical reaction occurring during this neutralisation reaction is as follows
$5 \mathrm{C}_2 \mathrm{O}_4^{2-}+\underset{\text { (Coloured) }}{2 \mathrm{MnO}_4^{-}}+16 \mathrm{H}^{+} \longrightarrow \underset{\text { (Colourless) }}{2 \mathrm{Mn}^{2+}}+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2$
When orange solution containing $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ ion is treated with an alkali, a yellow solution is formed and when $\mathrm{H}^{+}$ions are added to yellow solution, an orange solution is obtained. Explain why does this happen?
When orange solution containing $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ ion is treated with an alkali, a yellow solution of
$\mathrm{CrO}_4^{2-}$ is obtained. On the same way, $\underset{\substack{\text { Dischromate } \\ \text { (orange) }}}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}} \xrightarrow[\mathrm{H}^{+}]{\mathrm{OH}^{-}} \underset{\begin{array}{c}\text { Chromate } \\ \text { (yellow) }\end{array}}{\mathrm{CrO}_4{ }^{2-}}$
when $\mathrm{H}^{+}$ions are added to yellow solution, an orange solution is obtained due to interconversion.
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 $(\mathrm{pH}=7)$
$$\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \longrightarrow \underset{\text { (Brown ppt) }}{\mathrm{MnO}_2}+4 \mathrm{OH}^{-}$$