(i) LANTHANOIDS SHOWING $+4$ AND $+2$ OXIDATION STATES

The characteristic and most stable oxidation state of ALL lanthanoids is $+3$.
A few show $+2$ or $+4$ as well, and they do so when the resulting ion attains an
EMPTY $(4f^0)$, HALF-FILLED $(4f^7)$ or COMPLETELY FILLED $(4f^{14})$ configuration.

(I) $+4$ OXIDATION STATE : CERIUM (Ce)
$$Ce = [Xe]4f^1 5d^1 6s^2 \quad \Longrightarrow \quad Ce^{4+} = [Xe]4f^0$$
The stable noble gas $4f^0$ configuration is why $Ce^{4+}$ exists. $Ce^{4+}$ (as ceric
ammonium sulphate) is a well known oxidising agent in volumetric analysis; it is,
however, a strong oxidant and reverts to the more stable $+3$ state.
(Terbium, Tb, and praseodymium, Pr, also show $+4$.)

(II) $+2$ OXIDATION STATE : EUROPIUM (Eu)
$$Eu = [Xe]4f^7 6s^2 \quad \Longrightarrow \quad Eu^{2+} = [Xe]4f^7$$
The extra-stable HALF-FILLED $4f^7$ configuration is why $Eu^{2+}$ exists. $Eu^{2+}$ is
a good reducing agent, being readily oxidised to $Eu^{3+}$.
(Ytterbium, Yb $\rightarrow 4f^{14}$, and samarium, Sm, also show $+2$.)

(ii) WHY TRANSITION METALS ACT AS GOOD CATALYSTS

Reason 1 - VARIABLE OXIDATION STATES.
Because the $(n-1)d$ and $ns$ orbitals are close in energy, transition metals can readily
gain or lose electrons and exist in several oxidation states. They can therefore form
UNSTABLE INTERMEDIATE COMPOUNDS with the reactants, providing an ALTERNATIVE REACTION
PATH OF LOWER ACTIVATION ENERGY. The intermediates break down to give the products and
regenerate the catalyst.
Example: In the Contact process, $V_2O_5$ oxidises $SO_2$ to $SO_3$ and is itself
reduced to $V_2O_4$, which is then re-oxidised by air.
$$SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4; \qquad 2V_2O_4 + O_2 \rightarrow 2V_2O_5$$

Reason 2 - LARGE SURFACE AREA WITH FREE VALENCIES (adsorption).
Transition metals in the finely divided state offer a large surface with free valencies
(unpaired $d$ electrons). Reactant molecules are ADSORBED on this surface, which
(a) increases their local concentration and
(b) WEAKENS THE BONDS in the adsorbed molecules,
so the reaction proceeds faster.
Examples: Fe in Haber's process, Ni in the hydrogenation of oils, Pt in the Ostwald
process.

(iii) WHY Cr HAS A HIGHER MELTING POINT THAN Mn

Step 1 - What determines the melting point of a metal.
The melting point depends on the STRENGTH OF THE METALLIC BOND, which in turn depends on
the NUMBER OF UNPAIRED ELECTRONS available for interatomic (metallic) bonding. The
greater the number of unpaired electrons, the stronger the bonding and the higher the
melting point.

Step 2 - Write the configurations.
$$Cr\,(Z=24) = [Ar]3d^5 4s^1$$
$$Mn\,(Z=25) = [Ar]3d^5 4s^2$$

Step 3 - Count the electrons available for bonding.
In CHROMIUM, all five $3d$ electrons AND the single $4s$ electron are UNPAIRED, giving
SIX unpaired electrons available for metallic bonding. The metallic bonding is therefore
very strong.
In MANGANESE, the $4s$ orbital is FULLY PAIRED $(4s^2)$, and the $3d^5$ set is exactly
HALF-FILLED, a stable and symmetrical arrangement whose electrons are reluctant to
participate in bonding. Fewer electrons are effectively available.

Step 4 - Conclusion.
Chromium has stronger metallic bonding than manganese, so
$$\text{m.p. of Cr } (2130\ K) > \text{m.p. of Mn } (1518\ K)$$
In fact Cr has the highest melting point in the 3d series, while Mn shows an
anomalous dip.

(iv) ACIDIC $KMnO_4$ SOLUTION ON STANDING

Step 1 - What is observed.
On standing, an acidified solution of potassium permanganate slowly DECOMPOSES.
The permanganate ion oxidises WATER ITSELF, liberating OXYGEN GAS, while $Mn(VII)$ is
reduced to brown $Mn(IV)$. The characteristic pink/purple colour FADES and a BROWN
PRECIPITATE (or turbidity) of $MnO_2$ appears.

Step 2 - The equation involved.
$$\boxed{4MnO_4^-(aq) + 4H^+(aq) \longrightarrow 4MnO_2(s) + 3O_2(g) + 2H_2O(l)}$$
(Check: Mn goes from $+7$ to $+4$, gaining $3e^-$ each, $4 \times 3 = 12\,e^-$; oxygen
of water goes from $-2$ to $0$, losing $12\,e^-$ in all. Balanced.)

Step 3 - What this type of reaction is called.
It is an AUTOCATALYTIC REACTION (a self-catalysed reaction).
The $MnO_2$ produced acts as a CATALYST for the further decomposition of the
permanganate. So the reaction is slow to begin with, and then speeds up as more and more
$MnO_2$ accumulates - the classic signature of autocatalysis.

Step 4 - Practical consequence.
This is exactly why a standard solution of $KMnO_4$ cannot be kept for long: it must be
freshly prepared, boiled, filtered through sintered glass to remove $MnO_2$, and then
re-standardised before use in volumetric analysis.