Q.Explain with the help of balanced chemical equation, why the solution of Ce(IV) is prepared in acidic medium.

Lanthanides almost universally show the +3 oxidation state as their most stable/common state.

Lanthanides have the general configuration [Xe]4f^(1-14) 5d^(0-1) 6s2. They readily lose the two 6s electrons and one further electron (a 5d or 4f electron) to give the +3 ion, which is the most stable and common oxidation state across the whole series. A few show +2 or +4 in addition (e.g. Eu2+, Ce4+) whe …

Cerium readily shows +4 in addition to the usual +3 lanthanoid state, because losing its single 4f electron gives it the extra-stable empty-4f configuration of Xe (like the noble-gas core).

Most lanthanoids show the characteristic +3 oxidation state as their most stable state. However, a few show an additional +2 or +4 state when it leads to a specially stable ([Xe], f$^0$, f$^7$, or f$^{14}$) electron configuration.

Cerium (Ce, Z = 58) has configuration [Xe] 4f$^1$ 5d$^1$ 6s$^2$. Losing all four of these outer electrons (to give Ce$^{4+}$) achieves the very stable empty-4f ([Xe]) configuration, so Ce$^{4+}$ is unusually stable and readily formed — cerium is the classic lanthanoid example of the +4 state (used as an oxidising agent, e.g. ceric ammonium sulfate/nitrate in volumetric analysis).

OR: Mn$^{3+}$ vs Cr$^{3+}$ paramagnetism

• Cr$^{3+}$: [Ar] 3d$^3$ → configuration $t_{2g}^3$, giving 3 unpaired electrons. …

Because the 4f orbitals are shielded by the filled $5s^25p^6$ subshells and contribute little to bonding, lanthanoids characteristically lose the two $6s$ electrons and one $4f/5d$ electron to give the common, most stable $+3$ oxidation state.

Although a few lanthanoids also show $+2$ (e.g. $Eu^{2+}$, $Yb^{2+}$, where a half-filled or fully-filled 4f subshell gives extra stability) or $+4$ (e.g. $Ce^{4+}$, $Tb^{4+}$, similarly stabilised by an empty or half-filled 4f configuration), these are exceptions restricted to specific elements. Across the whole lanthanoid series, $+3$ is the state shown by every member and is by far the most common and characteristic oxidation st …

All lanthanoids show a common, most stable oxidation state of +3, corresponding to the loss of the two $6s$ electrons and one $5d$ (or $4f$) electron.

Lanthanoids have the general electronic configuration $[Xe]4f^{1-14}5d^{0-1}6s^2$. The $+3$ oxidation state is the most common and most stable across the whole series because it arises from the loss of the two outer $6s$ electrons plus one inner electron, giving a reasonably stable configuration; it is also the state in which most lanthanoid compounds and minerals occur. A few lan …

Eu²⁺ is stabilised by a half-filled 4f⁷ configuration, so europium — not cerium — shows the +2 oxidation state.

Among the lanthanides, oxidation states other than the usual +3 arise when they lead to an especially stable f-subshell configuration (empty, half-filled, or fully filled).

Europium (Eu, Z=63): neutral Eu has configuration $[Xe]4f^7 6s^2$. Losing 2 electrons (the 6s pair) gives $Eu^{2+}$ with configuration $4f^7$ — a half-filled f-subshell, which is exceptionally stable. This extra stability makes $Eu^{2+}$ a well-known, relatively stable oxidation state for europium (alongside the normal +3).

…