The D and F-Block Elements

Ce⁴⁺ has a 4f⁰ configuration (no unpaired electrons) and Yb²⁺ has a 4f¹⁴ configuration (all f‐orbitals filled). Both ions are diamagnetic.

$2{\mathrm{KMnO}}_4 \to {\mathrm{K}}_2{\mathrm{MnO}}_4 + {\mathrm{MnO}}_2 + {\mathrm{O}}_2$

${\mathrm{K}}_2{\mathrm{MnO}}_4$ will form on heating ${\mathrm{KMnO}}_4$ which has green colour.

Ce⁴⁺ has a 4f⁰ configuration (no unpaired electrons) and Yb²⁺ has a 4f¹⁴ configuration (all f‐orbitals filled). Both ions are diamagnetic.

$\left[\mathrm{Xe}\right]4{\mathrm{f}}^{\mathit{7}}5{\mathrm{d}}^{\mathit{1}}6{\mathrm{s}}^{\mathit{2}}$

$\left[\mathrm{Xe}\right]4{\mathrm{f}}^{\mathit{7}}5{\mathrm{d}}^{\mathit{1}}6{\mathrm{s}}^{\mathit{2}}$

$\begin{array}{lll}{\mathrm{Ti}}^{3+}& \begin{array}{ll}:& 3{\mathrm{d}}^1\end{array}& \mathrm{n}=1\\ {\mathrm{Cr}}^{2+}& \begin{array}{ll}:& 3{\mathrm{d}}^4\end{array}& \mathrm{n}=4\\ {\mathrm{Mn}}^{2+}& \begin{array}{ll}:& 3{\mathrm{d}}^5\end{array}& \mathrm{n}=5\\ {\mathrm{Fe}}^{2+}& \begin{array}{ll}:& 3{\mathrm{d}}^6\end{array}& \mathrm{n}=4\\ {\mathrm{Sc}}^{3+}& \begin{array}{ll}:& 3{\mathrm{d}}^0\end{array}& \mathrm{n}=0\end{array}$

The number of unpaired electron (n) in 3d orbital is same for $C{r}^{2+}$ and $F{e}^{2+}$. They will have the same spin-only magnetic moment.

$\mu =\sqrt{n(n+2) } =\sqrt{4(4+2)} =\sqrt{24} =4.90 BM$

$\begin{array}{lll}{\mathrm{Ti}}^{3+}& \begin{array}{ll}:& 3{\mathrm{d}}^1\end{array}& \mathrm{n}=1\\ {\mathrm{Cr}}^{2+}& \begin{array}{ll}:& 3{\mathrm{d}}^4\end{array}& \mathrm{n}=4\\ {\mathrm{Mn}}^{2+}& \begin{array}{ll}:& 3{\mathrm{d}}^5\end{array}& \mathrm{n}=5\\ {\mathrm{Fe}}^{2+}& \begin{array}{ll}:& 3{\mathrm{d}}^6\end{array}& \mathrm{n}=4\\ {\mathrm{Sc}}^{3+}& \begin{array}{ll}:& 3{\mathrm{d}}^0\end{array}& \mathrm{n}=0\end{array}$

The same number of unpaired electron (n) in 3d orbital is same for $C{r}^{2+}$ and $F{e}^{2+}$. They will have same spin-only magnetic moment.

Higher‐oxidation‐state lanthanide ions such as Ce⁴⁺ and Tb⁴⁺ readily get reduced to more stable Ln³⁺ states, making them oxidizing agents. The other listed ions are either already in their more stable +3 state (La³⁺, Lu³⁺) or not favored to act as oxidizers (Eu²⁺, Yb²⁺).

MnO₄²⁻ : the oxidation state of Mn: $x +4(-2) =-2 \Rightarrow x-8 =-2 \Rightarrow x=+6$

MnO₄⁻ : the oxidation state of Mn: $x+4(-2) =-1 \Rightarrow x=+7$

MnO₂: the oxidation state of Mn: $x+2(-2) =0 \Rightarrow x=+4$

Oxidation states appearing in the reaction are:
+4, +6, +7

Now, let's see which one is not exhibited:

A. +6 → exhibited

B. +2 → no

C. +4 → exhibited

D. +7 → exhibited

E. +3 → no

Higher‐oxidation‐state lanthanide ions such as Ce⁴⁺ and Tb⁴⁺ readily get reduced to more stable Ln³⁺ states, making them oxidizing agents. The other listed ions are either already in their more stable +3 state (La³⁺, Lu³⁺) or not favored to act as oxidizers (Eu²⁺, Yb²⁺).

The value of magnetic moment is calculated using the formula:

$\mathrm{\mu }= \sqrt{\mathrm{n}(\mathrm{n}+ 2)} \mathrm{B}.\mathrm{M}.$

where n = number of unpaired electrons.

Zn²⁺: [Ar] 3d¹⁰ → n = 0 → μ = 0 BM

Ti²⁺: [Ar] 3d² → n = 2 → μ = √8 ≈ 2.83 BM

Different values.

Cr²⁺: [Ar] 3d⁴ → n = 4 → μ = √24 ≈ 4.90 BM
Fe²⁺: [Ar] 3d⁶ → n = 4 → μ = √24 ≈ 4.90 BM
Same magnetic moment.

Ti³⁺: [Ar] 3d¹ → n = 1 → μ = √3 ≈ 1.73 BM
Cu²⁺: [Ar] 3d⁹ → n = 1 → μ = √3 ≈ 1.73 BM
Same magnetic moment.

V²⁺: [Ar] 3d³ → n = 3 → μ = √15 ≈ 3.87 BM
Cu⁺: [Ar] 3d¹⁰ → n = 0 → μ = 0 BM
Different values.