The D and F-Block Elements

$$\begin{gathered} E{u}^{2+}:\left[Xe\right]4f^{7}6s^0 \\ G{d}^{3+}:\left[Xe\right]4f^{7}5d^{0}6s^0 \\ E{u}^{3+}:\left[Xe\right]4f^{6}6s^0 \\ T{b}^{3+}:\left[Xe\right]4f^{8}6s^0 \\ S{m}^{2+}:\left[Xe\right]4f^{6}6s^0 \end{gathered}$$

Hence, only $E{u}^{2+}$ and $G{d}^{3+}$ have $4f^7$ configuration.

${\mathrm{Cr}}^{2+}, {\mathrm{Cu}}^{2+},$both are blue in colour.

The ionization energy depends upon the size and effective nuclear charge. when we move down the group I.E decreases and across the period the I.E increases.

The order of I.E is given as Pb>Tl >Al>In (Tl has more I.E than Al due to poor shielding of f orbital electrons)

The most stable oxidation state of Pb is +2, due to the inert pair effect, and In has a stable oxidation state is +3 .

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

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.

Among the given species, MnO₄⁻ has the highest standard reduction potential (E° ≈ +1.52 V), making it the strongest oxidizing agent. Comparatively, Cr₂O₇²⁻ (E° ≈ +1.33 V), VO₂⁺ (E° ≈ +1.00 V), and TiO₂ are weaker oxidizers. Answer: MnO₄⁻.

During the preparation of potassium permanganate from manganese dioxide, potassium manganate is formed when manganese dioxide reacts with potassium hydroxide. Potassium manganate reacts with HCl to give potassium permanganate.

$$\begin{gathered} 2{\mathrm{MnO}}_2+{\mathrm{O}}_2+4\mathrm{KOH} \to 2{\mathrm{K}}_2{\mathrm{MnO}}_4 +2{\mathrm{H}}_2\mathrm{O} \\ 3{\mathrm{K}}_2{\mathrm{MnO}}_4 +4\mathrm{HCl} \to 2{\mathrm{KMnO}}_4 +{\mathrm{MnO}}_2+4\mathrm{KCl} +2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

During the preparation of potassium permanganate from manganese dioxide, potassium manganate is formed when manganese dioxide reacts with potassium hydroxide. Potassium manganate reacts with HCl to give potassium permanganate.

$$\begin{gathered} 2{\mathrm{MnO}}_2+{\mathrm{O}}_2+4\mathrm{KOH} \to 2{\mathrm{K}}_2{\mathrm{MnO}}_4 +2{\mathrm{H}}_2\mathrm{O} \\ 3{\mathrm{K}}_2{\mathrm{MnO}}_4 +4\mathrm{HCl} \to 2{\mathrm{KMnO}}_4 +{\mathrm{MnO}}_2+4\mathrm{KCl} +2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

$$\begin{gathered} \mathrm{Gd} \left(64\right) =\left[\mathrm{Xe}\right] 4{\mathrm{f}}^{7}5{\mathrm{d}}^{1}6{\mathrm{s}}^2 \\ {\mathrm{Gd}}^{3+} = \left[\mathrm{Xe}\right] 4{\mathrm{f}}^7 \end{gathered}$$.

Both ${\mathrm{KMnO}}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ can oxidise ${\mathrm{I}}^{-}$to ${\mathrm{I}}_2$ in acidic medium.
Both ${\mathrm{KMnO}}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ can oxidize ${\mathrm{S}}^{2-}$ (sulhphide ions) to sulphur S in acidic medium.
Both ${\mathrm{KMnO}}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ easily oxidize ${\mathrm{Fe}}^{2+}$ to ${\mathrm{Fe}}^{3+}$ in acidic medium.

$Mn{O}_2+2KOH+KN{O}_3 \to K_{2}Mn{O}_4 +KN{O}_2+H_{2}O$

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 most stable oxidation state of Lanthanides is $+3.T{b}^{4+}$ is $\left[Xe\right]4f^7$. It tends to revert to the more stable oxidation state of +3 by the gain of an electron That is why $T{b}^{4+}$ is strongest oxidizing agent.

$V_2{O}_5$ reacts with alkalies to give $V{O}_4^{3-}$ In $V{O}_4^{3-}$ ion, vanadium is in +5 oxidation state.

The molecular structure of chlorine trifluoride ($Cl{F}_3$ ) is T- , following the VSEPR theory.
• Central atom: Chlorine (Cl)
• Total valence electrons: 7 + (3 × 7) = 28
• Bonded pairs: 3 fluorine atoms
• Lone pairs on Cl: 2
• Most stable structure: Trigonal bipyramidal electron geometry
• Fluorine atoms occupy axial positions.
• Two lone pairs occupy equatorial positions.
$V^{+3}, N{i}^{+2} and T{i}^{+2}$ have 2 unpaired electrons each.

${\mathrm{I}}^{-}+2{\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{IO}}_3^{-}+2{\mathrm{MnO}}_2+2{\mathrm{OH}}^{-}$

Among the given oxides, V₂O₅ (vanadium(V) oxide) is amphoteric. When it reacts with acids, it forms VO₂⁺ (dioxovanadium ion).

• The oxidation state of vanadium in V₂O₅ is +5.
• The oxidation state of vanadium in VO₂⁺ is also +5.

Since there is no change in oxidation state, the oxidation state remains +5 throughout the reaction.

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

$$\begin{gathered} \mathrm{Gd} \left(64\right) =\left[\mathrm{Xe}\right] 4{\mathrm{f}}^{7}5{\mathrm{d}}^{1}6{\mathrm{s}}^2 \\ {\mathrm{Gd}}^{3+} = \left[\mathrm{Xe}\right] 4{\mathrm{f}}^7 \end{gathered}$$.

$\left[\mathrm{Ar}\right]3\mathrm{d}6$

The correct answer is: K₂CrO₄ and K₂Cr₂O₇.

Reaction explanation:

K₂Cr₂O₇ reacts with KOH to form K₂CrO₄ (A) in an alkaline medium.

K₂CrO₄ converts back to K₂Cr₂O₇ (B) in an acidic medium (H₂SO₄).
Thus, [A] = K₂CrO₄ and [B] = K₂Cr₂O₇.

${\mathrm{V}}^{2+}, {\mathrm{Cr}}^{3+} ,{\mathrm{Mn}}^{3+}$ show violet color in aqueous solution.

Pd belongs to 5th period,whereas Os, Ir and Pt belong to 6th period

Both ${\mathrm{Cr}}^{2+}\mathrm{and} {\mathrm{Cu}}^{2+}$are blue in colour

Cerium has only 4 electrons in its outermost shell. Hence, it can only gain electrons and not lose extra electrons and act as an oxidizing agent in its +4 oxidation state

$$\begin{gathered} \mathrm{Ce} =\left[\mathrm{Xe}\right] 4{\mathrm{f}}^{1}5{\mathrm{d}}^{1}6{\mathrm{s}}^2 \\ \mathrm{Hence} , {\mathrm{Ce}}^{4+} \mathrm{acts} \mathrm{as} \mathrm{strongest} \mathrm{oxidizing} \mathrm{agent} \\ {\mathrm{Ce}}^{4+} =\left[\mathrm{Xe}\right] 4{\mathrm{f}}^{0}5{\mathrm{d}}^{0}6{\mathrm{s}}^0 \end{gathered}$$

During the preparation of potassium permanganate from manganese dioxide, potassium manganate is formed when manganese dioxide reacts with potassium hydroxide. Potassium manganate reacts with HCl to give potassium permanganate.

$$\begin{gathered} 2{\mathrm{MnO}}_2+{\mathrm{O}}_2+4\mathrm{KOH} \to 2{\mathrm{K}}_2{\mathrm{MnO}}_4 +2{\mathrm{H}}_2\mathrm{O} \\ 3{\mathrm{K}}_2{\mathrm{MnO}}_4 +4\mathrm{HCl} \to 2{\mathrm{KMnO}}_4 +{\mathrm{MnO}}_2+4\mathrm{KCl} +2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$