Some Important Compounds of Transition Elements — Revision Notes

⚡ 30-Second Revision

$KMnO_4$ (Potassium Permanganate):**

* Oxidation state of Mn: +7 * Color: Dark purple * Preparation: From $MnO_2$ (pyrolusite) \($MnO_2 \rightarrow K_2MnO_4 \rightarrow KMnO_4$\) * Reduction in acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (n-factor = 5, purple to colorless) * Reduction in neutral/weakly alkaline medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2(s) + 4OH^-$ (n-factor = 3, purple to brown ppt) * Reduction in strongly alkaline medium: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$ (n-factor = 1, purple to green) * Structure of $MnO_4^-$: Tetrahedral * Self-indicator: Yes

$K_2Cr_2O_7$ (Potassium Dichromate):**

* Oxidation state of Cr: +6 * Color: Orange-red * Preparation: From $FeCr_2O_4$ (chromite) \($FeCr_2O_4 \rightarrow Na_2CrO_4 \rightarrow Na_2Cr_2O_7 \rightarrow K_2Cr_2O_7$\) * Reduction in acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ (n-factor = 6, orange to green) * Chromate-Dichromate equilibrium: $2CrO_4^{2-} (yellow) + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (orange) + H_2O$ * Structure of $Cr_2O_7^{2-}$: Two $CrO_4$ tetrahedra sharing an oxygen * Self-indicator: No (requires external indicator)

2-Minute Revision

For NEET, focus on the two key transition metal compounds: Potassium Permanganate ($KMnO_4$) and Potassium Dichromate ($K_2Cr_2O_7$). Both are potent oxidizing agents. $KMnO_4$ is purple, with Mn in a +7 oxidation state.

Its oxidizing power and reduction product vary with pH: in acidic media, it forms colorless $Mn^{2+}$ (n-factor 5); in neutral/weakly alkaline, brown $MnO_2$ (n-factor 3); and in strongly alkaline, green $MnO_4^{2-}$ (n-factor 1).

It's a self-indicator. $K_2Cr_2O_7$ is orange, with Cr in a +6 oxidation state. In acidic media, it reduces to green $Cr^{3+}$ (n-factor 6). Remember the pH-dependent interconversion between yellow chromate ($CrO_4^{2-}$) and orange dichromate ($Cr_2O_7^{2-}$).

Unlike $KMnO_4$, $K_2Cr_2O_7$ needs an external indicator for titrations. Understand their preparation from pyrolusite ($MnO_2$) and chromite ($FeCr_2O_4$) respectively. Key structures are tetrahedral $MnO_4^-$ and two shared tetrahedra for $Cr_2O_7^{2-}$.

5-Minute Revision

To ace questions on important transition element compounds, concentrate on $KMnO_4$ and $K_2Cr_2O_7$. These are crucial for their strong oxidizing capabilities.

Potassium Permanganate ($KMnO_4$):

Properties: — Dark purple solid, forms purple solutions. Mn is in +7 oxidation state.
Preparation: — From pyrolusite ore ($MnO_2$). First, $MnO_2$ is fused with KOH and an oxidizing agent (like $O_2$) to form green potassium manganate ($K_2MnO_4$). Then, $K_2MnO_4$ is oxidized (chemically or electrolytically) to $KMnO_4$. Example: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$; $MnO_4^{2-} \rightarrow MnO_4^- + e^-$.
Redox Reactions (pH-dependent):

* Acidic: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (Purple to colorless, n-factor = 5) * Neutral/Weakly Alkaline: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2(s) + 4OH^-$ (Purple to brown precipitate, n-factor = 3) * Strongly Alkaline: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$ (Purple to green, n-factor = 1)

Structure: — $MnO_4^-$ is tetrahedral.
Indicator: — Acts as a self-indicator in titrations.

Potassium Dichromate ($K_2Cr_2O_7$):

Properties: — Orange-red crystalline solid, forms orange solutions. Cr is in +6 oxidation state.
Preparation: — From chromite ore ($FeCr_2O_4$). First, $FeCr_2O_4$ is fused with $Na_2CO_3$ and air to form yellow sodium chromate ($Na_2CrO_4$). Then, $Na_2CrO_4$ is acidified to form orange sodium dichromate ($Na_2Cr_2O_7$). Finally, $Na_2Cr_2O_7$ is treated with $KCl$ to precipitate $K_2Cr_2O_7$. Example: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$; $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$.
Redox Reactions: — Primarily used in acidic medium.

* Acidic: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ (Orange to green, n-factor = 6)

Chromate-Dichromate Equilibrium: — $2CrO_4^{2-} (yellow) + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (orange) + H_2O$. Acidification favors dichromate, basification favors chromate.
Structure: — $Cr_2O_7^{2-}$ consists of two $CrO_4$ tetrahedra sharing a common oxygen atom (bent $Cr-O-Cr$ bridge).
Indicator: — Requires an external indicator (e.g., diphenylamine) for titrations.

Key Differences: Remember $KMnO_4$ is a self-indicator, while $K_2Cr_2O_7$ is not. Their n-factors in acidic medium are 5 and 6 respectively. Pay attention to the specific color changes and reduction products for each compound under different conditions. Practice balancing redox equations and solving titration numericals.

Prelims Revision Notes

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**Potassium Permanganate ($KMnO_4$):**

* Formula: $KMnO_4$ * Color: Dark purple solid, purple solution. * Oxidation State of Mn: +7 (highest). * Preparation: From pyrolusite ore ($MnO_2$). * Step 1: Fusion of $MnO_2$ with KOH and $O_2$ (or $KNO_3$) to form potassium manganate ($K_2MnO_4$, green).

$2MnO_2 + 4KOH + O_2 \xrightarrow{\text{heat}} 2K_2MnO_4 + 2H_2O$ * Step 2: Oxidation of $K_2MnO_4$ to $KMnO_4$ (purple) using $Cl_2$, $CO_2$, or electrolytically. $3K_2MnO_4 + 2CO_2 \rightarrow 2KMnO_4 + MnO_2 + 2K_2CO_3$ $MnO_4^{2-} \xrightarrow{\text{anode oxidation}} MnO_4^- + e^-$ * Oxidizing Action (pH-dependent): * Acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (n-factor = 5, purple to colorless) * Neutral/Weakly Alkaline medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2(s) + 4OH^-$ (n-factor = 3, purple to brown ppt) * Strongly Alkaline medium: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$ (n-factor = 1, purple to green) * Structure: Tetrahedral $MnO_4^-$.

* Indicator: Self-indicator in titrations.

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**Potassium Dichromate ($K_2Cr_2O_7$):**

* Formula: $K_2Cr_2O_7$ * Color: Orange-red solid, orange solution. * Oxidation State of Cr: +6 (highest). * Preparation: From chromite ore ($FeCr_2O_4$). * Step 1: Fusion of $FeCr_2O_4$ with $Na_2CO_3$ and $O_2$ to form sodium chromate ($Na_2CrO_4$, yellow).

$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \xrightarrow{\text{heat}} 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$ * Step 2: Acidification of $Na_2CrO_4$ to form sodium dichromate ($Na_2Cr_2O_7$, orange). $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$ * Step 3: Conversion to $K_2Cr_2O_7$ by adding $KCl$ (due to lower solubility).

$Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7(s) + 2NaCl$ * Oxidizing Action: Primarily in acidic medium. * Acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ (n-factor = 6, orange to green) * Chromate-Dichromate Equilibrium: pH-dependent interconversion.

$2CrO_4^{2-} (yellow) + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (orange) + H_2O$ * Structure: Two $CrO_4$ tetrahedra sharing an oxygen atom (bent $Cr-O-Cr$ bridge). * Indicator: Requires an external indicator (e.

g., diphenylamine) for titrations.

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Key Differences: — $KMnO_4$ is self-indicating, $K_2Cr_2O_7$ is not. Different n-factors in acidic medium (5 for $KMnO_4$, 6 for $K_2Cr_2O_7$). Different pH dependence for reduction products.

Vyyuha Quick Recall

For $KMnO_4$ and $K_2Cr_2O_7$ redox products and colors:

Purple MnO$_4^-$ (Permanganate) Acidic: Colorless Mn$^{2+}$ (5e- gain) Neutral: Brown MnO$_2$ (3e- gain) Alkaline: Green MnO$_4^{2-}$ (1e- gain)

Orange Cr$_2$O$_7^{2-}$ (Dichromate) Acidic: Green Cr$^{3+}$ (6e- gain) Yellow CrO$_4^{2-}$ (Chromate) in Basic medium

Mnemonic: People Might Always Choose Mango Nectar Because Mango Always Gives More Organic Compounds And Good Chromium Yields Colorful Beauty.