Consider the following statements about manganate and permanganate ions. Identify the correct statements: (A) The geometry of both manganate and permanganate ions is tetrahedral. (B) The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively. (C) Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product. (D) Manganate ion is paramagnetic and permanganate ions is diamagnetic. (E) Acidified permanganate ion reduces oxalate, nitrite and iodide ions. Choose the correct answer from the options given below:

Related Formula

Core Logic

Manganate ion (MnO₄²⁻$\mathrm{MnO}_4^{2-}$), where Mn is in +6 oxidation state, is unstable in acidic medium and undergoes disproportionation. It oxidizes to Permanganate (MnO₄^-$\mathrm{MnO}_4^-$, +7 state) and reduces to Manganese dioxide (MnO₂$\mathrm{MnO}_2$, +4 state).

Pattern Recognition

Manganate (green, +6) disproportionates in acid to Permanganate (purple, +7) and MnO₂$MnO_2$ (brown/black precipitate, +4).

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Core Logic

The first reaction is the classical Chromyl Chloride Test: 4NaCl + K₂Cr₂O₇ + 6H₂SO₄ arrow 2CrO₂Cl₂ (A) + 2KHSO₄ + 4NaHSO₄ + 3H₂O$4\mathrm{NaCl} + \mathrm{K_2Cr_2O_7} + 6\mathrm{H_2SO_4} \rightarrow 2\mathrm{CrO_2Cl_2} (\text{A}) + 2\mathrm{KHSO_4} + 4\mathrm{NaHSO_4} + 3\mathrm{H_2O}$ Product A is Chromyl chloride (CrO₂Cl₂$\mathrm{CrO_2Cl_2}$), a red-orange gas.

When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B): CrO₂Cl₂ (A) + 4NaOH arrow Na₂CrO₄ (B) + 2NaCl + 2H₂O$\mathrm{CrO_2Cl_2} (\text{A}) + 4\mathrm{NaOH} \rightarrow \mathrm{Na_2CrO_4} (\text{B}) + 2\mathrm{NaCl} + 2\mathrm{H_2O}$

Acidifying the sodium chromate solution with H₂SO₄$H_2SO_4$ and adding H₂O₂$H_2O_2$ yields a deep blue solution of Chromium(VI) peroxide, CrO₅$CrO_5$ (C): Na₂CrO₄ (B) + H₂SO₄ + 2H₂O₂ arrow CrO₅ (C) + Na₂SO₄ + 3H₂O$\mathrm{Na_2CrO_4} (\text{B}) + \mathrm{H_2SO_4} + 2\mathrm{H_2O_2} \rightarrow \mathrm{CrO_5} (\text{C}) + \mathrm{Na_2SO_4} + 3\mathrm{H_2O}$

Structure of CrO₅$CrO_5$:

• It has a butterfly structure.
• Number of peroxy units (O₂²⁻$O_2^{2-}$), X = 2$X = 2$.
• Total number of oxygen atoms, Y = 5$Y = 5$.
• Oxidation state of Cr, Z = +6$Z = +6$.

Sum: X + Y + Z = 2 + 5 + 6 = 13$X + Y + Z = 2 + 5 + 6 = 13$.

Step 1: Final Calculation

X + Y + Z = 13$X + Y + Z = 13$

Pattern Recognition

Chromyl chloride test arrow$\rightarrow$ CrO₂Cl₂$CrO_2Cl_2$ (red gas). Absorbed in NaOH arrow$\rightarrow$ Na₂CrO₄$Na_2CrO_4$ (yellow). Tested with H₂O₂/H^+$H_2O_2/H^+$ arrow$\rightarrow$ CrO₅$CrO_5$ (butterfly structure, blue, two peroxy links, Cr in +6).

Chapter Mix

Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

Core Logic

• A is correct: Mn₂O₇$\text{Mn}_2\text{O}_7$ features Mn in +7 state (its highest oxidation state).
• B is correct: Oxygen stabilizes high oxidation states via multiple bonding.
• C is incorrect: Mn₂O₇$\text{Mn}_2\text{O}_7$ is a covalent green oil/oxide, not ionic.
• D is correct: Structure consists of two MnO₄$\text{MnO}_4$ tetrahedra sharing one bridging oxygen atom (O₃Mn-O-MnO₃$\text{O}_3\text{Mn}-\text{O}-\text{MnO}_3$).

Step 1: Final Conclusion

Statements A, B and D are correct, matching option (2).

Pattern Recognition

Sees: Properties and bonding of transition metal oxides like Mn₂O₇$\text{Mn}_2\text{O}_7$. Trap: Assuming high oxidation state oxides of transition metals are ionic.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

Related Formula

Core Logic

The first-row transition metal that does not liberate H₂$H_{2}$ gas from dilute HCl is Copper (Cu), because its standard reduction potential is positive (E°Cu²⁺/Cu = +0.34 V$E^{\circ}_{Cu^{2+}/Cu} = +0.34\text{ V}$).

When CuSO₄$CuSO_{4}$ is treated with excess KCN, it forms a very stable soluble cyano complex:

2Cu(CN)₂ arrow 2CuCN + (CN)₂$2Cu(CN)_{2} \rightarrow 2CuCN + (CN)_{2}$ CuCN + 3KCN arrow K₃[Cu(CN)₄]$CuCN + 3KCN \rightarrow K_{3}[Cu(CN)_{4}]$

The complex ion [Cu(CN)₄]³⁻$[Cu(CN)_{4}]^{3-}$ is highly stable (a perfect complex). When H₂S$H_{2}S$ is passed through this solution, it does not yield sufficient Cu⁺$Cu^{+}$ ions to exceed the solubility product (Kₛₚ$K_{sp}$) of Cu₂S$Cu_{2}S$.

Step 1: Final Conclusion

Since no copper sulphide precipitates, the amount of MS formed is 0 moles.

Pattern Recognition

Cu and Cd separation: Cu²⁺$Cu^{2+}$ forms a very stable cyanide complex that does not precipitate with H₂S$H_2S$, whereas Cd²⁺$Cd^{2+}$ forms a less stable complex that does precipitate as CdS$CdS$.

Chapter Mix

Class 12 Chemistry: d and f Block Elements Class 12 Chemistry: Coordination Compounds

Related Formula

Core Logic

Step 1: Compare IE₁$IE_1$:

Hence, IE₁(Cr) < IE₁(Mn)$IE_1(\text{Cr}) < IE_1(\text{Mn})$ (Statement-I is TRUE).

Step 2: Compare IE₂$IE_2$ and IE₃$IE_3$:

Hence, IE₃(Cr) < IE₃(Mn)$IE_3(\text{Cr}) < IE_3(\text{Mn})$. Thus Statement-II is FALSE.

Pattern Recognition

Sees: Ionization enthalpy comparison of Cr and Mn. Shortcut: Stable 3d⁵$3d^5$ configuration in Cr^+$\text{Cr}^+$ makes IE₂$IE_2$ very high, whereas 3d⁵$3d^5$ in Mn²⁺$\text{Mn}^{2+}$ makes IE₃$IE_3$ of Mn higher than Cr.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements Class 11 Chemistry: Classification of Elements and Periodicity in Properties