KMnO₄$KMnO_4$ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y$X + Y$ is ______.
Numerical Answer Type:
Enter a numerical valueAnswer: 10 to 10+4 marks
Solution & Explanation
Core Logic
Let's resolve both components step by step:
Finding X: In an acidic medium, the permanganate ion (KMnO₄$KMnO_4$, where Mn$Mn$ is in the +7$+7$ state) is reduced to the divalent manganese cation (Mn²⁺$Mn^{2+}$, state +2$+2$):
X = 7 - 2 = 5$X = 7 - 2 = 5$
Finding Y: During qualitative salt analysis, the acetate ion reacts with neutral ferric chloride to produce a characteristic blood-red coordination solution. Boiling this solution throws down a brown-red precipitate of basic ferric acetate, [Fe(OH)₂(CH₃COO)]$[Fe(OH)_2(CH_3COO)]$. In this complex, Iron retains its +3$+3$ oxidation state:
Fe³⁺ [Ar] 3d⁵ 4s⁰ Number of d-electrons (Y) = 5$$Fe^{3+} \implies [Ar] 3d^5 4s^0 \implies \text{Number of d-electrons } (Y) = 5$$
Summing the values yields:
X + Y = 5 + 5 = 10$$X + Y = 5 + 5 = 10$$
Pattern Recognition
This problem elegantly links standard redox transitions with qualitative inorganic salt tests. Remember that throughout the basic ferric acetate precipitation test, Iron remains steadily in its ferric +3$+3$ (d⁵$d^5$) core configuration.
Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Inorganic Qualitative Analysis
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
Q75jee_main_2026_21_jan_morningChromyl Chloride Test and Chromate Chemistry
Consider the following reactions:
NaCl + K₂Cr₂O₇ + H₂SO₄ arrow A + KHSO₄ + NaHSO₄ + H₂O$NaCl + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \rightarrow A + KHSO_{4} + NaHSO_{4} + H_{2}O$A + NaOH arrow B + NaCl + H₂O$A + NaOH \rightarrow B + NaCl + H_{2}O$B + H₂SO₄ + H₂O₂ arrow C + Na₂SO₄ + H₂O$B + H_{2}SO_{4} + H_{2}O_{2} \rightarrow C + Na_{2}SO_{4} + H_{2}O$
In the product 'C', 'X' is the number of O₂²⁻$O_{2}^{2-}$ units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X + Y + Z$X + Y + Z$ is ____.
Numerical Answer.Answer: 13 to 13
Solution
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$CrO_5$:
Chromyl Chloride Test diagram for Q75 - JEE Main 2026 Morning
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 testarrow$\rightarrow$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
Q66jee_main_2026_21_jan_eveningOxides of Manganese and Properties
Given below are some of the statements about Mn$\text{Mn}$ and Mn₂O₇$\text{Mn}_2\text{O}_7$. Identify the correct statements:
A. Mn forms the oxide Mn₂O₇$\text{Mn}_2\text{O}_7$ in which Mn is in its highest oxidation state.
B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn.
C. Mn₂O₇$\text{Mn}_2\text{O}_7$ is an ionic oxide.
D. The structure of Mn₂O₇$\text{Mn}_2\text{O}_7$ consists of one bridged oxygen.
Choose the correct answer from the options given below:
A.(1) A, B, C and D$(1) \text{ A, B, C and D}$
B.(2) A, B and D Only$(2) \text{ A, B and D Only}$
C.(3) A, C and D Only$(3) \text{ A, C and D Only}$
D.(4) A, B and C Only$(4) \text{ A, B and C Only}$
Solution
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
Q58jee_main_2026_22_january_morningReactions of Transition Metals
A first row transition metal (M) does not liberate H₂$H_{2}$ gas from dilute HCl. 1 mol of aqueous solution of MSO₄$MSO_{4}$ is treated with excess of aqueous KCN and then H₂S(g)$H_{2}S(g)$ is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is ____ mol.
A.2$\text{2}$
B.1$\text{1}$
C.3$\text{3}$
D.0$\text{0}$
Solution
Related Formula
Cu²⁺ + 4CN⁻ arrow [Cu(CN)₄]³⁻ (after redox with CN^-)$$Cu^{2+} + 4CN^{-} \rightarrow [Cu(CN)_{4}]^{3-} \quad (\text{after redox with } CN^-)$$
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:
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
Q54jee_main_2026_22_january_eveningIonization Enthalpy Trends in Transition Metals
Given below are two statements:
Statement-I: The first ionization enthalpy of Cr is lower than that of Mn.
Statement-II: The second and third ionization enthalpies of Cr are higher than those of Mn.
In the light of the above statements, choose the correct answer from the options given below:
Cr: 4s¹ Removal of single 4s electron requires less energy than removing 4s² in Mn.$$\text{Cr}: 4s^1 \implies \text{Removal of single } 4s \text{ electron requires less energy than removing } 4s^2 \text{ in Mn.}$$
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$IE_3$:
Cr^+ = 3d⁵ stable half-filled configuration d⁵, so IE₂(Cr) > IE₂(Mn)$$\text{Cr}^+ = 3d^5 \implies \text{stable half-filled configuration } d^5, \text{ so } IE_2(\text{Cr}) > IE_2(\text{Mn})$$For IE₃, Mn²⁺ = 3d⁵ removing electron from stable 3d⁵ in Mn²⁺ requires more energy than Cr²⁺ (3d⁴).$$\text{For } IE_3, \text{Mn}^{2+} = 3d^5 \implies \text{removing electron from stable } 3d^5 \text{ in Mn}^{2+} \text{ requires more energy than Cr}^{2+} (3d^4).$$
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
More The d- and f-Block Elements Questions — jee_main_2025_04_april_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
