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 ____.

Core Logic

An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons). Let's write the electronic configuration for the elements in question.

Step 1: Checking configurations

Cerium (Ce$Ce$, Z=58): [Xe] 4f¹ 5d¹ 6s²$[Xe] 4f^1 5d^1 6s^2$ arrow Ce⁴⁺: [Xe] 4f⁰$\rightarrow Ce^{4+}: [Xe] 4f^0$ (0 unpaired electrons arrow$\rightarrow$ Diamagnetic)

Lanthanum (La$La$, Z=57): [Xe] 4f⁰ 5d¹ 6s²$[Xe] 4f^0 5d^1 6s^2$ arrow La³⁺: [Xe] 4f⁰$\rightarrow La^{3+}: [Xe] 4f^0$ (0 unpaired electrons arrow$\rightarrow$ Diamagnetic)

Pattern Recognition

Ions with an empty f-subshell (f⁰$f^0$, e.g., La³⁺, Ce⁴⁺$La^{3+}, Ce^{4+}$) or a completely filled f-subshell (f¹⁴$f^{14}$, e.g., Lu³⁺, Yb²⁺$Lu^{3+}, Yb^{2+}$) are invariably diamagnetic.

Chapter Mix

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

Core Logic

Let's determine the electronic configuration for each species by first writing the neutral atom's configuration, and then removing electrons starting from the outermost 4s$4s$ orbital.

(A) Cr$Cr$ (Z=24): [Ar] 3d⁵ 4s¹$[Ar] 3d^5 4s^1$ arrow$\rightarrow$ Cr²⁺$Cr^{2+}$: [Ar] 3d⁴$[Ar] 3d^4$ (B) Mn$Mn$ (Z=25): [Ar] 3d⁵ 4s²$[Ar] 3d^5 4s^2$ arrow$\rightarrow$ Mn^+$Mn^+$: [Ar] 3d⁵ 4s¹$[Ar] 3d^5 4s^1$ (C) Ni$Ni$ (Z=28): [Ar] 3d⁸ 4s²$[Ar] 3d^8 4s^2$ arrow$\rightarrow$ Ni²⁺$Ni^{2+}$: [Ar] 3d⁸$[Ar] 3d^8$ (D) V$V$ (Z=23): [Ar] 3d³ 4s²$[Ar] 3d^3 4s^2$ arrow$\rightarrow$ V^+$V^+$: [Ar] 3d³ 4s¹$[Ar] 3d^3 4s^1$

Step 1: Match execution

A arrow$\rightarrow$ III B arrow$\rightarrow$ II C arrow$\rightarrow$ I D arrow$\rightarrow$ IV

Chapter Mix

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

Core Logic

(A) Mn₂O₇$Mn_2O_7$ is a covalent oxide and exists as a green oil at room temperature. (Correct) (B) V₂O₄$V_2O_4$ dissolves in acids to give VO²⁺$VO^{2+}$ (vanadyl) salts, not VO₂²⁺$VO_2^{2+}$. (Incorrect) (C) CrO$CrO$ has chromium in the +2$+2$ oxidation state. Lower oxidation state metal oxides are typically basic in nature. (Correct) (D) V₂O₅$V_2O_5$ is an amphoteric oxide; it reacts with both acids as well as bases. (Incorrect)

Step 1: Final Selection

Only statements A and C are correct, which corresponds to option (2).

Chapter Mix

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

Related Formula

Core Logic

This reaction represents the Chromyl Chloride test used to detect the presence of chloride ions. When potassium dichromate is heated with a metal chloride in concentrated sulfuric acid, red vapors of chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$) are evolved.

Step 1: Oxidation State Calculation

In chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$): Let the oxidation state of Chromium be x$x$. Oxygen is typically -2$-2$ and Chlorine is -1$-1$.

x - 4 - 2 = 0$x - 4 - 2 = 0$ x = +6$x = +6$ Thus, the oxidation state of Chromium in the product is 6$6$.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements Class 11 Chemistry: Practical Chemistry

Step 1: Statement A Analysis

CrO₄²⁻$CrO_4^{2-}$ (chromate ion) is tetrahedral, not square planar. Statement A is incorrect.

Step 2: Statement B Analysis

2Na₂CrO₄ + 2H^+ arrow Na₂Cr₂O₇ + 2Na^+ + H₂O$2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$. Dichromates are indeed prepared from chromates. Statement B is correct.

Step 3: Statement C Analysis

The green manganate ion (MnO₄²⁻$MnO_4^{2-}$) has manganese in the +6 oxidation state (3d¹$3d^1$). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect.

Step 4: Statement D Analysis

Dark green coloured K₂MnO₄$K_2MnO_4$ undergoes disproportionation in neutral or acidic media to yield permanganate (MnO₄^-$MnO_4^-$) and manganese dioxide (MnO₂$MnO_2$). Statement D is correct.

Step 5: Statement E Analysis

Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct.

Final Conclusion

The correct statements are B, D, and E.

Chapter Mix

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