Consider the following reactions:
NaCl + K_2Cr_2O_7 + H_2SO_4 rightarrow A + KHSO_4 + NaHSO_4 + H_2O$NaCl + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \rightarrow A + KHSO_{4} + NaHSO_{4} + H_{2}O$A + NaOH rightarrow B + NaCl + H_2O$A + NaOH \rightarrow B + NaCl + H_{2}O$B + H_2SO_4 + H_2O_2 rightarrow C + Na_2SO_4 + H_2O$B + H_{2}SO_{4} + H_{2}O_{2} \rightarrow C + Na_{2}SO_{4} + H_{2}O$
In the product 'C', 'X' is the number of O_2^2-$O_{2}^{2-}$ 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 Type:
Enter a numerical valueAnswer: 13 to 13+4 marks
Solution & Explanation
### Core Logic
The first reaction is the classical **Chromyl Chloride Test**:
4mathrmNaCl + mathrmK_2Cr_2O_7 + 6mathrmH_2SO_4 rightarrow 2mathrmCrO_2Cl_2 (textA) + 2mathrmKHSO_4 + 4mathrmNaHSO_4 + 3mathrmH_2O$4\mathrm{NaCl} + \mathrm{K_2Cr_2O_7} + 6\mathrm{H_2SO_4} \rightarrow 2\mathrm{CrO_2Cl_2} (\text{A}) + 2\mathrm{KHSO_4} + 4\mathrm{NaHSO_4} + 3\mathrm{H_2O}$
Product A is Chromyl chloride (mathrmCrO_2Cl_2$\mathrm{CrO_2Cl_2}$), a red-orange gas.
When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B):
mathrmCrO_2Cl_2 (textA) + 4mathrmNaOH rightarrow mathrmNa_2CrO_4 (textB) + 2mathrmNaCl + 2mathrmH_2O$\mathrm{CrO_2Cl_2} (\text{A}) + 4\mathrm{NaOH} \rightarrow \mathrm{Na_2CrO_4} (\text{B}) + 2\mathrm{NaCl} + 2\mathrm{H_2O}$
Acidifying the sodium chromate solution with H_2SO_4$H_2SO_4$ and adding H_2O_2$H_2O_2$ yields a deep blue solution of Chromium(VI) peroxide, CrO_5$CrO_5$ (C):
mathrmNa_2CrO_4 (textB) + mathrmH_2SO_4 + 2mathrmH_2O_2 rightarrow mathrmCrO_5 (textC) + mathrmNa_2SO_4 + 3mathrmH_2O$\mathrm{Na_2CrO_4} (\text{B}) + \mathrm{H_2SO_4} + 2\mathrm{H_2O_2} \rightarrow \mathrm{CrO_5} (\text{C}) + \mathrm{Na_2SO_4} + 3\mathrm{H_2O}$
Structure of CrO_5$CrO_5$:
Chromyl Chloride Test diagram for Q75 - JEE Main 2026 Morning
- It has a butterfly structure.
- Number of peroxy units (O_2^2-$O_2^{2-}$), 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 testrightarrow$\rightarrow$CrO_2Cl_2$CrO_2Cl_2$ (red gas). Absorbed in NaOH rightarrow$\rightarrow$Na_2CrO_4$Na_2CrO_4$ (yellow). Tested with H_2O_2/H^+$H_2O_2/H^+$rightarrow$\rightarrow$CrO_5$CrO_5$ (butterfly structure, blue, two peroxy links, Cr in +6).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 11 Chemistry: Redox Reactions
In acidic medium, K_2Cr_2O_7$K_2Cr_2O_7$ shows oxidising action as represented in the half reaction
Cr_2O_7^2- + XH^+ + Ye^- rightarrow 2A + ZH_2O$$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$$
X, Y, Z and A are respectively are:
A.8, 6, 4 text and Cr_2O_3$8, 6, 4 \text{ and } Cr_2O_3$
B.14, 7, 6 text and Cr^3+$14, 7, 6 \text{ and } Cr^{3+}$
C.8, 4, 6 text and Cr_2O_3$8, 4, 6 \text{ and } Cr_2O_3$
D.14, 6, 7 text and Cr^3+$14, 6, 7 \text{ and } Cr^{3+}$
Solution
### Core Logic
The balanced half-reaction for the dichromate ion acting as an oxidising agent in an acidic medium is:
Cr_2O_7^2- + 14H^+ + 6e^- rightarrow 2Cr^3+ + 7H_2O$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$
### Step 1: Compare with Given Equation
Comparing this with the given equation Cr_2O_7^2- + XH^+ + Ye^- rightarrow 2A + ZH_2O$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$:
X = 14$X = 14$Y = 6$Y = 6$Z = 7$Z = 7$A = Cr^3+$A = Cr^{3+}$
### Pattern Recognition
In acidic medium, dichromate (Cr_2O_7^2-$Cr_2O_7^{2-}$) always requires 14H^+$14H^+$ to balance 7O$7O$ atoms, forming 7H_2O$7H_2O$. Chromium reduces from +6 to +3 state, taking 6e^-$6e^-$ overall.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Class 11 Chemistry: Redox Reactions
Q73jee_main_2024_29_january_eveningLanthanoid Oxidation States
Which of the following acts as a strong reducing agent? (Atomic number : Ce = 58, Eu = 63, Gd = 64, Lu = 71)
A.mathrmLu^3+$\mathrm{Lu}^{3+}$
B.mathrmGd^3+$\mathrm{Gd}^{3+}$
C.mathrmEu^2+$\mathrm{Eu}^{2+}$
D.mathrmCe^4+$\mathrm{Ce}^{4+}$
Solution
### Related Formula
textElectronic configuration of mathrmEu = [mathrmXe] 4f^7 6s^2$$\text{Electronic configuration of } \mathrm{Eu} = [\mathrm{Xe}] 4f^7 6s^2$$
### Core Logic
The most common and stable oxidation state for lanthanoids is +3$+3$. In the case of Europium:
mathrmEu^2+ = [mathrmXe] 4f^7$$\mathrm{Eu}^{2+} = [\mathrm{Xe}] 4f^7$$
This configuration possesses a highly stable half-filled f$f$-subshell. However, because the +3$+3$ state is universally favored by thermodynamics in solution, textEu^2+$\text{Eu}^{2+}$ readily undergoes oxidation to lose one more electron:
mathrmEu^2+ rightarrow mathrmEu^3+ + 1e^-$$\mathrm{Eu}^{2+} \rightarrow \mathrm{Eu}^{3+} + 1e^-$$
By releasing an electron to stabilize into the +3$+3$ state, it behaves as a potent reducing agent.
### Step 1: Evaluation
Conversely, textCe^4+$\text{Ce}^{4+}$ acts as a powerful oxidizing agent to return to +3$+3$, while textLu^3+$\text{Lu}^{3+}$ and textGd^3+$\text{Gd}^{3+}$ are already perfectly configured at their native stable limits.
### Pattern Recognition
Europium(II$II$) has a stable half-filled f^7$f^7$ configuration, yet easily loses an electron to attain the highly stable +3$+3$ state typical of lanthanoids, making it a strong reducing agent.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q75jee_main_2024_29_january_eveningProperties of Zinc, Cadmium and Mercury
Which of the following statements are correct about Zn, Cd and mathrmHg$\mathrm{Hg}$ ?
A. They exhibit high enthalpy of atomization as the d-subshell is full.
B. Zn and Cd do not show variable oxidation state while Hg shows +mathrmI$+\mathrm{I}$ and +mathrmII$+\mathrm{II}$.
C. Compounds of Zn, Cd and Hg are paramagnetic in nature.
D. Zn, Cd and Hg are called soft metals.
Choose the most appropriate from the options given below:
A. B, D only
B. B, C only
C. A, D only
D. C, D only
Solution
### Related Formula
textGeneral configuration of Group 12: (n-1)d^10 ns^2$$\text{General configuration of Group 12: } (n-1)d^{10} ns^2$$
### Core Logic
Analyzing each statement based on inorganic chemistry principles:
* **Statement A is false**: Because their d$d$-subshell is completely full (d^10$d^{10}$), these elements do not form strong metallic bonds. As a result, they exhibit the *lowest* enthalpy of atomization in their respective periods.
* **Statement B is true**: textZn$\text{Zn}$ and textCd$\text{Cd}$ show only a stable +2$+2$ oxidation state, whereas textHg$\text{Hg}$ exhibits variable states forming both +1$+1$ (as textHg_2^2+$\text{Hg}_2^{2+}$) and +2$+2$.
* **Statement C is false**: With a fully paired d^10$d^{10}$ subshell, their compounds lack unpaired electrons and are explicitly diamagnetic.
* **Statement D is true**: Due to weak metallic bonds, these elements have low melting points and are classified as soft metals.
### Step 1: Selection Verification
Statements B and D are true, matching choice (1).
### Pattern Recognition
Group 12 metals have a full d^10$d^{10}$ subshell, leading to exceptionally weak metallic bonding, low enthalpies of atomization, and diamagnetic characteristics.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q75jee_main_2024_27_jan_morningQualitative Analysis of Lead
Yellow compound of lead chromate gets dissolved on treatment with hot textNaOH$\text{NaOH}$ solution. The product of lead formed is a :
A. Tetraanionic complex with coordination number six
B. Neutral complex with coordination number four
C. Dianionic complex with coordination number six
D. Dianionic complex with coordination number four
Solution
### Related Formula
Dissolution reaction pathway:
textPbCrO_4 + 4textNaOH (hot excess) rightarrow textNa_2[textPb(OH)_4] + textNa_2textCrO_4$$\text{PbCrO}_4 + 4\text{NaOH (hot excess)} \rightarrow \text{Na}_2[\text{Pb(OH)}_4] + \text{Na}_2\text{CrO}_4$$
### Core Logic
The reaction yields sodium tetrahydroxoplumbate(II), [textPb(OH)_4]^2-$[\text{Pb(OH)}_4]^{2-}$. The charge of the complex species is -2$-2$ (dianionic), and it binds 4 hydroxo coordination ligands, matching a coordination number of four.
### Chapter Mix
Class 12 Chemistry: d-and f-Block Elements
Class 12 Chemistry: Coordination Compounds
Q78jee_main_2024_27_jan_morningChromyl Chloride Test
textNaCl$\text{NaCl}$ reacts with conc. H_2SO_4$H_2SO_4$ and K_2Cr_2O_7$K_2Cr_2O_7$ to give reddish fumes (B), which react with textNaOH$\text{NaOH}$ to give yellow solution (C). (B) and (C) respectively are;
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