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
### Core Logic
The noble gas configuration of Xenon (Z=54$Z=54$) provides the primary core layout. For Neodymium (Z=60$Z=60$), the 6 remaining valence electrons distribute into the inner 4textf$4\text{f}$ orbital subshell rather than filling the 5textd$5\text{d}$ subshell due to shielding effects. This results in an absolute atomic ground state electronic configuration of text[Xe] 4textf^4 6texts^2$\text{[Xe]} 4\text{f}^4 6\text{s}^2$.
### Pattern Recognition
Lanthanide filling sequences generally bypass 5d$5d$ progression except for specific exceptions (La, Gd, Lu).
### Chapter Mix
Class 12 Chemistry: d-and f-Block Elements
Q63jee_main_2024_29_jan_morningPotassium Dichromate and Chromyl Chloride Test
In chromyl chloride test for confirmation of Cl^-$Cl^-$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10\%$10\%$H_2O_2$H_2O_2$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is
A.+6$+6$
B.+5$+5$
C.+10$+10$
D.+3$+3$
Solution
### Core Logic
The reaction sequence for the chromyl chloride test is:
Cl^- + K_2Cr_2O_7 + H_2SO_4 rightarrow CrO_2Cl_2$$Cl^- + K_2Cr_2O_7 + H_2SO_4 \rightarrow CrO_2Cl_2$$
The chromyl chloride gas is then passed through a basic medium (like NaOH$NaOH$) to form a yellow solution of chromate ions:
CrO_2Cl_2 xrightarrowtextBasic medium CrO_4^2- + Cl^-$$CrO_2Cl_2 \xrightarrow{\text{Basic medium}} CrO_4^{2-} + Cl^-$$
Acidification of the yellow CrO_4^2-$CrO_4^{2-}$ solution followed by the addition of H_2O_2$H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO_5$CrO_5$).
CrO_4^2- xrightarrow[textyellow solution,
1. textAcidification CrO_5 text (blue compound)$$CrO_4^{2-} \xrightarrow[\text{yellow solution},
{1. \text{Acidification}} CrO_5 \text{ (blue compound)}$$
### Step 1: Oxidation State Calculation
Potassium Dichromate and Chromyl Chloride Test diagram for Q63 - JEE Main 2024 Morning
The structure of chromium pentoxide (CrO_5$CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O^2-$O^{2-}$) and four peroxide oxygens (O_2^2-$O_2^{2-}$). Therefore, there are 2 peroxo linkages.
Let the oxidation state of Chromium be x$x$.
x + 1(-2) + 4(-1) = 0$$x + 1(-2) + 4(-1) = 0$$x - 2 - 4 = 0$x - 2 - 4 = 0$x = +6$x = +6$
Thus, the oxidation state of Cr in CrO_5$CrO_5$ is +6$+6$.
### Pattern Recognition
A classic oxidation state trap. Calculating simply via formula CrO_5$CrO_5$ yields x - 10 = 0 implies x = +10$x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 11 Chemistry: Redox Reactions
### Core Logic
Potassium permanganate (KMnO_4$KMnO_4$) is a strong oxidizing agent. When heated to 513mathrmK$513\mathrm{K}$, it undergoes thermal decomposition to give potassium manganate (K_2MnO_4$K_2MnO_4$), manganese dioxide (MnO_2$MnO_2$), and oxygen gas (O_2$O_2$).
The balanced chemical equation is:
2KMnO_4 xrightarrowDelta K_2MnO_4 + MnO_2 + O_2$$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$$
### Step 1: Final Identification
The products formed along with O_2$O_2$ are K_2MnO_4$K_2MnO_4$ (green) and MnO_2$MnO_2$ (black).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
In alkaline medium. MnO_4^-$MnO_4^-$ oxidises I^-$I^-$ to
A.IO_4^-$IO_4^-$
B.IO^-$IO^-$
C.I_2$I_2$
D.IO_3^-$IO_3^-$
Solution
### Core Logic
The behavior of the permanganate ion (MnO_4^-$MnO_4^-$) varies with the pH of the medium.
In a faintly alkaline or neutral medium, MnO_4^-$MnO_4^-$ oxidizes iodide (I^-$I^-$) completely to iodate (IO_3^-$IO_3^-$) while getting reduced to manganese dioxide (MnO_2$MnO_2$).
The balanced ionic equation is:
2MnO_4^- + H_2O + I^- rightarrow 2MnO_2 + 2OH^- + IO_3^-$$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$$
### Pattern Recognition
Rule of thumb for I^-$I^-$ oxidation by KMnO_4$KMnO_4$:
In acidic medium: I^- rightarrow I_2$I^- \rightarrow I_2$
In alkaline/neutral medium: I^- rightarrow IO_3^-$I^- \rightarrow IO_3^-$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q70jee_main_2024_30_january_eveningProperties of Transition Metal Compounds
The orange colour of K_2Cr_2O_7$K_2Cr_2O_7$ and purple colour of KMnO_4$KMnO_4$ is due to
B.textd rightarrow textd transition in KMnO_4 text and charge transfer transitions in K_2textCr_2textO_7$\text{d} \rightarrow \text{d transition in KMnO}_4 \text{ and charge transfer transitions in K}_2\text{Cr}_2\text{O}_7$
C.textd rightarrow textd transition in K_2textCr_2textO_7 text and charge transfer transitions in KMnO_4.$\text{d} \rightarrow \text{d transition in K}_2\text{Cr}_2\text{O}_7 \text{ and charge transfer transitions in KMnO}_4.$
D.textd rightarrow textd transition in both.$\text{d} \rightarrow \text{d transition in both.}$
Solution
### Core Logic
In K_2Cr_2O_7$K_2Cr_2O_7$, Chromium is in the +6$+6$ oxidation state, which means its electronic configuration is d^0$d^0$. Since there are no d-electrons, d-d$d-d$ transitions cannot occur. The orange color is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium.
Similarly, in KMnO_4$KMnO_4$, Manganese is in the +7$+7$ oxidation state, which also corresponds to a d^0$d^0$ configuration. Again, no d-d$d-d$ transitions are possible. The intense purple color is due to ligand-to-metal charge transfer (LMCT) from oxygen to manganese.
### Step 1: Final Conclusion
Both compounds owe their colors to charge transfer transitions.
### Pattern Recognition
Compounds of transition metals in their highest oxidation states (where they have d^0$d^0$ configurations, like Cr^+6$Cr^{+6}$, Mn^+7$Mn^{+7}$, V^+5$V^{+5}$) are deeply colored primarily due to Charge Transfer spectra, NOT d-d transitions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d and f Block Elements
More d and f Block Elements Questions — jee_main_2026_21_jan_morning
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