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
Keywords:#Chromyl chloride test#CrO5 structure#JEE Main 2026 Morning Q75#d and f Block Elements JEE Main 2026
More d and f Block Elements Previous-Year Questions
Q63jee_main_2026_21_jan_morningCompounds of Transition Elements
MnO_4^2-$MnO_{4}^{2-}$, in acidic medium, disproportionates to :
A.Mn_2O_7text and MnO_2$Mn_{2}O_{7}\text{ and }MnO_{2}$
B.mathrmMnO_4^-text and MnO$\mathrm{MnO}_4^-\text{ and }MnO$
C.mathrmMnO_4^-text and mathrmMnO_2$\mathrm{MnO}_4^-\text{ and }\mathrm{MnO}_2$
D.mathrmMn_2mathrmO_7text and MnO$\mathrm{Mn}_{2}\mathrm{O}_{7}\text{ and }MnO$
Solution
### Related Formula
3mathrmMnO_4^2- + 4mathrmH^+ rightarrow 2mathrmMnO_4^- + mathrmMnO_2 + 2mathrmH_2mathrmO$$3\mathrm{MnO}_4^{2-} + 4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^- + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$$
### Core Logic
Manganate ion (mathrmMnO_4^2-$\mathrm{MnO}_4^{2-}$), where Mn is in +6 oxidation state, is unstable in acidic medium and undergoes disproportionation.
It oxidizes to Permanganate (mathrmMnO_4^-$\mathrm{MnO}_4^-$, +7 state) and reduces to Manganese dioxide (mathrmMnO_2$\mathrm{MnO}_2$, +4 state).
### Pattern Recognition
Manganate (green, +6) disproportionates in acid to Permanganate (purple, +7) and MnO_2$MnO_2$ (brown/black precipitate, +4).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q39jee_main_2025_03_april_eveningOxidation States and Stability of Transition Metals
Given below are two statements:
Statement I: mathrmCrO_3$\mathrm{CrO}_3$ is a stronger oxidizing agent than mathrmMoO_3$\mathrm{MoO}_3$.
Statement II: mathrmCr(VI)$\mathrm{Cr(VI)}$ is more stable than mathrmMo(VI)$\mathrm{Mo(VI)}$.
In the light of the above statements, choose the correct answer from the options given below :
A. Statement I is false but Statement II is true
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false
Solution
### Related Formula
In transition metal groups:
- Stability of higher oxidation states increases down the group:
textStability: mathrmCr(VI) < mathrmMo(VI) < mathrmW(VI)$$\text{Stability: } \mathrm{Cr(VI)} < \mathrm{Mo(VI)} < \mathrm{W(VI)}$$
- Oxidizing power is inversely proportional to the stability of the high oxidation state.
### Core Logic
Statement I Analysis:
- Since mathrmCr(VI)$\mathrm{Cr(VI)}$ is less stable than mathrmMo(VI)$\mathrm{Mo(VI)}$, chromium is easily reduced from +6$+6$ to +3$+3$, making mathrmCrO_3$\mathrm{CrO}_3$ a much stronger oxidizing agent than mathrmMoO_3$\mathrm{MoO}_3$. Statement I is True.
### Step 1: Analyze Statement II
- Statement II asserts that mathrmCr(VI)$\mathrm{Cr(VI)}$ is more stable than mathrmMo(VI)$\mathrm{Mo(VI)}$. As we go down a transition metal group, the higher oxidation states become increasingly stable due to better shielding of the core electrons and relativistic effects. Hence, mathrmMo(VI)$\mathrm{Mo(VI)}$ is more stable than mathrmCr(VI)$\mathrm{Cr(VI)}$. Statement II is False.
### Step 2: Conclusion
Therefore, Statement I is True but Statement II is False, matching Option (2).
### Pattern Recognition
For d-block elements, higher oxidation states are more stable down the group (e.g., mathrmMo(VI)$\mathrm{Mo(VI)}$ and mathrmW(VI)$\mathrm{W(VI)}$ are very stable and non-oxidizing, whereas mathrmCr(VI)$\mathrm{Cr(VI)}$ is unstable and strongly oxidizing). This is the exact opposite of p-block elements where the inert pair effect makes lower oxidation states more stable down the group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q49jee_main_2025_03_april_eveningEnthalpy of Atomisation and Magnetic Properties
Among, mathrmSc$\mathrm{Sc}$, mathrmMn$\mathrm{Mn}$, mathrmCo$\mathrm{Co}$ and mathrmCu$\mathrm{Cu}$, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ________ BM (in nearest integer).
Numerical Answer.Answer: 4 to 4
Solution
### Related Formula
Spin-only magnetic moment (mu$\mu$) is given by:
mu = sqrtn(n+2)mathrm~BM$$\mu = \sqrt{n(n+2)}\mathrm{~BM}$$
where n$n$ is the number of unpaired d-electrons.
### Core Logic
Enthalpies of atomization of the given 3d transition elements (in mathrmkJ/mol$\mathrm{kJ/mol}$):
- Scandium (mathrmSc$\mathrm{Sc}$): 326$326$
- Manganese (mathrmMn$\mathrm{Mn}$): 281$281$
- Cobalt (mathrmCo$\mathrm{Co}$): 425$425$
- Copper (mathrmCu$\mathrm{Cu}$): 339$339$
Thus, Cobalt (mathrmCo$\mathrm{Co}$) has the highest enthalpy of atomization.
### Step 1: Determine unpaired electrons in mathrmCo^2+$\mathrm{Co}^{2+}$
Electronic configuration of Cobalt (Z=27$Z=27$):
mathrmCo: [mathrmAr] 3d^7 4s^2$$\mathrm{Co}: [\mathrm{Ar}] 3d^7 4s^2$$
For divalent Cobalt ion (mathrmCo^2+$\mathrm{Co}^{2+}$):
mathrmCo^2+: [mathrmAr] 3d^7$$\mathrm{Co}^{2+}: [\mathrm{Ar}] 3d^7$$
In the d-subshell (five orbitals):
- Three orbitals are paired, and three are unpaired (n=3$n=3$).
### Step 2: Calculate spin-only magnetic moment
mu = sqrt3(3+2) = sqrt15 approx 3.87mathrm~BM$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\mathrm{~BM}$$
Rounding to the nearest integer gives 4$4$.
### Pattern Recognition
Enthalpy of atomization generally peaks near the middle of transition series due to maximum metallic bonding. However, mathrmMn$\mathrm{Mn}$ (3d^5 4s^2$3d^5 4s^2$) is an anomaly with an exceptionally low value (281mathrm~kJ/mol$281\mathrm{~kJ/mol}$) due to its highly stable half-filled d^5$d^5$ subshell configuration which reduces electron delocalization in metallic bonding.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q36jee_main_2025_07_april_morningEnthalpy of Atomisation
The number of valence electrons present in the metal among mathrmCr$\mathrm{Cr}$, mathrmCo$\mathrm{Co}$, mathrmFe$\mathrm{Fe}$ and mathrmNi$\mathrm{Ni}$ which has the lowest enthalpy of atomisation is:
A. 8
B. 9
C. 6
D. 10
Solution
### Core Logic
Let's look at the enthalpy of atomisation values for the given 3d transition metals:
- **Chromium (mathrmCr$\mathrm{Cr}$)**: 397 text kJ mol^-1$397 \text{ kJ mol}^{-1}$
- **Iron (mathrmFe$\mathrm{Fe}$)**: 416 text kJ mol^-1$416 \text{ kJ mol}^{-1}$
- **Cobalt (mathrmCo$\mathrm{Co}$)**: 425 text kJ mol^-1$425 \text{ kJ mol}^{-1}$
- **Nickel (mathrmNi$\mathrm{Ni}$)**: 430 text kJ mol^-1$430 \text{ kJ mol}^{-1}$
Among the choices, **Chromium (mathrmCr$\mathrm{Cr}$)** has the lowest enthalpy of atomisation (397 text kJ mol^-1$397 \text{ kJ mol}^{-1}$), due to a highly stable half-filled d-subshell configuration which leads to weaker metallic bonding relative to the other metals listed.
The valence electronic configuration of mathrmCr$\mathrm{Cr}$ is:
mathrmCr = [mathrmAr] 3mathrmd^5 4mathrms^1$$\mathrm{Cr} = [\mathrm{Ar}] 3\mathrm{d}^5 4\mathrm{s}^1$$
Total valence electrons = 5 + 1 = 6$= 5 + 1 = 6$.
### Pattern Recognition
In transition metals, manganese (Mn$Mn$) has the absolute lowest enthalpy of atomisation in the 3d series because of its completely half-filled d^5$d^5$ and completely filled s^2$s^2$ stability. Since Mn$Mn$ is not in the list, Chromium ("Cr"$"Cr"$) is next, having 6$6$ valence electrons.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
More d and f Block Elements Questions — jee_main_2026_21_jan_morning
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