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 — Page 4
Q37jee_main_2025_24_jan_morningPreparation and Properties of Potassium Permanganate
Preparation of potassium permanganate from mathrmMnO_2$\mathrm{MnO}_2$ involves two step process in which the 1^textst$1^{\text{st}}$ step is a reaction with KOH and mathrmKNO_3$\mathrm{KNO}_3$ to produce
### Related Formula
2MnO_2 + 4KOH + O_2 xrightarrowKNO_3 2K_2MnO_4 + 2H_2O$$2MnO_2 + 4KOH + O_2 \xrightarrow{KNO_3} 2K_2MnO_4 + 2H_2O$$
### Core Logic
The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore (MnO_2$MnO_2$). Fusing the solid reactant directly along an alkaline base payload (KOH$KOH$) combined explicitly with an oxidizing carrier (KNO_3$KNO_3$) yields the intermediate green product, **potassium manganate** (K_2MnO_4$K_2MnO_4$).
### Pattern Recognition
Step 1 yields the +6$+6$ green compound (K_2MnO_4$K_2MnO_4$); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple +7$+7$ agent (KMnO_4$KMnO_4$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q27jee_main_2025_28_jan_eveningOxides and Oxoanions of Transition Metals
The amphoteric oxide among V_2O_3$V_{2}O_{3}$, V_2O_4$V_{2}O_{4}$ and V_2O_5$V_{2}O_{5}$ upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :
A.+3$+3$
B.+7$+7$
C.+5$+5$
D.+4$+4$
Solution
### Related Formula
Oxidation state equation for an oxoanion VO_4^3-$VO_4^{3-}$:
x + 4(-2) = -3$x + 4(-2) = -3$
### Core Logic
Among the given oxides of Vanadium:
- V_2O_3$V_2O_3$ is basic.
- V_2O_4$V_2O_4$ is less basic / amphoteric.
- V_2O_5$V_2O_5$ is predominantly amphoteric (reacts with both acids and alkalies).
When V_2O_5$V_2O_5$ reacts with an alkali, it forms the orthovanadate ion (VO_4^3-$VO_4^{3-}$).
### Step 1: Finding the Oxidation State
In VO_4^3-$VO_4^{3-}$ ion:
x - 8 = -3 implies x = +5$$x - 8 = -3 \implies x = +5$$
Thus, the oxidation state of Vanadium in the resulting oxide anion is +5$+5$.
### Pattern Recognition
As the oxidation state of a transition metal increases, its oxide shifts from basic to amphoteric to acidic. V_2O_5$V_2O_5$ has the highest oxidation state (+5$+5$) here and dissolves in alkali to retain its +5$+5$ oxidation state in VO_4^3-$VO_4^{3-}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Q46jee_main_2025_28_jan_eveningMagnetic Properties and Oxidation States
The spin only magnetic moment (mu)$(\mu)$ value (B.M.) of the compound with strongest oxidising power among Mn_2O_3$Mn_{2}O_{3}$, TiO and VO is ______ B.M. (Nearest integer).
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
Spin-only magnetic moment expression:
mu = sqrtn(n+2)mathrm\ B.M.$$\mu = \sqrt{n(n+2)}\mathrm{\ B.M.}$$
### Core Logic
Evaluating the oxidation states and stability profiles:
- In TiO$TiO$: Ti^2+$Ti^{2+}$
- In VO$VO$: V^2+$V^{2+}$
- In Mn_2O_3$Mn_2O_3$: Mn^3+$Mn^{3+}$Mn^3+$Mn^{3+}$ possesses a very high reduction potential (E^circ_Mn^3+/Mn^2+ = +1.57mathrm\ V$E^\circ_{Mn^{3+}/Mn^{2+}} = +1.57\mathrm{\ V}$), making it an exceptionally strong oxidizing agent because it easily gains an electron to form stable Mn^2+$Mn^{2+}$ (d^5$d^5$ configuration).
### Step 1: Calculate the Magnetic Moment of Mn(III)
Electronic configuration of Mn^3+$Mn^{3+}$:
Mn^3+ = [Ar]3d^4 implies n = 4text unpaired electrons$$Mn^{3+} = [Ar]3d^4 \implies n = 4\text{ unpaired electrons}$$
Calculating the spin-only magnetic moment:
mu = sqrt4(4+2) = sqrt24 approx 4.89mathrm\ B.M.$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89\mathrm{\ B.M.}$$
### Step 2: Rounding to Nearest Integer
Rounding 4.89mathrm\ B.M.$4.89\mathrm{\ B.M.}$ to the nearest integer gives 5$5$.
### Pattern Recognition
High reduction potentials are strongly tied to manganese in its +3$+3$ oxidation state. To quickly estimate magnetic moments, remember that a system with n$n$ unpaired electrons always results in a value of 'n.textsomething$n.\text{something}$' B.M. Thus, 4$4$ unpaired electrons rightarrow 4.89mathrm\ B.M.$\rightarrow 4.89\mathrm{\ B.M.}$, which rounds up to 5$5$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Qjee_main_2025_29_jan_morningMelting Points of Transition Elements
The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :
### Formulas Used
Melting point trends in 3d$3d$, 4d$4d$, and 5d$5d$ series transition metals depend on the extent of metallic bonding and d$d$-electron participation.
### Core Logic
According to NCERT transition element periodic trends:
* **3d$3d$ Series (mathrmMn$\mathrm{Mn}$ vs mathrmFe$\mathrm{Fe}$)**: Manganese (mathrmMn$\mathrm{Mn}$, 3d^5 4s^2$3d^5 4s^2$) has an abnormally low melting point compared to Iron (mathrmFe$\mathrm{Fe}$, 3d^6 4s^2$3d^6 4s^2$) because its stable, half-filled d^5$d^5$ configuration holds d$d$-electrons more tightly, reducing their participation in metallic bonding rightarrow mathbfmathrmMn < mathrmFe$\rightarrow \mathbf{\mathrm{Mn} < \mathrm{Fe}}$.
* **4d$4d$ Series (mathrmTc$\mathrm{Tc}$ vs mathrmRu$\mathrm{Ru}$)**: Technetium (mathrmTc$\mathrm{Tc}$, 4d^5 5s^2$4d^5 5s^2$) similarly shows a dip in melting point compared to Ruthenium (mathrmRu$\mathrm{Ru}$, 4d^7 5s^1$4d^7 5s^1$) due to the stable 4d^5$4d^5$ configuration rightarrow mathbfmathrmTc < mathrmRu$\rightarrow \mathbf{\mathrm{Tc} < \mathrm{Ru}}$.
* **5d$5d$ Series (mathrmRe$\mathrm{Re}$ vs mathrmOs$\mathrm{Os}$)**: Rhenium (mathrmRe$\mathrm{Re}$, 5d^5 6s^2$5d^5 6s^2$) has optimal interatomic interaction and a higher melting point than Osmium (mathrmOs$\mathrm{Os}$, 5d^6 6s^2$5d^6 6s^2$) rightarrow mathbfmathrmOs < mathrmRe$\rightarrow \mathbf{\mathrm{Os} < \mathrm{Re}}$.
Combining these trends yields: **mathrmMn < mathrmFe$\mathrm{Mn} < \mathrm{Fe}$, mathrmTc < mathrmRu$\mathrm{Tc} < \mathrm{Ru}$, and mathrmOs < mathrmRe$\mathrm{Os} < \mathrm{Re}$**
### Pattern Recognition
Stable half-filled d^5$d^5$ configurations in 3d$3d$ (mathrmMn$\mathrm{Mn}$) and 4d$4d$ (mathrmTc$\mathrm{Tc}$) restrict d$d$-electron delocalization, creating characteristic dips in melting point curves compared to adjacent metals.
**Correct Option:** **(C)**
Qjee_main_2025_29_jan_morningPreparation and Properties of Potassium Dichromate
The molar mass of the water insoluble product formed from the fusion of chromite ore mathrm(FeCr_2O_4)$\mathrm{(FeCr_2O_4)}$ with mathrmNa_2mathrmCO_3$\mathrm{Na}_2\mathrm{CO}_3$ in presence of mathrmO_2$\mathrm{O}_2$ is ________ mathrmg \, mol^-1$\mathrm{g \, mol^{-1}}$.
Numerical Answer.Answer: 160 to 160
Solution
### Related Formula
textBalanced fusion reaction process description$$\text{Balanced fusion reaction process description}$$
### Core Logic
Write the balanced chemical equation for the industrial preparation stage of chromate salts:
4mathrmFeCr_2O_4 + 8mathrmNa_2CO_3 + 7mathrmO_2 rightarrow 8mathrmNa_2CrO_4 + 2mathrmFe_2O_3 + 8mathrmCO_2$$4\mathrm{FeCr_2O_4} + 8\mathrm{Na_2CO_3} + 7\mathrm{O_2} \rightarrow 8\mathrm{Na_2CrO_4} + 2\mathrm{Fe_2O_3} + 8\mathrm{CO_2}$$
Evaluating the solubilities of the products:
* mathrmNa_2CrO_4$\mathrm{Na_2CrO_4}$ is highly soluble in water.
* mathrmFe_2O_3$\mathrm{Fe_2O_3}$ (Iron(III) oxide) is water-insoluble.
Molar Mass of mathrmFe_2O_3$\mathrm{Fe_2O_3}$:
M = (2 cdot 55.85) + (3 cdot 16.0) simeq (2 cdot 56) + (3 cdot 16) = 112 + 48 = 160 mathrm~g/mol$$M = (2 \cdot 55.85) + (3 \cdot 16.0) \simeq (2 \cdot 56) + (3 \cdot 16) = 112 + 48 = 160 \mathrm{~g/mol}$$
### Pattern Recognition
Transition metal oxides in high oxidation states with minimal ionic breakdown parameters reliably act as insoluble precipitates in water.
### 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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