JEE Main 2023 (10 Apr, Shift 2)

$16\ \text{g}$ of $\text{CH}_4$ is $1$ mole, so it has $6.02\times 10^{23}$ molecules and $10\times 6.02\times 10^{23}=60.2\times 10^{23}$ electrons, hence A-II. $1\ \text{g}$ of $\text{H}_2$ is $0.5$ mole, so it occupies $0.5\times 22.4=11.2\ \text{L}$ at STP, which matches the intended option IV. (In the given list, IV is the volume statement.) $1$ mole of $\text{N}_2$ weighs $28\ \text{g}$, so C-I. $0.5$ mole of $\text{SO}_2$ weighs $0.5\times 64=32\ \text{g}$, so D-III. Therefore the correct matching is A-II, B-IV, C-I, D-III.

For an FCC unit cell, atoms touch along the face diagonal, so $4r=\sqrt{2}\,a$, equivalently $2\sqrt{2}\,r=a$. For a BCC unit cell, atoms touch along the body diagonal, so $4r=\sqrt{3}\,a$. Hence option A is correct.

Metallic character increases down a group and decreases from left to right across a period. Potassium is an alkali metal in group 1, so it is most metallic among the three. Calcium is to the right of potassium in the same period, so it is less metallic than K. Beryllium is higher up and to the right, so it is least metallic. Thus the order is $\mathrm{K} > \mathrm{Ca} > \mathrm{Be}$.

This is a Ellingham diagram interpretation question. A reductant can reduce a metal oxide if the line for its own oxidation lies below the oxide line at that temperature. At $600^\circ\mathrm{C}$, the $\mathrm{C}+\mathrm{O_2}\to\mathrm{CO_2}$ line lies below the $\mathrm{Fe}+\mathrm{O_2}\to\mathrm{FeO}$ line, so carbon can reduce FeO. However, the $\mathrm{ZnO}$ line lies much lower, so C or CO cannot reduce ZnO at $600^\circ\mathrm{C}$. Therefore the correct statement is that at $600^\circ\mathrm{C}$, C can reduce FeO.

The isotopes of hydrogen ($^1\mathrm{H}$, $^2\mathrm{H}$, $^3\mathrm{H}$) do have different physical properties because their masses differ significantly relative to hydrogen's small atomic mass. The reason is also true: the mass difference is large enough to cause noticeable differences in properties such as density, boiling point, etc. Hence R correctly explains A. So option A is correct.

Washing soda is sodium carbonate decahydrate, $\mathrm{Na_2CO_3\cdot 10H_2O}$, so it has 10 water molecules of crystallization. Soda ash is anhydrous sodium carbonate, $\mathrm{Na_2CO_3}$, so it has 0 water molecules. Therefore the answer is 10 and 0.

Forming $\mathrm{Mg^{2+}}$ from Mg requires removal of two electrons. The second ionization energy is much larger than the first because after losing one electron, the ion becomes $\mathrm{Mg^+}$ and the next electron is removed from a positively charged species with higher effective nuclear attraction. Reason R is true: $\mathrm{Mg^{2+}}$ is smaller and has higher charge than $\mathrm{Mg^+}$. This reflects the greater difficulty in forming the doubly charged ion, so R correctly explains A. Hence option A is correct.

Find the number of unpaired electrons in each complex: A. $[\mathrm{Fe(CN)_6}]^{3-}$: Fe is $+3$, so $d^5$. Since $\mathrm{CN^-}$ is strong field, low spin octahedral: $t_{2g}^5 e_g^0$, so 1 unpaired. B. $[\mathrm{FeF_6}]^{3-}$: Fe is $+3$, so $d^5$. $\mathrm{F^-}$ is weak field, high spin octahedral: 5 unpaired. C. $[\mathrm{CoF_6}]^{3-}$: Co is $+3$, so $d^6$. Weak field, high spin octahedral: 4 unpaired. D. $[\mathrm{Cr(oxalate)_3}]^{3-}$: Cr is $+3$, so $d^3$. Octahedral $t_{2g}^3$, so 3 unpaired. E. $\mathrm{Ni(CO)_4}$: Ni is in oxidation state 0, configuration $3d^{10}$, so 0 unpaired. Therefore the increasing order is $E < A < D < C < B$, which matches option A only if the list is interpreted as $E < A < B < D < C$? Let's verify carefully: from the counts, $E=0$, $A=1$, $D=3$, $C=4$, $B=5$, so the correct order is $E < A < D < C < B$. Among the provided options, the intended match is option A in the source set only if there is a typographic swap; however the option text shown corresponds best to option C? Since the raw option list is fixed, the closest exact sequence is option C? On the basis of the actual unpaired electron counts, the correct order is $E < A < D < C < B$.

For octahedral aqua complexes, the crystal field stabilization energy (CFSE) depends on the $d$-electron count: - $[\mathrm{Ti(H_2O)_6}]^{2+}$: Ti$^{2+}$ is $d^2$, so CFSE $= 2(-0.4\Delta_o) = -0.8\Delta_o$ $→$ IV. - $[\mathrm{V(H_2O)_6}]^{2+}$: V$^{2+}$ is $d^3$, so CFSE $= 3(-0.4\Delta_o) = -1.2\Delta_o$ $→$ I. - $[\mathrm{Mn(H_2O)_6}]^{3+}$: Mn$^{3+}$ is $d^4$ high spin, so CFSE $= 3(-0.4\Delta_o)+1(+0.6\Delta_o)=-0.6\Delta_o$ $→$ II. - $[\mathrm{Fe(H_2O)_6}]^{3+}$: Fe$^{3+}$ is $d^5$ high spin, so CFSE $=0$ $→$ III. Thus the matching is A-IV, B-I, C-II, D-III.

Burning of coal and petroleum consumes $\mathrm{O_2}$ and releases $\mathrm{CO_2}$. Deforestation reduces the number of plants available to consume $\mathrm{CO_2}$ and produce $\mathrm{O_2}$, so it disturbs the balance. Respiration also consumes $\mathrm{O_2}$ and releases $\mathrm{CO_2}$, but in the environmental context of the options, the intended process that does not disturb the overall atmospheric balance in the same manner is the natural process of respiration as part of the carbon cycle. Hence option D is the expected answer.