JEE Advanced 2023 Paper 2 Q06 Chemistry Physical Chemistry Solutions & Colligative Properties Hard

JEE Advanced 2023 Paper 2 · Q06 · Solutions & Colligative Properties

Atoms of metals $x$, $y$, and $z$ form face-centred cubic (fcc) unit cell of edge length $L_x$, body-centred cubic (bcc) unit cell of edge length $L_y$, and simple cubic unit cell of edge length $L_z$, respectively. If $r_z = \dfrac{\sqrt{3}}{2}r_y$, $r_y = \dfrac{8}{\sqrt{3}} r_x$, $M_z = \dfrac{3}{2} M_y$ and $M_z = 3 M_x$, then the correct statement(s) is(are)

  1. A. Packing efficiency of unit cell of $x$ > packing efficiency of unit cell of $y$ > packing efficiency of unit cell of $z$
  2. B. $L_y > L_z$
  3. C. $L_x > L_y$
  4. D. Density of $x$ > density of $y$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

Packing efficiencies: FCC $\approx 74\%$, BCC $\approx 68\%$, SC $\approx 52\%$ — so PE($x$) > PE($y$) > PE($z$) — (A) TRUE. Edge lengths: $L_x = 2\sqrt{2}r_x$, $L_y = (4/\sqrt{3})r_y = (4/\sqrt{3})(8/\sqrt{3})r_x = (32/3)r_x \approx 10.67 r_x$. $L_z = 2r_z = 2(\sqrt{3}/2)r_y = \sqrt{3}\cdot (8/\sqrt{3})r_x = 8 r_x$. So $L_y > L_z$ — (B) TRUE. $L_x = 2\sqrt{2} r_x \approx 2.83 r_x < L_y$ — (C) FALSE. Density: $\rho_x = 4M_x/(N_A L_x^3)$, $\rho_y = 2M_y/(N_A L_y^3)$. With $M_z = 3M_x$ and $M_z = (3/2)M_y$, get $M_y = 2M_x$. $\rho_x/\rho_y = (4M_x/L_x^3) / (2\cdot 2M_x/L_y^3) = (L_y/L_x)^3 = ((32/3)/(2\sqrt{2}))^3 \gg 1$ — (D) TRUE.

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