Solid State JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Solid State, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 15 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   The density (in $\mathrm{g} \mathrm{cm}^{-3}$ ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is $\qquad$ . Use: Atomic mass of metal $=105.6 \mathrm{amu}$ and Avogadro's constant $=6 \times 10^{23} \mathrm{~mol}^{-1}$
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   Answer: 11

   $\begin{aligned} & \text { CCP : } \\ & (\mathrm{Z}=4, \mathrm{a}=400 \mathrm{pm}, \mathrm{M}=105.6 \mathrm{~g} / \mathrm{mol}) \\ & \text { Density }=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}} \\ & =\frac{4 \times 105.6}{\left(400 \times 10^{-10}\right)^3 \times 6 \times 10^{23}}=11.00 \mathrm{~g} / \mathrm{cm}^3\end{aligned}$
2. Q2JEE Advanced Adv 2023 (Paper 2)
   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) [Given: ${M}_{x},{M}_{y}$, and ${M}_{z}$ are molar masses of metals $x,y$, and $z$, respectively. ${r}_{x},{r}_{y}$, and ${r}_{z}$ are atomic radii of metals $x,y$, and $z$, respectively.]
   1. A.Packing efficiency of unit cell of $x\gt$ Packing efficiency of unit cell of $y\gt$ Packing efficiency of unit cell of $z$
   2. B.${L}_{y}\gt {L}_{z}$
   3. C.${L}_{x}\gt {L}_{y}$
   4. D.Density of $x\gt$ Density of $y$
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   Answer: A,B,D

   Metal $x$ forms FCC (edge length ${L}_{x}$ ) Metal y forms BCC (edge length ${L}_{y}$ ) Metal $z$ forms SC (edge length ${L}_{z}$ ) Given ${r}_{z}=\dfrac{\sqrt{3}}{2}{r}_{y}$ and ${r}_{y}=\dfrac{8}{\sqrt{3}}{r}_{x}$ $∴{r}_{z}=\dfrac{\sqrt{3}}{2}\times \dfrac{8}{\sqrt{3}}{r}_{x}=4{r}_{x}$ ${M}_{z}=\dfrac{3}{2}{M}_{y}{M}_{z}=3{M}_{x}$ $∴{M}_{y}=2{M}_{x}$ Packing efficiency $FCC\gt BCC\gt SC$ Packing efficiency unit cell $x\gt y\gt z$ In FCC unit cell:- atoms along the face diagonals are in contact. $∴\sqrt{2}{L}_{x}=4{r}_{x}\Rightarrow {L}_{x}=2\sqrt{2}{r}_{x}$ In BCC unit cell: atoms along the body diagonal are $∴\sqrt{3}{L}_{y}=4{r}_{y}\Rightarrow {L}_{y}=\dfrac{4}{\sqrt{3}}{r}_{y}=\dfrac{4}{\sqrt{3}}\times \dfrac{8}{\sqrt{3}}{r}_{x}=\dfrac{32}{3}{r}_{x}$ ${L}_{y}=\dfrac{32}{3}{r}_{x}$ In SC unit cell, atoms along the edge are in contact $∴{L}_{z}=2{r}_{z}$ $=2\times 4{r}_{x}=8{r}_{x}$ ${L}_{x}=2\sqrt{2}{r}_{x}$ ${L}_{y}=\dfrac{32}{3}{r}_{x}$ ${L}_{z}=8{r}_{x}$ $∴{L}_{y}\gt {L}_{z}\gt {L}_{x}$ Density of $x$ (Number of atoms of $x$ per unit cell $(z)=4)$ ${d}_{x}=\dfrac{{zM}_{x}}{{\left({L}_{x}\right)}^{3}{N}_{A}}=\dfrac{4\times {M}_{x}}{{\left(2\sqrt{2}{r}_{x}\right)}^{3}\times {N}_{A}}$ $=\dfrac{4{M}_{x}}{16\sqrt{2}{r}_{x}^{3}{N}_{A}}=\dfrac{{M}_{x}}{4\sqrt{2}{r}_{x}^{3}{N}_{A}}$ Density of $y$ : (Number of atoms of y per unit cell $(z)=2$ ) ${d}_{y}=\dfrac{{zM}_{y}}{{\left({L}_{y}\right)}^{3}{N}_{A}}=\dfrac{2\times 2{M}_{x}}{{\left(\dfrac{32}{3}{r}_{x}\right)}^{3}{N}_{A}}=\dfrac{108{M}_{x}}{32768{r}_{x}^{3}{N}_{A}}$ $∴$ Density of $x\gt$ density of $y$.
3. Q3JEE Advanced Adv 2022 (Paper 2)
   Atom $X$ occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in $$) of the resultant solid is closest to
   1. A.$25$
   2. B.$35$
   3. C.$55$
   4. D.$75$
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   Answer: (B)

   Atom $X$ occupies FCC lattice sites as well as alternate tetrahedral voids of FCC. In FCC, tetrahedral voids are $8$ (in a unit cell) Hence Atom $X$ in a unit cell (FCC lattice sites) $=8\times \dfrac{1}{8}+6\times \dfrac{1}{2}=4$ Atom $X$ in a unit cell (in T.V.) $=\dfrac{1}{2}\times 8=4$ Total atom$X$ in one unit cell $=8$ For relation between a and r, since Tetrahedral void forms at $\dfrac{1}{4}$th of body diagonal, $\dfrac{a\sqrt{3}}{4}=2r$ $a=\dfrac{8r}{\sqrt{3}}$ Packing efficiency $=\dfrac{8\times \dfrac{4}{3}{\pi r}^{3}}{{a}^{3}}\times 100=\dfrac{8\times \dfrac{4}{3}{\pi r}^{3}}{{\left(\dfrac{8r}{\sqrt{3}}\right)}^{3}}\times 100=35$
4. Q4JEE Advanced Adv 2018 (Paper 1)
   Consider an ionic solid $MX$, with $NaCl$ structure. Construct a new structure $\left(Z\right)$, whose unit cell is constructed from the unit cell of $MX$, following the sequential instructions given below. Neglect the charge balance. $(i)$ Remove all the anions $\left(X\right)$, except the central one. $(ii)$ Replace all the face centered cations $\left(M\right)$, by anions $\left(X\right)$. $(iii)$ Remove all the corner cations $\left(M\right)$. $(iv)$ Replace the central anion $\left(X\right)$, with cation $\left(M\right)$. The value of $\left(\dfrac{numberofanions}{numberofcations}\right)$ in $Z$ is _____.
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   Answer: 3

   ${X}^{-}\Rightarrow O.V$ ${M}^{+}\Rightarrow FCC$ ${M}^{+}$ $X^-$ (i) 4 1 (ii) 4 – 3 3 + 1 (iii) 4 – 3 – 1 3 + 1 (iv) 1 3 $Z=\dfrac{3}{1}=3$.
5. Q5JEE Advanced Adv 2017 (Paper 1)
   A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of $400pm$ . If the density of the substance in the crystal is $8g{cm}^{-3},$ then the number of atoms present in $256g$ of the crystal $isN\times {10}^{24}$ . The value of $N$is
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   Answer: 2

   Formula of density $=\dfrac{Z\times M}{{N}_{A}\times {a}^{3}}$ For FCC unit cell $Z,=4$ Edge length, $a=4\times {10}^{-8}cm$ $M=\dfrac{d\times {N}_{A}\times {a}^{3}}{Z}=\dfrac{8\times 6\times {10}^{23}\times 64\times {10}^{-24}}{4}g/mol$ Number of atoms $=\dfrac{wt\left(g\right)}{molarmass}\times {N}_{A}=\dfrac{256\times 10\times 6\times {10}^{23}}{8\times 6\times 16}=2\times {10}^{24}$ Hence, the required value of $N=2$