Q.$2 \times 10^{8}$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

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1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $r$ is the radius of a metal atom, then the total volume of atoms present in a body centered cubic cell of a metal is (A) $\dfrac{4}{3}\pi r^3$ (B) $\dfrac{8}{3}\pi r^3$ (C) $\dfrac{16}{3}\pi r^3$ (D) $\dfrac{20}{3}\pi r^3$

   ›Reveal solutionSolution

   A bcc unit cell holds exactly 2 atoms; multiplying that by the volume of a single sphere ($\tfrac43\pi r^3$) gives $\tfrac83\pi r^3$.

   Concept and Intuition

   Counting "effective" atoms per unit cell means accounting for how much of each atom actually lies inside the cell's geometric boundary, since atoms are shared between adjacent cells at corners, edges, and faces. For bcc, corner atoms are shared among 8 cells (each contributes $\tfrac18$) and the single body-centre atom is wholly inside (contributes 1) — no face-centred atoms are present in bcc.

   Step-by-Step Solution

   1. Count atoms per bcc unit cell: 8 corner atoms, each shared among 8 unit cells, contribute $8\times\tfrac18=1$ atom; plus 1 body-centre atom fully inside, contributing 1 atom.
   2. Total atoms per unit cell $= 1+1 = 2$.
   3. Volume of a single spherical atom of radius $r$: $\dfrac{4}{3}\pi r^3$.
   4. Total volume occupied by atoms in the cell $= 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$.

   Common Mistakes …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A compound is formed by elements $A$ and $B$. Element $B$ forms $ccp$ structure. Atoms of $A$ occupy $x$ % of tetrahedral voids. All the octahedral voids are vacant. Compound formula is $A_2B_3$. What is $x$ %? (A) 50 (B) 25 (C) 33.3 (D) 66.6

   ›Reveal solutionSolution

   With B in ccp (4 atoms/cell, 8 tetrahedral voids) and stoichiometry $A_2B_3$ fixing A at $8/3$ atoms/cell, A occupies $1/3 = 33.3\%$ of the tetrahedral voids.

   Concept and Intuition

   In a ccp (fcc) lattice, the number of atoms per unit cell is 4. The number of tetrahedral voids is twice the number of atoms (= 8), and the number of octahedral voids equals the number of atoms (= 4). Given a compound's formula, the ratio of A to B atoms per unit cell must match the formula ratio, which lets us back out how many A atoms are present, and hence what fraction of the available voids they occupy.

   Step-by-Step Solution

   1. B forms ccp $\Rightarrow$ atoms of B per unit cell $= 4$.
   2. Number of tetrahedral voids $= 2 \times 4 = 8$; number of octahedral voids $= 4$ (all stated to be vacant — A occupies none of these).
   3. Compound formula is $A_2B_3$, so the required mole ratio is $A:B = 2:3$.
   4. With $B = 4$: $A = \dfrac{2}{3}\times 4 = \dfrac{8}{3} \approx 2.667$ atoms per unit cell. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A metal $X$ of atomic mass 75 u forms a cubic lattice of edge length 5Å. If the density of the lattice is 2 g $\mathrm{cm^{-3}}$, the radius (in Å) of the metal atom is ($N = 6 \times 10^{23}\ \mathrm{mol^{-1}}$) (A) 1.083 (B) 4.330 (C) 2.165 (D) 6.495

   ›Reveal solutionSolution

   This tests density-to-Z identification followed by the touching-atoms radius formula; the lattice is BCC ($Z=2$), giving $r \approx 2.165$ Å.

   Concept and Intuition

   The density of a crystal lattice relates the number of atoms per unit cell ($Z$), molar mass ($M$), Avogadro's number ($N_A$), and the unit cell volume ($a^3$) via $\rho = \dfrac{ZM}{N_A a^3}$. Once $Z$ is identified (1 = simple cubic, 2 = BCC, 4 = FCC), the atomic radius follows from the geometric touching condition specific to that lattice type.

   Step-by-Step Solution

   1. Convert edge length: $a = 5\ \text{Å} = 5\times10^{-8}\ \text{cm}$, so $a^3 = 125\times10^{-24}\ \text{cm}^3 = 1.25\times10^{-22}\ \text{cm}^3$.
   2. Apply the density formula: $\rho = \dfrac{ZM}{N_A a^3} \Rightarrow 2 = \dfrac{Z(75)}{(6\times10^{23})(1.25\times10^{-22})}$.
   3. Compute the denominator: $(6\times10^{23})(1.25\times10^{-22}) = 75$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Atoms of element B form hcp lattice and atoms of element A occupy $\frac{2}{3}$rd of the tetrahedral voids. The formula of the compound formed by the elements A and B is (A) $A_3B_4$ (B) $A_4B_3$ (C) $AB_4$ (D) $A_4B$

   ›Reveal solutionSolution

   Using the standard result "tetrahedral voids = 2 × number of close-packed atoms" and the given 2/3 occupancy, the A:B ratio works out to 4:3, so the formula is $A_4B_3$.

   Concept and Intuition

   In any close-packed structure (hcp or ccp) built from $N$ atoms, there are exactly $N$ octahedral voids and $2N$ tetrahedral voids per $N$ atoms — this is a fixed geometric fact of close packing, independent of whether the packing is hcp or ccp. Here, element B forms the hcp lattice itself, so if there are $N$ atoms of B, there are $2N$ tetrahedral voids available. Element A occupies $\frac{2}{3}$ of these voids, so the number of A atoms is $\frac{2}{3}\times 2N = \frac{4N}{3}$.

   Step-by-Step Solution

   1. Let the number of B atoms (forming the hcp lattice) = $N$.
   2. Tetrahedral voids available = $2N$ (standard close-packing result).
   3. A occupies $\tfrac{2}{3}$ of the tetrahedral voids: number of A atoms $= \tfrac{2}{3}\times 2N = \tfrac{4N}{3}$. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The fraction of the total volume occupied by the atoms present in a face centred cubic unit cell is (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{3\sqrt2}$ (C) $\dfrac{\pi\sqrt3}{8}$ (D) $\dfrac{\pi}{2\sqrt2}$

   ›Reveal solutionSolution

   The packing fraction of an FCC unit cell is the standard result $\pi/(3\sqrt2) \approx 0.74$, the highest possible for identical spheres.

   Concept and Intuition

   In an FCC unit cell, atoms touch along the face diagonal, so $4r = \sqrt2\,a$, i.e. $r = \dfrac{\sqrt2\,a}{4}$. There are 4 atoms per FCC unit cell ($8\times\tfrac18$ at corners $+\ 6\times\tfrac12$ at face centres $=4$).

   Step-by-Step Solution

   1. Relation between edge length and radius: $a = 2\sqrt2\,r$.
   2. Volume of unit cell: $a^3 = (2\sqrt2\,r)^3 = 16\sqrt2\,r^3$.
   3. Volume occupied by 4 spheres: $4\times\dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A crystal lattice has A (cations), B (cations) and O (anions). Atoms of O form hcp lattice. Atoms of A occupy 25% of octahedral voids and atoms of B occupy 25% of tetrahedral voids. What is the molecular formula of the crystal lattice? (A) $ABO_2$ (B) $A_2BO_4$ (C) $AB_2O_4$ (D) $ABO_3$

   ›Reveal solutionSolution

   Using the standard hcp void ratios (octahedral voids = N, tetrahedral voids = 2N for N anions), 25% occupancy of octahedral by A and 25% of tetrahedral by B gives the ratio A:B:O = 1:2:4, i.e. $AB_2O_4$.

   Concept and Intuition

   In any close-packed lattice (hcp or ccp) built from N anions (here O), the number of octahedral voids equals N, and the number of tetrahedral voids equals 2N. This ratio (voids : anions = 1 : 2 for octahedral, 2 : 1 for tetrahedral) is what lets us find the empirical formula purely from the fractional occupancy, without needing to know the actual unit cell size.

   Step-by-Step Solution

   1. Let O form the hcp lattice with N atoms of O per reference unit → octahedral voids = N, tetrahedral voids = 2N.
   2. A occupies 25% of octahedral voids: number of A = $0.25 \times N = 0.25N$.
   3. B occupies 25% of tetrahedral voids: number of B = $0.25 \times 2N = 0.5N$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.An element of molar mass $250\ g\,mol^{-1}$ crystallizes in a simple cubic lattice. The density of the unit cell is $7.2\ g\,cm^{-3}$. What is the radius of atom (in Å) of that element? ($N_0 = 6.02 \times 10^{23}mol^{-1}$) (A) $3.86$ (B) $1.93$ (C) $7.72$ (D) $5.79$

   ›Reveal solutionSolution

   This tests the density–unit cell relation for a simple cubic lattice, followed by the touching-atoms geometry along the cube edge. The atomic radius works out to about 1.93 Å.

   Concept and Intuition

   The density formula connects the mass in one unit cell (Z formula units × molar mass, divided by Avogadro's number) to the volume of the unit cell ($a^3$). Once the edge length $a$ is known, the geometric arrangement of the lattice tells us how atoms touch (for simple cubic: along the cell edge), which relates $a$ to the atomic radius $r$.

   Step-by-Step Solution

   1. For a simple cubic (SC) unit cell, the number of atoms per unit cell is $Z = 1$ (only corner atoms contribute, $8 \times \tfrac{1}{8} = 1$).
   2. Use the density formula: $\rho = \dfrac{Z M}{N_0 a^3}$, so $a^3 = \dfrac{Z M}{N_0 \rho}$.
   3. Substitute: $a^3 = \dfrac{1 \times 250}{6.02\times10^{23} \times 7.2} = \dfrac{250}{4.3344\times10^{24}} = 5.767\times10^{-23}\ cm^3$.
   4. Take the cube root: $a = (5.767\times10^{-23})^{1/3}$. Writing $5.767\times10^{-23} = 57.67 \times 10^{-24}$, and $\sqrt[3]{57.67}\approx 3.867$, we get $a \approx 3.867\times10^{-8}\ cm$. …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Which of the following is not correct about the hexagonal close packing? (A) The co-ordination number is 12 (B) Packing efficiency in it is 74% (C) Tetrahedral voids of the second layer are covered by the spheres of the third layer (D) In this arrangement, spheres of the fourth layer are aligned with those of the first layer

   ›Reveal solutionSolution

   Hexagonal close packing repeats every TWO layers (ABAB…), so the fourth layer must align with the second layer, not the first — that is the one false statement here. The answer is (D).

   Concept and Intuition

   Both hcp and ccp are built by stacking close-packed 2-D layers, but they differ in how the third layer is placed. Starting from layers A (first) and B (second, offset into A's triangular voids), the third layer can either sit directly above A (repeating the pattern every 2 layers: ABAB… → hcp) or shift again to a new position C (repeating every 3 layers: ABCABC… → ccp). In hcp, the spheres of the third layer occupy the tetrahedral voids of the second layer that lie directly over the first layer's spheres. This gives coordination number 12 and packing efficiency 74% in both hcp and ccp, since local sphere-packing density is identical either way.

   Step-by-Step Solution

   1. (A) Coordination number in hcp is 12 (6 in-plane neighbours + 3 above + 3 below) — this is correct, not the answer.
   2. (B) Packing efficiency of hcp (and ccp) is 74% — correct, not the answer. …

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9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A compound is formed by A (cations), B(cations) and O(anions). Atoms of O form ccp lattice. Atoms of A occupy 50% of octahedral voids and atoms of B occupy 25% of tetrahedral voids. What is the molecular formula of the compound? (A) $AB_2O_4$ (B) $AB_2O_2$ (C) $ABO_3$ (D) $ABO_2$

   ›Reveal solutionSolution

   In a ccp arrangement of $O^{2-}$ (4 per cell, with 4 octahedral and 8 tetrahedral voids), 50% octahedral occupancy by A and 25% tetrahedral occupancy by B give 2 A, 2 B, and 4 O per cell — formula $ABO_2$.

   Concept and Intuition

   In any ccp (cubic close-packed / face-centred cubic) lattice of $N$ effective atoms/ions per unit cell, there are exactly $N$ octahedral voids and $2N$ tetrahedral voids. For $O^{2-}$ forming the ccp lattice, $N = 4$ per unit cell (the standard fcc count: $8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4$). So there are 4 octahedral voids and 8 tetrahedral voids available for the cations to occupy.

   Step-by-Step Solution

   1. Number of $O^{2-}$ per unit cell (ccp) = 4.
   2. Number of octahedral voids = 4 (equal to the number of lattice points); A occupies 50% of these: $0.50 \times 4 = 2$ atoms of A. …

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10. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
    Q.An oxide of metal, M crystallizes in a hexagonal close packed array of oxide ions. Two out of every three octahedral holes are occupied by metal ions. The correct formula of metal oxide is (A) $M_2O_3$ (B) $MO$ (C) $M_3O_4$ (D) $MO_2$

    ›Reveal solutionSolution

    One octahedral hole exists per anion in hcp; filling 2/3 of them with M gives M:O = 2:3, i.e. $M_2O_3$.

    Concept and Intuition

    In any close-packed (hcp or ccp) arrangement of $n$ anions, there are exactly $n$ octahedral holes and $2n$ tetrahedral holes. The stoichiometry of an ionic solid built on a close-packed anion lattice is fixed purely by what fraction of these holes the cation occupies — no need to know unit cell details explicitly.

    Step-by-Step Solution

    1. Let the number of oxide ions ($O^{2-}$) be $n$ (forming the hcp lattice).
    2. Number of octahedral holes = $n$ (one per anion in any close-packed lattice).
    3. Metal ions occupy $\dfrac{2}{3}$ of these holes: number of $M$ ions $= \dfrac{2}{3}n$.
    4. Ratio $M:O = \dfrac{2}{3}n : n = 2:3$. …

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11. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.The number of unit cells in 5.85 g of cube shaped ideal crystal of NaCl (Z=4) is $x \times 10^{21}$. The value of $x$ is (At. Wt.: Na = 23 u, Cl = 35.5 u), (N = $6 \times 10^{23}\ mol^{-1}$) (A) 10 (B) 18 (C) 15 (D) 20

    ›Reveal solutionSolution

    Convert the given mass of NaCl to moles, then formula units, then divide by $Z=4$ (formula units per fcc unit cell) to get the number of unit cells; $x = 15$.

    Concept and Intuition

    NaCl crystallises in a face-centred cubic lattice with $Z = 4$ formula units of NaCl per unit cell. To find the number of unit cells in a given mass of crystal, first convert mass → moles → number of formula units (using Avogadro's number), then divide by $Z$ since each unit cell "contains" 4 formula units worth of ions.

    Step-by-Step Solution

    1. Molar mass of NaCl $= 23 + 35.5 = 58.5$ g/mol.
    2. Moles of NaCl $= \dfrac{5.85}{58.5} = 0.1$ mol.
    3. Number of NaCl formula units $= 0.1 \times (6\times10^{23}) = 6\times10^{22}$. …

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12. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
    Q.Metals A and B crystallize in "simple cubic" and "face centered cubic" lattice respectively. The number of metal atoms A and B per unit cell are respectively (A) 2, 4 (B) 1, 4 (C) 2, 1 (D) 1, 2

    ›Reveal solutionSolution

    Counting shared lattice points per unit cell: simple cubic contributes 1 atom, face-centred cubic contributes 4 atoms.

    Concept and Intuition

    In crystallography, an atom sitting at a corner, face, or edge of a unit cell is shared with neighbouring cells, so it only contributes a fraction of itself to that particular cell: a corner atom contributes $1/8$ (shared among 8 cells), a face-centred atom contributes $1/2$ (shared between 2 cells), and an edge atom contributes $1/4$. Counting all contributions per cell gives the effective number of atoms per unit cell for each lattice type.

    Step-by-Step Solution

    1. Simple cubic (SC): only 8 corner atoms, each contributing $1/8$. Total $= 8\times\tfrac18 = 1$ atom/cell.
    2. Face-centred cubic (FCC): 8 corner atoms ($8\times\tfrac18=1$) plus 6 face-centred atoms, one on each face, each shared between 2 cells ($6\times\tfrac12=3$).
    3. Total for FCC $= 1+3 = 4$ atoms/cell. …

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