Question.
Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are [NEET 2021]
(a) 8,4
(b) 6,12
(c) 2,1
(d) 12,6
Answer/Explanation
Ans. (d)
Number of octahedral and tetrahedral voids are equal to $\mathrm{N}$ and $2 \mathrm{~N}$ respectively, where $N$ is the number of atoms in unit cell.
Number of atoms per unit cell in hexagonal primitive unit cell=6.
Number of tetrahedral voids $=2 \mathrm{~N}=2 \times 6=12$.
Number of octahedral voids $=N=6$.
Question.
The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cell is [NEET 2021]
(a)7
(b)5
(c)2
(d)3
Answer/Explanation
Ans. (d)
Body centred unit cell exists in three systems, i.e. cubic, orthorhombic and tetragonal.
Question.
An element has a body centered cubic (bcc) structure with a cell edge of $288 \mathrm{pm}$. The atomic radius is [NEET (Sep.) 2020]
(a) $\frac{\sqrt{2}}{4} \times 288 \mathrm{pm}$
(b) $\frac{4}{\sqrt{3}} \times 288 \mathrm{pm}$
(c) $\frac{4}{\sqrt{2}} \times 288 \mathrm{pm}$
(d) $\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}$
Answer/Explanation
Ans. (d)
In bcc crystal,
$\because \quad 4 r=\sqrt{3} a ; r=\frac{\sqrt{3} a}{4}=\frac{\sqrt{3} \times 288}{4} \mathrm{pm}$
where, r = radius of atoms,
a= edge length of the unit cell.
Question.
A compound is formed by cation $C$ and anion $A$. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75\% of octahedral voids. The formula of the compound is [NEET (National) 2019]
(a) $\mathrm{C}_3 \mathrm{~A}_2$
(b) $\mathrm{C}_3 \mathrm{~A}_4 $
(c) $\mathrm{C}_4 \mathrm{~A}_3 $
(d) $\mathrm{C}_2 \mathrm{~A}_3$
Answer/Explanation
Ans. (b)
Anions (A) form hexagonal close packed (hcp) lattice, so
Number of anions $(A)=6$
Number of octahedral voids $=$ Number of atoms in the close packed structure $=6$. Cations (C) occupy $75 \%$ of octahedral voids, so number of cations $(C)=6 \times \frac{75}{100}$ $=6 \times 3 / 4=9 / 2$
$\therefore$ The formula of compound $=C_{9 / 2} A_6$ $=C_9 A_{12}=C_3 A_4$ Thus, option (b) is correct.
Question.
Iron exhibits bcc structure at room temperature. Above $900^{\circ} \mathrm{C}$, it transforms to fcc structure. The ratio of density of iron at room temperature to that at $900^{\circ} \mathrm{C}$ (assuming molar mass and atomic radii of iron remains constant with temperature) is [NEET 2018]
(a) $\frac{3 \sqrt{3}}{4 \sqrt{2}}$
(b) $\frac{4 \sqrt{3}}{3 \sqrt{2}}$
(c) $\frac{\sqrt{3}}{\sqrt{2}}$
(d) $\frac{1}{2}$
Answer/Explanation
Ans. (a)
Density of unit celld $=\frac{Z \times M}{N_A \times a^3}$
where, $Z=$ Number of atoms per unit cell $M=$ Molar mass
$a^3=$ Volume of unit cell $[a=$ edge length $]$
$N_A=$ Avogadro’s number $=6.022 \times 10^{23}$
For bcc, $Z=2$, radius $(r)=\frac{\sqrt{3} a}{4}$ $a=\frac{4 r}{\sqrt{3}}$
For fcc, $Z=4, r=\frac{a}{2 \sqrt{2}} \Rightarrow a=2 \sqrt{2} r$
According to question
$
\frac{d_{\text {room temp. }}}{d_{900^{\circ} \mathrm{C}}}=\frac{\left(\frac{Z M}{N_A a^3}\right)_{\mathrm{bcc}}}{\left(\frac{Z M}{N_A a^3}\right)_{\mathrm{fcc}}}
$
On substituting the given values, we get $\frac{d_{\text {room temp. }}}{d_{900^{\circ} \mathrm{C}}}=\frac{2 \times M}{N_A \times\left(\frac{4 r}{\sqrt{3}}\right)^3} / \frac{4 \times M}{N_A \times(2 \sqrt{2} r)^3}$ $[\because$ Given, Mand r of iron remains constant with temperature ]
$
\begin{aligned}
& =\frac{2 \times 3 \sqrt{3}}{64 r^3} \times \frac{16 \sqrt{2} r^3}{4} \\
\frac{d_{\mathrm{bcc}}}{d_{f c c}} & =\frac{3}{4} \sqrt{\frac{3}{2}}
\end{aligned}
$
Question.
The ionic radii of $A^{+}$and $B^{-}$ions are $0.98 \times 10^{-10} \mathrm{~m}$ and $1.81 \times 10^{-10} \mathrm{~m}$. The coordination number of each ion in $A B$ is [NEET 2016, Phase I]
(a) 4
(b) 8
(c) 2
(d) 6
Answer/Explanation
Ans. (d)
Given, ionic radius of cation $\left(\mathrm{A}^{+}\right)=0.98$ $\times 10^{-10} \mathrm{~m}$
Ionic radius of anion $\left(B^{-}\right)=1.81 \times 10^{-10} \mathrm{~m}$
$\therefore$ Coordination number of each ion in $A B$ $=$ ?
Now, we have
$
\begin{aligned}
\text { Radius ratio } & =\frac{\text { Radius of cation }}{\text { Radius of anion }} \\
& =\frac{0.98 \times 10^{-10} \mathrm{~m}}{1.81 \times 10^{-10} \mathrm{~m}} \\
& =0.541
\end{aligned}
$
If radius ratio range is in between $0.441$ $0.732$, ion would have octahedral structure with coordination number ‘six’.
Question.
In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion $\left(\mathrm{Ca}^{2+}\right)$ and fluoride ion $\left(\mathrm{F}^{-}\right)$are [NEET 2016, Phase II]
(a) 4 and 2
(b) 6 and 6
(c) 8 and 4
(d) 4 and 8
Answer/Explanation
Ans. (c)
In $\mathrm{CaF}_2$ (Fluorite structure), $\mathrm{Ca}^{2+}$ ions are arranged in ccp arrangement $\mathrm{Ca}^{2+}$ ions are present at all corners and at the centre of each face of the cube) while ${ }^{-}$ ions occupy all the tetrahedral sites.
From the above figure, you can clearly see that coordination number of $\mathrm{F}^{-}$is 4 while that of $\mathrm{Ca}^{2+}$ is 8 .
Question.
The vacant space in bee lattice cell is [CBSE AIPMT 2015]
(a)26%
(b)48%
(c)23%
(d)32%
Answer/Explanation
Ans. (d)
∴ Packing efficiency in bee lattice= 68%.
∴ Vacant space in bee lattice = 100 -68 =32%
Question.
A given metal crystallises out with a cubic structure having edge length of $361 \mathrm{pm}$. If there are four metal atoms in one unit cell, what is the radius of one atom?
[CBSE AIPMT 2015]
(a) $40 \mathrm{pm}$
(b) $127 \mathrm{pm}$
(c) $80 \mathrm{pm}$
(d) $108 \mathrm{pm}$
Answer/Explanation
Ans. (b)
Given, edge length $=361 \mathrm{pm}$
Four metal atoms in one unit cell
i.e. effective number in unit cell $(z)=4$
(given)
$\therefore$ It is a FCC structure
$\therefore$ Face diagonal $=4 r$
$
\begin{aligned}
\sqrt{2} a & =4 r \\
r & =\frac{\sqrt{2} \times 361}{4} \\
& =127 \mathrm{pm}
\end{aligned}
$
Question.
Lithium metal crystallises in a body centred cubic (bcc) crystal. If the length of the side of the unit cell of lithium is $351 \mathrm{pm}$, the atomic radius of the lithium will be [CBSE AIPMT 2009]
(a) $240.8 \mathrm{pm}$
(b) $151.8 \mathrm{pm}$
(c) $75.5 \mathrm{pm}$
(d) $300.5 \mathrm{pm}$
Answer/Explanation
Ans. (b)
In case of body centred cubic (bcc) crystal,
$
a \sqrt{3}=4 r
$
Given, edge length, $a=351 \mathrm{pm}$
Hence, atomic radius of lithium,
$
\begin{aligned}
r & =\frac{a \sqrt{3}}{4}=\frac{351 \times 1.732}{4} \\
& =151.98 \mathrm{pm}
\end{aligned}
$
Question.
Copper crystallises in a face centred cubic (fcc) lattice with a unit cell length of $361 \mathrm{pm}$. What is the radius of copper atom in pm? [CBSE AIPMT 2009]
(a)128 pm
(b)157 pm
(c)181 pm
(d) 108 pm
Answer/Explanation
Ans. (a)
In case of face centred cubic (fee) lattice,
$
\text { radius }=\frac{\sqrt{2} a}{4}
$
$\therefore$ Radius of copper atom (fcc lattice)
$
=\frac{\sqrt{2} \times 361}{4}=128 \mathrm{pm}
$
Question.
Percentage of free space in body centred cubic (bcc) unit cell is [CBSE AIPMT 2008]
(a) $30 \%$
(b) $32 \%$
(c) $34 \%$
(d) $28 \%$
Answer/Explanation
Ans. (b)
In bcc unit cell, the number of atoms $=2$ Thus, volume of atoms in unit cell (v) $=2 \times \frac{4}{3} \pi r^3$
For bcc structure $(r)=\frac{\sqrt{3}}{4} a$ $(V)=2 \times \frac{4}{3} \pi\left(\frac{\sqrt{3}}{4} a\right)^3=\frac{\sqrt{3}}{8} \pi a^3$
Volume of unit cell $(V)=a^3$
Percentage of volume occupied by unit cell
$
\begin{aligned}
& =\frac{\text { Volume of the atoms in unit cell }}{\text { Volume of unit cell }} \\
& =\frac{\frac{\sqrt{3}}{8} \pi a^3}{a^3} \times 100=\frac{\sqrt{3}}{8} \pi \times 100=68 \%
\end{aligned}
$
Hence, the free space in bcc unit cell $=100-68=32 \%$
Question.
If ‘ $a$ ‘ stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively, [CBSE AIPMT 2008]
(a) $\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a$
(b) $\frac{1}{2} a: \sqrt{3} a: \frac{1}{\sqrt{2}} a$
(c) $\frac{1}{2} a: \frac{\sqrt{3}}{2} a: \frac{\sqrt{2}}{2} a$
(d) $1 a: \sqrt{3} a: \sqrt{2} a$
Answer/Explanation
Ans. (a)
If $a=$ edge length of cubic systems
For simple cubic structure, radius $=\frac{a}{2}$
For body centred cubic structure, radius
$
=\frac{\sqrt{3}}{4} a
$
For face centred cubic structure, radius
$
=\frac{a}{2 \sqrt{2}}
$
Hence, the ratio of radii
$
=\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a
$
Question.
Which one of the following statements is an incorrect? [CBSE AIPMT 2008]
(a) The fraction of the total volume occupied by the atoms in a primitive cell is $0.48$
(b) Molecular solids are generally volatile
(c) The number of carbon atoms in an unit cell of diamond is 4
(d) The number of Bravais lattices in which a crystal can be categorised is 14
Answer/Explanation
Ans. (a)
Volume of atoms in a unit cell $(V)=\frac{4}{3} \pi r^3$
For primitive cell, $r=\frac{a}{2}$
$
V=\frac{4}{3} \pi\left(\frac{a}{2}\right)^3=\frac{\pi a^3}{6}
$
Volume of the unit cell $(V)=a^3$
Thus, total volume occupied by the atoms
$
\begin{aligned}
& =\frac{\text { Volume of the atoms in unit cell }}{\text { Volume of unit cell }} \\
& =\frac{\pi a^3}{6} \times \frac{1}{a^3}=\frac{\pi}{6}=0.52=100-0.52=0.48
\end{aligned}
$
Question.
The fraction of total volume occupied by the atoms present in a simple cube is [CBSE AIPMT 2007]
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3 \sqrt{2}}$
(c) $\frac{\pi}{4 \sqrt{2}}$
(d) $\frac{\pi}{4}$
Answer/Explanation
Ans. (a)
For simple cube,
$\operatorname{Radius}(r)=\frac{a}{2} \quad[a=$ edge length $]$
Volume of the atom $=\frac{4}{3} \pi\left(\frac{a}{2}\right)^3$
$
\therefore \text { Packing fraction }=\frac{\frac{4}{3} \pi\left(\frac{a}{2}\right)^3}{a^3}=\frac{\pi}{6}
$
Question.
In a face centred cubic (fcc) lattice, a unit cell is shared equally by how many unit cells? [CBSE AIPMT 2005]
(a) 8
(b) 4
(c) 2
(d) 6
Answer/Explanation
Ans. (d)
In a face centred cubic (fcc) lattice, a unit cell is shared equally by six unit cells.
Question.
A compound formed by elements $X$ and $Y$ crystallises in a cubic structure in which the $X$-atoms are at the corners of a cube and the $Y$-atoms are at the face centres. The formula of the compound is [CBSE AIPMT 2004]
(a) $X Y_3$
(b) $X_3 Y$
(c) $X Y$
(d) $X Y_2$
Answer/Explanation
Ans. (a)
In unit cell, $X$-atoms at the corners
$ =\frac{1}{8} \times 8=1 $$ $Y$-atoms at the face centres $=\frac{1}{2} \times 6=3$ Ratio of $X$ and $Y=1: 3$. Hence, formula is $X Y_3$.
Question.
Zn converts from its melted state to its solid state, it has hcp structure, then find out the number of nearest atoms. [CBSE AIPMT 2001]
(a) 6
(b) 8
(c) 12
(d) 4
Answer/Explanation
Ans. (c)
HCP is a closed packed arrangement, in which the unit cell is hexagonal and coordination number is 12 .
Question.
A compound formed by elements $A$ and $B$ crystallises in the cubic structure, where $A$ atoms are present at the corners of a cube and $B$ atoms are present at the face centres. The formula of the compound is [CBSE AIPMT 2000]
(a) $\mathrm{A}_2 \mathrm{~B}_2$
(b) $A B_3$
(c) $A B$
(d) $A_3 B$
Answer/Explanation
Ans. (b)
A-atoms are present at the corners of a cube. So, the number of $A$-atoms per unit cell $=8 \times \frac{1}{8}=1$
Similarly, B-atoms are present at face centres of a cube.
So, the number of B-atoms per unit cell
$
=6 \times \frac{1}{2}=3
$
Hence, the formula of compound is $A B_3$.
Question.
The edge length of face centred unit cubic cell is $508 \mathrm{pm}$. If the radius of the cation is $110 \mathrm{pm}$, the radius of the anion is [CBSE AIPMT 1998]
(a) $288 \mathrm{pm}$
(b) $398 \mathrm{pm}$
(c) $144 \mathrm{pm}$
(d) $618 \mathrm{pm}$
Answer/Explanation
Ans. (c)
Edge length $(0)=2 r^{+}+2 r^{-}$
$
\begin{aligned}
a & =2\left(r^{+}+r^{-}\right) \\
a & =508 \mathrm{pm} \\
r^{+} & =110 \mathrm{pm} \\
\frac{508}{2} & =r^{+}+r^{-} \\
254=110 & +r^{-} \\
r^{-}= & 254-110=144 \mathrm{pm}
\end{aligned}
$
Question.
The intermetallic compound LiAg crystallises in cubic lattice in which both lithium and silver have coordination number of eight. The crystal class is [CBSE AIPMT 1997]
(a) simple cube
(b) body centred cube
(c) face centred cube
(d) None of the above
Answer/Explanation
Ans. (b)
In body centered cubic, each atom/ion has a coordination number of 8 .
Question.
The edge length of a centred unit cubic cell is $508 \mathrm{pm}$. If the radius of the cation is $100 \mathrm{pm}$, the radius of the anion is [CBSE AIPMT 1996]
(a) $288 \mathrm{pm}$
(b) $398 \mathrm{pm}$
(c) $154 \mathrm{pm}$
(d) $618 \mathrm{pm}$
Answer/Explanation
Ans. (c)
For centred unit cell,
$
\begin{aligned}
& 2\left(r^{+}+r^{-}\right)=a \\
& r^{+}=\text {radii of cation } \\
& r^{-}=\text {radii of anion } \\
& 2\left(100+r^{-}\right)=508 \\
& 100+r^{-}=\frac{508}{2} \\
& r^{-}=\frac{508}{2}-100 \\
&=254-100 \\
&=154 \mathrm{pm}
\end{aligned}
$
Question.
In the fluorite structure, the coordination number of $\mathrm{Ca}^{2+}$ ion is [CBSE AIPMT 1993]
(a) 4
(b) 6
(c) 8
(d) 3
Answer/Explanation
Ans. (c)
In fluorite structure each $\mathrm{Ca}^{2+}$ ion is surrounded by eight $\mathrm{F}^{-}$ions. Thus, the coordination number of $\mathrm{Ca}^{2+}$ is eight.
Question.
The number of atoms contained in a fcc unit cell of a monoatomic substance is [CBSE AIPMT 1993]
(a) 1
(b) 2
(c) 4
(d) 6
Answer/Explanation
Ans. (c)
Face centred cubic is also called cubic close packed arrangement. It has points at all the corners as well as at the centre of each of the six faces.
The number of atoms present at corners per unit cell $=8 \times \frac{1}{8}=1$.
The number of atoms present at faces per unit cell $=6 \times \frac{1}{2}=3$
$\therefore$ Total number of atoms in ccp or fcc arrangement $=1+3=4$