Chemistry

Solid State and Crystal Structures

151 Questions

Solid state chemistry focuses on the structural properties of crystalline solids. Questions cover Bravais lattices, unit cells, and close packed structures. This topic is essential for chemistry papers in UPSC and state PSC examinations.

Bravais lattice typesUnit cell calculationsClose packed structuresCrystal densityBragg's LawWurtzite structure

Solid State and Crystal Structures Questions

Multiple choice imperfections in solids solid state the solid state chemistry

Give the correct order of initials T (true) or F (False) for following statements:
I. In an anti-fluorite structure, anions form FCC lattice and cations occupy all tetrahedral voids.
II. If the radius of cations and anions are $0.2\ \mathring A$ and $0.95\ \mathring A$, then co-ordination number of cation in the crystal is $4$.
III. An ion is transferred from a lattice site to an interstitial position in Frenkel defect.
IV. Density of crystal always increases due to substitutional impurity defect.

  1. TFFF

  2. FTTF

  3. TFFT

  4. TFTF
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

I. In an anti-flourite structure, anions form FCC lattice and cations occupy all tetrahedral voids.


II. Ratio of radius $= \dfrac{0.2}{0.95} = 0.21$ which lies between $0.155-0.225$ and so, it has co-ordination number of $3$.


III. Frenkel defect is a defect which is created when an ion leaves its appropriate site in the lattice and occupies an interstitial site. A hole or vacancy is thus produced in the lattice.

IV. The density of crystal may increase/decrease due to substitutional impurity defect.


Hence, the correct option is $\text{D}$

Multiple choice imperfections in solids solid state the solid state chemistry

Frenkel defect is found in crystals in which the radius ratio is:

  1. low
  2. $1.3$
  3. $1.5$

  4. slightly less than unity
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Frenkel defect is exhibited in ionic compounds in which the radius ratio is low. The cations and anions differ much in their sizes and the ions have low co-ordination numbers. However, in Schottky defect the size of cation and anion

are almost same, ex- KCl, AgBr.
Hence, the correct option is $\text{A}$

Multiple choice imperfections in solids solid state the solid state chemistry

In a solid lattice, the cation has left a lattice site and is located at an interstitial position. The lattice defect is known as____________.

  1. Interstitial defect

  2. Valency defect

  3. Frenkel defect

  4. Schottky defect
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Frenkel defect is a defect which is created when an ion leaves its appropriate site in the lattice and occupies an interstitial site. A hole or vacancy is thus produced in the lattice.
The electroneutrality of the crystal is maintained since the number of positive and negative ions is the same. Since positive ions are small in size, they usually leave their positions in the lattice and occupy interstitial positions.

Multiple choice imperfections in solids solid state the solid state chemistry

What type of crystal defect is indicated in the diagram below?

$Na^{+}$ $Cl^{-}$ $Na^{+}$ $Cl^{-}$ $Na^{+}$ $Cl^{-}$
$Cl^{-}$ $Cl^{-}$ $Na^{+}$ $Na^{+}$
$Na^{+}$ $Cl^{-}$ $Cl^{-}$ $Na^{+}$ $Cl^{-}$
$Cl^{-}$ $Na^{+}$ $Cl^{-}$ $Na^{+}$ $Na^{+}$

  1. Frenkel defect

  2. Schottky defect

  3. Interstitial defect

  4. Frenkel and Schottky defects
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Schottky defect is indicated in the diagram as equal number of sodium cation and chloride anion are missing.
In Frenkel defect, a cation or anion leaves its regular site and occupies interstitial position.

Multiple choice imperfections in solids solid state the solid state chemistry

Which kind of defect is shown by the given crystal?
${ K }^{ + }$  ${ Cl }^{ - }$   ${ K }^{ + }$   ${ Cl }^{ - }$   ${ K }^{ + }$     ${ Cl }^{ - }$
${ Cl }^{ - }$  $\Box $      ${ Cl }^{ - }$    ${ K }^{ + }$   $\Box $       ${ K }^{ + }$
${ K }^{ + }$   ${ Cl }^{ - }$   $\Box $     ${ Cl }^{ - }$     ${ K }^{ + }$    ${ Cl }^{ - }$ 
${ Cl }^{ - }$   ${ K }^{ + }$   ${ Cl }^{ - }$   ${ K }^{ + }$   $\Box $        ${ K }^{ + }$   

  1. Schottky defect

  2. Frenkel defect

  3. Schottky Frenkel defects

  4. Substitution disorder
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the given crystal equal number of cations and anions are missing (two ${K}^{+}$ and two ${Cl}^{-}$) from their normal lattice sites and the crystal maintains electrical neutrality. Hence, this is Schottky defect.

Multiple choice imperfections in solids solid state the solid state chemistry

In a crystal, at $827^{\circ}C$, one out of $10^{10}$ lattice site is found to be vacant, while in the same solid, one out of $2 \times 10^9$ lattice site is found to be vacant at $927^{\circ}C$. What is the enthalpy of vacancy formation in kJ/mol unit?

  1. $76.8$
  2. $176.8$
  3. $33$
  4. $23$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\cfrac {^nV}{N}=Ae^{-\Delta H _V/RT} \longrightarrow (x)$

$\left(\cfrac {^nV}{N}\right) _{827^oC}=\cfrac {1}{10^{10}} \longrightarrow (1)$
$\left(\cfrac {^nV}{N}\right) _{927^oC}= \cfrac {1}{2 \times 10^9} \longrightarrow (2)$                                   $\therefore T _1= 827+273=1100K ; T _2=927+273=1200K$

$\Rightarrow \cfrac {(2)}{(1)}= \cfrac {1/2\times 10^9}{1/10^{10}}= \cfrac {10^ {10}}{2 \times 10^9}= 0.5 \times 10=5$
$x \Rightarrow 2.303 \log 5= - \cfrac {\Delta H _V}{R} \left(\cfrac {1}{T _1}-\cfrac {1}{T _2}\right)$
$\therefore \Delta H _V= 2.303 \times \log 5 \times 8.314 \times \left(\cfrac {1100\times 1200}{100}\right)$
               $=1.766 \times 10^5 J$
               $=176.6kJ/mol$ $unit$ .

Multiple choice imperfections in solids solid state the solid state chemistry

Calcium crystallizes in a face centred cubic unit cell with a $=0.556$ nm. Calculate the density if it contains:

(i) $0.1\%$ Frenkel defect
(ii) $0.1\%$ Schottky defect

  1. $1.546, 1.546$
  2. $1.546, 1.544$
  3. $15.46, 15.46$
  4. $21.76, 41.66$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1.$ In frenkel defect there is no change in original density

$\rho=\cfrac{Z\times M _A}{N _A\times a^3}$
$\implies \rho=\cfrac{4\times 40}{6.023\times 10^{23}\times 171.8\times 10^{-24}}$
$\implies \rho=1.546$ $g/cm^3$

$2.$ In schottky defect $Z _{eff}$ changes so density changes
$Z _{new}=4-\cfrac{4\times 0.1}{100}=3.996$
$\rho =\cfrac { Z\times M _{ A } }{ N _{ A }\times a^{ 3 } } $
$\implies \rho =\cfrac { 3.996\times 40 }{ 6.023\times 10^{ 23 }\times 171.8\times 10^{ -24 } } $
$\implies \rho =1.544$$g/cm^{ 3 }$

Multiple choice imperfections in solids solid state the solid state chemistry

A strong current of trivalent gaseous boron passed through a germanium crystal decreases the density of the crystal due to part replacement of germanium by boron and due to interstitial vacancies created by missing Ge atoms. In one such experiment, one gram of germanium is taken and the boron atoms are found to be $150$ ppm by weight when the density of the Ge crystal decreases by $4\%$. Calculate the percentage of missing vacancies due to germanium which are filled up by boron atoms. Atomic weight of Ge $= 72.6$ amu and $B = 11$ amu.

  1. $2.4\, \%$
  2. $1.2\, \%$
  3. $6.6\, \%$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As given,
$1$ g Germanium $\equiv  150$ ppm of Boron $=1.5 \times 10^{-4}$ g of Boron
$4\%$ decrease in density of Ge occurs in one experiment due to replacement of Ge by B.

The weight becomes $0.96$ g (Ge + B).
Hence, $0.04$ g is missing.
$0.04$ missing means $(0.04-0.00015)$ g Ge missing $= \dfrac{(0.04-0.00015)}{72.61}$ moles
Ge missing $=$ $5.488\times 10^{-4}$ moles
Boron replaced $=$ $1.5\times 10^{-4} g = \dfrac{(1.5\times 10^{-4})}{10}$ moles
Equivalently replaced $= \dfrac 34\times $ $\dfrac{1.5\times 10^{-4}}{10}$ moles of Ge ( Valency of B is $3$ and Ge is $4$.)
$=0.1125\times 10^{-4}$ moles of Ge
Percentage of missing vacancies filled by B atoms $=\dfrac {0.1125\times 100}{5.488}$ $=2.05\%$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

The formula of calgon is:

  1. $Na _{2}\left [ Na _{4}\left ( PO _{3} \right ) _{6} \right ]$
  2. $Na _{4}\left [ Na _{2}\left ( PO _{3} \right ) _{6} \right ]$
  3. $Na _{4}\left [ Na _{2}\left ( PO _{3} \right ) _{3} \right ]$
  4. $Na _{2}\left [ Na _{4}\left ( PO _{3} \right ) _{4} \right ]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The formula of calgon is $Na _2[Na _4(PO _3) _6]$. Calgon is a trade name of a complex salt, sodium hexametaphosphate $(NaPO _3) _6$. It is used for softening hard water. 
Calgon ionizes to give a complex anion.
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

The distance between two consecutive atoms of the crystal lattice is $1.227\overset {\circ}{A}$. The maximum order of diffraction of electrons accelerated through $10^{4}$ volt will be:

  1. $10$
  2. $\dfrac {1}{10}$
  3. $100$
  4. $\dfrac {1}{100}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The de Broglie wavelength lambda = h / sqrt(2 * m * e * V). For V = 10^4 V, lambda is approx 0.1227 Angstroms. Bragg's law is 2 * d * sin(theta) = n * lambda. For maximum order n, sin(theta) = 1, so n = 2 * d / lambda = 2 * 1.227 / 0.1227 = 20. The calculation suggests 20, but 10 is the closest option.

Multiple choice

The space lattice of graphite is

  1. Cubic

  2. Tetragonal

  3. Rhombic

  4. Hexagonal
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Graphite consists of layers of carbon atoms arranged in a hexagonal pattern, which defines its hexagonal crystal system.