Tag: imperfections in solids

Questions Related to imperfections in solids

Which of the following is incorrect about interstitial compounds?

  1. Very reactive

  2. High metallic conductivity

  3. Very hard

  4. High melting point


Correct Option: A
Explanation:

Interstital compound are almost inert so (A) is wrong statement.

Which forms interstitial compounds?

  1. $Fe$

  2. $Co$

  3. $Ni$

  4. All


Correct Option: D
Explanation:

A interstitial compound has ions or atoms of a nonmetal occupy interstitial positions in a metal lattice. All of the three transition metal mentioned above have the tendency to form interstitial compounds.

Addition of non-metals like B and C to the interstitial sites of a transition metal results the metal :

  1. of more ductability

  2. of less ductability

  3. Less malleable

  4. of more hardness


Correct Option: B,C,D
Explanation:

Addition of non-metals like B and C to the interstitial sites of a transition metal results in the  more hardness so it will be less ductile and less malleable.
Hence options B,C & D are correct.

Find the solid which has brilliant blue color.
  1. $Fe _2O _3$

  2. $CuSO _4$

  3. $NaBr$

  4. $KCl$

  5. $Ca(OH) _2$


Correct Option: B
Explanation:

Copper sulphate $\displaystyle CuSO _4$ has brilliant blue color.
$Cu(II)$ ions with an outer electronic configuration of  $\displaystyle 3d^{9}$ have one unpaired electron. This results in blue colour which is due to $d-d$ transitions.

What type of stoichiometric defect is shown by $ZnS$?

  1. Schottky defect

  2. Frenkel defect

  3. Both Frenkel and Schottky defect

  4. Non-stoichiometric defect


Correct Option: B
Explanation:
The defect arises when some of the ions of the lattice occupy interstitial sites leaving lattice sites vacant is known as Frenkel defect. This defect is generally found in ionic crystals where anion is much larger in size than the cation.
The atomic sizes of $Zn$ and $S$ are different. So, we can say that they show Frenkel defect.
So, option $B$ is correct.

Silver halides generally show ?

  1. Schottky defect

  2. Frenkel defect

  3. both Frenkel and Schottky defects

  4. cation excess defect


Correct Option: C
Explanation:

Silver halides show both Frenkel and Schotkky defects. For Frenkel defect the reason is that there is size difference between the sizes of silver and halide.

Schottky defect is possible because silver halides are highly ionic.
So, correct answer is option C.

Assertion: In any ionic solid $[MX]$ with schottky defects, the numbers of positive and negative ions are same.
Reason: Equal number of cation and anion vacancies are present.

  1. Both assertion and reason are true and reason is the correct explanation of assertion.

  2. Both assertion and reason are true and reason is not the correct explanation of assertion.

  3. Assertion is true but reason is false.

  4. Both assertion and reason are false.


Correct Option: A
Explanation:

Schottky Defect: This defect occurs when oppositely charged ions leave their lattice site creating vacancies in such a way that electrical neutrality of crystal is maintained. It is generally seen in highly ionic compounds where a difference in size of cation and anion is small. 

So, an equal number of cations and anions vacancies are present in the crystals with Schottky Defect.

Which of the following defects is also known as dislocation defect?

  1. Frenkel defect

  2. Schottky defect

  3. Non-stoichimetric

  4. Simple interstitial defect


Correct Option: A
Explanation:

Frenkel defect is one in which atom is displaced from its lattice point to the interstitial site, creating a vacancy at the lattice point. Here, since dislocation of atom lattice point happens. So, it is also called as dislocation defect.

Fill in the blanks by choosing an appropriate option. A sample of ferrous oxide has actual formula $Fe {0.93}O _{1.00}.$ In this sample __(i) fraction of metal ions are $Fe^+$ ions. ____(ii) type of non-stoichiometric defect is present in this sample. 

  1. (i) 0.849 (ii) Metal deficiency

  2. (i) 0.790 (ii) Metal deficiency

  3. (i) 0.145 (ii) Metal excess

  4. (i) 0.93 (ii) Vacancy defect


Correct Option: A
Explanation:

Given, sample of ferrous oxide is $Fe _{0.93}O _{1.00}$.

Let number of $Fe^{2+}$ ions is $x$ then number of $Fe^{3+}$ ions is $(0.93-x)$

Now, for electrical neutrality, $(+2)x + (+3)(0.93-x)=+2$

$\Rightarrow 2x+2.79 – 3x = 2$

$\Rightarrow x=0.79$

So, number of $Fe^{2+}$ ions $=0.79$

Number of $Fe^{3+}$ ions $=0.14$

So, fraction of $Fe^{2+}$ ions $= \cfrac {0.79}{0.93}=0.849$

$FeO$ has non stochiometric metal deficiency defect in which number of $Fe^{2+}$ ions are lesser than $O^-$ ions as compared to stochiometric formula. The neutrality is maintained to variable oxidation state possessing capability of $Fe$.