
A non-stoichiometric compound \[F{e_7}{S_8}\] consist of iron in both \[F{e^{2 + }}\]and \[F{e^{3 + }}\] form and sulphur is present as sulphide ions. Calculate cation vacancies as a percentage of \[F{e^{2 + }}\] initially present in the sample. (Express your answer as \[\dfrac{{100}}{x}\% \] and you should write x as answer)
Answer
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Hint: Calculate the total charge on both Fe and S. Afterwards calculate the ratio of \[F{e^{2 + }}\] and \[F{e^{3 + }}\] and use the formula to find the percentage.
Complete step by step answer:
Non-stoichiometric compounds are chemical compounds, almost always solid inorganic compounds, having elemental composition whose proportions cannot be represented by a ratio of small natural numbers.
A non-stoichiometric compound \[F{e_7}{S_8}\] consist of iron in both \[F{e^{2 + }}\]and \[F{e^{3 + }}\] form and sulphur is present as sulphide ions \[{S^{2 - }}\].
Let the ratio of \[F{e^{2 + }}\]and
\[F{e^{3 + }}\] be x:y
\[
F{e_7}{S_8} \\
x + y = 7 \\
\]
Balancing the charges of \[8{S^{ - 2}}\]=16
\[
F{e_7}^{ + 16} \\
2x + 3y = 16 \\
\]
The values of x and y are:
\[
x = 5 \\
y = 2 \\
\]
So the given non-stoichiometric compound \[F{e_7}{S_8}\] has 5, 2 and 8 number of \[F{e^{2 + }}\], \[F{e^{3 + }}\] and \[{S^{2 - }}\] respectively .
To calculate the cation vacancies as a percentage of \[F{e^{2 + }}\]:
Hence the formula is:
\[
Cation{\text{ vacancies = }}\dfrac{{F{e^{2 + }} \times 100}}{{total{\text{ }}Fe}} \\
= \dfrac{{3 \times 100}}{8} \\
= 37.5\% \\
\]
Additional information: A cation missing from its lattice site is called cation vacancies . To maintain electrical neutrality, one of the nearest metal ions acquires two positive charges. This type of defect occurs in compounds where metal can exhibit variable valency. e.g., Transition metal compounds like NiO, FeO, FeS etc.
Cations exhibit frenkel defects generally.
Frenkel defects are mostly exhibited in ionic solids where the smaller ion (usually the cation) is dislocated.
Note:
The question has asked the cation vacancies as a percentage of \[F{e^{2 + }}\]. Kindly note to take \[F{e^{2 + }}\]only in the formula of cation vacancies.
Complete step by step answer:
Non-stoichiometric compounds are chemical compounds, almost always solid inorganic compounds, having elemental composition whose proportions cannot be represented by a ratio of small natural numbers.
A non-stoichiometric compound \[F{e_7}{S_8}\] consist of iron in both \[F{e^{2 + }}\]and \[F{e^{3 + }}\] form and sulphur is present as sulphide ions \[{S^{2 - }}\].
Let the ratio of \[F{e^{2 + }}\]and
\[F{e^{3 + }}\] be x:y
\[
F{e_7}{S_8} \\
x + y = 7 \\
\]
Balancing the charges of \[8{S^{ - 2}}\]=16
\[
F{e_7}^{ + 16} \\
2x + 3y = 16 \\
\]
The values of x and y are:
\[
x = 5 \\
y = 2 \\
\]
So the given non-stoichiometric compound \[F{e_7}{S_8}\] has 5, 2 and 8 number of \[F{e^{2 + }}\], \[F{e^{3 + }}\] and \[{S^{2 - }}\] respectively .
To calculate the cation vacancies as a percentage of \[F{e^{2 + }}\]:
Hence the formula is:
\[
Cation{\text{ vacancies = }}\dfrac{{F{e^{2 + }} \times 100}}{{total{\text{ }}Fe}} \\
= \dfrac{{3 \times 100}}{8} \\
= 37.5\% \\
\]
Additional information: A cation missing from its lattice site is called cation vacancies . To maintain electrical neutrality, one of the nearest metal ions acquires two positive charges. This type of defect occurs in compounds where metal can exhibit variable valency. e.g., Transition metal compounds like NiO, FeO, FeS etc.
Cations exhibit frenkel defects generally.
Frenkel defects are mostly exhibited in ionic solids where the smaller ion (usually the cation) is dislocated.
Note:
The question has asked the cation vacancies as a percentage of \[F{e^{2 + }}\]. Kindly note to take \[F{e^{2 + }}\]only in the formula of cation vacancies.
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