Question

If AgCl is doped with \( 1 \times 10^{-4} \) mole percent of CdCl\(_2\), the number of cation vacancies (in mol\(^{-1}\)) is

• A. \(6.023 \times 10^{19} \)
• B. \(6.023 \times 10^{21} \)
• C. \(6.023 \times 10^{17} \)
• D. \(6.023 \times 10^{23} \)

Hint:

When doping an ionic crystal with an impurity of a different valence: Cationic doping:} If a cation of higher valence (e.g., \(M^{2+}\)) replaces a cation of lower valence (e.g., \(A^+\)), cation vacancies are created to maintain electrical neutrality. For each \(M^{2+}\) incorporated, one \(A^+\) vacancy is formed. Anionic doping:} Similarly, if an anion of higher valence replaces one of lower valence, anion vacancies are created. Conversion:} Convert the given percentage or mole fraction of dopant to moles of dopant, then use the stoichiometry of vacancy formation to find moles of vacancies, and finally multiply by Avogadro's number to get the number of vacancies.

Answer 1

The Correct Option is C

Step 1: Understand the doping process and defect formation.
AgCl is an ionic compound where Ag\(^{+}\) ions occupy cation sites. CdCl\(_2\) is doped into AgCl. When a Cd\(^{2+}\) ion (from CdCl\(_2\)) replaces an Ag\(^{+}\) ion in the AgCl lattice, the charge balance needs to be maintained.
An Ag\(^{+}\) ion has a +1 charge. A Cd\(^{2+}\) ion has a +2 charge.
When one Cd\(^{2+}\) ion substitutes for one Ag\(^{+}\) ion, there is an excess of +1 positive charge at that site. To maintain electrical neutrality in the crystal, an additional positive charge equivalent to +1 must be removed. This is achieved by the creation of a cation vacancy, meaning another Ag\(^{+}\) ion leaves its lattice site.
Therefore, for every one Cd\(^{2+}\) ion incorporated into the AgCl lattice, one Ag\(^{+}\) cation vacancy is created.
Step 2: Calculate the mole fraction of CdCl\(_2\) doping.
The doping concentration is given as \(1 \times 10^{-4}\) mole percent.
"Mole percent" means out of 100 moles.
So, \(1 \times 10^{-4}\) mole percent of CdCl\(_2\) means that there are \(1 \times 10^{-4}\) moles of CdCl\(_2\) for every 100 moles of AgCl (or total solution/mixture).
To express this as a mole fraction per mole of AgCl:
Mole fraction of CdCl\(_2\) = \( \frac{1 \times 10^{-4} \, \text{mol}}{100 \, \text{mol}} = 1 \times 10^{-6} \) mol of CdCl\(_2\) per mole of AgCl.
Step 3: Calculate the moles of cation vacancies per mole of AgCl.
As established in Step 1, for every mole of CdCl\(_2\) doped, one mole of cation vacancies is created.
Therefore, if there are \(1 \times 10^{-6}\) moles of CdCl\(_2\) doped per mole of AgCl, then there will be \(1 \times 10^{-6}\) moles of cation vacancies per mole of AgCl.
Step 4: Convert moles of vacancies to the number of vacancies.
To find the actual number of cation vacancies, multiply the moles of vacancies by Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)).
Number of cation vacancies = (Moles of vacancies) \(\times\) Avogadro's number
Number of cation vacancies = \( (1 \times 10^{-6} \, \text{mol}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \)
Number of cation vacancies = \( 6.022 \times 10^{(-6 + 23)} \)
Number of cation vacancies = \( 6.022 \times 10^{17} \)
This value matches option (3).
The final answer is \( \boxed{6.022 \times 10^{17}} \).