Understanding Doping in Ionic Solids

When an ionic solid like sodium chloride (NaCl) is doped with an impurity, it can create point defects in the crystal lattice. NaCl is an ionic compound with Na$^+$ cations and Cl$^-$ anions arranged in a face-centered cubic structure. Doping involves substituting some of the host ions with impurity ions. To maintain electrical neutrality in the overall crystal, the substitution must be balanced, often leading to the formation of vacancies or interstitial defects.

Cation Vacancies from Doping with BCl$_3$

In this problem, NaCl is doped with BCl$_3$. NaCl contains Na$^+$ ions (charge +1). Boron is in Group 13 and typically forms B$^{3+}$ ions. If B$^{3+}$ ions replace Na$^+$ ions in the lattice, there is a charge difference:

Charge introduced by B$^{3+}$ = $+3$

Charge of replaced Na$^+$ = $+1$

Net excess positive charge per substitution = $+3 - (+1) = +2$

To maintain electrical neutrality, this excess positive charge must be balanced by a deficiency of positive charge or an excess of negative charge. In the case of doping NaCl with a higher-valent cation, this balance is typically achieved by creating cation vacancies. Each Na$^+$ vacancy has an effective charge of $-1$ relative to the lattice (since a $+1$ ion is missing from a neutral site).

To balance the $+2$ excess charge introduced by one B$^{3+}$ ion replacing an Na$^+$ ion, we need to create two cation vacancies (each with an effective charge of $-1$).

Thus, for every B$^{3+}$ ion that substitutes for an Na$^+$ ion, two Na$^+$ vacancies are created.

Calculating the Number of Dopant Ions

The concentration of BCl$_3$ doping is given as $10^{-3}$ mol%. This means that for every 100 moles of NaCl, there are $10^{-3}$ moles of BCl$_3$.

We need to find the number of cation vacancies per mole of NaCl. So, let's determine the amount of BCl$_3$ per mole of NaCl:

Amount of BCl$_3$ per mole of NaCl = $\frac{10^{-3} \text{ mol}}{100 \text{ mol of NaCl}} \times 1 \text{ mol of NaCl} = 10^{-5}$ moles of BCl$_3$.

Assuming that all the BCl$_3$ contributes B$^{3+}$ ions to the lattice, the number of B$^{3+}$ ions per mole of NaCl is equal to the number of moles of BCl$_3$ multiplied by Avogadro's number ($N_A$).

$N_A = 6.023 \times 10^{23}$ mol$^{-1}$.

Number of B$^{3+}$ ions per mole of NaCl = (Moles of BCl$_3$) $\times N_A$

Number of B$^{3+}$ ions = $(10^{-5} \text{ mol}) \times (6.023 \times 10^{23} \text{ ions/mol})$

Number of B$^{3+}$ ions = $6.023 \times 10^{18}$ ions

Calculating the Number of Cation Vacancies

As discussed earlier, for every one B$^{3+}$ ion incorporated into the lattice replacing an Na$^+$ ion, two cation vacancies are created to maintain charge neutrality.

Number of cation vacancies = $2 \times$ (Number of B$^{3+}$ ions)

Number of cation vacancies = $2 \times (6.023 \times 10^{18})$

Number of cation vacancies = $12.046 \times 10^{18}$

This is the number of cation vacancies created per mole of NaCl doped with $10^{-3}$ mol% of BCl$_3$ (assuming B$^{3+}$ replaces Na$^+$).

Summary of Calculation

• Doping concentration = $10^{-3}$ mol%
• Moles of dopant (BCl$_3$) per mole of NaCl = $10^{-5}$ mol
• Number of dopant ions (B$^{3+}$) per mole of NaCl = $10^{-5} \times 6.023 \times 10^{23} = 6.023 \times 10^{18}$ ions
• Number of cation vacancies per dopant ion = 2 (to balance charge of B$^{3+}$ replacing Na$^+$)
• Total cation vacancies per mole of NaCl = $2 \times 6.023 \times 10^{18} = 12.046 \times 10^{18}$ vacancies

The final answer is $\boxed{12.046 \times 10^{18}}$.