Q84 E
Question
The lattice energy of \({\rm{KF}}\) is \({\rm{794kJ/mol}}\), and the interionic distance is \({\rm{269 pm}}\). The \({\rm{Na - F}}\) distance in \({\rm{NaF}}\), which has the same structure as \({\rm{KF}}\), is \({\rm{231 pm}}\). Which of the following values is the closest approximation of the lattice energy of \({\rm{NaF}}\): \({\rm{682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol,}}\) or \({\rm{3175 kJ/mol}}\)? Explain your answer.
Step-by-Step Solution
VerifiedThe value for lattice energy for \(NaF\) is \({\rm{924 kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}\) and \({\rm{924 kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}\) closely approximates to the result.
The lattice energy is the change in energy that occurs when one mole of a crystalline ionic compound is formed from its constituent ions, which are considered to be in a gaseous state at the start. It's a metric for the forces that bind ionic solids together.
Formula to calculate lattice energy is –
\({\rm{U = C}}\left( {\frac{{{{\rm{Z}}^{\rm{ + }}}{{\rm{Z}}^{\rm{ - }}}}}{{{{\rm{R}}_{\rm{o}}}}}} \right)......(1)\)
Where \({{\rm{R}}_{\rm{o}}}\), is the interatomic distance.
As far as charges are considered, this is same in both \({\rm{KF}}\) and \({\rm{NaF}}\). Major difference is expected to be in interatomic distance i.e., \({\rm{2}}{\rm{.69}}\mathop {\rm{A}}\limits^{\rm{o}} \) vs \({\rm{2}}{\rm{.31}}\mathop {\rm{A}}\limits^{\rm{o}} \).
From the data for \({\rm{LiF}}\), with \({{\rm{Z}}^{\rm{ + }}}{{\rm{Z}}^{\rm{ - }}}{\rm{ = - 1}}\),
\(\begin{align}C &= \frac{{U{R_o}}}{{{Z^ + }{Z^ - }}}\\ &= \frac{{794 \times 2.69}}{{ - 1}}\\ &= - 2135{\rm{ }}kJ\mathop A\limits^o mo{l^{ - 1}}\end{align}\).
Then, it is obtained that –
\(\begin{align}{U_{NaF}} &= \frac{{ - 2135 \times - 1}}{{2.31\mathop A\limits^o }}\\ &= 924{\rm{ }}kJmo{l^{ - 1}}\end{align}\)
Therefore, value for lattice energy is obtained as \({\rm{924 kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}\).