I cam across a question in an IB HL chemistry paper which reads:
Which of the following statements help(s) to explain why the lattice enthalpy value for lithium fluoride (+1022 kj/mol) is less than that of calcium fluoride (+2602 kj/mol)?
I) The ionic radius of lithium is less than that of calcium.
II) The ionic charge of lithium is less than that of calcium.
I understand that the greater the charge, the greater the bond strength.
But surely, MgF2, for example, would have a greater lattice enthalpy than CaF2 because it has a smaller ionic radius?
Which of the following statements help(s) to explain why the lattice enthalpy value for lithium fluoride (+1022 kj/mol) is less than that of calcium fluoride (+2602 kj/mol)?
I) The ionic radius of lithium is less than that of calcium.
II) The ionic charge of lithium is less than that of calcium.
I understand that the greater the charge, the greater the bond strength.
But surely, MgF2, for example, would have a greater lattice enthalpy than CaF2 because it has a smaller ionic radius?
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Hi, thank your for your question.
I believe this question is only asking evidence to support the fact that LiF's lattice enthalpy value is less than that of CaF2. Therefore, if radius decreases, then the bond strength should increase, leading to a greater lattice enthalpy value. This answer doesn't support the statement in the question.
Also, there is one thing you should know about the F=kqq/r2 law. Charge always dominates over the distance. So CaF2 will be stronger than NaF because of the 2+ charge from Ca.
You are right, for MgF2 has a greater lattice enthalpy than CaF2 because both of them has the same kqq value. Now this is where the distance thing comes in. MgF2 has a smaller distance so its bond is stronger.
The MgF2 thing cannot be applied to LiF because the charge is 1 on Li. The condition is totally different. Since I said that charge always dominates over distance, CaF2 is stronger.
Anyway, as a conclusion, you can't pick I) because first of all it doesn't explain or support the statement in the question.
Hope that helped.
I believe this question is only asking evidence to support the fact that LiF's lattice enthalpy value is less than that of CaF2. Therefore, if radius decreases, then the bond strength should increase, leading to a greater lattice enthalpy value. This answer doesn't support the statement in the question.
Also, there is one thing you should know about the F=kqq/r2 law. Charge always dominates over the distance. So CaF2 will be stronger than NaF because of the 2+ charge from Ca.
You are right, for MgF2 has a greater lattice enthalpy than CaF2 because both of them has the same kqq value. Now this is where the distance thing comes in. MgF2 has a smaller distance so its bond is stronger.
The MgF2 thing cannot be applied to LiF because the charge is 1 on Li. The condition is totally different. Since I said that charge always dominates over distance, CaF2 is stronger.
Anyway, as a conclusion, you can't pick I) because first of all it doesn't explain or support the statement in the question.
Hope that helped.