Question

circle the compound in each pair with the greater lattice energy (in magnitude). then justify your answer.
a. nacl or kcl
b. lif or licl
c. mg(oh)2 or mgo
d. fe(oh)2 or fe(oh)3
e. nacl or na2o
f. mgo or bas

Explanation:

Step1: Recall lattice - energy factors

Lattice energy is directly proportional to the product of ion charges and inversely proportional to the distance between ions (ionic radii).

Step2: Analyze pair A

For NaCl and KCl, the anions are the same ($Cl^-$). Cation charges are the same ($+ 1$ for $Na^+$ and $K^+$). But ionic radius of $Na^+

Step3: Analyze pair B

For LiF and LiCl, cations are the same ($Li^+$). Anion charges are the same ($ - 1$). Ionic radius of $F^-

Step4: Analyze pair C

For $Mg(OH)_2$ and MgO, in MgO, $Mg^{2 + }$ and $O^{2 - }$ have charges of + 2 and - 2 respectively. In $Mg(OH)_2$, $Mg^{2+}$ has charge + 2 and $OH^-$ has charge - 1. Also, $O^{2 - }$ is smaller than $OH^-$. So, MgO has greater lattice energy.

Step5: Analyze pair D

For $Fe(OH)_2$ and $Fe(OH)_3$, $Fe^{3+}$ has a higher charge than $Fe^{2+}$ in $Fe(OH)_2$. Anions are the same ($OH^-$). So, $Fe(OH)_3$ has greater lattice energy.

Step6: Analyze pair E

For NaCl and $Na_2O$, in $Na_2O$, $Na^+$ has charge + 1 and $O^{2 - }$ has charge - 2. In NaCl, $Na^+$ has charge + 1 and $Cl^-$ has charge - 1. Also, $O^{2 - }$ is smaller than $Cl^-$ in terms of ionic radius. So, $Na_2O$ has greater lattice energy.

Step7: Analyze pair F

For MgO and BaS, charges on ions are + 2 and - 2 for both compounds. But ionic radius of $Mg^{2+}

Answer:

A. NaCl
B. LiF
C. MgO
D. $Fe(OH)_3$
E. $Na_2O$
F. MgO