Answer:
a. KCl, c. BaCl2 and e. LiF.
Step-by-step explanation:
Hello,
In this case, we can identify the ionic compounds by verifying the difference in the electronegativity between the bonding compounds when it is 1.7 or more (otherwise it is covalent) as shown below:
a. KCl: 3.0-0.9=2.1 -> Ionic.
b. C2H4: 2.5-2.1=0.4 -> Covalent.
c. BaCl2: 3.0-0.8=2.2 -> Ionic.
d. SiCl4: 3.0-1.8=1.2 -> Covalent.
e. LiF: 4.0-1.0=3.0 -> Ionic.
Therefore, ionic compounds are a. KCl, c. BaCl2 and e. LiF.
Regards.