Answer
a) Pure covalent
b) Ionic
c) Polar covalent
Work Step by Step
a) The elctronegativity difference between $I$ and $I$ is zero. So, the bond is pure covalent
b) The elctronegativity difference between $Cs(0,7)$ and $Br(2.8)$ is $2.1$. So, the bond is ionic.
c) The elctronegativity difference between $P(2.1)$ and $O(3.5)$ is $1.4$. So, the bond is polar covalent
(See table 9.1 and figure 9.8)
