POI is a bent molecule similar to water. How are the two molecules different it terms of bond angles and what causes this? phosphorus oxyiodide ($\ce{POI}$) has the VSPER AXE formula of $\ce{AX2E}$ while water, $\ce{H2O}$ has the AXE code of $\ce{AX2E2}$. According to my work sheet $\ce{POI}$ should have a $120^\circ$ bond angle while $\ce{H2O}$ should have a bond angle of $104.5^\circ$. 

Does this mean that due to the extra lone pair it creates less resistance resulting in a greater angle? Also, does it having three different elements in the equation do anything?
Does this mean that:
  
  
    Due to the extra lone pair $\color{red}{\text{it}}$ $\color{limegreen}{\text{creates less resistance}}$ resulting in $\color{blue}{\text{a greater angle}}$?
  


More like:

Due to the extra lone pair $\color{red}{\text{in } \ce{H2O} \text{ the pendant atoms}}$ $\color{limegreen}{\text{experience more repulsion}}$ resulting in $\color{blue}{\text{a smaller angle}}$ $\color{magenta}{\text{compared to }\ce{POI}}$? 




  Also, does it having three different elements in the equation do anything?


No, I think the three different elements in the question were there to evaluate if you were:


estute to the fact that oxygen forms a double bond thus
phosphorous has 1 unbonded pair of electrons (not 2) and
if the double bond would hang you up in obtaining an answer.