Answer
a) C $\lt$ N $\lt$ O
b) Se $\lt$ S $\lt$ Cl
c) Sn $\lt$ Ge $\lt$ Si
d) Ti $\lt$ Ge $\lt$ S
Work Step by Step
In order to find relative electronegativity, you need to remember the periodic trend. It increases across a period and decreases down a group.
For part "A" the elements are arranged across the first period. This means that the electronegativities will increase as the elements increase in atomic number across the period, yielding C $\lt$ N $\lt$ O.
For part B, the elements follow two trends. An easy way to remember this trend is that as you get closer to fluorine, the electronegativities increase. Se is farthest from F and Cl is the closest, giving the result Se $\lt$ S $\lt$ Cl.
For part C, you must recall the group trend. As electronegativity goes down a group, it decreases. This yields the result Sn $\lt$ Ge $\lt$ Si.
For part D, remember that the elements closer to Fluorine have higher electronegativities. This means that S will have the highest and Ti will have the lowest, giving us the answer: Ti $\lt$ Ge $\lt$ S
