# Order the following bonds by increasing polarity? "K"-"F", "C"-"C", "C"-"N", "C"-"O", "C"-"F"

Jun 21, 2017

I'll give you the general steps and outline how to do it, but this is something you should do yourself so that you get how to do it.

1. Look up the electronegativities of each atom in your textbook.
2. Find the difference in electronegativity for each atom in the bond.
3. Larger difference = more polar (thus, the more electronegative atom hogs the electron density more) and vice versa.

You should come up with the following electronegativities (listed above the atoms):

• $\stackrel{2.5}{\text{C""-"stackrel(4.0)"F}}$, " "|DeltaEN| = ???
• $\stackrel{2.5}{\text{C""-"stackrel(3.0)"N}}$, " "|DeltaEN| = ???
• $\stackrel{0.8}{\text{K""-"stackrel(4.0)"F}}$, " "|DeltaEN| = ???
• $\stackrel{2.5}{\text{C""-"stackrel(2.5)"C}}$, $\text{ } | \Delta E N | = 0$
• $\stackrel{2.5}{\text{C""-"stackrel(3.5)"O}}$, " "|DeltaEN| = ???

At this point, it is just subtraction and organization by number magnitude. Again, larger $| \Delta E N |$ $=$ more polar.

Highlight below after finishing to check your answer...

${\overbrace{\textcolor{w h i t e}{\stackrel{0.8}{\text{K""-"stackrel(4.0)"F"))^(DeltaEN = 3.2) > overbrace(color(white)(stackrel(2.5)"C""-"stackrel(4.0)"F"))^(DeltaEN = 1.5) > overbrace(color(white)(stackrel(2.5)"C""-"stackrel(3.5)"O"))^(DeltaEN = 1.0) > overbrace(color(white)(stackrel(2.5)"C""-"stackrel(3.0)"N"))^(DeltaEN = 0.5) > overbrace(color(white)(stackrel(2.5)"C""-"stackrel(2.5)"C}}}}}^{\Delta E N = 0}$