Order the following bonds by increasing polarity? #"K"-"F"#, #"C"-"C"#, #"C"-"N"#, #"C"-"O"#, #"C"-"F"#
1 Answer
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.
- Look up the electronegativities of each atom in your textbook.
- Find the difference in electronegativity for each atom in the bond.
- 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)"C""-"stackrel(4.0)"F"# ,#" "|DeltaEN| = ???# #stackrel(2.5)"C""-"stackrel(3.0)"N"# ,#" "|DeltaEN| = ???# #stackrel(0.8)"K""-"stackrel(4.0)"F"# ,#" "|DeltaEN| = ???# #stackrel(2.5)"C""-"stackrel(2.5)"C"# ,#" "|DeltaEN| = 0# #stackrel(2.5)"C""-"stackrel(3.5)"O"# ,#" "|DeltaEN| = ???#
At this point, it is just subtraction and organization by number magnitude. Again, larger
Highlight below after finishing to check your answer...
#overbrace(color(white)(stackrel(0.8)"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"))^(DeltaEN = 0)#
