Question

Use the chart to determine which is an example of a highly polar covalent bond. Ca – Cl Fe – O H – S F – F Done Elements Electronegativity aluminum (Al) 1.61 calcium (Ca) 1 carbon (C) 2.55 chlorine (Cl) 3.16 fluorine (F) 4 hydrogen (H) 2.2 iron (Fe) 1.83 nitrogen (N) 3.04 oxygen (O) 3.44 phosphorus (P) 2.19 potassium (K) 0.82 silicon (Si) 1.9 sulfur (S) 2.58

Answer 1

The most polar bond has the greatest difference in electronegativity.  Thus, we calculate the differences for each bond:

Ca-Cl: 3.16-1=2.16

Fe-O 3.44-1.83=1.61

H-S 2.58-2.2=0.38

F-F=4-4=0

The Ca-Cl bond has the greatest difference in electronegativity and is therefore the most polar.

Answer 2

Answer: Fe - O

Explanation:

1) Start by showing the chart information in a more understandable way:

Elements ---------- Electronegativity

aluminum (Al) ----- 1.61

calcium (Ca) ------- 1

carbon (C) ---------- 2.55

chlorine (Cl) -------- 3.16

fluorine (F) --------- 4

hydrogen (H) ------ 2.2

iron (Fe) ------------- 1.83

nitrogen (N) --------- 3.04

oxygen (O) ----------- 3.44

phosphorus (P) ----- 2.19

potassium (K) -------- 0.82

silicon (Si) ------------- 1.9

sulfur (S) --------------- 2.58

2) Find the electronegativity differences for the atoms in each compound:

i) Ca – Cl: 3.16 - 1 = 2.16

ii) Fe – O: 3.44 - 1.83 = 1.61

iii) H – S: 2.58 - 2.2 = 0.38

iii= F – F = 4 - 4 = 0

3) Analysis

i) The higher the electronegativity difference between two non-metal elements, the the more the polar the covalent bond. Therefore, the bond H - S is the most polar among the non-metal - non-metal bonds listed. But 0.38 is a small difference, so this is not very polar.

ii) For metal - non-metal bonds, when the difference in electronegativities is too high (close to or greater than 2.0) the ionic character is dominant, and so you cannot classify the bond as polar but as ionic. Therefore, Ca - Cl with 2.16 electronegativity difference is a ionic bond.

iii) The bond in Fe – O has electronegativty difference of 1.61, so it is yet covalent, and highly polar.