Electronegativity / predicting structure / ranking order

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The discussion focuses on calculating electronegativity differences for three compounds: CaBr2, Na3N, and CH4, with results showing values of 1.9, 2.1, and 0.3, respectively. Based on these differences, CaBr2 and Na3N are identified as ionic compounds, while CH4 is classified as a molecular compound with a non-polar covalent bond. The ranking of ionic character places Na3N first, followed by CaBr2, and CH4 last. The conversation emphasizes that ionic bonds typically form between metals and non-metals, particularly with halogens. Overall, the thread illustrates the relationship between electronegativity and bond type in chemical compounds.
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Homework Statement



a) Calculate the differences in electronegativity between the elements in each of these compounds.
I)CaBr2
II) Na3N
III) CH4
b) Predict whether each of these compounds would be an ionic or a molecular compound and justify each prediction.
c) Rank the bonds in these compounds in order of decreasing ionic character. Where do we always find compounds containing metals, in this ranking order?

Homework Equations


none.

The Attempt at a Solution



a)
i) CaBr_2
∆EN=2.9-1.0
=1.9
ii) Na_3 N
∆EN=3.0-0.9
=2.1
iii) CH_4
∆EN=2.5-2.2
=0.3
b)
i) CaBr_2
∆EN>1.7 ∴The compound is ionic and has a non-polar covalent bond
ii) Na_3 N
∆EN>1.7 ∴The compound is ionic and has a non-polar covalent bond
iii) CH_4
1.7>∆EN>0 ∴The compound is molecular and has a polar covalent bond
c)
Rank 1: Na_3 N Most Ionic Character
Rank 2: CaBr_2 2nd most Ionic Character
Rank 3: CH_4 Least Ionic Character
We find metals in compounds that form Ionic bonds when non-metals meet metals Ions form leading to one atom completely taking the electrons from the other atom since one of the atoms want to take atoms to complete their octet. This usually occurs with halogens in group 17 since they desperately want to complete their valence shell since every atom wants to be like the noble gases.
 
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Ionic compounds have ionic bonds, not covalent bonds. The electonegativity difference between C and H is small enough that most would consider the C-H bond to be a non-polar covalent bond (as opposed to something like an O-H bond or N-H bond, which would be considered polar covalent).

In terms of electonegativity difference (from largest difference to smallest difference), you would have:
Ionic > polar covalent > non-polar covalent
 
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I made some adjustments according to your comments:

b)
i) CaBr_2
∆EN>1.7 ∴The compound has an ionic bond
ii) Na_3 N
∆EN>1.7 ∴The compound has an ionic bond
iii) CH_4
0.4>∆EN>0 ∴The compound is molecular and has a non polar covalent bond
c)
Rank 1: Na_3 N Most Ionic Character
Rank 2: CaBr_2 2nd most Ionic Character
Rank 3: CH_4 Least Ionic Character ( non polar covalent bond)

What do you think? did I address the issues, sorry I'm still learning I'm very new to chemistry.
 
Looks good.
 
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