Order the following compounds according to the increasing ionic c...

Question

Order the following compounds according to the increasing ionic character of their bonds: CCl4, BaCl2, TiCl3, ClO2.

Accepted Answer

welcome back everyone. We need to in order of increasing ionic character of the bonds, sort the following compounds. We're going to begin by recalling that. The greater the election negativity difference of the compound that corresponds to a greater ionic character of the compounds bonds. So beginning with our first compound silicon die bromide or sorry, sulfur di bromide we have in our textbooks the electro negativity value for our atom of roaming Which we would see is a value of 2.8. And noting down the election negativity value. Force atom of sulfur in our textbooks we see a value of 2.5. So for the election negativity difference we would have 2.8 minus 2.5. Leaving us with a difference of 0.3. Moving on to our second compound, we have scanned iem three bromide Where we would note down the election negativity value for an atom of scandia in our textbooks we would see that that value is 1.3. So taking the election negativity difference, we have the election negativity value for bromine which we stated earlier was 2.8 subtracted from 1.3 and this difference gives us 1.5. So moving on to our fourth compound, we have strong tm to bromide where noting down the election negativity value for an atom of strontium in our textbooks we see a value of one and so taking the election negativity difference, we have the election negativity value of bromine being 2.8 minus one and this leaves us with a difference of one point oh and lastly for our fourth compound, we have silicon tetra bromide. We're noting down our election negativity value for an atom of silicon in our textbooks we see a value of 1.8. So finding the election negativity difference, we would take 2.8 for bromine minus 1.8 for silicon and this difference is going to leave us with one point oh And just to make a correction here are difference for our election negativity value for strontium bromide should be 1.8. So apologies for this mistake here and now we want to correct our difference for or sorry. Now we want to rank our compounds by increasing ionic character. So we're going to begin with the compound that had the lowest ionic character or lowest election negativity difference And that would be our value of .3 for sulfur di bromide. So this is the lowest ionic character compound based on that election activity difference. And this would be listed again first in our ranking. And we would say that next. In our ranking, we have the second lowest value which corresponded to silicon touch. A bromide with an election negativity difference of 1.0 and that would be listed as we stated. Second in our ranking because it has greater ionic character than or sulfur di bromide. Moving up next in our ranking, we have the third smallest difference in electro negativity which is associated with skandia. Um three bromide which is a value of 1.5. And so that would be listed again third in our ranking because Skandia three bromide has a greater ionic character than silicon touch, A bromide and sulfur di bromide. And first or listed fourth in our ranking. Sorry, We have our last compound which associated with the greatest election activity difference being our compound strontium to bromide with election activity difference of 1.8. And so this would be the compound with the greatest ionic character. Whereas as we stated initially, sulfur di bromide is the compound with the lowest ionic character because it had the smallest election negativity difference. And so for our final answer, we have the below ranking that we've written out to complete this example. I hope everything I explained was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

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