Consider the following bonds: Ca and O, C and O, K and O, Oand O,...

Question

Consider the following bonds: Ca and O, C and O, K and O, Oand O, and N and O. (6.7)d. Arrange the covalent bonds in order of decreasing polarity.

Accepted Answer

Welcome back everyone. We need to list the bonds in order of increasing bond polarity, silicon and bromine, rubidium and bromine. Bromine and bromine phosphorus and bromine and strontium and bromine. We're going to begin by recalling that an increase of bond polarity is related to a increase in our election negativity difference of our bond. So this will be attributed by the difference of election negativity of our atoms making up our bond beginning with our first bond silicon bonded to bromine. Noting down the electron negativity value of silicon from our textbooks will find a value of 1.8 for bromine, we'll find a value of 2.8. So calculating our electron negativity difference of our bond, we would take the higher value for bromine 2.8 minus the lower value for silicon 1.8. And we'd find a difference of 1.0. Our next bond is our rubidium bromine bond. In which noting down the electron negativity value of rubidium from our textbooks will find a value of .8. So calculating the electron negativity difference, we take the higher value for bromine 2.8 minus the lower value for rubidium .8 and we'll find a difference equal to 2.0. Next, we have our bromine bromine bond And calculating the election negativity difference. We'd take 2.8 -2.8, which would give us a difference of zero. So far we can rank our bromine bromine bond With the lowest elect ne difference of zero as our least polar bond. Next, we have our phosphorus bromine bond. Noting down the electron negativity value for phosphorus from our textbooks, we'll find a value of 2.1. So calculating the electron negativity difference, we take 2.8 for bromine, the higher value minus our lower value 2.1 for phosphorus. And we'll find an electron negativity difference equal to .7. Next, we have our strontium bromine bond. Noting down our electron negativity value for strontium from our textbooks, we'll find a value of about one. So calculating our electron negativity difference, we would take our higher value for bromine 2.8 minus our lower value for Stron T M one. And we'll find an electron negativity difference of 1.8. Now, based on these calculations, Our highest election negativity difference was 1.8 attributed to our strong and roaming bond. So we would label it as the most polar and we would define the following ranking where, as we stated, our lowest election negativity difference was zero for our bromine bromine bond, which is going to be the least polar. And ranked last or lowest. Our second Lowest election negativity difference is attributed to our phosphorus roaming bond of .7, which is ranked second to last. Our third lowest election negativity difference was attributed to our silicon bromine bond with a value of 1.0, which is ranked third lowest. So we have our silicon bromine bond, then we have second lowest, our second lowest calculation for the electron negativity difference, Which was actually 1.8. So just to make a correction, our most polar bond was our calculation for rubidium bromine Which had a election negativity difference of 2.0 our strong and bromine bond, as we just stated, has a election negativity difference of 1.8. So that is what's listed second In our ranking and then listed first in our ranking as the most polar bond. We have our greatest election activity difference of 2. attributed to our rubidium bromine bond, which is listed first in our ranking again as the most polar. And this is in order of increasing bond polarity, which increases in the direction of our rubidium bromine bond being the most polar. And again, our bromine bromine bond being the least polar. So this ranking is our final answer in increasing bond polarity. I hope that this made sense. And let us know if you have any questions

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Consider the following bonds: Ca and O, C and O, K and O, Oand O, and N and O. (6.7)d. Arrange the covalent bonds in order of decreasing polarity.

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