Use your knowledge of electronegativity to pick out the most p...

Question

Use your knowledge of electronegativity to pick out the most polar bond in the list below.

Accepted Answer

Welcome back. Everyone predict which of the following pairs of elements will form the most polar bond. We've got choice, a silicon bonded to bromine, choice B phosphorus bond to chlorine, choice C carbon bonded to fluorine or choice D nitrogen bonded to oxygen. Let's recall that the difference in electron negativity between two atoms tells us the bond classification or bond polarity. And so recall that the greater the electron activity difference correlates to a higher polarity of that bond between the atoms sharing them. And so going to choice a, we're going to note down the electron activity values of each of the atoms making up the bond. And so we have silicon which on our product tables recall has an electron negativity value of 1.8. Whereas Bromine recall has an electron activity value of 2.8. And so calculating delta en or the difference in elect negativities, we would start with the larger value for bromine 2.8 subtracted by the smaller value for silicon of 1.8. And from that difference, we would find a difference in electron negativity for that bond of one point. Oh, we're going to compare this result to each of our other answer choices to determine the most polar bond moving on to choice. B recall that phosphorus on our product table has an electron negativity value of 2.1. Whereas chlorine recall on our periodic table has an electron negativity value of 3.0. Calculating the difference in electron negativity between these atoms. We have delta en starting with the higher value for chlorine. We have 3.0 subtracted by the smaller value for phosphorus of 2.1. And from this difference, we would find a difference in electron negativity equal to a value of 0.9. And so far because choice B with the bond of phosphorus to chlorine has a lower electron negativity difference of just 0.9. We would rule out choice B because it is a less polar bond than the silicon bromine bond. In choice A with an electron negativity difference of 1.0 moving on to, to C we've got carbon which on our periodic tables has an electron negativity value of 2.5. And fluorine recall is the most electron electron negative atom sorry, corresponding to a value of 4.0. Calculating the difference in electron negativity delta en we would take the higher value for fluorine of 4.0 And we would subtract by the value for carbon of 2.5. And this will result in the electron negativity difference equal to a value of 1.5 And so far that makes the carbon fluorine bond more polar than the silicon bromine bond in choice A. So we can now rule out choice. A. Now moving on to choice d the nitrogen oxygen bond beginning with the electron negativity value of nitrogen recall that it's a value of 3.0 on the periodic table. And then for oxygen recall that it has an electron negativity value of 3.5. So calculating the difference in electron negativity delta en we would take the higher value of 3.5 subtracted by the smaller value of 3.0. And from this difference, we would find a result of 0.5. And this would be the smallest elect negativity difference of all of the bonds we've analyzed. So we can easily rule out choice D that leaves us with choice C as the only correct choice. And our final answer which allows us to confirm that the carbon fluorine bond represents a pair of elements that form the most polar bond with their election negativity difference of 1.5. I hope that this made sense. And let us know if you have any questions.

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Use your knowledge of electronegativity to pick out the most polar bond in the list below.