Video Transcript
Which of the following diatomic molecules is likely to have the greatest difference in electronegativity between its atoms? (A) O2, (B) CO, (C) HF, (D) HBr, or (E) CsF.
This is a question about electronegativity, which is defined as the tendency of an atom to attract a bonding pair of electrons. Each of the five answer choices is a different diatomic molecule, meaning a molecule made up of two atoms bonded to one another. The question is asking us to find the diatomic molecule that has the greatest difference in electronegativity between the two atoms. Each element has its own electronegativity value. If we have a diatomic molecule where one atom has a very high electronegativity, while the other atom has a very low electronegativity, that molecule has a large difference in electronegativity between its atoms.
Essentially, our task is to compare the electronegativities of these elements and find the diatomic molecule that has the atom with the highest electronegativity and the lowest electronegativity. We can solve this problem two different ways. First, we can use the periodic trends of the periodic table to determine which elements are more or less electronegative than other elements. The second method for solving this problem uses the Pauling scale, which links the electronegativities of atoms to specific numbers. After quantifying the electronegativities, we can then compare them.
Let’s start with method one. Here, we have a periodic table with each of the six elements that appear in the answer choices highlighted in pink. In order to answer this question, we need to know the periodic trends in electronegativity. In other words, how does electronegativity change as we move side to side on the periodic table as well as up and down on the periodic table? The horizontal trend for electronegativity is that it increases as we move to the right across a period in the periodic table.
To understand this trend, we can compare two elements from either side of the periodic table. Elements to the right of the periodic table, like fluorine, have nearly full outer electron shells. They’re much more likely to attract a bonding pair of electrons in order to try to fill that electron shell. On the other hand, elements to the left of the periodic table, like cesium, have fewer valence electrons. Cesium has a nearly empty shell and is much more likely to give up that electron than attract additional electrons. The vertical trend is that electronegativity increases as we move up a group.
Elements further up in a group have fewer electron shells. Because there are fewer electron shells, the nucleus of the atom is closer to the bonding pair of electrons that the atom interacts with. So there’s a stronger attraction between the nucleus and the bonding pair of electrons. Now that we know the periodic trends for electronegativity, we can begin to compare the electronegativities of these elements.
Since electronegativity increases up and to the right, fluorine, the atom that’s the furthest up and to the right of the group, is the most electronegative element we see here. In fact, it’s the most electronegative element on the entire periodic table. On the other hand, cesium in the bottom-left corner of the periodic table is the least electronegative element of the bunch. It’s also the least electronegative element on the entire periodic table. Looking at our answer choices, one stands out. Cesium fluoride has both the most electronegative atom and the least electronegative atom. Therefore, it must have the greatest difference in electronegativity between its atoms. Choice (E), cesium fluoride, is the correct answer.
As we mentioned earlier, another way to solve this problem or a way to confirm our answer from the first method is to use the Pauling electronegativity scale. On this scale, each element on the left is associated with a unitless number on the right. The higher the number, the greater the electronegativity of that element. Once again, we can see that fluorine is the most electronegative element with a Pauling electronegativity of 3.98. Cesium is the least electronegative element with a Pauling electronegativity of 0.79. Sometimes, we may want a quantified difference in electronegativity instead of just a relative comparison. In those cases, we can use the formula Δ𝐸𝑁 equals the absolute value of 𝐸 one minus 𝐸 two.
Δ𝐸𝑁 represents the difference in electronegativity. 𝐸 one and 𝐸 two represent the Pauling electronegativities of the two atoms in the bond. If we wanted to calculate the electronegativity difference for cesium fluoride, we would use 3.98 for fluorine and 0.79 for cesium. Carrying out the arithmetic, we get 3.19, the electronegativity difference for cesium fluoride. We use the absolute value bars in this equation to make sure that our end result is a positive number. It also ensures that we can include the Pauling electronegativity values of each of the two elements in either order and still end up with the same answer. Cesium fluoride’s electronegativity difference of 3.19 is indeed the highest electronegativity difference for any of these diatomic molecules.
Note that since oxygen gas is a diatomic molecule that involves two atoms of the same element, there is no difference in electronegativity between its atoms. Its Δ𝐸𝑁 is zero. Whether we look at the trends in the periodic table or at the Pauling electronegativity scale, our answer is the same. The diatomic molecule that is likely to have the greatest difference in electronegativity between its atoms, that’s choice (E), cesium fluoride.
