Consider the six second- and third-row elements in groups 4A–6A of the periodic table:

Possible structures for the binary fluorides of each of these elements in its highest oxidation state are shown below.

(a) Identify the nonfluorine atom in each case, and write the molecular formula of each fluoride.

Video transcript

Hello, everyone today. With the following problem, consider the elements listed in the periodic table. The binary bromine formed by these elements in their highest oxidation state could have one of the following structures identify the non bromine atom and molecular formula in each possible structure. So first, we will start with our carbon. Now, we note that carbon is a group four A element and its highest oxidation state is positive four. And so that means that it can accommodate up to four hay lines. So it would form a structure sea or carbon followed by four, bro means moving on to our nitrogen. Nitrogen is a group of five A element and it has the highest oxidation state of positive five. However, due to its small tonic radius and high electronegativity, it can actually only accommodate three hides. So it would be given the following structure of nitrogen followed by three, bro means and just so we can keep track, we can see here that our carbon would have the structure of A and A nitrogen that has three, bro means would have the structure of B moving on to our oxygen. Oxygen is a group six A element. And so its highest oxidation state is positive six. However, just like nitrogen, it has a small atomic radius in a high electron negativity. And so this time, it can only accommodate up to two ha lines. It is also known that oxygen has an oxidation number of minus two. And so it would actually only form with two ha lines or accommodate up to two ha lines giving us the following formula which corresponds to eat. Now, we have our Germanium, which is going to be a group four, a element as well just as carbon and it will have the highest oxidation state of positive of four. And we would say that it could accommodate up to 4h lines. So we would have the Germanium followed by four bro memes which would have a structure similar to a, then we have our arsenic and arsenic is a group five A element and its highest oxidation is positive five. But because it has a larger atomic radius and a lower electron negativity instead of it only accommodating three, it can accommodate up to five hides. So we give the formula, you know, the arsenic followed by five, bro means and so this gives us the structure of C and then last but not least we have selenium, which is a group six A element which gives its highest oxidation number or state of positive six. And just like arsenic, it has a larger atomic radius and a lower electron negativity, meaning it could accommodate six H lights giving us the formula of selenium followed by six bro means or the choice of answer or the structure of D. And so if we look at our answer choices, we set that our structures for our day included our carbon and our germanium. So if we look at our answer choices, answer choice A is incorrect as it does not have this. If we look at answers, choice B, this also does not have this. If we look at answer choice C, we see that carbon and bromine both have the structure of A. As we said, if we look at nitrogen, it has a structure of B, oxygen has the structure of E arsenic. Has a structure of C and selenium has a structure of D, meaning that our correct answer here is answer to AC and with that, we have solved the problem overall, I hope this helped. And until next time.