Consider the Lewis structure for the polyatomic oxyanion shown here, where X is an element from the third period (Na - Ar). By changing the overall charge, n, from 1- to 2- to 3- we get three different polyatomic ions. For each of these ions (b) determine the formal charge of the central atom, X;

Question

Accepted Answer

Hey everyone, we're given the following generic lewis structure. For a poly atomic oxy an ion where X is a period for element we're told that three different poly atomic ions are formed when the overall charge and is changed to one minus two minus or three minus. Were asked to determine the identity and calculate the formal charge of the central atom X for each ion formed. Now to answer this question, let's first go ahead and determine the total number of valence electrons. Our structure has Looking at our structure, we have 10 loan pairs so we can go ahead and multiply this by two since one lone pair consists of two valence electrons And this will get us to 20 valence electrons. Next we can see that we have three bonds and we're going to multiply this by two valence electrons as well. Since one bond consists of two valence electrons. This will get us to a total of six Adding these two values up. We get 26 valence electrons. Now to determine the number of valence electrons our X will give we can go ahead and subtract the number of valence electrons our oxygen gives. So we know that oxygen is in our group six. And since we have three of our oxygen's, we're going to multiply three times six to get 18. And when we subtract these two values we get a total of eight valence electrons. So this is the total number of valence electrons. Our X will give, not accounting for our overall charge. So let's go ahead and account for our overall charge. Starting with N equals one minus. This means that we added one electron. So to determine the number of valence electrons, our X is going to give We can go ahead and take that number eight And subtract one valence electron and this will get us to a total of seven. So this means Our X is going to be in group seven a and it's going to be in period four. So this means that we have browning. Now let's go ahead and do this with N equals two minus again. Taking the same steps. We're going to take that value of eight and subtract two valence electrons. This will get us to a total of six, which means that our period four element is going to be in our group six A. So we have selenium lastly looking at N equals three minus, We're going to take that value of eight again And we're going to subtract three. This will get us to a total of five valence electrons. So this means our element is going to be in group five A period four and that's going to be our arsenic. Now let's go ahead and calculate our formal charge and we know our formula is our group number minus the sum of our bonds plus our non bonding electrons. So calculating the formal charge for bro mean we're going to take that group number of seven And we're going to subtract the sum of three plus two since we have three bonds and two non bonding electrons. This will get us to a formal charge of plus two. Now looking at selenium, we're going to take that group number of six and we're going to subtract that sum of three plus two, which will get us to a value of plus one. Next looking at arsenic, we're going to take its group number of five and subtract the sum of three plus two, which will get us to a formal charge of zero. So this is going to be our final answers for this question Now, I hope that made sense. And let us know if you have any questions.

Consider the Lewis structure for the polyatomic oxyanion shown here, where X is an element from the third period (Na - Ar). By changing the overall charge, n, from 1- to 2- to 3- we get three different polyatomic ions. For each of these ions (b) determine the formal charge of the central atom, X;

Verified Solution

Key Concepts

Lewis Structures

Formal Charge

Polyatomic Ions