QUESTION ABOUT IONIC AND ATOMIC RADII

[E_Villavicencio 2N]

Question number 93 in the book says:
Below is pictured the reaction between an atom of sodium and an atom of chlorine. Identify each element and the ions formed and explain your reasoning.

The picture shows the atoms A (SMALLER) and B(LARGER), and then the ions C (LARGER) and D(SMALLER)

Taking into account the trends of atomic and ionic size, I would say that A is Chlorine and B is Sodium (because the size decreases from left to right and Cl is on the right), and then C is Cl(-) and D is Na(+) (because Cl is gaining and electron and therefore becomiong bigger).
However the solutions manual shows totally the opposite. In it, A is sodium, B is Chlorine, C is Na+ and D is Cl-

Can somebody explain that to me please?

[Amy_Bugwadia_3I]

You are correct! Maybe you were looking at the wrong problem's answer? At the back of my book, it says the following for the answer for 2.93: "A  Cl; B  Na; C  Cl- ; D  Na+", which is the answer that fits your initial explanation. You're right about the fact that Cl has a smaller atomic radius than Na, because Cl is in the same period as Na but is further to the right (it has more electrons and thus they are held tighter and closer to the nucleus than they are in Na). When Cl gains an electron and becomes the ion Cl-, it does become bigger, just as Na+ is smaller than Na because it loses its one valence electron and thus doesn't have as many subshells.

[Emily_Lenh2A]

On Lavelle's website, he pointed out and posted errors from the solution manual. Problem 2.93 happens to be one of the questions that have an error in the solution manual.
Here is the link: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Solution_Manual_Errors_6Ed.pdf
Hope this helps!