LeAirraBullingor2k
Posts: 77
Joined: Wed Mar 13, 2019 12:15 am

Hello, I understand that the trend of ionic radius increases as you move to the left and down a column but I'm not understanding how to order ions with different charges.

For example, S2-, Cl-, P3-. If they were neutrally charged, it would be Cl, S, and P in order of increasing ionic radius but how do you take into account the different negative charges?

Nyari Muchaka_Discussion 4A
Posts: 53
Joined: Thu Sep 19, 2019 12:17 am

All three of those atoms essentially copy the orientation of Argon as they have gained electrons in order to satisfy the octet rule. However, since they have gained electrons but keep the same number of protons, the positive charge is spread out more over the atom with each additional negative charge. Since the protons are unable to hold onto the extra electrons as well, the ionic radius of the atom increases. P3- would have the largest radius as it had the most electrons added and thus the charge throughout the atom is the weakest. S2- is the second largest, and Cl- the smallest as it only had one electron added and thus the positive charge had to be spread out less.

LeAirraBullingor2k
Posts: 77
Joined: Wed Mar 13, 2019 12:15 am

Nyari Muchaka_Discussion 4A wrote:All three of those atoms essentially copy the orientation of Argon as they have gained electrons in order to satisfy the octet rule. However, since they have gained electrons but keep the same number of protons, the positive charge is spread out more over the atom with each additional negative charge. Since the protons are unable to hold onto the extra electrons as well, the ionic radius of the atom increases. P3- would have the largest radius as it had the most electrons added and thus the charge throughout the atom is the weakest. S2- is the second largest, and Cl- the smallest as it only had one electron added and thus the positive charge had to be spread out less.

So because they're isoelectronic, you determine the size solely on added electrons?

Alex Tchekanov Dis 2k
Posts: 118
Joined: Sat Aug 24, 2019 12:16 am

Because they are isoelectronic, the shielding effect will be different and the strength of the positive nuclear pull will be less on the P atom because the proton to electron ratio is lower. And thus, because there is a weaker relative positive charge pulling on the electrons in the atoms with fewer protons, the electrons will be able to move further away from the nucleus, thus having a greater radius.

905416023
Posts: 54
Joined: Thu Jul 25, 2019 12:17 am

Are there exceptions to the ionic radius trend? Like any elements that don't follow it?

Nyari Muchaka_Discussion 4A
Posts: 53
Joined: Thu Sep 19, 2019 12:17 am

LeAirraBullingor3k wrote:
Nyari Muchaka_Discussion 4A wrote:All three of those atoms essentially copy the orientation of Argon as they have gained electrons in order to satisfy the octet rule. However, since they have gained electrons but keep the same number of protons, the positive charge is spread out more over the atom with each additional negative charge. Since the protons are unable to hold onto the extra electrons as well, the ionic radius of the atom increases. P3- would have the largest radius as it had the most electrons added and thus the charge throughout the atom is the weakest. S2- is the second largest, and Cl- the smallest as it only had one electron added and thus the positive charge had to be spread out less.

So because they're isoelectronic, you determine the size solely on added electrons?

Yes. Look at the ratio between protons and electrons. More electrons to protons, means a larger radius as you have a weaker nuclear force.