Which is the larger species in each pair? a. Sr or Sr 2+ b. N or N 3- c. Ni or Ni 2+ d. S 2- or Ca 2+

1. The size of an atom or ion is determined by the number of electron shells it has and the effective nuclear charge. The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. The more protons in the nucleus, the stronger the pull on the electrons and the smaller the atom or ion.. 2. For Sr and Sr 2+ , Sr 2+ has lost two electrons compared to Sr. This means it has a higher effective nuclear charge and fewer electron shells, making it smaller.. 3. For N and N 3- , N 3- has gained three electrons compared to N. This means it has a lower effective nuclear charge and more electron shells, making it larger.. 4. For Ni and Ni 2+ , Ni 2+ has lost two electrons compared to Ni. This means it has a higher effective nuclear charge and fewer electron shells, making it smaller.. 5. For S 2- and Ca 2+ , S 2- has gained two electrons compared to Ca 2+ which has lost two electrons. This means S 2- has a lower effective nuclear charge and more electron shells, making it larger.

Which is the larger species in each pair? a. Sr or Sr 2+ b. N or N 3- c. Ni or Ni 2+ d. S 2- or Ca 2+

Hey everyone in this example, we need to identify the elements or ion that is larger in size for each of the given sets. Before we do so, we want to recall our facts on atomic size corresponding to our ionic charge and our total electrons. So what we want to recall is that the greater the amount of electrons in an atom will correspond to a larger atomic radius for atom. The second thing we want to recall is that the more positive our ion that's going to relate to a loss of electrons in the atom and the more negative our ion will relate to our atom gaining electrons. And lastly, we want to go ahead and recall that when we have so electronic atoms or ions, we're going to be recalling that this means that they have the same number of electrons, whether they're an ion or an atom, a neutral atom. And so we're going to be determining which is greater in size by referring to the electronic charge on the ion. And so the more negative the charge, for example, something with a negative three charge that will correspond to increasing size. So a -3 ion is going to be greater in size by its atomic radius compared to a ion with a minus two charge, then compared to an ion with a minus one charge and then so on and so forth as we go down the spectrum All the way to positive three. And when we get to positive three, for example, we would go ahead and say that we are decreasing significantly in our atomic size. So now that we have this laid out, let's go ahead and determine our answers. So for part A we're comparing the sulfur atom to a sulfide an ion Again, we want to recognize that our sulfide an ion is more negative. It has a 2 - charge, meaning it has two extra electrons compared to the neutral atom of Sulfur. And so we would say that because this is more negative, it therefore has more electrons. And so therefore our sulfide an ions radius is going to be larger than our sulfur atoms radius. So we would go ahead and box in as our answer. The sulfide an ion because it's larger in size. Moving onto choice B. We have the nitride an ion versus the nitride cat ion. Now in this case we're gonna want to go ahead and recall again that the more negative our charge that is associated with a gain of electrons. And so the more electrons we have attached to our atom, we have a greater atomic radius meaning a larger atoms size. And so in this case we have a negative charge on our nitride an ion. So we would say that the an ion is negative and has an extra electron. And so therefore our night rides radius is going to be greater than the nitride cat ion radius which is missing an electron because it has that plus charge. So we would go ahead and box in the night. Tried an ion. Moving onto choice see we're comparing a neutral barium atom to a caddy on B A. Two plus again, we recognize that the neutral barium atom is not missing two electrons like the barium kati on here because we recall that the positive charge means that it's missing two electrons. And so the barium cat ion has more electrons. And so therefore sorry, the barium neutral atom has more electrons. So we would box that one in because we would say that therefore barium radius is going to be greater than that of the barium cat ions radius which is missing electrons. Two electrons specifically. And lastly we have the copper an ion compared to our copper cat ion. And so again we're missing two electrons on our copper cat ion whereas our copper and ion has an extra electron and so we have more electrons on our copper and ion and so are copper and ions radius is going to be greater than that of our copper cat ions radius. So we would go ahead and box in our copper an ion as our final choice to complete this example as our answer for the largest or the larger atomic radius in the set. So if you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice video

Ch.9 - Periodic Properties of the Elements

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Tro 6th Edition

Ch.9 - Periodic Properties of the Elements

Problem 74

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Chapter 9, Problem 74

Which is the larger species in each pair? a. Sr or Sr²⁺ b. N or N^3-c. Ni or Ni²⁺ d. S^2-or Ca²⁺

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1. The size of an atom or ion is determined by the number of electron shells it has and the effective nuclear charge. The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. The more protons in the nucleus, the stronger the pull on the electrons and the smaller the atom or ion.

2. For Sr and Sr2+, Sr2+ has lost two electrons compared to Sr. This means it has a higher effective nuclear charge and fewer electron shells, making it smaller.

3. For N and N3-, N3- has gained three electrons compared to N. This means it has a lower effective nuclear charge and more electron shells, making it larger.

4. For Ni and Ni2+, Ni2+ has lost two electrons compared to Ni. This means it has a higher effective nuclear charge and fewer electron shells, making it smaller.

5. For S2- and Ca2+, S2- has gained two electrons compared to Ca2+ which has lost two electrons. This means S2- has a lower effective nuclear charge and more electron shells, making it larger.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ionic Radius

Ionic radius refers to the size of an ion in a crystal lattice. Cations, which are positively charged ions, are generally smaller than their neutral atoms due to the loss of electrons, leading to increased effective nuclear charge on the remaining electrons. Conversely, anions, which are negatively charged ions, are larger than their neutral atoms because the addition of electrons increases electron-electron repulsion, causing the electron cloud to expand.

Charge and Size Relationship

The charge of an ion significantly affects its size. When an atom loses electrons to form a cation, it becomes smaller due to the reduced electron-electron repulsion and increased attraction between the nucleus and the remaining electrons. In contrast, when an atom gains electrons to form an anion, it becomes larger because the added electrons increase repulsion among them, leading to a larger ionic size.

Comparative Analysis of Ions

To determine which species is larger in a pair of ions, one must compare their ionic charges and the corresponding neutral atoms. Generally, for a given element, the neutral atom is larger than its cation, while the anion is larger than the neutral atom. Thus, when comparing pairs like Sr and Sr²⁺ or N and N³⁻, the neutral species will typically be the larger one, while the anion will be larger than the neutral atom.