Arrange this isoelectronic series in order of increasing atomic radius: Se^2- , Sr²⁺ , Rb⁺ , Br^- .

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Accepted Answer

All right. Hello everyone. So for this question, let's go ahead and arrange this isoelectronic series. In order of increasing atomic radius, we have selenium two negative strontium, two positive rubidium positive and bromine negative or in other words, bromide. But underneath the series itself, we have four different answer choices labeled a through d arranging them in a different order. Now, just for the record increasing atomic radius means we are to arrange these ions in the isoelectronic series in order from smallest to largest atomic radius. So with that in mind, right, recall first and foremost that the atomic and ionic radius refers to the distance between the nucleus of a given atom or ion and its outermost electron shell. Now, generally speaking, the ionic radius of an atom or an ion is going to increase as the number of electrons increases. However, notice the ions that were provided in this series, they were specifically grouped together as an isoelectronic series. The prefix iso refers to something that is similar about more than one object and electronic referring to the number of electrons. All this is to say that isoelectric species are atoms and or ions that have the same number of electrons. So in this case, right, we cannot compare all of the ions provided by the number of electrons considering they have the same exact amount. So in this case, we're going to have to refer to the number of protons in each species because earlier, we talked about how the number of electrons increases as the ionic radius increases. But with protons, that trend is actually the opposite because in this case, almost spelled that wrong. But it just so happens that the fewer protons you have in a given nucleus, the larger that radius happens to be that atomic or ionic radius. So this means we have to compare the number of protons for all of these ions. Now recall that the number of protons in a given nucleus corresponds to the atomic number of that element. So in this case, we have selenium strong team rubidium and brome. So now let's go ahead and identify their atomic numbers. Selenium to start off has an atomic number of 34 which means that it has 34 protons inside of its nucleus. Strontium has 38 rubidium has 37 and bromine has 35 right. So in this case, because sodium has the smallest or the fewest protons of the isoelectronic series, it is also going to have the largest atomic radius. So in this case, selenium two negative would be the last own on my list considering it is largest. Now the second largest would then have to be Brolin with 35 protons inside of its nucleus, which means bromide or BR negative is the second largest afterwards is rubidium with 37 protons. And finally, the smallest of the isoelectronic series would be strontium two positive with 38 protons inside of its nucleus. And there you have it. So from smallest to largest in terms of atomic radius, we have strontium two positive followed by rubidium positive, followed by bromide, followed by selenium two negative, which is the largest and this corresponds to option c in the multiple choice. So that is going to be our final answer and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Arrange this isoelectronic series in order of increasing atomic radius: Se^2- , Sr²⁺ , Rb⁺ , Br^- .

Verified Solution

Key Concepts

Isoelectronic Species

Atomic Radius Trends

Effective Nuclear Charge (Z_eff)

Arrange this isoelectronic series in order of increasing atomic radius: Se2- , Sr2+ , Rb+ , Br- .

Verified Solution

Key Concepts

Isoelectronic Species

Atomic Radius Trends

Effective Nuclear Charge (Z_eff)

Arrange this isoelectronic series in order of increasing atomic radius: Se^2- , Sr²⁺ , Rb⁺ , Br^- .