What tripositive ion has the electron configuration 3Kr4 4d3?

Question

Accepted Answer

Hello everyone today. We have the following problem. The following electron configuration belongs to what pens a positive ion and then we have argon and brackets there. So the element in question here is are gone And Argon has 18 electrons per the periodic table. Pence a positive just refers to an ion has a positive five charge. Hence Penta meaning five in positive meaning positive. And so we know that the number of protons that are possible are equal to the number of electrons plus some charge. Because essentially in a neutral species, the number of protons equals the number of electrons. However, we have a charged species here and so the number of protons is going to be equal to however many electrons there are plus whatever charge is in the way. And so we know That we have 18 electrons and our charge is five meaning The number of protons that we must have is 23. If we go to the periodic table, the number of protons corresponds to the atomic number of an element. And so therefore The element with an atomic number of 23 is going to be vanadium and thus we're going to have V five plus as our final answer to represent that overall, I hope it has helped. And until next time

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