Write the electron configuration for each ion. c. Sr 2+ d. Co 3+ e. Cu 2+

Hello. Everyone in this video. We're giving all these ions and you're running the condensed ground state electron configuration for all of these. And we're seeing which one of these ions will have electron configurations identical to their inert gas. So we'll first start off by writing the noble gas configuration with the same thing as the condense ground state electron configuration. And then we'll take into account the ion part of the ion. So let's start off with a let's start off by identifying what I have here on this bond, this puree table. So I can color coded on my blocks because the blocks are where we need to consider. We're writing any electron configuration. So in yellow I have my S block in blue. I have my P block in great. I have my D. Block and in pink I have my F. Block. It's also labeled the rose so 23456 and seven rows. So when I'm doing my electric configuration I'm kind of thinking, I'm thinking that we're reading a page in the book. I was kind of reading and reading and reading until we get to our desired element In our case. 1st. First up is going to be our a. There's R. S. C. Element. So I'm gonna go ahead and try and find that my pureed table. So my sc element is going to be right over here and this blue outline. All right. So we're doing the noble gas configuration. We need to find the noble gas that's closest to here. Instead of finding it forwards, we're going backwards, we find a forest. it doesn't make any sense because we never read any of these here because we're going to stop right up here. So we're gonna go ahead and go backwards. The backwards one is going to be argon. Argon is going to be our noble guest that we're selecting. So for eight we're starting off with A. R continuing to read, let's see, we're going here and we're reading this now. So we're at the fourth row, so four were in this yellow color, which we said to be the S block. Then we're reading 1, 2 elements of that block. So to moving on to this D block, that's a little bit different. The D block is always going to be the road number minus one for what we write on the electron configuration. So we think it's stuff far through but 4 -1 is three. And of course from the D Block and we're only reading one element. So I put a little one. Now, considering our ion parts now, it's a three plus charge. So we're losing three Electrons, Sir Losing two and 1. So speed left with just the argon. All right now, moving on to the B is going to be our L I plus R. L. I element is going to be right over here and again going backwards to find a normal guest is going to be helium. So indicate the H E in brackets. Now reading down just get two right here, we're on the second row In our S. lock and we're reading one element. So back here we have L. I. Plus one charge. So plus one charge means we're losing one electron. So just be two S one that we're losing hour left with just the H. E. Moving on, we're going to our seat, there's 02 minus oxygen is going to be right over here now, reading backwards back at helium. So that's going to be our noble guests now, reading, reading, Going to read these two. So that's going to be second row and R. S block with two elements that we're reading. Moving across. Purim table, we get to this blue color which is our P block same row. So that's going to be are too P. And how many are you reading? We're reading 123 and four, so have four. All right. No, let's see here. So we have our two minus charge, meaning that we're actually going to add two electrons. So add two electrons here. Then we're actually beginning the configuration of H. E to us too. Two P six. Taking a look at this noble gas configuration is the same thing as neon, isn't it? Let's see where I had here. We're reading two more and it is neon. So we actually have neon All right next we're moving on to D and we have our f minus the F element is going to be right next oxygen. So going backwards, of course, we grab back to helium turnover gas will be at helium because it's right next door to this oxygen that we just wrote for. We're basically adding one. So from here we're adding one instead of two. So have to us too. And two P five. Taking a look at the question again, we're looking for f minus. So we're adding one electron and again, we get back because five plus one is six. We get back to here, which is just the neon. Right? So N E again. All right, And let's go back to the blue color. So we're moving on to E now and E S C A two plus C A is right over here. So reading backwards, we're going to this A R double guess then from A R. Reading down, Reading just these two elements in the 4th row. Right? So it's going to be again in the 4th row in our s block. And two Going back to this question is a 2-plus ion. Mean that two electrons are being lost. So we'll say bye bye to those and we're just left with the A. R. Now, the last element, go go back to a pink color is our F. It's going to be V two plus. Now the V element is going to be right over here to writing noble gas configuration going backwards or again at a R. That's our noble guest that we're going to choose. Now, continuing the reading, we get to read these two, we're just going to be in the fourth row, the sp and we need to elements and moving on to this D block again to four minus one is three. Again, we're in the D block And we're reading 1 2, 3 elements. So three. And because we have the two plus charge, we're losing two electrons. So we're actually losing the four S two because we will always lose electrons from the higher quantum number four versus 34. S hired. This goes, bye bye. And we're left with skin this up. We're left with ARN three D 3. So if we see a pattern here, the only islands that have similar or identical electron configurations too, and they're inert gas is going to be a B c d E and F is the only one that's not. So our final answer for this problem, there's going to be the B c C and E. And those are going to be our final answer brothers question

Ch.8 - Periodic Properties of the Elements

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Ch.8 - Periodic Properties of the Elements

Problem 65c,d,e

Chapter 8, Problem 65c,d,e

Write the electron configuration for each ion. c. Sr²⁺ d. Co³⁺ e. Cu²⁺

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Identify the atomic number of the neutral atom for each element: Sr (38), Co (27), Cu (29).

Write the electron configuration for the neutral atom: Sr: [Kr] 5s^2, Co: [Ar] 3d^7 4s^2, Cu: [Ar] 3d^10 4s^1.

For Sr^2+, remove 2 electrons from the outermost shell of Sr: Remove from 5s orbital.

For Co^3+, remove 3 electrons starting from the outermost shell of Co: Remove from 4s and then 3d orbitals.

For Cu^2+, remove 2 electrons starting from the outermost shell of Cu: Remove from 4s and then 3d orbitals.

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Key Concepts

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

Electron Configuration

Electron configuration describes the distribution of electrons in an atom's orbitals. It is represented using a notation that indicates the energy levels and sublevels occupied by electrons. Understanding electron configuration is essential for predicting an element's chemical behavior and reactivity.

Ionic Charge and Electron Loss

Ionic charge refers to the net charge of an ion, which results from the loss or gain of electrons. For cations, such as Sr2+, Co3+, and Cu2+, electrons are lost, leading to a positive charge. The number of electrons lost is crucial for determining the correct electron configuration of the ion.

Transition Metals and Electron Configuration

Transition metals often exhibit unique electron configurations due to their partially filled d orbitals. The electron configuration of transition metal ions can differ from their neutral atoms, as they may lose electrons from both the s and d orbitals. This is particularly relevant for ions like Co3+ and Cu2+, where understanding the specific electron loss is key to writing their configurations.