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8. Give the ground-state electron configuration of (a) F; (b) Ga"+ , (c) Pb?+ 9. Give the ground-state electron configuration along with total number of unpair...

Question

8. Give the ground-state electron configuration of (a) F; (b) Ga"+ , (c) Pb?+ 9. Give the ground-state electron configuration along with total number of unpaired electrons of (a) Sc, (b) V3t (c) Mn?+ , (d) Cr2t , (e) Cu; 10. Give the ground-state electron configuration of (a) Rh' ; (b) Eu' , (c) Mo6+ . Show orbital diagram to find the number of unpaired electrons. 11. Consider the process of shielding in atoms, using Be as an example. What is being shielded? What is it being shiel

8. Give the ground-state electron configuration of (a) F; (b) Ga"+ , (c) Pb?+ 9. Give the ground-state electron configuration along with total number of unpaired electrons of (a) Sc, (b) V3t (c) Mn?+ , (d) Cr2t , (e) Cu; 10. Give the ground-state electron configuration of (a) Rh' ; (b) Eu' , (c) Mo6+ . Show orbital diagram to find the number of unpaired electrons. 11. Consider the process of shielding in atoms, using Be as an example. What is being shielded? What is it being shielded from? What is doing the shielding? 12. In general, ionization energies increase across a period from left to right. Explain why the second ionization energy of chromium is higher; not lower; than that of manganese 13 . Half filled orbitals are more stable then partially filled orbitals_ Explain this using exchange energy concept. Elaborate it using a figure. 14. Define Madelung $ rule and explain the reason 15. Calculate the value of effective nuclear charge for the outer most valence electron in Ni2+?



Answers

The orbital occupancies for the d orbitals of several complex ions are diagrammed below.
(a) Which diagram corresponds to the orbital occupancy of the cobalt ion in $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-2} ?$
(b) If diagram D depicts the orbital occupancy of the cobalt ion in $\left[\mathrm{CoF}_{6}\right]^{n}$ , what is the value of $n ?$
(c) $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is paramagnetic and $\left[\mathrm{Ni (\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic. Which diagrams correspond to the orbital occupancies of the nickel ions in these species?
(d) Diagram $\mathrm{C}$ shows the orbital occupancy of $\mathrm{V}^{2+}$ in the octahedral complex $\mathrm{VL}_{6}$ Can you determine whether L is a strong- or weak-field ligand? Explain.

In this problem, we are going to be considering lithium. We are asked to do several things, the first of which is to write the electron configuration for lithium. That's super simple. It's element number three. So it's one is too to s one as part of a. We are also asked to calculate the effect of nuclear charge and the effective nuclear charge is going to be equal to our nuclear charge minus our core electrons. Which will be 3 -2 equals one. So that's our effective nuclear charge B were given the following information. The energy of an electron in a one electron atom or ion equals -2.18 times 10 To the -18 jewels times Z squared over N squared where Z. Is the nuclear charge. And actually they use the effect of nuclear charge on this one. And this one was answered in your text, I'm going to make a note that they used the effective equals one and N is the principal quantum number, which is to. So in order to calculate what we're asked to do is calculate or estimate the first ionization energy of lithium. In order to do that, we simply took 218 times 10 to the -18 jewels times one squared over two squared. And when I did this I got 545 times 10 to the -19. And that would have been jules per atom. Is the amount of heat required 5.45 Times 10 to the -19 jules per adam to convert that to jules per mole. Was that jules per adam? There's 6.022 Times 10 to the 23rd atoms per mole. And I got for this just cleared my calculator 3.28 Times 10 to the 5th jules per mole. and of course that will equal times 10 to the Second, killer jewels per mole. That is our estimated first ionization energy. You see if I got that right? Yes. Okay, we're going to compare the result with the value on Table 7.4. So that is part C. Compare 328 killer jewels per mole to the value. It said that we got on our next one. Got to see where I wrote that down quickly here. Okay, So on our table. Sorry, it took so long to get that there. We got 520 killer goals per mole and we can see that um that is substantially higher estimate for Z effect. Hang on just a second here. Not going to start this whole thing over, but I'm going to go double check something super fast. Make sure I've got the right values here cheating a little bit but I'm looking in the back of the book. 110. Yes, we're correct. So, um, our estimated value is indeed lower than the text. First ionization energy. Our our estimate for the z effective was a lower limit Pepsi. It didn't account for our core electrons which don't perfectly shield The two S Electron. Okay, there. I feel better about that now. And our last part of this question is D Indy says what value of the effective nuclear charge gives the proper value for ionization energy, Which we said was 520. Killing joules per mole. And we are going to probably use some Slater's rules on here. I think I used a quick Slater's rule on here. Yeah, I did. So I used a quick slater's rule. If we want to get 520 killer joules per mole, and we're going to have that equaling 2.18 times 10 to the -18 jewels over X squared and four. So that was that number squared. We're going to get an ex squared here, or X is going to be About 1.26 as our effective nuclear charge. Now, if I use Slater's rules to figure out what are effectively nuclear charge would be, I'm going to get when I've got one S 2 to S one. I have two electrons That I'm going to multiply by zero, Equals 1.70. So I'm going to take 3 -1.70 Equals 1.30. And compare that to the 1-6. We just calculated that will be very close and notice that that number is a little bit higher because the electrons in my one s don't perfectly shield As indicated by multiple them, multiplying them each by 0.85 instead of multiplying by them by one. And I think that's it.

Here we are continuing to look at electron configurations. So, for example, we've got Philip Krypton followed by 45 and we are populating our not degenerate orbital's. So here we have high spins, we've got 12345. So we jump into the hospital before returning back to the T2, 500 electrons. Next we have M3 plus to fill up to Krypton, followed by four D 3. So again, we have two e in the T two Orbital's and so we have three unpowered electron here, all in the T two. Finally, we have Ceo three plus when we fill up to argon, Followed by three D 6, draw our orbital's E and ET two, where we have 123456, where we have five, where we have four and paired electrons 1, 2, 3, 4.

The electronic configuration of titanium to be positive. It's equipped to configuration followed by four. The I am. The energy level diagram can be represented us. Does the state war with him and this is A. G or become Yeah, it has a total of for you on beard electrons. 2nd 1 is more libidinal. Mhm Sorry Mm. or three positive. The electronic configuration will be krypton 40 three. The energy level type from can be 12, three. There's a state to this e.g.. It has a total of three um Beard electrons and option C. We have cobalt three positive great clothes arkan Followed by three D six. Yeah. One fully beard two. Okay, so mhm Yeah, it has a total of four on beard electrons.

So we were looking at chemistry on trends to be specific. So this links into our period a city which is also covered in this podcast series, where we're looking at how elements are placed on the periodic table to reveal their period of city to us and the trends and the properties that they possess, which allows us to determine how exactly these elements might work in chemical reactions. So the smallest atomic radius in the Group 16 is oxygen. The largest atomic radius in the sixth period is freedom. Oh, a moving on to the sad point. We're looking for the metal with the smallest atomic radius in Assad period. That has the element at the left hand side. So this is aluminum. The the highest organization energy and Group 14 that is carbon lowest in the first period has the element at the left side off the period that is all seen on so the most metallic element in the 15th period that is bismuth the I. So next we have the highest energy level field in the fourth period, which is crypt on element with dick following configuration and E three asked to three piece two is silicon element with the configuration KR 5 to 46 is Are you? I went with the configuration. Ir three d three is titanium element in the first period with the pseudo noble gas configuration with plus three is cadmium element in the fourth period that forms three bliss. Die magnetic ion be element in the fourth period that forms a two plus iron with half filled the orbital is manganese on n heaviest land Tonight is why be element in the third period. That's like I saw electronics with a are one in the tu minus is so for that air alkaline as metal where the Catalan is isil Electronic with K R is strong tea. Um lastly the matter Lloyd from group five A with the most acidic oxide is arsenic.


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