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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 105a

One of the molecular orbitals of the H2- ion is sketched below:

(a) Is the molecular orbital a s or p MO? Is it bonding or antibonding?

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Hello everyone. So in this video we're given this diagram right over here are trying to determine if the molecular orbital is sigma or pi molecular orbital. I'm trying to classify if it's going to be bonding or anti bonding. So let's go ahead and write the molecular orbital that we are familiar with. So let's go ahead and maybe scroll down a little bit. So we have a little bit more room. All right, so let's go ahead and draw that out. Starting off at the bottom, we have our sigma two people right on top. Or let's go ahead and actually save some space. What we'll do is write it on the side. So sigma to people. Then on top we have our sigma star. Toupee right on top we have pi two P On top of that we have our Sigma two p. Then on top we have our pi star to pee. And finally on the very top we have our sigma star to peep. So what these little stars mean is just the anti bonding. Alright, so we know That we have the C2 plus carry on. Let's go ahead and rewrite that because we don't really see it here. So we have our C two plus ion. So we know that carbon because there's a C here. So carbon has four valence electrons. And since we can use the P R a trend carbon is in Group 48. So that's why we know that there's four valence electrons. And because in this molecule here we have two atoms of carbon. Then we'll double the amount of valence electrons. So four times two is eight. So we have a total of eight valence electrons. But then we do have this plus ion and that plus ion just means or the charge just means that we lose one electron. So inform our eight, we'll subtract one and we'll get a total Of seven valence electrons because we're -1. Okay, so filling out this molecular orbital diagram right over here that we just drew out. Keeping in mind of course our hunt's role to fill from the bottom to the top. Let's go ahead and see which our seventh valence electron will go ahead and land on. I'm gonna go ahead and determine if it's going to be bonding or anti bonding and it's going to be prior sigma. So starting from the bottom then we have 12, 34, five. So again we have 1234, 56 and seven. So as you can see our seventh valence electron has landed right over here and our pie And it's going to have no start right because stars anti bonding, there's no start meaning that it is bonding and you can see in the beginning orbital that usually molecular orbital start with R. S orbital. However, the bonding and the anti bonding of R. S orbital cancel each other out. So that's why we can go ahead and start with our p orbital instead. So because of an overlap or orbital overlap, it has to be bonding and it's good to see here is also bonding. So just say maybe like part a this is going to be bonding and then because the overlap is sideways as well as the molecular orbital diagram here that it's landed on pi, so we'll have the pi molecular orbital. So to answer the question, determine if the molecular herbal is sigma or pi it is Pie I was trying to cast if I if it's bonding or anti bonding, it is bonding. So those two will be my final answer for this question. Thank you all so much for watching.
Related Practice
Textbook Question

The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero.

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Textbook Question

The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? Would you expect the molecule to be planar for both Lewis structures?

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Textbook Question

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as p2p9p2p*, s2s9s2s*, and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (c) The electronic absorption spectrum of the N2 molecule has the lowest energy peak at 170 nm. To what orbital transition does this correspond?

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Textbook Question

One of the molecular orbitals of the H2- ion is sketched below: (d) Compared to the H¬H bond in H2, the H¬H bond in H2- is expected to be which of the following: (i) Shorter and stronger, (ii) longer and stronger, (iii) shorter and weaker, (iv) longer and weaker, or (v) the same length and strength?

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Textbook Question

Place the following molecules and ions in order from smallest to largest bond order: N22+, He2+, Cl2 H2-, O22-.

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Textbook Question
Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMOLUMO gap; see Exercise 9.104). Suppose you have two samples, one is lycopene which is responsible for the red color in tomato, and the other is curcumin which is responsible for the yellow color in turmeric. Which one has the larger HOMO-LUMO gap?
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