Search found 30 matches
- Sat Dec 08, 2018 8:57 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: A X5 E
- Replies: 6
- Views: 689
Re: A X5 E
It would be square pyramidal
- Sat Dec 08, 2018 2:41 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Electronegativity & Acid Strength
- Replies: 2
- Views: 557
Re: Electronegativity & Acid Strength
Acidity increases with the increase of electronegativity.
- Fri Dec 07, 2018 9:01 pm
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Review questions
- Replies: 1
- Views: 297
Re: Review questions
H2Se has a higher boiling point because it has a larger London dispersion force. You can figure this out because Se is larger than S, and the bigger the molecule, the bigger the dispersion force. The larger the intermolecular forces, the higher the boiling point will be.
- Thu Nov 29, 2018 9:28 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Trigonal Bipyramidal
- Replies: 6
- Views: 738
Re: Trigonal Bipyramidal
Due to the high repulsive energy that lone pairs have, you have to think of the best possible place to put it so that the molecule is the most stable. Thus, when placing lone pairs, putting it in the equitorial position provides the most stability.
- Thu Nov 29, 2018 9:23 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR (lone pairs taking axial vs equatorial positions)
- Replies: 2
- Views: 476
Re: VSEPR (lone pairs taking axial vs equatorial positions)
When there is a lone pair, the molecule has to be the in the best optimum position available so that the electrons are equally spread apart. So for example, when there is a lone pair in a AX3E2, the best optimum place for the lone pair is in the equitorial axes.
- Thu Nov 29, 2018 9:01 pm
- Forum: Trends in The Periodic Table
- Topic: Ionization Energy
- Replies: 2
- Views: 1239
Re: Ionization Energy
This is beacause an atom that has half-filled or full orbitals are more stable. For oxygen, its electron configuration is 1s2 2s2 2p4, and nitrogen's is 1s2 2s2 2p3. Nitrogen has half-filled 2p orbitals, so losing an electron would cause it to become more unstable. However, oxygen can become more st...
- Sun Nov 25, 2018 7:56 pm
- Forum: Properties of Electrons
- Topic: excitation
- Replies: 2
- Views: 587
Re: excitation
When an electron absorbs energy, it goes into an excited state. When it returns to its unexcited state, the electron loses energy by emitting it through light.
- Sat Nov 17, 2018 5:29 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Seesaw and T-Shape
- Replies: 1
- Views: 269
Re: Seesaw and T-Shape
For seesaw, you have approximately 90,120, and 180 degrees. For T-shape, you would have 90 degrees and 180 degrees.
- Thu Nov 15, 2018 11:26 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: bond angle
- Replies: 4
- Views: 515
Re: bond angle
It should be 90 and 180 degree
- Thu Nov 15, 2018 11:23 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lewis and VSEPR for I3-
- Replies: 3
- Views: 3125
Re: Lewis and VSEPR for I3-
The VSEPR model should be linear. As for the Lewis structure, you should have one of the iodine being a central atom with one bond to each iodine atom. The central atom has 3 lone pairs, and the other two iodine atoms also have 3 lone pairs.
- Thu Nov 15, 2018 11:17 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Length
- Replies: 6
- Views: 628
Re: Bond Length
Yes, bond length and bond strength usually have a correlation. The longer the bond, the weaker the bond; The shorter the bond, the stronger the bond.
- Thu Nov 15, 2018 11:14 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Repulsion Strength
- Replies: 7
- Views: 706
Re: Repulsion Strength
It means that the VSEPR model can give you the quantitative shapes of the molecules like tetrahedral or trigonal planar or something. However, when there are lone pairs involved, the VSEPR model cannot give you an exact measurement of the bond angles due to the higher electron repulsion, thus it can...
- Thu Nov 15, 2018 11:04 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Effect of Molecular Shape on Lewis Structure
- Replies: 3
- Views: 361
Re: Effect of Molecular Shape on Lewis Structure
Yeah, I don't you have to take molecular shapes into consideration when drawing Lewis Structures. However, I do think you should be able to recognize molecular shape when looking at a Lewis Structure from the electron density.
- Thu Nov 15, 2018 11:00 pm
- Forum: Ionic & Covalent Bonds
- Topic: H bonds and melting points
- Replies: 4
- Views: 643
Re: H bonds and melting points
Hydrogen bonds have higher melting and boiling point because they have some of the highest intermolecular forces, so a large amount of energy is needed to separate them.
- Sun Nov 04, 2018 5:27 pm
- Forum: Trends in The Periodic Table
- Topic: Which Ion is larger?
- Replies: 2
- Views: 294
Re: Which Ion is larger?
S^-2 is bigger because it has more electrons that is free to move around more.
- Sun Nov 04, 2018 11:31 am
- Forum: Quantum Numbers and The H-Atom
- Topic: Drawing orbitals
- Replies: 4
- Views: 1669
Re: Drawing orbitals
I'm not 100% sure, but I think you would draw the shape of the p orbital, but would draw one on the x-axis, one on the y-axis, and one on the z-axis.
- Sun Nov 04, 2018 11:25 am
- Forum: Electronegativity
- Topic: What are the trends useful for?
- Replies: 12
- Views: 1012
Re: What are the trends useful for?
Can someone explain why we use [Ar] You can use Atomic radius to also tell ionization energy. The general trend for atomic radius is that it decreases from left to right across the period, which means the nucleus has a greater pull on the electrons, causing their radius to be smaller. So, ionizatio...
- Sat Nov 03, 2018 4:31 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Ionization Energies
- Replies: 4
- Views: 406
Re: Ionization Energies
Ionization energy has a general trend that increases left to right across and decreases down a group. Ionization energy is the energy required to remove an electron from an atom, so to answer your question, yes.
- Sun Oct 28, 2018 4:10 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Copper and Chromium Exception
- Replies: 1
- Views: 259
Copper and Chromium Exception
Can someone explain to me the exception for copper and chromium? I heard it in the lecture, but I had trouble understanding it. Please clarify this for me.
- Sun Oct 28, 2018 4:07 pm
- Forum: Trends in The Periodic Table
- Topic: Ionization energy and electron affinity
- Replies: 1
- Views: 298
Re: Ionization energy and electron affinity
Ionization energy is the energy required to remove the outermost bound electron. Ionization energy increases from left to right on the periodic table. Electron affinity is kind of like the opposite. It is the energy change that occurs when a neutral atom attracts an electron to become a negative ion...
- Sun Oct 28, 2018 3:44 pm
- Forum: Ionic & Covalent Bonds
- Topic: Hydrogen Bonds
- Replies: 5
- Views: 499
Re: Hydrogen Bonds
One of the biggest reasons is that hydrogen bonds are intermolecular forces, which happens between two atoms in the same molecule. A hydrogen bond is merely an electromagnetic attraction, not really a bond.
- Sat Oct 20, 2018 3:47 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 2.43 Electron Configurations
- Replies: 1
- Views: 188
Re: 2.43 Electron Configurations
Yeah I believe electron configuration won't be on test 2. Lavelle said the test will be on information up until Friday's lecture, and we haven't learned electron configuration yet.
- Sat Oct 20, 2018 3:41 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Prob 1.15 6th Ed.
- Replies: 2
- Views: 293
Re: Prob 1.15 6th Ed.
First, you have to recognize that ultraviolet light is included in the Lyman series, which means its initial n is n=1. This can be found in your book. Then, you can use the Rydberg equation: v=R((1/n1^2)-(1/n2^2) Now that we now that its initial is n=1, you plug in 1 in n1. v=R((1/1^2)-(1/n2^2) In t...
- Sat Oct 20, 2018 3:17 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Homework Question 1.D13
- Replies: 2
- Views: 244
Re: Homework Question 1.D13
The possible values are 0,1,2,3,4,5,6
The allowed values of L are 0,1,2,...,n-1.
The allowed values of L are 0,1,2,...,n-1.
- Thu Oct 11, 2018 7:33 pm
- Forum: Properties of Light
- Topic: Homework for Week 2 [ENDORSED]
- Replies: 3
- Views: 268
Re: Homework for Week 2 [ENDORSED]
I think it depends on your TA, but I heard in a lecture and Lavelle said that you can turn in hw from the fundamental chapter if you wish to review more. I did some of the fundamental problems in my hw for week 2 as well because I needed the practice for the test.
- Thu Oct 11, 2018 7:30 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: G.21
- Replies: 2
- Views: 795
Re: G.21
First, you convert 0.500 g of KCL, K2s, K3Po4 into moles. Then, you multiply the moles by the mole ratio between the moles of potassium and the compound. 0.500g KCl= .0067 mol *1 0.500g K2S= .0045 mol*2 0.500g K3PO4= .0023 mol*3 Then, you just add the results for the total amount of moles of potassi...
- Thu Oct 11, 2018 7:04 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect
- Replies: 2
- Views: 323
Re: Photoelectric Effect
The photoelectric effect affects only metals, but yes to answer your question, it can affect everyday objects. Technically, it will lose mass, but it is of negligible quantity.
- Thu Oct 04, 2018 4:28 pm
- Forum: SI Units, Unit Conversions
- Topic: Formula Units
- Replies: 1
- Views: 254
Formula Units
Can someone tell me how to do E9 (b)? I have 7th edition.
- Wed Oct 03, 2018 4:57 pm
- Forum: Balancing Chemical Reactions
- Topic: Stoichiometric coefficients
- Replies: 3
- Views: 206
Re: Stoichiometric coefficients
Yeah I think it's fine. I personally don't do it because it's more confusing for me.
- Wed Oct 03, 2018 4:50 pm
- Forum: SI Units, Unit Conversions
- Topic: state of compounds
- Replies: 3
- Views: 450
Re: state of compounds
yeah, I think you would have to state the state of the compound when doing chemical equations. Sometimes you can tell what state it's in by the context of the question. However, there are solubility rules you can memorize if you search them up online.