to know how many p orbitals there are count the regions of electron density.
1 region: s
2 regions: sp
3 regions: sp2
4 regions: sp3
5 regions: sp3d
Search found 30 matches
- Tue Dec 04, 2018 3:02 pm
- Forum: Hybridization
- Topic: P orbital
- Replies: 7
- Views: 650
- Tue Dec 04, 2018 3:00 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: polarity
- Replies: 11
- Views: 854
Re: polarity
A molecule that is a dipole (meaning positively charged on one side and negatively charged on the other) is polar. A molecule that is nonpolar is, as previously stated, symmetrical.
- Tue Dec 04, 2018 2:57 pm
- Forum: Naming
- Topic: Naming a Compound
- Replies: 2
- Views: 404
Naming a Compound
What are the rules to naming a compound and in what order do you write the element names?
- Thu Nov 29, 2018 11:52 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX2E2
- Replies: 14
- Views: 4584
Re: AX2E2
Molecules with the molecular formula AX2E2 have four regions of electron density so its electron geometry would be tetrahedral. Its molecular geometry would be bent with a bond angle of <<109.5.
- Thu Nov 29, 2018 11:41 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Radicals
- Replies: 2
- Views: 346
Re: Radicals
Typically, questions will not give you a radical. Most of the time the questions will take away or add an electron so that all electrons are either bonded or in a lone pair.
- Thu Nov 29, 2018 11:38 am
- Forum: Biological Examples
- Topic: CH 17 6TH EDITION HW 17.31d
- Replies: 4
- Views: 434
Re: CH 17 6TH EDITION HW 17.31d
sodium bisoxalato(diaqua)ferrate(III)
sodium: Na
bisoxalato: (C2O4)2
diaqua: (OH2)2
ferrate(III): Fe
Na[Fe(OH2)2(C2O4)2]
sodium: Na
bisoxalato: (C2O4)2
diaqua: (OH2)2
ferrate(III): Fe
Na[Fe(OH2)2(C2O4)2]
- Tue Nov 20, 2018 5:21 pm
- Forum: Lewis Structures
- Topic: Midterm question
- Replies: 8
- Views: 690
Re: Midterm question
Since resonance is the blending of structures, each N-O bond is an average of the double and single bond shown on the lewis structure. Therefore you would take the average of the two bonds. (140pm+120pm)/2 = 130pm
- Tue Nov 20, 2018 5:16 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Hund's rule vs Pauli's
- Replies: 3
- Views: 1105
Re: Hund's rule vs Pauli's
Pauli exclusion principle:
1. no more than 2e- per orbital
2. if two electrons are in the same orbital then spin paired
Hund's Rule:
1. due to electron repulsion electrons in the same shell occupy different orbitals
2. with parallel spin
1. no more than 2e- per orbital
2. if two electrons are in the same orbital then spin paired
Hund's Rule:
1. due to electron repulsion electrons in the same shell occupy different orbitals
2. with parallel spin
- Tue Nov 20, 2018 5:11 pm
- Forum: Sigma & Pi Bonds
- Topic: What makes up a sigma bond?
- Replies: 1
- Views: 255
Re: What makes up a sigma bond?
A sigma bond consists of two orbitals, each with 1e- interacting end-to-end to for 1sigma bond. Sigma bonds allow bound atoms to rotate, resulting sigma bonds have e- density with cylindrical symmetry around internuclear axis.
- Thu Nov 15, 2018 9:27 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: 6th edition 4.67
- Replies: 2
- Views: 506
Re: 6th edition 4.67
The higher valence shell, the lower the ionization energy. This is because the more inner shells there are, the more the valence electrons get shielded from the electrostatic attraction of the nucleus. Electrons close to the nucleus have a higher ionization energy than electrons further away.
- Thu Nov 15, 2018 9:25 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge
- Replies: 10
- Views: 1672
Re: Formal Charge
If an element is in row 3+ then it can have an expanded octet. However, formal charge is still important because it makes sure your lewis structure is in its most stable form, making it easier for you to identify the shape of the molecule.
- Thu Nov 15, 2018 9:18 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Trigonal Planar
- Replies: 5
- Views: 481
Re: Trigonal Planar
Whether a structure has a single bond or multiple bonds do not matter in case of determining bond angles. Instead, look at the number of regions of electron density.
- Thu Nov 08, 2018 1:49 pm
- Forum: Resonance Structures
- Topic: Resonance Structures
- Replies: 3
- Views: 505
Re: Resonance Structures
Resonance structures are generally more stable since they are lower in energy.
- Thu Nov 08, 2018 1:45 pm
- Forum: Lewis Structures
- Topic: Mystery Element
- Replies: 4
- Views: 552
Re: Mystery Element
Typically, the center element would be the element that is the least electronegative. Or sometimes, the question will give you a specific element to use.
- Thu Nov 08, 2018 1:33 pm
- Forum: Octet Exceptions
- Topic: Octet
- Replies: 6
- Views: 826
Re: Octet
Exceptions to the octet rule include:
1. If the element is the the 3rd or more period, it can have more than an octet since it can use the d-subshell.
2. Radicals: H, He, Li, Be can have less than an octet
1. If the element is the the 3rd or more period, it can have more than an octet since it can use the d-subshell.
2. Radicals: H, He, Li, Be can have less than an octet
- Fri Nov 02, 2018 11:33 am
- Forum: Lewis Structures
- Topic: Formal Charge Tricks
- Replies: 5
- Views: 3605
Re: Formal Charge Tricks
To find the formal charge use the formula FC = V (L+ S/2).
Where FC=formal charge
V= number of valence electrons
L= number of lone electrons that arent bonded
S= number of shared electrons that are being bonded
Where FC=formal charge
V= number of valence electrons
L= number of lone electrons that arent bonded
S= number of shared electrons that are being bonded
- Fri Nov 02, 2018 11:29 am
- Forum: Photoelectric Effect
- Topic: Photoelectric Experiment - Chem Test 2
- Replies: 3
- Views: 554
Re: Photoelectric Experiment - Chem Test 2
To find 5 moles of ejected electron, you would multiply your answer by plancks constant and multiply that by 5.
- Fri Nov 02, 2018 11:22 am
- Forum: Octet Exceptions
- Topic: The center atom
- Replies: 20
- Views: 3315
Re: The center atom
In this case, oxygen is in the middle since the formal charge with oxygen in the middle is 0; whereas if chlorine is the the middle the formal charge would be +1.
- Thu Oct 25, 2018 11:12 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: 1D.23 b and d, 7th edition
- Replies: 1
- Views: 323
Re: 1D.23 b and d, 7th edition
the ml value corresponds to the orientation of the orbital,
for example, the p orbital can have an ml value -1, 0, +1
which corresponds with the px, py, and pz orbitals.
Therefore, when the ml value is specified, there can only be one orbital.
for example, the p orbital can have an ml value -1, 0, +1
which corresponds with the px, py, and pz orbitals.
Therefore, when the ml value is specified, there can only be one orbital.
- Thu Oct 25, 2018 11:03 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Units
- Replies: 8
- Views: 721
Re: Units
The Heisenberg Uncertainty Equation is deltap x delta x = h/4pi,
where deltap =m x deltav
m= mass (kilograms)
deltav = distance(meters)/time(second)
deltax = distance(meters)
h= planck's constant (kg x m^2/s)
where deltap =m x deltav
m= mass (kilograms)
deltav = distance(meters)/time(second)
deltax = distance(meters)
h= planck's constant (kg x m^2/s)
- Thu Oct 25, 2018 9:49 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: electron config. 3d and 4s
- Replies: 5
- Views: 1657
Re: electron config. 3d and 4s
3d is a lower energy level because it is in the 3rd shell, whereas the 4s electrons are in the 4th shell. This means that the 4s electrons are outer electrons, further from the nucleus, and are shielded by the 3d electrons. Also, when removing an electron, the electron would be removed from the 4s o...
- Fri Oct 19, 2018 10:50 am
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: H bar
- Replies: 5
- Views: 593
Re: H bar
He has the equation sheet posted on his website with all the equations he's giving us for the tests. In this case the equation on the sheet is (delta p)(delta x)>=h/(4pi), so you dont need to know the value of hbar since hbar =h/2pi.
- Fri Oct 19, 2018 10:44 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Nodal Planes
- Replies: 3
- Views: 493
Re: Nodal Planes
Nodes are regions of electron absense, since psi=0 at those regions. So, in the cone (nodal shape) there aren't any electrons there.
- Fri Oct 19, 2018 10:40 am
- Forum: DeBroglie Equation
- Topic: De Broglie equation
- Replies: 3
- Views: 656
Re: De Broglie equation
c=(lambda)(frequency) and E=h(frequency)
Therefore, E=(hc)/lambda
ALSO, E=mc^2, where m*c=p, so E=pc
Thus, pc=(hc)/lambda,
Therefore, lambda=h/p, and p=momentum(m*v)
So, lambda=h/(m*v)
Therefore, E=(hc)/lambda
ALSO, E=mc^2, where m*c=p, so E=pc
Thus, pc=(hc)/lambda,
Therefore, lambda=h/p, and p=momentum(m*v)
So, lambda=h/(m*v)
- Thu Oct 11, 2018 9:15 pm
- Forum: Photoelectric Effect
- Topic: photoelectric effect
- Replies: 2
- Views: 403
Re: photoelectric effect
One of the equations that Lavelle wrote on his board was that E(photon)-E(energy to remove e)=E(excess). So basically, the total energy of the photon is equal to the energy required to remove an electron (threshold energy) plus the kinetic energy of the photon (excess energy). Another equation that ...
- Thu Oct 11, 2018 8:53 pm
- Forum: Properties of Light
- Topic: HW 1A.9 [ENDORSED]
- Replies: 4
- Views: 322
Re: HW 1A.9 [ENDORSED]
In order to find the energy, use the formula E=hv, since the frequency is given.
- Thu Oct 11, 2018 3:05 pm
- Forum: Properties of Electrons
- Topic: Help with question 1.13
- Replies: 2
- Views: 226
Re: Help with question 1.13
for the first part, use the rydberg formula where v=R( (1/n sub1^2)-(1/n sub2^2)) and plug that into c=lambda*v. Therefore you get c=lambda*(R( (1/n sub1^2)-(1/n sub2^2))). 3.00x10^8=(3.29x10^15)*((1/4^2)-(1/2^2))*lambda. lambda = 4.86x10^-7m = 486nm for part b, this belongs to the balmer series sin...
- Wed Oct 03, 2018 1:41 pm
- Forum: Empirical & Molecular Formulas
- Topic: Confusion on number of moles
- Replies: 4
- Views: 388
Re: Confusion on number of moles
Depending on the chemical equation, CuCl2H24H20 and Cl2 probably had a 1:2 molar ratio, respectively. Thus, you would have to multiply, .0417molsCuCl2H24H20 x (2molCl2/1molCuCl2H24H20), leaving you with .0834molsCl2.
- Wed Oct 03, 2018 1:34 pm
- Forum: Empirical & Molecular Formulas
- Topic: Writing formula from diagram
- Replies: 2
- Views: 337
Re: Writing formula from diagram
For exercises F1 I just counted the number of carbons, hydrogens, and oxygens that I saw on the diagram, using the given code. For F1 I got C10H16O.
- Wed Oct 03, 2018 1:27 pm
- Forum: Limiting Reactant Calculations
- Topic: Post-assessment problem, neutralizing acids
- Replies: 2
- Views: 270
Re: Post-assessment problem, neutralizing acids
For this problem, what I did was to first convert 1kg CaCO3 into moles of CaCO3 using its molar mass. 1000g/100.09g = 9.99molCaCO3. I assumed that CaCO3 was the limiting reagent since that was the only given information. Then, I converted 9.99molCaCO3 to moles of CO2. Since the balanced equation sho...