Chemistry
Conventions
Given some element \(\mathrm{X}\)
$$\begin{equation}
\begin{split}
\ce{^{A}_{Z}X}
\end{split}
\end{equation}$$
$$\begin{equation}
\begin{split}
\ce{X-A}
\end{split}
\end{equation}$$
Where
- \(A = \text{neutrons + protons}\)
- \(Z = \text{protons}\)
- Cations
- Positively (＋) Charged
- Anions
- Negatively (−) Charged
Mnemonic
Cations are Pawsitive
Conventions on homework
When a question says, determine the energy of 1 ㏖ of photons, the unit will be ᴶ/㏖.
Units
$$\begin{equation}
\begin{split}
1 \mol &= \sci{6.022}{23}
\end{split}
\end{equation}$$
$$\begin{equation}
\begin{split}
\small 1 \;\mathrm{amu} &\approx \;\text{protons} + \;\text{neutrons}
\end{split}
\end{equation}$$
SI Prefixes
| Value | Prefix | Symbol |
|---|---|---|
| \(10^{1}\) | deca | \(\mathrm{da}\) |
| \(10^{2}\) | hecto | \(\mathrm{h}\) |
| \(10^{3}\) | kilo | \(\mathrm{k}\) |
| \(10^{6}\) | mega | \(\mathrm{M}\) |
| \(10^{9}\) | giga | \(\mathrm{G}\) |
| \(10^{12}\) | tera | \(\mathrm{T}\) |
| Value | Prefix | Symbol |
|---|---|---|
| \(10^{-1}\) | deci | \(\mathrm{d}\) |
| \(10^{-2}\) | centi | \(\mathrm{c}\) |
| \(10^{-3}\) | mili | \(\mathrm{m}\) |
| \(10^{-6}\) | micro | \(\mathrm{\mu}\) |
| \(10^{-9}\) | nano | \(\mathrm{n}\) |
| \(10^{-12}\) | pico | \(\mathrm{p}\) |
Classification of Matter
Overview
Mixtures
Heterogeneous mixture
Where the prefix Hetero- means different
Homogeneous mixture
Where the prefix Homo- means same
Quantum Mechanics
The Electromagnetic Spectrum
Overview
Formulas
$$\begin{equation}
\begin{split}
\mathrm{f} = \nu &= \frac{c}{\lambda}
\end{split}
\end{equation}$$
$$\begin{equation}
\begin{split}
\mathrm{E}
&= \mathrm{h}\times\mathrm{f}\\
&= \mathrm{h}\frac{\mathrm{c}}{\lambda}
\end{split}
\end{equation}$$
$$\begin{equation}
\begin{split}
\mathrm{T} &= \frac{1}{f}
\end{split}
\end{equation}$$
$$\begin{equation}
\begin{split}
\lambda &= \frac{c}{f}
\end{split}
\end{equation}$$
Values
| Name | Symbol | Unit | Description | Range |
|---|---|---|---|---|
| Wavelength | \(\mathrm{\lambda}\) | Any unit for distance | Distance between two analogous points | Always Positive |
| Frequency | \(f\) or \(\nu\) (nu) | \(㎐ = \frac{\text{1 cycle}}{\text{second}}\)\(\mathrm{s}^{-1}\) | Number of cycles | Always Positive |
| Energy | \(\mathrm{E}\) | \(\mathrm{J}\) (joule) | Amount of energy (\(\mathrm{E}\)) in a light packet |
Constants
| Name | Symbol | Unit | Value |
|---|---|---|---|
| Speed of Light | \(\mathrm{c}\) | \(\v\) | \(\mathrm{c} = \sci{3.00}{8}\v\) |
| Planck's constant | \(\mathrm{h}\) |
|
\(\mathrm{h} = \sci{6.626}{-34}\;\mathrm{J}\cdot\mathrm{s}\) |
Other Formulas
de Broglie Relation
$$\begin{equation}
\begin{split}
\lambda &= \frac{\mathrm{h}}{\mathrm{m}\mathrm{v}}\\
\text{where}&\\
m &= \text{mass}\\
v &= \text{velocity} \neq \nu\;\text{(nu)}
\end{split}
\end{equation}$$
Heisenberg's Uncertainty Principle
$$\begin{equation}
\begin{split}
\Delta{x} \times \mathrm{m} \Delta{v} \geq \frac{\mathrm{h}}{4\pi}
\end{split}
\end{equation}$$
Where
- \(\Delta{x}\) is the uncertainty in position.
- \(\Delta{v}\) is the uncertainty in velocity.
- \(\mathrm{m}\) is the mass of the particle.
- \(\mathrm{h}\) is the plank's constant.
In general it states that the more you know about an electrons position, the less you know about it's velocity.
Energy of an Electron in an Orbital with Quantum Number \(\mathrm{n}\) in a Hydrogen Atom
$$\begin{equation}
\begin{split}
E_n &= \sci{-2.18}{-18} \mathrm{J} \left(\frac{1}{n^2}\right)
\end{split}
\end{equation}$$
Change in Energy That Occurs in an Atom When It Undergoes a Transition between Levels
\(n_{\small\text{initial}}\) and \(n_{\small\text{final}}\)
$$\begin{equation}
\begin{split}
\Delta{E} &= \sci{-2.18}{-18} \mathrm{J} \left(\frac{1}{n^2_f} - \frac{1}{n^2_i}\right)
\end{split}
\end{equation}$$
- If \(\Delta{E}\) is negative, energy is being released.
- If \(\Delta{E}\) is positive, energy is being absorbed.
Electron Configuration
Traditional Chart
Better Method
Examples
Electron configuration for \(_{26}\mathrm{Fe}\)
$$\begin{equation}
\begin{split}
\ce{\underbrace{1s^2 2s^2 2p^6 3s^2 3p^6}_{\smallText{Equal to Argon}} 4s^2 3d^6}
&= \ce{[Ar] 4s^2 3d^6}
\end{split}
\end{equation}$$
Since the electron configuration for Argon is
$$\begin{equation}
\begin{split}
\ce{[Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6}
\end{split}
\end{equation}$$
Electron configuration for \(_{26}\mathrm{Fe}^{+2}\)
Beginning with the electron configuration for \(_{26}\mathrm{Fe}\)
$$\begin{equation}
\begin{split}
\underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}_{\smallText{Equal to Argon}}\;
\overbrace{4s^2\; 3d^6}^{\mathclap{
\begin{gathered}
\smallText{Which should we remove 2 $\mathrm{e}^{-}$ from?}\\
\smallText{($4s^2$ or $3d^6$)?}\\
\end{gathered}
}}
&= \ce{[Ar]
\underbrace{4s^2\; 3d^6}_{\mathclap{
\begin{gathered}
\smallText{Which should we remove 2 $\mathrm{e}^{-}$ from?}\\
\smallText{($4s^2$ or $3d^6$)?}\\
\end{gathered}
}}
}
\end{split}
\end{equation}$$
Remove the electrons from the term with the higher electron state. Warning! Do not just remove the electrons from the rightmost term since the rightmost term may be a lower electron state. For instance given \(4s^2\; 3d^6\)
- \(4s^2\) is in a higher electron state
- \(3d^6\) is in a lower electron state
As shown
$$\begin{equation}
\begin{split}
\underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}\;
\overbrace{4s^{\left(2 - 2\right)}}^{ \mathclap{
\begin{gathered}
\smallText{higher state}\\
\smallText{remove $2\mathrm{e^{-}}$}
\end{gathered}
}}\;
\underbrace{3d^6}_{ \mathclap{\smallText{lower state}}}\;
&= [\mathrm{Ar}]\;
\overbrace{4s^{\left(2 - 2\right)}}^{ \mathclap{
\begin{gathered}
\smallText{higher state}\\
\smallText{remove $2\mathrm{e^{-}}$}
\end{gathered}
}}\;
\underbrace{3d^6}_{ \mathclap{\smallText{lower state}}}\\\\
\underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}\;
\underbrace{3d^6}_{ \mathclap{\smallText{unchanged}}}\;
&= [\mathrm{Ar}]\;
\underbrace{3d^6}_{ \mathclap{\smallText{unchanged}}}
\end{split}
\end{equation}$$
Therefore the electron configuration for \(_{26}\mathrm{Fe}^{+2}\) is:
$$\begin{equation}
\begin{split}
\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^6}
&= \ce{[Ar] 3d^6}
\end{split}
\end{equation}$$
Electron configuration for \(_{24}\mathrm{Cr}\)
It would appear that the electron configuration for \(_{24}\mathrm{Cr}\) would be
$$\begin{equation}
\begin{split}
\overbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}^{\smallText{Equal to Argon}}\;
\underbrace{4s^2\; 3d^4}_{ \mathclap{\smallText{this is wrong!}}}\;
&= [\mathrm{Ar}]\;
\underbrace{4s^2\; 3d^4}_{ \mathclap{\smallText{this is wrong!}}}\;
\end{split}
\end{equation}$$
But this is wrong! It's actually
$$\begin{equation}
\begin{split}
\overbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}^{\smallText{Equal to Argon}}\;
\underbrace{4s^1\; 3d^5}_{ \mathclap{\smallText{notice the superscripts}}}\;
&= [\mathrm{Ar}]\;
\underbrace{4s^1\; 3d^5}_{ \mathclap{\smallText{notice the superscripts}}}\;
\end{split}
\end{equation}$$
How-tos
What are the valence electrons?
Given
$$\begin{equation}
\begin{split}
\ce{1s^2 2s^2 2p^6 3s^2 3p^4}
\end{split}
\end{equation}$$
The valance electrons will be the ones in the highest energy state. Therefore
$$\begin{equation}
\begin{split}
\overbrace{1s^2\; 2s^2\; 2p^6}^{\mathclap{
\smallText{Core electrons}
}}\;
\underbrace{3s^2\;3p^4}_{\mathclap{
\begin{gathered}
\smallText{Highest state}\\
\smallText{Therefore these are the valence electrons}
\end{gathered}
}}
\end{split}
\end{equation}$$
Therefore there are \(6\) valence electrons.
Given
$$\begin{equation}
\begin{split}
\ce{[Ne] 3s^2 3p^2}
\end{split}
\end{equation}$$
The valance electrons will be the ones in the highest energy state. Therefore
$$\begin{equation}
\begin{split}
[\mathrm{Ne}]
\underbrace{3s^2 3p^2}_{\mathclap{
\begin{gathered}
\smallText{Highest state}\\
\smallText{Therefore these are the valence electrons}
\end{gathered}
}}
\end{split}
\end{equation}$$
Therefore there are \(4\) valence electrons.
Quantum Numbers
Overview
| Symbol | Description |
|---|---|
| \(\mathrm{n}\) | The principle quantum number |
| \(l\) | The angular momentum quantum number |
| \(\mathrm{m}_1\) | The magnetic quantum number |
| \(\mathrm{m}_s\) | The spin quantum number |
The Principle Quantum Number (\(\mathrm{n}\))
| Value of \(\mathrm{n}\) | Value of \(l\) | Orbital Sublevel |
|---|---|---|
| \(\mathrm{n} = 1\) | \(l = 0\) | \(1\mathrm{s}\) |
| \(\mathrm{n} = 2\) | \(l = 0\)\(l = 1\) | \(2\mathrm{s}\)\(2\mathrm{p}\) |
| \(\mathrm{n} = 3\) | \(l = 0\)\(l = 1\)\(l = 2\) | \(3\mathrm{s}\)\(3\mathrm{p}\)\(3\mathrm{d}\) |
| \(\mathrm{n} = 4\) | \(l = 0\)\(l = 1\)\(l = 2\)\(l = 3\) | \(4\mathrm{s}\)\(4\mathrm{p}\)\(4\mathrm{d}\)\(4\mathrm{f}\) |
Angular Momentum Quantum Number
| Value | Result |
|---|---|
| \(l = 0\) | \(\mathrm{s}\) |
| \(l = 1\) | \(\mathrm{p}\) |
| \(l = 2\) | \(\mathrm{d}\) |
| \(l = 3\) | \(\mathrm{f}\) |
| Value of \(l\) | Value of \(\mathrm{m_l}\) |
|---|---|
| \(l = 0\) | \(\mathrm{m_l} = 0\) |
| \(l = 1\) | \(\mathrm{m_l} = -1\)\(\mathrm{m_l} = 0\)\(\mathrm{m_l} = 1\) |
| \(l = 2\) | \(\mathrm{m_l} = -2\)\(\mathrm{m_l} = -1\)\(\mathrm{m_l} = 0\)\(\mathrm{m_l} = 1\)\(\mathrm{m_l} = 2\) |
| \(l = 2\) | \(\mathrm{m_l} = -3\)\(\mathrm{m_l} = -2\)\(\mathrm{m_l} = -1\)\(\mathrm{m_l} = 0\)\(\mathrm{m_l} = 1\)\(\mathrm{m_l} = 2\)\(\mathrm{m_l} = 3\) |
Summary
$$\begin{equation}
\begin{split}
l \leq \mathrm{n} - 1
\end{split}
\end{equation}$$
$$\begin{equation}
\begin{split}
-l \leq \mathrm{m_l} \leq l
\end{split}
\end{equation}$$
Useful Formulas
The equation for a maximum number of electrons a given energy level can hold given some value for \(n\)
$$\begin{equation}
\begin{split}
\text{max}_{\mathrm{e}^{-}} &= 2n^2
\end{split}
\end{equation}$$
How many orbitals are possible given some value for \(n\)
$$\begin{equation}
\begin{split}
\text{max}_{\smallText{orbitals}} &= n^2
\end{split}
\end{equation}$$
Examples
Light
Interference and Diffraction
Constructive Interference
If two waves of equal amplitude are in phase when they interact—that is, they align with overlapping crests—a wave with twice the amplitude results. This is called constructive interference.
Destructive Interference
If two waves are completely out of phase when they interact—that is, they align so that the crest from one overlaps with the trough from the other—the waves cancel by destructive interference.
