# Question #8f16a

Jun 14, 2016

(1)

Chromium(III) has an oxidation number of $\text{III}$. Alone, this doesn't tell us enough information, but oxide is the name of the ${\text{O}}^{2 -}$ anion. So, we know that $\text{Cr}$ has to have a positive oxidation state with a magnitude of $3$.

Therefore, we can say that chromium(III) oxide contains the ${\text{Cr}}^{\textcolor{b l u e}{3 +}}$ and ${\text{O}}^{\textcolor{red}{2 -}}$ ions. To balance out the charges and generate a compound:

• The magnitude of the anion charge becomes the subscript of the cation.
• The magnitude of the cation charge becomes the subscript of the anion.

With ${\text{Cr}}^{\textcolor{b l u e}{3 +}}$ and ${\text{O}}^{\textcolor{red}{2 -}}$, we have...

$\to$ ${\text{Cr"_color(red)(2)"O}}_{\textcolor{b l u e}{3}}$

(2)

• Option A gives us $\left(n , l , {m}_{l}\right) = \left(2 , 0 , 0\right)$. This is possible, since it correlates with a $2 s$ orbital, for which $n = 2$ and $l = 0$. By definition, if $l = 0$, ${m}_{l} = \left\{0\right\}$, which makes sense because there exists only one $2 s$ orbital.

• Option B is also possible, since $\left(n , l , {m}_{l}\right) = \left(2 , 1 , - 1\right)$, and if $l = 1$, ${m}_{l} = \left\{0 , \pm 1\right\}$. This is simply one of the $2 p$ orbitals, since for $l = 1$, we are looking at a $p$ orbital.

• Option C is again possible. For $n = 3$ instead of $2$, we just now have a $3 p$ orbital. A $2 p$ orbital exists, therefore a $3 p$ orbital exists.

• Option D, where $\left(n , l , {m}_{l}\right) = \left(1 , 1 , 1\right)$ couldn't work, because $n = 1$. By definition, $l = 0 , 1 , . . . , n - 1$. Since $n - 1 = 0$, this is self-contradictory. When $l = 0$, $l$ cannot equal $1$.

Hence, option D gives the uh... correctly incorrect configuration.

(3)

Recall the equation for the de Broglie wavelength (not to be confused with the Compton wavelength):

$\setminus m a t h b f \left({\lambda}_{\text{dB}} = \frac{h}{p} = \frac{h}{m v}\right)$

You were already given the mass $m$ and the speed $v$, and you should have Planck's constant $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ on-hand or memorized. $p = m v$ is the classical momentum from physics.

Hence, you can solve for $\lambda$ (without a calculator) to get:

$\textcolor{b l u e}{\lambda} = \left(6.626 \times {10}^{- 34} \text{J"cdot"s")/("2.0 kg"xx"50 m/s}\right)$

$= \left(6.626 \times {10}^{- 34} \cancel{\text{kg")cdot"m"^cancel(2)"/"cancel("s"^2)cdotcancel("s"))/("2.0" cancel("kg")xx50 cancel("m")"/"cancel("s}}\right)$

$= \frac{6.626 \times {10}^{- 34}}{100} \text{m}$

$= \textcolor{b l u e}{6.626 \times {10}^{- 36} \text{m}}$

You can figure out how to round this to find the answer in your choices.

(4)

Nitrogen is on row $2$, column $15$. Starting from $\text{He}$'s known configuration of $1 {s}^{2}$, we add on the $2 s$ and $2 p$ electrons.

$1 {s}^{2} \implies \text{He}$ configuration
$1 {s}^{2} 2 {s}^{1} \implies \text{Li}$ configuration
$1 {s}^{2} 2 {s}^{2} \implies \text{Be}$ configuration
$1 {s}^{2} 2 {s}^{2} 2 {p}^{1} \implies \text{B}$ configuration
$1 {s}^{2} 2 {s}^{2} 2 {p}^{2} \implies \text{C}$ configuration
$1 {s}^{2} 2 {s}^{2} 2 {p}^{3} \implies \text{N}$ configuration

So that would explain why all configurations given as answer choices have exactly $7$ electrons.

Now, recall:

• Aufbau principle: electrons fill orbitals in order from lowest to highest energy.
• Hund's rule: Electrons generally fill each orbital one electron at a time (in order to maximize the number of parallel electron spins).
• Pauli Exclusion Principle: Each orbital can only contain two electrons, and they must be different spins. One must be $+ \frac{1}{2}$ (spin-up) if the other is $- \frac{1}{2}$ (spin-down).

Hence, only option D is correct.

Option A is less stable because two electrons are paired in the $2 p$ orbitals when that creates coulombic repulsions and destabilizes the configuration. It goes against the notions of Hund's rule.

Option B is impossible, because the second electron spin is violating the Pauli Exclusion Principle in the $2 s$ orbital. If two electrons have the same spin in the same orbital, they disappear.

Option C is impossible for the same reason, except the $1 s$ orbital is problematic.