Question

Question #8f16a

Answer 1

(1)

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

Therefore, we can say that chromium(III) oxide contains the `"Cr"^(color(blue)(3+))` and `"O"^(color(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 `"Cr"^(color(blue)(3+))` and `"O"^(color(red)(2-))`, we have...

`->` `"Cr"_color(red)(2)"O"_color(blue)(3)`

(2)

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

• Option B is also possible, since `(n,l,m_l) = (2,1,-1)`, and if `l = 1`, `m_l = {0, pm1}`. This is simply one of the `2p` 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 `3p` orbital. A `2p` orbital exists, therefore a `3p` orbital exists.

• Option D, where `(n,l,m_l) = (1,1,1)` 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):

`\mathbf(lambda_("dB") = h/p = h/(mv))`

You were already given the mass `m` and the speed `v`, and you should have Planck's constant `h = 6.626xx10^(-34) "J"cdot"s"` on-hand or memorized. `p = mv` is the classical momentum from physics.

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

`color(blue)(lambda) = (6.626xx10^(-34) "J"cdot"s")/("2.0 kg"xx"50 m/s")`

`= (6.626xx10^(-34) cancel("kg")cdot"m"^cancel(2)"/"cancel("s"^2)cdotcancel("s"))/("2.0" cancel("kg")xx50 cancel("m")"/"cancel("s"))`

`= (6.626xx10^(-34))/(100) "m"`

`= color(blue)(6.626xx10^(-36) "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 `"He"`'s known configuration of `1s^2`, we add on the `2s` and `2p` electrons.

`1s^2 => "He"` configuration
`1s^2 2s^1 => "Li"` configuration
`1s^2 2s^2 => "Be"` configuration
`1s^2 2s^2 2p^1 => "B"` configuration
`1s^2 2s^2 2p^2 => "C"` configuration
`1s^2 2s^2 2p^3 => "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 `+1/2` (spin-up) if the other is `-1/2` (spin-down).

Hence, only option D is correct.

Option A is less stable because two electrons are paired in the `2p` 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 `2s` orbital. If two electrons have the same spin in the same orbital, they disappear.

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