Question #8f16a
1 Answer
(1)
Chromium(III) has an oxidation number of
Therefore, we can say that chromium(III) oxide contains the
- 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(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
Hence, you can solve for
#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
#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
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
Option B is impossible, because the second electron spin is violating the Pauli Exclusion Principle in the
Option C is impossible for the same reason, except the