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 selfcontradictory. 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# (spinup) if the other is#1/2# (spindown).
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