Question #8f16a

1 Answer
Jun 14, 2016

(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.