Kaplan FL 1 PS #33

[MedChallenge]

I thought I knew this topic pretty well since it's a fundamental, but I'm just stumped. Would someone please explain this question to me? Btw, this is from the old-school paper version of Kaplan's FLs...don't know if it's the same for the CBT version.

33. ​

​

​

​

Given the order in which orbitals are filled, which
molecule is a triplet in its ground state?​

A.​

​

​

​

​

H2​

B.​

​

​

​

​

O2​

C.​

​

​

​

​

N2​

D. F2

33. B​

The answer to this question is found in the passage:You are given the order of orbital filling
in paragraph three and are told in the last paragraph what a triplet state is. Remember that when
there are orbitals of equal energy, each one will be half-filled before any of the orbitals is completely
filled, and all the half-filled orbitals will have electrons with the same spin. Looking at
the answer choices for the question, you can see that they are all diatomic; thus, there will be
two electrons in a hydrogen molecule, fourteen in nitrogen, sixteen in oxygen and eighteen in
fluorine. The passage does not describe how to fill the​

n = 1 shell, but from knowing that the​

n​

​

= 2 shell has the sigma 2s bonding orbitals and one sigma 2s antibonding orbital, there will be
sigma bonding and antibonding orbitals in the n = 1 shell, and these will also hold four electrons.
Therefore, molecular hydrogen will fill half the n = 1 shell with the total of two electrons
that it has and there will be zero electrons in its second shell. Working in a similar way you will
have ten electrons in the second shell for molecular nitrogen, twelve for oxygen, and fourteen
for fluorine. When you fill in the molecular orbitals, you will find that nitrogen just fills the pi
2x and 2y bonding orbitals. Since there are no unpaired electrons, molecular nitrogen can’t be a
triplet. Molecular oxygen has two extra electrons, which go into the pi 2x and 2y antibonding
orbitals. In other words it has unpaired electrons with the same spin. If you’re not sure, check
the final choice: In fluorine, these pi 2x and 2y antibonding orbitals are filled with the two extra
electrons, so fluorine is not a triplet. Thus, it turns out that the oxygen molecule is a triplet in its​

ground state, so choice B is the correct answer.

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[angkim]

I thought I knew this topic pretty well since it's a fundamental, but I'm just stumped. Would someone please explain this question to me? Btw, this is from the old-school paper version of Kaplan's FLs...don't know if it's the same for the CBT version.

33. ​

​

​

​

Given the order in which orbitals are filled, which
molecule is a triplet in its ground state?​

A.​

​

​

​

​

H2​

B.​

​

​

​

​

O2​

C.​

​

​

​

​

N2​

D. F2

33. B​

The answer to this question is found in the passage:You are given the order of orbital filling
in paragraph three and are told in the last paragraph what a triplet state is. Remember that when
there are orbitals of equal energy, each one will be half-filled before any of the orbitals is completely
filled, and all the half-filled orbitals will have electrons with the same spin. Looking at
the answer choices for the question, you can see that they are all diatomic; thus, there will be
two electrons in a hydrogen molecule, fourteen in nitrogen, sixteen in oxygen and eighteen in
fluorine. The passage does not describe how to fill the​

n = 1 shell, but from knowing that the​

n​

​

= 2 shell has the sigma 2s bonding orbitals and one sigma 2s antibonding orbital, there will be
sigma bonding and antibonding orbitals in the n = 1 shell, and these will also hold four electrons.
Therefore, molecular hydrogen will fill half the n = 1 shell with the total of two electrons
that it has and there will be zero electrons in its second shell. Working in a similar way you will
have ten electrons in the second shell for molecular nitrogen, twelve for oxygen, and fourteen
for fluorine. When you fill in the molecular orbitals, you will find that nitrogen just fills the pi
2x and 2y bonding orbitals. Since there are no unpaired electrons, molecular nitrogen can’t be a
triplet. Molecular oxygen has two extra electrons, which go into the pi 2x and 2y antibonding
orbitals. In other words it has unpaired electrons with the same spin. If you’re not sure, check
the final choice: In fluorine, these pi 2x and 2y antibonding orbitals are filled with the two extra
electrons, so fluorine is not a triplet. Thus, it turns out that the oxygen molecule is a triplet in its​

ground state, so choice B is the correct answer.

I thought this question required going back to the passage. End of paragraph 3 says that "for a diatomic molecule, orbitals int he n=2 energy level are fille as follows..." so the first 5 orbitals are equal in energy so then one electron would fill up each, with up to 10 e in this level. Then the energy level with pi* 2px and 2py are filled up to 4 e. Then the last sigma*2pz, up to 2 electrons. From the answer explanation, you gotta count for the 1s orbital of 4 e. so yeah, if you filled up the orbitals like above for each diatomic molecule, the O2 has only 2 electrons to fill the pi*2px and pi*2py and the 2 electrons are unpaired. I hope this helps!

I'm going over all the Kaplan diagnostics too. So hope we can help each other! I've been posting on Kaplan MCAT forum as well. Good luck!

 

[justhanging]

I thought I knew this topic pretty well since it's a fundamental, but I'm just stumped. Would someone please explain this question to me? Btw, this is from the old-school paper version of Kaplan's FLs...don't know if it's the same for the CBT version.

33. ​

Given the order in which orbitals are filled, which
molecule is a triplet in its ground state?​

A.​

H2​

B.​

O2​

C.​

N2​

D. F2

33. B​

The answer to this question is found in the passage:You are given the order of orbital filling
in paragraph three and are told in the last paragraph what a triplet state is. Remember that when
there are orbitals of equal energy, each one will be half-filled before any of the orbitals is completely
filled, and all the half-filled orbitals will have electrons with the same spin. Looking at
the answer choices for the question, you can see that they are all diatomic; thus, there will be
two electrons in a hydrogen molecule, fourteen in nitrogen, sixteen in oxygen and eighteen in
fluorine. The passage does not describe how to fill the​

n = 1 shell, but from knowing that the​

n​

= 2 shell has the sigma 2s bonding orbitals and one sigma 2s antibonding orbital, there will be
sigma bonding and antibonding orbitals in the n = 1 shell, and these will also hold four electrons.
Therefore, molecular hydrogen will fill half the n = 1 shell with the total of two electrons
that it has and there will be zero electrons in its second shell. Working in a similar way you will
have ten electrons in the second shell for molecular nitrogen, twelve for oxygen, and fourteen
for fluorine. When you fill in the molecular orbitals, you will find that nitrogen just fills the pi
2x and 2y bonding orbitals. Since there are no unpaired electrons, molecular nitrogen can’t be a
triplet. Molecular oxygen has two extra electrons, which go into the pi 2x and 2y antibonding
orbitals. In other words it has unpaired electrons with the same spin. If you’re not sure, check
the final choice: In fluorine, these pi 2x and 2y antibonding orbitals are filled with the two extra
electrons, so fluorine is not a triplet. Thus, it turns out that the oxygen molecule is a triplet in its​

ground state, so choice B is the correct answer.

Fill up the p orbitals for oxygen and you'll notice that it has a +1 spin.

4 electrons in p orbitals, 3 spin up and one spin down. Each spin up is worth +1/2 and spin down -1/2. Thats how you get the +1 spin. The plus one spin gives the triplet state (2*spin# + 1 = triplet state). I am sorry that I didn't use the explanation from the answer.