Which of these crystal-field splitting diagrams represents:

a. a weak-field octahedral complex of Fe³⁺ ,

b. a strong-field octahedral complex of Fe³⁺ 

c. a tetrahedral complex of Fe³⁺

d. a tetrahedral complex of Ni²⁺ (The diagrams do not indicate the relative magnitudes of ∆. ) [Find more in Section 23.6.]

Video transcript

Hello, everyone today. With the following problem match the crystal field splitting diagrams to the following descriptions. First, we have a square planar complex of platinum two plus and then we have a tetrahedral complex of platinum two plus. So first let's identify the electron configurations to determine the number of D electrons. So neutral platinum as the following electron configuration, more specifically the noble gas electron configuration where if we refer to the noble gas, that is just before we get to platinum, it will be xenon. So you put xenon in brackets and then we enter the F orbital. So you have 4 F-14 electrons and then we return back to the D orbitals for five D nine. And then we're in the S orbital for six S one. And so for platinum to lose two electrons from the positive two charge, it would lose them from the highest energy orbitals and the highest energy orbitals are six S and five D. Thus, when it loses two electrons, so this platinum two plus will lose two electrons. We will still have the same noble gas in brackets. We will still have the 4 F-14. However, we would have five D eight as we lost one electron from the six S and one electron from five D nine. So there are a tool of V eight electrons there. Now looking at our crystal field splitting diagrams, we have eight electrons. So our only options are one in three, looking at square planar, this has the shape of diagram at three. So this platinum for one will be diagram three which makes the diagram for the second one tetrahedral diagram. One. Now we have three and four which three being a weak field octahedral complex of Cobalt three plus and four being a strong field octahedral complex of Cobalt three plus. So as we do with platinum, we will determine the electron the noble gas electron configuration first for neutral Cobalt, which the most recent noble gas just before cobalt is argon. So we put that in brackets followed by the D orbital which is 3d 7 and then the S orbital four S two. Now for Cobalt to lose three electrons, hence the three plus charge, it needs to lose them from the highest energy orbital which is four S and three D. Thus, if we write the electron config the noble gas electron configuration for cobalt three plus, we will still have the argon in brackets. But this time, we will lose two electrons from the forest two and then one from the 3d 7 giving us 3d 6. So we have essentially six electrons in the outer shell. Now between the 2nd and 4th diagram, having six electrons diagram, two showcases the weak field. Since in this weak field, we fill up the electrons first before pairing. So we fill up all orbitals before pairing. So our for part three, the answer will be the second diagram. And then for part four, it will be the fourth diagram which a strong field wherein the pair electrons will happen first in the lower energy orbitals. And so with that, we have solved the problem overall, I hope this helped. And until next time.