simple AAMC GC ques (VSPER)

[Neplina94]

Which cation is most likely to be found in place of Fe(II) in the square planar binding domain of hemoglobin?

The options were Mg 2+, Li+, Co 2+ and Na+.

It says the answer is Co 2+ because it can support square planar binding because it is a transition metal.

My question: Is there a general rule that I can follow to make these easier?

Like do i assume only transition metals can support square planar binding. What is a good rule of thumb to know which groups/types of atoms support which binding?

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[aldol16]

Like do i assume only transition metals can support square planar binding. What is a good rule of thumb to know which groups/types of atoms support which binding?

Easiest way to do this question is to look for the closest thing on the periodic table. Fe2+ is a transition metal with an oxidation state of +2. Only one answer choice fits that.

Generally speaking, the majority of things that adopt square planar geometries are going to be transition-metal complexes. Now, things like XeF4 can also do square planar but in terms of biology, most of what you're going to see square planar are coordination complexes of transition metals like iron.

 

[Hi_I'mPaul]

Which cation is most likely to be found in place of Fe(II) in the square planar binding domain of hemoglobin?

The options were Mg 2+, Li+, Co 2+ and Na+.

It says the answer is Co 2+ because it can support square planar binding because it is a transition metal.

My question: Is there a general rule that I can follow to make these easier?

Like do i assume only transition metals can support square planar binding. What is a good rule of thumb to know which groups/types of atoms support which binding?

If you know the hybridization of square planar is sp^3d^2 (at a minimum you should definitely know that it has a steric number of 6 with 2 lone pairs), then the only answer choice that contains a d orbital is Cobalt. You can also use logic while looking at the periodic table, as what aldol16 pointed out, but I am a fan of using logic combined with the underlying rationale! Good luck ~

 

[Neplina94]

If you know the hybridization of square planar is sp^3d^2 (at a minimum you should definitely know that it has a steric number of 6 with 2 lone pairs), then the only answer choice that contains a d orbital is Cobalt. You can also use logic while looking at the periodic table, as what aldol16 pointed out, but I am a fan of using logic combined with the underlying rationale! Good luck ~

Thank you, one more question of this, square planar can support 6 bonds therefore do i interpret sp3d2 as, 1 bond corresponding to s, 3 corresponding to p3 and 2 corresponding to d2? So by that reasoning trigonal planar is sp2 (support 3 bonds) , and tetrahedral is sp3? Am i thinking of this properly, that the number next to each orbital (s,p,d) also represents the number of bonds in the atom on that orbital? Thank you!

 

[Hi_I'mPaul]

Thank you, one more question of this, square planar can support 6 bonds therefore do i interpret sp3d2 as, 1 bond corresponding to s, 3 corresponding to p3 and 2 corresponding to d2? So by that reasoning trigonal planar is sp2 (support 3 bonds) , and tetrahedral is sp3? Am i thinking of this properly, that the number next to each orbital (s,p,d) also represents the number of bonds in the atom on that orbital? Thank you!

Correct!!! just count the number of molecules attached to the central atom for hybridization which can help you solve problems like this one. I'll offer a few more examples:

So for BH3, count the number of molecules attached to boron which is 3, so (s,p,p) which equates to sp^2 or trigonal planar. XeF4 is tricky because it actually contains 2 lone pairs so you must count them for a total of 6 "things" attached to Xe... therefore, (s,p,p,p,d,d) which is equal to the sp3d2. Lastly, I'll do trigonal bipyramidal which an example would be something like PCl5. with 5 things attached it will be (s,p,p,p,d), so the hybridization will be sp3d. Good work!! Now, just do some research on orbital vs molecular geometry on the MCAT and you should be solid on these types of questions.

 

[Neplina94]

Correct!!! just count the number of molecules attached to the central atom for hybridization which can help you solve problems like this one. I'll offer a few more examples:

So for BH3, count the number of molecules attached to boron which is 3, so (s,p,p) which equates to sp^2 or trigonal planar. XeF4 is tricky because it actually contains 2 lone pairs so you must count them for a total of 6 "things" attached to Xe... therefore, (s,p,p,p,d,d) which is equal to the sp3d2. Lastly, I'll do trigonal bipyramidal which an example would be something like PCl5. with 5 things attached it will be (s,p,p,p,d), so the hybridization will be sp3d. Good work!! Now, just do some research on orbital vs molecular geometry on the MCAT and you should be solid on these types of questions.

Wonderful, thank you!