Question

Critical thinking problem: The iron in hemoglobin in our red blood cells is used to bind and carry dioxygen throughout the body. Iron is one of the atoms that can form hybrid orbitals using its d-shell electrons, which allows for a greater number of sigma bonds than just four. When iron is only bound to protein, it forms five sigma bonds with one lone pair. The iron atom is pulled down below the central square plane due to one of the bonds with the hemoglobin protein. The thick lines indicate bonds pointing out of the screen, while dashed lines indicate bonds pointing into the screen, so as to suggest 3-D structure.

When iron binds to dioxygen, it now has six sigma bonds. The geometry changes, even though the hybridization orbitals do not change. The dashed lines are still present, as seen from the top-down view, but the thick lines hide them in the side-on view.

Why would the molecular geometry for iron change in this case? Would this have anything to do with the ability of hemoglobin to bind oxygen tightly?

Answer 1

The change in molecular geometry of iron when it binds to dioxygen allows for a tighter binding of oxygen by hemoglobin.

Step-by-step explanation:

When iron binds to dioxygen, it forms six sigma bonds instead of the previous five sigma bonds when it is only bound to protein. The change in molecular geometry is due to the fact that the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. This change in geometry allows for stronger binding of oxygen to the iron, leading to a tighter binding of oxygen by hemoglobin.

Answer 2

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Contents Home Bookshelves Physical & Theoretical Chemistry Supplemental Modules (Physical and Theoretical Chemistry) Electronic Structure of Atoms and Molecules Expand/collapse global location

Predicting the Hybridization of Simple Molecules

Last updatedAug 16, 2020

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Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).1This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. An innovative method proposed for the determination of hybridization state on time economic ground 2,3,4.

Prediction of sp, sp2, sp3 Hybridization state

We Know, hybridization is nothing but the mixing of orbital’s in different ratio to form some newly synthesized orbitals called hybrid orbitals. The mixing pattern is as follows:

s + p (1:1) - sp hybrid orbital; s + p (1:2) - sp2 hybrid orbital ; s + p (1:3) - sp3 hybrid orbital

Formula used for the determination of sp, sp2 and sp3 hybridization state:

Power on the Hybridization state of the central atom = (Total no of σ bonds around each central atom -1)

All single (-) bonds are σ bond, in double bond (=) there is one σ and 1π, in triple bond (≡) there is one σ and 2π. In addition to these each lone pair (LP) and Co-ordinate bond can be treated as one σ bond subsequently.

Eg.:

a. In NH3: central atom N is surrounded by three N-H single bonds i.e. three sigma (σ) bonds and one lone pair (LP) i.e. one additional σ bond. So, in NH3 there is a total of four σ bonds [3 bond pairs (BPs) + 1 lone pair (LP)] around central atom N. Therefore, in this case power of the hybridization state of N = 4-1 = 3 i.e. hybridization state = sp3.

b. In H2O: central atom O is surrounded by two O-H single bonds i.e. two sigma (σ) bonds and two lone pairs i.e. two additional σ bonds. So, altogether in H2O there are four σ bonds (2 bond pairs + 2 lone pairs) around central atom O, So, in this case power of the hybridization state of O = 4-1 =3 i.e. hybridization state of O in H2O = sp3.

c. In H3BO3:- B has 3σ bonds (3BPs but no LPs) and oxygen has 4σ bonds (2BPs & 2LPs) so, in this case power of the hybridization state of B = 3-1 = 2 i.e. B is sp2 hybridized in H3BO3. On the other hand, power of the hybridization state of O = 4-1= 3 i.e. hybridization state of O in H3BO3 is sp3.

d. In I-Cl: I and Cl both have 4σ bonds and 3LPs, so, in this case power of the hybridization state of both I and Cl = 4 - 1 = 3 i.e. hybridization state of I and Cl both are sp3.

e. In CH2=CH2: each carbon is attached with 2 C-H single bonds (2 σ bonds) and one C=C bond (1σ bond), so, altogether there are 3 sigma bonds. So, in this case, power of the hybridization state of both C = 3-1 = 2 i.e. hybridization state of both C’s are sp2.

Prediction of sp3d, sp3d2, and sp3d3 Hybridization States