Why does nitrogen sp3 hybridize?

[Mayhem]

We know that carbon sp3 hybridizes to be able to form three $\sigma$-bonds; in its non-hybridized state, carbon only allows two bonds. A quick review of its electron configuration confirms this.

However, nitrogen has a a non-hybridized electron configuration of three half-filled 2p orbitals, which should (on paper) allow for three $\sigma$-bonds already.

Looking around, it was suggested that nitrogen utilizes sp3 hybridization to achieve an optimal trigonal planar geometry as it has a non-bonding lone pair when its formal charge is neutral. Is this all there is to it?

 

[etotheipi]

We know that carbon sp3 hybridizes to be able to form three σ-bonds; in its non-hybridized state, carbon only allows two bonds. A quick review of its electron configuration confirms this.

What do you mean by this? Hybridisation is a mathematical procedure by which $x$ atomic orbitals can be combined in such a way to give $x$ hybrid orbitals with a suitable geometry, for better describing the bonding (and not an actual physical state of the atom, or anything). If you mix the $2s$ and 3 x $2p$ AOs to produce 4 (not 3) tetrahedrally directed $sp^3$ HAOs, these can be used to describe 4 $\sigma$ bonds to other atoms.

Even before hybridisation, it's energetically favourable to uncouple the $2s^2$ valence electrons and promote one of these to the $2p_z$ to achieve a configuration of $2s^1, 2p_x^1, 2p_y^1, 2p_z^1$, which sort of explains why carbon is quadrivalent instead of divalent.

Looking around, it was suggested that nitrogen utilizes sp3 hybridization to achieve an optimal trigonal planar geometry as it has a non-bonding lone pair when its formal charge is neutral. Is this all there is to it?

The best way to describe the bonding, will depend on the geometry of the molecule as well as stability considerations. For a trigonal arrangement, it's best to think of it as $sp^2$ (not $sp^3$) hybridised; this has a trigonal arrangement and leaves a lone pair in the non-bonding $p$ orbital, which might for example conjugate with other $p$ orbitals in an extended $\pi$ system in a planar molecule [i.e. for example, an amide], which has a stabilising effect.

If the nitrogen is involved in bonding with tetrahedral geometry (e.g. ${NH_4}^{-}$), then naturally you'll want to view it in terms of $sp^3$ hybridisation.

I'm sure others (e.g. @TeethWhitener) will know an awful lot more about this than I do.

 

[Mayhem]

What do you mean by this? Hybridisation is a mathematical procedure by which $x$ atomic orbitals can be combined in such a way to give $x$ hybrid orbitals with a suitable geometry, for better describing the bonding (and not an actual physical state of the atom, or anything). If you mix the $2s$ and 3 x $2p$ AOs to produce 4 tetrahedrally directed $sp^3$ HAOs, these can be used to describe 4 $\sigma$ bonds to other atoms.

But, in any case the carbon still has 4 (not 2?) valence electrons, $2s^1, 2p_x^1, 2p_y^1, 2p_z^1$, so however you choose to describe the bonding (e.g. hybridisation of a certain number of these AOs, or another way of constructing suitable AOs from these), the carbon will still naturally want to form 4 bonds, no?

In its ground sate, carbon has [He] 2s $\uparrow \downarrow$ 2p $\uparrow$ 2p $\uparrow$ 2p __ electron configuration, so the third orbital can't participate in covalent bonding, and thus only two bonds are possible. It is through hybridization that a [He] 2s $\uparrow$ 2p $\uparrow$ 2p $\uparrow$ 2p $\uparrow$ electron configuration is achieved. That is, one electron from the 2s orbital is donated to the empty 2p orbital.

So you're right, just not for ground state carbon (you can verify this by looking at https://ptable.com/#Electrons). My confusion lies in the fact that nitrogen has a ground state configuration which does allow for three covalent bonds, but when it forms three bonds, in $\mathrm{NH_3}$ for example, the atom is said to be sp3 hybridized anyway. That's where my confusion lies.

 

[etotheipi]

My confusion lies in the fact that nitrogen has a ground state configuration which does allow for three covalent bonds, but when it forms three bonds, in $\mathrm{NH_3}$ for example, the atom is said to be sp3 hybridized anyway. That's where my confusion lies.

But, the 3 $2p$ orbitals on the nitrogen are all at right angles to each other, and the $2s$ orbital is spherically symmetric. So, this arrangement is unsuitable to describe either tetrahedral or trigonal planar geometry, for instance.

In $NH_3$, the $sp^3$ hybridisation ends up producing, from the original AOs, 4 tetrahedrally directed HAOs. Three of these are involved in $\sigma$ N-H bonding and one contains a lone pair. The tetrahedral geometry arising from the hybrid description is a better model of the actual molecule, since loosely speaking we'd expect the electron pairs constituting the bonds & the l.p. to be as far apart as possible.

 

[Mayhem]

But, the 3 $2p$ orbitals on the nitrogen are all at right angles to each other, and the $2s$ orbital is spherically symmetric. So, this arrangement is unsuitable to describe either tetrahedral or trigonal planar geometry, for instance.

I think this adequately answers my question, as this boils down to "it sp3 hybridizes for optimal geometry" (roughly speaking).

By the way, I think this is not quite precise. If you just promote one electron from the s to the empty p, that's an energy-requiring process and in the end you still have essentially exactly the same AOs, just with an excited state.

That's different to hybridisation, where you construct new orbitals with total energy 4E(sp3) equal to the sum of the previous energy E(2s) + 3E(2p), i.e. it's purely a mathematical redeployment.

I haven't looked into the physics of it, but we've been told something along the lines "hybridization naturally occurs when carbon is in the vicinity of each other". Don't ask me why. Maybe in the bigger picture, this is a more favorable state for the carbon atoms to be in, but for single carbon atoms, the GS electron configuration is the most stable? I don't know.

 

[etotheipi]

I haven't looked into the physics of it, but we've been told something along the lines "hybridization naturally occurs when carbon is in the vicinity of each other". Don't ask me why. Maybe in the bigger picture, this is a more favorable state for the carbon atoms to be in, but for single carbon atoms, the GS electron configuration is the most stable? I don't know.

This can't be right, because hybridisation is not something that "occurs". It's just a way of mathematically combining the functions describing the original atomic orbitals to produce a new set of functions which are better suited to describe the bonding.

In other words, how to describe the bonding is purely the choice of the chemist! And hybridisation as an operation has no physical meaning, it's not a "state" or anything.

 

[Mayhem]

This can't be right, because hybridisation is not something that "occurs". It's just a way of mathematically combining the functions describing the original atomic orbitals to produce a new set of functions which are better suited to describe the bonding.

In other words, how to describe the bonding is purely the choice of the chemist! And hybridisation as an operation has no physical meaning, it's not a "state" or anything.

Next you'll tell me that oxidation states is also just a construct!

It honestly doesn't surprise me, and unsurprisingly gen chem textbooks do not delve into the physics of it all. From what I understand, an orbital is a space of probability where an electron has a 90% chance of existing, and that seems like a mathematical construction. Is the underlying physics of hybridization really just a process that changes the geometry of this space of probability?

 

[etotheipi]

From what I understand, an orbital is a space of probability where an electron has a 90% chance of existing, and that seems like a mathematical construction. Is the underlying physics of hybridization really just a process that changes the geometry of this space of probability?

You have to be careful with the '90%' definition... technically, there are an infinite number of closed surfaces $\Omega$ in space in which the total probability is 90%!

You can think of it like, each orbital on an atom has a particular wavefunction $\psi_{n,l,m_l}(r, \theta, \phi)$ and we can always choose this in such a way that it is real valued (but you can also choose it to be complex-valued). When two atoms form a bond using $AO1$ on atom $1$ and $AO2$ on atom $2$, you get a bonding orbital by loosely adding in phase, $MO1 = AO1 + AO2$, and an antibonding orbital by loosely adding out of phase, $MO2 = AO1 - AO2$.

The problem is that, for anything other than very simple (e.g. diatomic) molecules, the geometry cannot be described using the raw AOs. So, one approach is to combine AOs on the same atom to produce a new set of wavefunctions associated with each atom, which have the correct geometry to now form bonds with other atoms. In that sense, it's a completely mathematical operation.

But even this is not the most accurate picture, since by hybridising you lose some information, and in reality electrons are not localised between atoms. The best way, in principle, is to use a computer to determine the accurate forms of the molecular orbitals which are often much more complicated.

Once again, I'm sure other members know a lot more about this than I do!

 

[Mayhem]

You have to be careful with the '90%' definition... technically, there are an infinite number of closed surfaces $\Omega$ in space in which the total probability is 90%!

You can think of it like, each orbital on an atom has a particular wavefunction $\psi_{n,l,m_l}(r, \theta, \phi)$ and we can always choose this in such a way that it is real valued (but you can also choose it to be complex-valued). When two atoms form a bond using $AO1$ on atom $1$ and $AO2$ on atom $2$, you get a bonding orbital by loosely adding in phase, $MO1 = AO1 + AO2$, and an antibonding orbital by loosely adding out of phase, $MO2 = AO1 - AO2$.

The problem is that, for anything other than very simple (e.g. diatomic) molecules, the geometry cannot be described using the raw AOs. So, one approach is to combine AOs on the same atom to produce a new set of wavefunctions associated with each atom, which have the correct geometry to now form bonds with other atoms. In that sense, it's a completely mathematical operation.

But even this is not the most accurate picture, since by hybridising you lose some information. The best way, in principle, is to use a computer to determine the accurate forms of the molecular orbitals which are often much more complicated.

Once again, I'm sure other members know a lot more about this than I do!

Ah, I had no idea about any of this. Seems like this is again one of those things where "I don't even know what I don't know".

Thanks, though.

 

[TeethWhitener]

I’m not sure I can add much to this discussion without completely hijacking the thread (and also I’m at work, so can’t spend too long here), but I just wanted to echo that it gets tricky when you’re trying to decide what’s “really” going on in these systems, as opposed to what simply fits our models (NB—this is true in all areas of science). Fundamentally, it’s a bunch of nuclei and electrons interacting, which is a big quantum mess. But maybe we can build a useful simplified model to explain some data.

As for the hybridization of nitrogen in, e.g., ammonia, I’ve never personally been wild about calling it sp³ hybridized, for the reason @Mayhem mentioned in post 3. If you’re going to use the argument that a carbon atom moves into an excited state with two half-open subshells whose energy is compensated by the formation of two extra covalent bonds, (which I happen to think is a nice heuristic), then you can’t draw the same analogy with a nitrogen atom. Instead, I tend to think of three H’s binding to the three orthogonal nitrogen p orbitals and being electrostatically repelled from one another until they settle into an equilibrium position. But if you use a modern LCAO-MO (basically building up your molecular orbitals by a linear combination of atomic orbitals) approach, you find that the “nitrogen atom” orbitals that are occupied in ammonia have significant s and p mixed character. So is it that the steric repulsion of the H atoms causes a rehybridization of the N atom electrons into a more sp³-like set of orbitals? Again, it’s really just a bunch of nuclei and electrons doing whatever it is they do.

 

[Mayhem]

Again, it’s really just a bunch of nuclei and electrons doing whatever it is they do.

This neatly summarizes why a first principles look on chemical interactions might not be feasible in practice.

 

[TeethWhitener]

This neatly summarizes why a first principles look on chemical interactions might not be feasible in practice.

I definitely don’t agree with this. We have plenty of techniques that are extremely good at modeling chemical behavior from first principles by solving the multielectron Schrodinger equation numerically. But sometimes you don’t want to wait for a full-CI calculation to finish, and sometimes you don’t need to: simple models can do a lot of good work before they fail. (And of course, when they fail is precisely when we learn something new.) So for instance, orbital hybridization does a pretty darn good job at describing molecular geometry and reactivity in a broad category of organic compounds. VSEPR theory does a pretty god job at describing p-block compound molecular geometry, etc.

 

[snorkack]

But, the 3 $2p$ orbitals on the nitrogen are all at right angles to each other, and the $2s$ orbital is spherically symmetric. So, this arrangement is unsuitable to describe either tetrahedral or trigonal planar geometry, for instance.

In $NH_3$, the $sp^3$ hybridisation ends up producing, from the original AOs, 4 tetrahedrally directed HAOs. Three of these are involved in $\sigma$ N-H bonding and one contains a lone pair. The tetrahedral geometry arising from the hybrid description is a better model of the actual molecule, since loosely speaking we'd expect the electron pairs constituting the bonds & the l.p. to be as far apart as possible.

In contrast, PH₃ has bond angles of 93,5 degrees. Close to unhybridized p orbitals.
Trigonal planar would be sp².

 

[TeethWhitener]

In contrast, PH3 has bond angles of 93,5 degrees. Close to unhybridized p orbitals.

Which is why I like the heuristic in post 10 better than a blanket statement like “trigonal pyramidal molecules have sp³-hybridized central atoms” that you find in some chem textbooks. Phosphorus has a much larger atomic radius than nitrogen, which means the H’s experience less steric repulsion in PH₃ than in NH₃, so you get closer to 90 degrees in PH₃.

 

[DrDu]

But if you use a modern LCAO-MO (basically building up your molecular orbitals by a linear combination of atomic orbitals) approach,

Well, MO theory was modern in maybe 1950 to -60, mainly because calculations were simpler and less time consuming on the first computers. Especially for compounds of non-metals, VB does often a better job, as it incorporates some electron interaction, which becomes more important for non-metals whose valence shells are more crowded than those of metallic elements.

 

[DrDu]

But, the 3 $2p$ orbitals on the nitrogen are all at right angles to each other, and the $2s$ orbital is spherically symmetric. So, this arrangement is unsuitable to describe either tetrahedral or trigonal planar geometry, for instance.

In $NH_3$, the $sp^3$ hybridisation ends up producing, from the original AOs, 4 tetrahedrally directed HAOs. Three of these are involved in $\sigma$ N-H bonding and one contains a lone pair. The tetrahedral geometry arising from the hybrid description is a better model of the actual molecule, since loosely speaking we'd expect the electron pairs constituting the bonds & the l.p. to be as far apart as possible.

This contains the implicit assumption, that orbitals have always to be directed such that the overlap becomes maximal. While this is a good rule of the thumb, a maximum always means that things change but little if we move slightly away from it (first derivative vanishes at the maximum). So, even if we assume p orbitals on nitrogen in NH3, the angle between the hydrogens may be larger than 90 degrees due to the nuclear repulsion without a considerable weakening of the bonds due to non-optimal overlap.
What has not been mentioned in this thread is the fact that the energetic separation of s and p orbitals increases from Carbon to Fluorine, so that hybridization becomes energetically increasingly unfavorable.
For third row elements and below, s and p orbitals are of vastly different size, so that hybridization also becomes inefficient.

 

[TeethWhitener]

Well, MO theory was modern in maybe 1950 to -60, mainly because calculations were simpler and less time consuming on the first computers. Especially for compounds of non-metals, VB does often a better job, as it incorporates some electron interaction, which becomes more important for non-metals whose valence shells are more crowded than those of metallic elements.

Maybe you’re more in the loop on modern valence bond theory than I am, but nearly all of the work I see done in computational chemistry nowadays is MO-based.

 

[DrDu]

Maybe you’re more in the loop on modern valence bond theory than I am, but nearly all of the work I see done in computational chemistry nowadays is MO-based.

Well, I'd rather say that DFT is the workhorse today. Personally, I think that VB is still an excellent way to think about chemical bonding and you can nowadays project multi configuration SCF wavefunction on VB structures to generate insight. MO and DFT are especially superior in terms of speed of ab initio calculation which lend to their triumph in the 50ies. But this does not necessarily mean that they are always better (or more "modern") wrt teaching.

 

[TeethWhitener]

Yes I should have said, between MO and VB, most computational work is MO. But you’re right, DFT is taking over everything now. And I agree that VB has real conceptual advantages over MO.