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In trialkylamines $\ce{NR3}$, the $\ce{R}$ alkyl group attached to nitrogen increases the electronegativity of the nitrogen atom. I also found in this question that this happens due to the repulsion between the lone pair of the nitrogen atom and the $\ce{R}$ group, which increases the bond angle and hence the s-character. (This has a direct link with Bent's rule.)

Why does the increase in bond angle increase the s-character of the compound? Please provide the orbital structures with the explanation.

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In symmetric trialkylamines, $\ce{NR3}$, the bond angle under consideration is $\ce{\angle (R'-N-R)}$, where $\ce{R' = R}$

In this answer, it has been proven using Coulson's Theorem, that for the bond angle between any two equivalent bonds, we can use the formula,

$$\cos \theta = \frac{s}{s-1}$$

A brief explanation of Coulson's Theorem and how it is used to prove this property follows.

Coulson's Theorem states that the bond angle between any two bonds can be written as

$$\cos \theta = -\frac{1}{\sqrt{\lambda_i\lambda_j}}$$

The way this is proved is as follows,

Any bond can be written to be a linear combination of $\mathrm s$ and $\mathrm p$, orbitals. Since $\mathrm s$ and $\mathrm p$ are two wavefunctions that are orthogonal to each other, we state that the hybridised orbital formed by their linear combination to be $\psi = N(\mathrm s +\sqrt{\lambda}\mathrm p)$, where $N$ is the normalization factor. In such a case, we say that the orbital in question is $\mathrm {sp}^{\lambda}$ hybridised. Therefore in our case we can say that the $\ce{N-R}$ bond and $\ce{N-R'}$ bond are $\psi_R = N_1(\mathrm s +\sqrt{\lambda_1}\mathrm p_1)$ and $\psi_{R'} = N_2(\mathrm s +\sqrt{\lambda_2}\mathrm p_2)$.

Now, we know that $\mathrm s$ has no direction and $\mathrm p$ is directional. Therefore, the angle between the $\mathrm p$-orbitals will be the bond angle $\theta$.

Also, the two bonds should not have anything in common. i.e. they should be orthogonal to each other. This implies that the inner product needs to be zero.

The inner product for two vectors a and b would be $\left \langle a | b \right \rangle = |a||b|\cos(\theta)$ which is equivalent to the dot product in this case.

Therefore, we get,

$$\left\langle N_1(\mathrm s + \sqrt{\lambda_1}\mathrm p_1)|N_2(\mathrm s + \sqrt{\lambda_2}\mathrm p_2)\right\rangle = 0$$

Simplifying and expanding, we get,

\begin{align} \left\langle N_1(\mathrm s + \sqrt{\lambda_1}\mathrm p_1)|N_2(\mathrm s + \sqrt{\lambda_2}\mathrm p_2)\right\rangle &= \left\langle \mathrm s|\mathrm s \right\rangle + \left\langle \mathrm s|\sqrt{\lambda_1}\mathrm p_1 \right\rangle + \left\langle \mathrm s|\sqrt{\lambda_{2}}\mathrm p_2 \right\rangle + \left\langle \sqrt{\lambda_1}\mathrm p_1|\sqrt{\lambda_2}\mathrm p_2 \right\rangle \\ 0&= 1 + 0+ 0+ \sqrt{\lambda_1\lambda_2}\cos(\theta) \end{align}

This is because $\left\langle \mathrm s|\mathrm s \right \rangle$ is basically the magnitude of $\mathrm s$ which is equal to one and $\mathrm s$ is orthogonal $ \mathrm p $ making those inner products zero.

Finally, we end up with Coulson's theorem which states (as mentioned before),

$$1 + \sqrt{\lambda_1\lambda_2}\cos (\theta) = 0$$

A more detailed and expansive proof has been given by Martin - マーチン here in How is Bent's rule consistent with LCAO MO theory?

A worked out example of how Coulson's theorem can be applied to find the hybridization of the carbons in cyclopropane here by ron

Hybridization is basically the mathematics of superposition of wave functions in order to generate hybridized orbitals and hence it can be explained in the best way mathematically.

Using the first equation mentioned under the assumption that $\lambda_1$ = $\lambda_2$, we can see that as the bond angle increases, the value of s-character will also increase.

A brief explanation of equation (1) follows,

From Coulson's theorem, we have,

$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$

Now, if both bonds are equivalent in nature, we get $\lambda_1 = \lambda_2 = \lambda$, therefore,

$$1 + \lambda^2\cos(\theta) = 0$$

This implies,

$$\cos \theta = \frac{-1}{\lambda^2}$$

Now for such a molecule the hybridisation would be sp$\lambda^2$ and so percentage s-character would be

$$\text{%s character} = \frac{1}{1 + \lambda^2}$$

Solving for $\lambda^2$, we get:

$$\lambda^2 = \frac{1-s}{s}$$

Substituting this in the equation for $\cos \theta$, we get:

$$\cos \theta = \frac{s}{s-1}$$

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  • $\begingroup$ Sorry to be nitpicking here: bond in question is sp𝜆 hybridised; that is not correct (or sloppy form). The bond orbital of the central atom is a spX hybrid orbital (if hybridisation is an applicable approximation). || write the σ-bond formed between them is also sloppy. There is not going to be a bond between orbitals at the same atom. $\endgroup$ Jun 8 at 21:11
  • $\begingroup$ Also, you use $\lambda_1$ and $\lambda_2$, which explicitly states that the two hybrid orbitals compared do not have to be equal. The special case you are mentioning would be $\lambda_1=\lambda_2$. $\endgroup$ Jun 8 at 21:17
  • $\begingroup$ Related: How is Bent's rule consistent with LCAO MO theory? (including full derivation of Coulson's Theorem). Why does cyclopropane react with bromine? Ron's answer has a worked example. $\endgroup$ Jun 8 at 21:35
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within hybrid theory, R groups carry non-bonding electrons (typically in s orbitals) beside the orbitals for bonding that they lend Nitrogen for use in hybridization, so now Nitrogen has more s orbitals to mix with their own p orbitals. It’s not so much this increase in s character (non-directional) rather than the decrease in p character (directional) that influence the bond angle because the p orbitals (like z, x, y axes) determine how hybrid orbitals point in space. Sure Nitrogen is still labeled sp3 hybridized in textbooks, but the ratio between s and p orbital contributions is no longer 25% as in the R=hydrogen case. It’s just a slightly higher percentage depending on how large the R group is.

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  • $\begingroup$ Hybridisation is not a theory; it is a mathematical interpretation within the molecular orbital theory. This post can otherwise benefit from shorter sentences, some paragraphing. $\endgroup$ Jun 8 at 21:41

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