Revealing strong interactions with the reduced density gradient: a benchmark for covalent, ionic and charge-shift bonds

Abstract

The visualization of covalent interactions has been a common practice in theoretical chemistry thanks to the electron localization function (ELF). More recently, the reduced density gradient (RDG) has been introduced as a tool for revealing non-covalent interactions. Along reactions, interactions change from weak to strong and vice versa. Thus, a tool enabling to analyze all of them simultaneously is fundamental for reactivity studies. This contribution is aimed at filling this gap within the NCI approach. We will highlight the ability of RDG to reveal also strong interactions and produce a benchmark that covers covalent, ionic and charge-shift bonds. The ability of the NCI descriptor to differentiate among them will be highlighted.

Appendix: Critical points of s

We now develop the gradient of s to shed some chemical insight into the critical points of s. CPs of s are given by the following equation:

$$\begin{aligned} \nabla s =\frac{1}{C_{s}}\sum _{i,j} \left [\frac{H_{ij}}{ \rho ^{4/3} |\nabla \rho |} - \frac{4}{3} \frac{|\nabla \rho |}{\rho ^{7/3}}\right ]\left(\frac{\partial \rho }{\partial x_{j}}\right ){\mathbf {u_{j}}} = 0, \end{aligned}$$

(9)

where $x_i$ stands for $\{x,y,z\}$, ${\mathbf {u_{i}}}$ for three unit vectors along directions x, y and z, and $H_{ij}$ for the electron density Hessian matrix elements $\frac{\partial ^2\rho }{\partial x_{i}x_{j}}$. Since at CPs of s $\nabla s$ must be zero along each direction x, y and z, the following three equalities must be satisfied:

$$\begin{aligned} \frac{1}{\rho ^{4/3}|\nabla \rho |} \Big [H_{11} + H_{12} + H_{13} \Big ] \Big (\frac{\partial \rho }{\partial x}\Big )= & {} \frac{4}{3} \frac{|\nabla \rho |}{\rho ^{7/3}} \Big (\frac{\partial \rho }{\partial x}\Big ), \end{aligned}$$

(10)

$$\begin{aligned} \frac{1}{ \rho ^{4/3}|\nabla \rho |} \Big [H_{12} + H_{22} + H_{23} \Big ] \Big (\frac{\partial \rho }{\partial y}\Big )= & {} \frac{4}{3} \frac{|\nabla \rho |}{\rho ^{7/3}} \Big (\frac{\partial \rho }{\partial y}\Big ), \end{aligned}$$

(11)

$$\begin{aligned} \frac{1}{ \rho ^{4/3}|\nabla \rho |} \Big [H_{13} + H_{32} + H_{33} \Big ] \Big (\frac{\partial \rho }{\partial z}\Big )= & {} \frac{4}{3} \frac{|\nabla \rho |}{\rho ^{7/3}} \Big (\frac{\partial \rho }{\partial z}\Big ). \end{aligned}$$

(12)

To connect CPs of s with those of the electron density and with the Laplacian of the electron density, we shall consider different cases:

Case 1. :

AIM-CPs: $\frac{\partial \rho }{\partial x_{i}}=0$ for all i. $|\nabla \rho |$ and $\frac{\partial \rho }{\partial x_{i}}$ are equal to zero along each direction. Thus, there is a division 0/0 for each left-hand side of Eqs. 10, 11, and 12. Since $\frac{\partial \rho }{\partial x_{i}}$ approaches to zero faster than $|\nabla \rho |$ , this limit is zero along each direction. Therefore, at critical points of $\rho$, $\nabla s$ tends to zero.

Case 2. :

Non-AIM-CPs: $|\nabla \rho | \ne 0$,

$$\begin{aligned} \Big [H_{11} + H_{12} + H_{13} \Big ] \Big (\frac{\partial \rho }{\partial x}\Big )= & {} \frac{4}{3} \frac{|\nabla \rho |^2}{\rho } \Big (\frac{\partial \rho }{\partial x}\Big ), \end{aligned}$$

(13)

$$\begin{aligned} \Big [H_{21} + H_{22} + H_{23} \Big ] \Big (\frac{\partial \rho }{\partial y}\Big )= & {} \frac{4}{3} \frac{|\nabla \rho |^2}{\rho } \Big (\frac{\partial \rho }{\partial y}\Big ), \end{aligned}$$

(14)

$$\begin{aligned} \Big [H_{31} + H_{32} + H_{33} \Big ] \Big (\frac{\partial \rho }{\partial z}\Big )= & {} \frac{4}{3} \frac{|\nabla \rho |^2}{\rho } \Big (\frac{\partial \rho }{\partial z}\Big ). \end{aligned}$$

(15)

Case 2.1. :

$\frac{\partial \rho }{\partial x_{i}} \ne 0$ for all i

$$\begin{aligned} H_{11} + H_{12} + H_{13}= & {} \frac{4}{3} \frac{|\nabla \rho |^2}{\rho }, \end{aligned}$$

(16)

$$\begin{aligned} H_{21} + H_{22} + H_{23}= & {} \frac{4}{3} \frac{|\nabla \rho |^2}{\rho }, \end{aligned}$$

(17)

$$\begin{aligned} H_{31} + H_{32} + H_{33}= & {} \frac{4}{3} \frac{|\nabla \rho |^2}{\rho } . \end{aligned}$$

(18)

.

Case 2.2. :

Mutual cancelation of all off-diagonal terms: $H_{12} + H_{13}= H_{21} + H_{23} = H_{31} + H_{32} = 0$

$$\begin{aligned} H_{11} = H_{22} = H_{33} = \frac{4}{3} \frac{|\nabla \rho |^2}{\rho }. \end{aligned}$$

(19)

It is worth noting that the trace of H, $H_{11} + H_{22} + H_{33}$, which is equal to the Laplacian of the electron density, $\nabla ^{2} \rho$, takes positive values and equal to $4 \frac{|\nabla \rho |^2}{\rho }$.

Case 2.3. :

H is diagonal. Then, all off-diagonal terms are zero and Equality 19 is obtained.