# Average electronegativity difference in ternary compounds

Villars and Hulliger [1, p. 300] define the average electronegativity difference of a ternary compound $$\ce{A_xB_yC_z}$$ as follows:

$$\bar{\Delta\chi} = 2x(\chi_\ce{A} - \chi_\ce{B}) + 2x(\chi_\ce{A} - \chi_\ce{C}) + 2y(\chi_\ce{B} - \chi_\ce{C}),$$

given that $$x \leq y \leq z$$ and $$x + y + z = 1.$$ Can someone give explanation to how it arises?

### Reference

1. Villars, P.; Hulliger, F. Structural-Stability Domains for Single-Coordination Intermetallic Phases. Journal of the Less Common Metals 1987, 132 (2), 289–315. https://doi.org/10.1016/0022-5088(87)90584-4.
• Is this formula heuristically determined ? I don't find any physical significance Commented May 29 at 15:31

Let us consider a binary compound of type $$AB_{n}$$, where A and B both are metals. On normalization it can be written as $$A_{x}B_{y}$$, where x and y satisfy the following conditions: $$$$x \leq y$$$$ $$$$x + y = 1$$$$ The weighted average of metallic electronegativity difference given by Villars, et.al is : $$$$\Delta\chi = 2x(\chi_{A} - \chi_{B})$$$$ where '$$\chi_{A}$$' and $$\chi_{B}$$' are electronegativity of 'A' and 'B' in Martynov-Batsanov scale respectively.

Derivation:

The terms 'x' and 'y' are function of n and is given as: $$$$x = \frac{1}{n+1}$$$$ $$$$y = \frac{n}{n+1}$$$$

where 'x' and 'y' satisfies the condition mentioned above.

In binary compound with chemical formula $$AB_{n}$$ we have 1 atom of A, n atoms of B and total of n+1 atoms.

1] The maximum number of pairs of A-B possible is : n

2] The total number of pairs possible with n+1 atoms is : $$\binom{n+1}{2} = \frac{n(n+1)}{2}$$

where, $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

The weighted average of s' is given as: $$$$_{w} = \frac{\Sigma_{i=1}^{t} w_{i}x_{i}}{\Sigma_{i=1}^{t} w_{i}}$$$$ In our case, the weight for A-B ($$w_{A-B}$$) pairs is given as: $$$$w_{A-B} = \frac{n}{\frac{n(n+1)}{2}} = \frac{2}{n+1}$$$$ The weight for other possible pairs ($$w_{other}$$) is given as: $$$$w_{other} = \frac{\binom{n}{2}}{\binom{n+1}{2}} = \left(\frac{\frac{n(n-1)}{2}}{\frac{n(n+1)}{2}}\right) = \frac{n-1}{n+1}$$$$ Therefore the weighted average of metallic electronegativity difference is given as: $$$$\Delta \chi = \frac{2}{n+1}(\chi_{A} - \chi_{B})= 2x(\chi_{A} - \chi_{B})$$$$