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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.
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  • $\begingroup$ Is this formula heuristically determined ? I don't find any physical significance $\endgroup$ Commented May 29 at 15:31

1 Answer 1

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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: \begin{equation} x \leq y \end{equation} \begin{equation} x + y = 1 \end{equation} The weighted average of metallic electronegativity difference given by Villars, et.al is : \begin{equation} \Delta\chi = 2x(\chi_{A} - \chi_{B}) \end{equation} 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: \begin{equation} x = \frac{1}{n+1} \end{equation} \begin{equation} y = \frac{n}{n+1} \end{equation}

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: \begin{equation} <s>_{w} = \frac{\Sigma_{i=1}^{t} w_{i}x_{i}}{\Sigma_{i=1}^{t} w_{i}} \end{equation} In our case, the weight for A-B ($w_{A-B}$) pairs is given as: \begin{equation} w_{A-B} = \frac{n}{\frac{n(n+1)}{2}} = \frac{2}{n+1} \end{equation} The weight for other possible pairs ($w_{other}$) is given as: \begin{equation} 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} \end{equation} Therefore the weighted average of metallic electronegativity difference is given as: \begin{equation} \Delta \chi = \frac{2}{n+1}(\chi_{A} - \chi_{B})= 2x(\chi_{A} - \chi_{B}) \end{equation}

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