Question

For an Na+—Cl- ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to EA = -1.436/r ER =(7.32 *10-6 )/r8 For these expressions, energies are expressed in electron volts per Na+—Cl- pair, and r is the distance in nanometers. The net energy EN is just the sum of the preceding two expressions

Answer 1

Answer:

Explanation:

$\text{The curve of the plot}$ $\mathbf{E_A,E_R, \ and \ E_N}$ $\text{can be seen in the attached diagram below}$

$\text{From the plot}$, $\mathbf{r_o = 0.2 4nm \ and \ E_o =-5.3 eV}$

$\mathbf{We \ knew \ that: E_N = E_A + E_R}$

$\mathtt{GIven \ E_A = \dfrac{-1.436}{r}\ \ \ , E_R = \dfrac{7.32 \times 10^{-6}}{r^n} \ \ and \ \ n=8 }$

$\mathtt{Then; E_N = -\dfrac{-1.436}{r}+ \dfrac{7.32\times 10^{-6}}{r^8}}$

$\mathtt{Also; r_o = \Big( \dfrac{A}{nB} \Big)^{\dfrac{1}{1-n}}} \\ \\ \mathtt{ r_o = \Big( \dfrac{1.986}{8 \times 7.32\times 10^{-6}} \Big)^{\dfrac{1}{1-8}}} \\ \\ \mathbf{r_o = 0.236 nm}$

$E_o = \dfrac{-1.436}{\Big[\dfrac{1.436}{8(732\times 10^{-6})}\Big]^{\dfrac{1}{1-8}}} + \dfrac{7.32 \times 10^{-6}}{\Big[ \dfrac{1.436}{8\times7.32 \times 10^{-6} } \Big]^{\dfrac{8}{1-8}}}$

$\mathbf{E_o = -5.32 \ eV}$