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

Step-by-step 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 = (-1.436)/(r)\ \ \ , E_R = (7.32 * 10^(-6))/(r^n) \ \ and \ \ n=8 }`

`\mathtt{Then; E_N = -(-1.436)/(r)+ (7.32* 10^(-6))/(r^8)}`

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

`E_o = \frac{-1.436}{\Big[(1.436)/(8(732* 10^(-6)))\Big]^{(1)/(1-8)}} + \frac{7.32 * 10^(-6)}{\Big[ (1.436)/(8*7.32 * 10^(-6) ) \Big]^{(8)/(1-8)}}`

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