Question

The atomic radii of a divalent cation and a monovalent anion are 0.19 nm and 0.126 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance

Answer 1

Step-by-step explanation:

Given that,

Radius of divalent cation,
`a_1=0.19\ nm`

Radius of monovalent anion,
`a_2=0.126\ nm`

Charge for divalent cation,
`Z_1=+2`

Charge for monovalent cation,
`Z_2=-1`

The sum of radii of both cations and anion is,
`a_o=0.19+0.126=0.316\ nm=0.316* 10^(-9)\ m`

(a) The force of attraction between two ions is given by :

`F=k(-Z_1Z_2e^2)/(a_o^2)`

`F=-9* 10^9* (2* (-1)* (1.6* 10^(-19))^2)/((0.316* 10^(-9))^2)`

`F=4.61* 10^(-9)\ N`

(b) Let F' is the force of repulsion at this same distance such that,

F + F' = 0

So,
`F'=-4.61* 10^(-9)\ N`

Hence, this is the required solution.