The force of attraction between a divalent cation and a divalent anion is 1.91 x 10-8 n. if the ionic radius of the cation is 0.074 nm, what is the anion radius?

The force of attraction (F) is given by the formula:

F = (1/4π∈r²)(Zc*e)(Za*e)(1)

where:

∈ = permittivity of free space = 8.85*10^-15 F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602*10^-19 C

r = interionic distance = rc + ra

where rc and ra are the radius of cation and anion respectively

F = 1.91 * 10^-8 N

therefore based on equation (1) we have:

1.91 *10^-8 = [1/4π(8.85*10^-15)r²](2*1.602*10^-19)²

r² = 0.04832 * 10^-15

r = 6.951 nm

Now: r = rc + ra

where rc = 0.074 nm

thus, ra = r - rc = 6.951 - 0.074 = 6.877 nm

sorry i dont quite understand, were you given more information

Step-by-step explanation: