Calculate the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, the centers of which are separated by a distance of 2.9 nm.

The attraction force between the cation and anion is

Solution:

According to the question:

Separation distance between the center of the charges, d = 2.9 nm =

Also, valence is the no. of electrons required to reach the stable configuration of octet in the outer most orbit or shell of an atom.

Thus

+ 2 means +2e = Q

-1 means - 1e = Q'

where

e = electronic charge =

The Coulombian Force is given as:

where

Now,

Here, the negative sign is indicative of the attractive nature of force between two oppositely charged particles.

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

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