The force of attraction between a divalent cation and a divalent anion is 1.12 x 10-8 n. if the ionic radius of the cation is 0.071 nm, what is the anion radius?

= 0.2162 nm

Explanation:

The force of attraction can be written as:

Where,  

K is the Coulomb's constant having value 9×10⁹ N. m²/C²

 is the charge on the cation

 is the charge on the anion

e is electronic charge =

r is the distance between the cation and anion and in bond, it is the sum of the ionic radius of cation and anion

So,

Given, F =

Applying in the equation,

Also, 1 m =

So, r = 0.2872 nm

Also,

= 0.071 nm

Thus,

= 0.2162 nm

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."