Question

By using Pauling's method calculate the ionic radii of K⁺ and Cl⁻ ions in the potassium chloride crystal. Given that `"d"_("K"^+) - "Cl"^-` = 3.14 Å

Answer 1

Given `"d"_("K"^+) - "Cl"^-` = 3.14 Å

`"r"_("K"^+)` = ?

`"r"_("Cl"^-)` = ?

i.e. `"r"_("K"^+) + "r"_("Cl"^-)` = 3.14 Å .....(1)

We know that,

`("r"_("K"^+))/("r"_("Cl"^-)) = ("Z"_"eff")_("Cl"^-)/("Z"_"eff")_("K"^+)`

`("Z"_"eff")_("Cl"^-)` = Z – S

= 17 – [(0.35 × 7) + (0.85 × 8) + (1 × 2)]

= 17 – 11.25

= 5.75

`("Z"_"eff")_("K"^+)` = Z – S

= 19 – [(0.35 × 7) + (0.85 × 8) + (1 × 2)]

= 19 – 11.25

= 7.75

∴ `("r"_(("K"^+)))/("r"_(("Cl"^-))) = ("Z"_"eff")_("Cl"^-)/("Z"_"eff")_("K"^+) = 5.75/7.75 = 0.74`

`"r"_(("K"^+))` = 0.74 `"r"_(("Cl"^-)`

Substitute the value of `"r"_(("K"^+))` in equation (1)

0.74 `"r"_(("Cl"^-)) + "r"_(("Cl"^-))` = 3.14 Å

1.74 `"r"_(("Cl"^-)) = (3.14 Å)/(1.74) = 1.81 Å`