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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
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Solution
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`
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