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Question

The distance between adjacent oppositely charged ions in rubidium chloride is 328.5 pm, in potassium chloride is 313.9 pm, in sodium bromide is 298.1 pm and in potassium bromide is 329.3 pm. The distance between adjacent oppositely charged ions in rubidium bromide is

Options

A.314.7 pm
B.338.5 pm
C.339.3 pm
D.343.9 pm

Solution

$r_{Rb^{+}} + r_{Cl^{-}} = 328.5\text{pm} = x$

$r_{K^{+}} + r_{Cl^{-}} = 313.9\text{pm} = y $$${r_{Na^{+}} + r_{Br^{-}} = 298.1\text{pm} = z }{r_{K^{+}} + r_{Br^{-}} = 329.3\text{pm=w} }{\therefore r_{Rb^{+}} + r_{Br^{-}} = x + w - y = 343.9\text{pm}}$$

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