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Solid State

Section: Chemistry

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Question : 11 of 19

Marks: +1, -0

The arrangement of X

^{-}

^{+}

ion in solid AX is given in the figure (not drawn to scale). If the radius of X

^{-}

is 250 pm. the radius of A

^{+}

[JEE Adv 2013 P1]

104 pm
125 pm
183 pm
57pm

Solution:

According to the given figure,

A^{+}

is present in the octahedral void of

X^{-}

.The limiting radius in octahedral void is related to the radius of the sphere as

r_{\text{void}}

= 0.414

r_{\text{sphere}}

r_{A^{+}}

= 0.414

r_{X^{-}}

= 0.414 × 250 pm = 103.5

≈ 104 pm

Hence (A) is correct.

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