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The arrangeme...

Question

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid AX is given in the figure (not drawn to scale). If the radius of $X^{-}$ is 250 pm, the radius of $A^{+}$ is

104 pm

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183 pm

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125 pm

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57 pm

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Solution

The correct option is A
104 pm

From figure,
$A^{\oplus}$ is present in the octahedral void of $X^{⊖}$
For octahedral void,
$\frac{r A^{\oplus}}{r X^{⊖}} = 0.414$
$\Rightarrow r_{A^{\oplus}} = 0.414 \times r_{X^{⊖}} = 0.414 \times 250 p m ≃ 104 p m$

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The arrangement of X− ions around A+ 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+ is

The correct option is A 104 pm From figure, A⊕ is present in the octahedral void of X⊖ For octahedral void, rA⊕rX⊖=0.414 ⇒rA⊕=0.414×rX⊖=0.414×250 pm≃104 pm

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid AX is given in the figure (not drawn to scale). If the radius of $X^{-}$ is 250 pm, the radius of $A^{+}$ is

The correct option is A
104 pm

From figure,
$A^{\oplus}$ is present in the octahedral void of $X^{⊖}$
For octahedral void,
$\frac{r A^{\oplus}}{r X^{⊖}} = 0.414$
$\Rightarrow r_{A^{\oplus}} = 0.414 \times r_{X^{⊖}} = 0.414 \times 250 p m ≃ 104 p m$