A compound AB has an NaCl like structure- The radius of A- and B- is 1-14 -5em-A and 1-86 -5em-A- What is the minimum distance between two A- ions in angstroms- Take -2- 1-41

The edge length is given by the expression, a= 2(r^+ + r^ ) nbsp; nbsp; = 2(1.14+1.86) nbsp; nbsp; = 6 .5em∘A The minimum distance between two catio ...

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Standard XII

Chemistry

Ionic Compounds and Radius Ratios

A compound AB...

Question

A compound AB has an $N a C l$ like structure. The radius of $A^{+}$ and $B^{-}$ is $1.14 \circ A$ and $1.86 \circ A$ . What is the minimum distance between two $A^{+}$ ions in angstroms? (Take $\sqrt{2}$ = 1.41)

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Solution

The edge length is given by the expression,
$a = 2 (r^{+} + r^{-})$
= 2(1.14+1.86)
= 6 $\circ A$
The minimum distance between two cations is given by $\frac{a}{\sqrt{2}}$
Substituting,
$\frac{6}{\sqrt{2}}$ = $3 \times \sqrt{2}$ = 4.23 $\circ A$

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A compound AB has an NaCl like structure. The radius of A+ and B− is 1.14 ∘A and 1.86 ∘A. What is the minimum distance between two A+ ions in angstroms? (Take √2= 1.41)

The edge length is given by the expression, a=2(r++r−) = 2(1.14+1.86) = 6 ∘A The minimum distance between two cations is given by a√2 Substituting, 6√2 = 3×√2 = 4.23 ∘A

A compound AB has an $N a C l$ like structure. The radius of $A^{+}$ and $B^{-}$ is $1.14 \circ A$ and $1.86 \circ A$ . What is the minimum distance between two $A^{+}$ ions in angstroms? (Take $\sqrt{2}$ = 1.41)

The edge length is given by the expression,
$a = 2 (r^{+} + r^{-})$
= 2(1.14+1.86)
= 6 $\circ A$
The minimum distance between two cations is given by $\frac{a}{\sqrt{2}}$
Substituting,
$\frac{6}{\sqrt{2}}$ = $3 \times \sqrt{2}$ = 4.23 $\circ A$