CsBr has a BCC structure- The length of edge is 4-3 A- The minimum distance between Cs -and Br -ion will be in angstrom-

For BCC: 2(r^++r^ ) = √(3) a (For minimum distance) or , 2(r^++r^ ) = √(3)× 4.3 or, (r^+ + r^ ) = √(3)×4.32 = 3.72 ∘A Hence, the distance between Cs^+ and Br^ ...

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Standard XII

Chemistry

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CsBr has a BC...

Question

$C s B r$ has a BCC structure. The length of edge is $4.3 \circ A$ . The minimum distance between $C s^{+}$ and $B r^{-}$ ion will be (in angstrom):

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Solution

For BCC:
$2 (r^{+} + r^{-}) = \sqrt{3} a$ (For minimum distance)
or , $2 (r^{+} + r^{-}) = \sqrt{3} \times 4.3$
or, $(r^{+} + r^{-}) = \sqrt{3} \times \frac{4.3}{2} = 3.72 \circ A$
Hence, the distance between $C s^{+}$ and $B r^{-}$ ions will be $3.72 \circ A$

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CsBr has a BCC structure. The length of edge is 4.3 ∘A. The minimum distance between Cs+ and Br− ion will be (in angstrom):

For BCC: 2(r++r−)=√3a (For minimum distance) or , 2(r++r−)=√3×4.3 or, (r++r−)=√3×4.32=3.72 ∘A Hence, the distance between Cs+ and Br− ions will be 3.72 ∘A

$C s B r$ has a BCC structure. The length of edge is $4.3 \circ A$ . The minimum distance between $C s^{+}$ and $B r^{-}$ ion will be (in angstrom):

For BCC:
$2 (r^{+} + r^{-}) = \sqrt{3} a$ (For minimum distance)
or , $2 (r^{+} + r^{-}) = \sqrt{3} \times 4.3$
or, $(r^{+} + r^{-}) = \sqrt{3} \times \frac{4.3}{2} = 3.72 \circ A$
Hence, the distance between $C s^{+}$ and $B r^{-}$ ions will be $3.72 \circ A$