Chemistry - Solid State Question with Solution | TestHub

ChemistrySolid StateGeneralEasy2 minPYQ_2020

ChemistryEasysingle choice

CsBr has bcc structure with edge length 4.3 $Å$ . The shortest inter ionic distance in between ${Cs}^{+}$ and $B r^{-}$ is –

Options:

Answer:

Solution:

In BCC structure the cations and anions touch along the body diagonal

$2 (r_{{cs}^{+}} + r_{{Br}^{-}}) = \sqrt{3} \times 4 \cdot 3$ (Body Diagonal $= \sqrt{3} a$ )

$r_{{cs}^{+}} + r_{{Br}^{-}} = \frac{1 \cdot 732 \times 4 \cdot 3}{2} = 3 \cdot 72 Å$

Stream:NTA_ABHYASSubject:ChemistryTopic:Solid StateSubtopic:General

⏱ 2mℹ️ Source: PYQ_2020

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