Question

$PbS$ has $aNaCl$ type structure.The shortest distance between $P{b}^{2+}$ ion and $S^{2-}$ion is $297$ pm.The edge length in cm of the unit cell will be :

• A. $1.24\times {10}^{-8}$
• B. $8.58\times {10}^{-8}$
• C. $5.9\times {10}^{-8}$
• D. $5.9\times {10}^{-10}$

Answer 1

The correct option is C $5.9\times {10}^{-8}$

In NaCl,
$a=2(r_{+}+r_{-})$
Given, $(r_{+}+r_{-})=297\text{pm}$
Hence $a=5.9\times {10}^{-8}cm$