Question

The edge length of the unit cell of KCl (fcc) is $6.28\mathring{A}$. Assuming anion-cation contact along the cell edge, calculate the radius (in $\mathring{A}$) of the potassium ion.

$[r_{C{l}^{-}}=1.8\mathring{A}]$

• A. $0.9$
• B. $1.3$
• C. $1.8$
• D. None of the above

Answer 1

The correct option is B $1.3$

KCl has same structure vas of NaCl.

As we know, for NaCl type structure,

$r^{+}+r^{-}=\frac{a}{2}$

$r^{+}=\frac{6.28}{2}-1.8=1.34\mathring{A}$

Hence, the correct option is $\text{B}$