Question

CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Calculate the inter-ionic distance (in pm) in CsCl:

• A. $141.42$
• B. $282.84$
• C. $346.41$
• D. $173.2$

Answer 1

The correct option is C $346.41$

In BCC arrangement, the inter-ionic distance is calculated by:
$a\sqrt{3}=2(r_{+}+r_{-})$

Here, given edge lenfgth =400 pm
$\frac{400\sqrt{3}}{2}=(r_{+}+r_{-})$
$(r_{+}+r_{-})=346.4\text{pm}$