$C s B r$ has bcc structure with edge lengths $4.3$ . The shortest inter ionic distance $(in pm)$ in between $C s^{+}$ and $B r^{-}$ is: (Given, $\sqrt{3} = 1.73$ )

Open in App

Solution

For a bcc structure,
$Atomic radius r = \frac{\sqrt{3}}{4} a = \frac{\sqrt{3}}{4} \times 4.3 = 1.86$
We know,
$r = half distance between two nearest neighbouring atoms$ .
$∴ Shortest inter ionic distance = 2 \times 1.86 = 3.72 pm$

Suggest Corrections

Similar questions

C s B r

4.3 pm

C s^{+}

B r^{-}

$C s B r$ has a BCC structure where the edge length is 4.3. The shortest inter ionic distance between $C s^{+}$ and $B r^{-}$ is:

C s B r

4.3 ˚ A

˚ A

C s^{+}

B r^{-}

C s B r

C s C l

6 pm

C s^{+}

B r^{-}

CsBr crystal has bcc structure. It has an edge length of $4.3 \circ A$ . The shortest interionic distance between $C s^{+}$ and $B r^{-}$ ions is

Join BYJU'S Learning Program