Question

An ionic solid has $C_{5}Cl$ structure. The length of body diagonal is $7.0\stackrel{\text{o}}{\text{A}}.$ The edge length of the cube and inter-ionic distance respectively are:

• A. $4.04\stackrel{\text{o}}{\text{A}},3.5\stackrel{\text{o}}{\text{A}}$
• B. $3.5\stackrel{\text{o}}{\text{A}},4.04\stackrel{\text{o}}{\text{A}}$
• C. $7.54\stackrel{\text{o}}{\text{A}},4.04\stackrel{\text{o}}{\text{A}}$
• D. $5.3\stackrel{\text{o}}{\text{A}},4.04\stackrel{\text{o}}{\text{A}}$

Answer 1

The correct option is A $4.04\stackrel{\text{o}}{\text{A}},3.5\stackrel{\text{o}}{\text{A}}$

In $CsCl$ type structure, $C{l}^{-}$ ions are at the corners of the cube and $C{s}^{+}$ ion is at the body center. So, it is body centered cubic unit cell.

Now, the atoms/ ion that they body diagonal will touch each other.

So, if $a$ is the edge length of the cube and $r_{-}$ is the radius of the $C{l}^{-}$ and $r_{+}$ is radius of $C{s}^{+}$

$\sqrt{3}a=2(r_{-}+r_{+})$

$\Rightarrow 2(r_{+}+r_{-})=7\mathring{A}$ (given)

$\Rightarrow r_{+}+r_{-}=\frac{7}{2}\mathring{A}=3.5\mathring{A}$

So, inter ionic distance $=\frac{\sqrt{3}}{2}a=r_{+}+r_{-}=3.5\mathring{A}$

And edge length of cube $=\frac{diagonalofcube}{\sqrt{3}}=\frac{7}{\sqrt{3}}\mathring{A}=4.04\mathring{A}$