Question

$N{H}_{4}Cl$ crystallizes in a body-centered cubic lattice with an edge length of unit cell equal to 387 pm. If the radius of the $C{l}^{-}$ ion is 181 pm, the radius for $N{H}_4^{+}$ ion is?

• A. 154.1 pm
• B. 92.6 pm
• C. 366.3 pm
• D. None of these

Answer 1

The correct option is A

154.1 pm

For bcc lattice, $2r_{N{H}_4^{+}}+2r_{C{l}^{-}}=a\sqrt{3}$
$(N{H}_4^{+}$ ion lies in the cubic void $C{l}^{-}$ ions)
Or, $r_{N{H}_4^{+}}+r_{C{l}^{-}}=\frac{a\sqrt{3}}{2}=\frac{387\times 1.732}{2}=335.1$
Or $r_{C{l}^{-}}=335.1-101.0=154.1pm$