$N H_{4} C l$ 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?

154.1 pm

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92.6 pm

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366.3 pm

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None of these

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Solution

The correct option is A
154.1 pm

For bcc lattice, $2 r_{N H_{4}^{+}} + 2 r_{C l^{-}} = a \sqrt{3}$
$(N H_{4}^{+}$ 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 $r_{C l^{-}} = 335.1 - 101.0 = 154.1 p m$

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

$N H_{4} C l$ 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?