$N H_{4} C l$ 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

154. 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. 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)
$o r,_{N H_{4}^{+}}^{r} + r_{C l^{-}} = \frac{a \sqrt{3}}{2} = \frac{387 \times 1.732}{2} = 335.1 o r, r_{C l} = 335.1 - 101.0 = 154.1 p m$

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Similar questions

$N H_{4} C l$ 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?

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387 p m

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$N H_{4} C l$ 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?

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N H_{4} C l

crystallises in a body-centered cubic lattice with an unit cell edge length of 387 pm. The distance between the oppositely charged ions in the lattice is:

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NH4Cl crystallizes in a body-centered cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl− ion is 181 pm, radius for NH+4 ion is

The correct option is A 154. pmFor bcc lattice, 2rNH+4+2rCl−=a√3 (NH+4ion lies in the cubic void Cl− ions) or, rNH+4+rCl−=a√32=387×1.7322=335.1or, rCl=335.1−101.0=154.1 pm

$N H_{4} C l$ 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

The correct option is A 154. 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)
$o r,_{N H_{4}^{+}}^{r} + r_{C l^{-}} = \frac{a \sqrt{3}}{2} = \frac{387 \times 1.732}{2} = 335.1 o r, r_{C l} = 335.1 - 101.0 = 154.1 p m$