$N H_{4} C l$ crystallizes in a body centred cubic lattice , with a unit cell distance of $387 p m$ . Calculate the radius of the $N H_{4}^{+}$ ion if the radius of the $C l^{-}$ ion is $181 p m$ .

608.67 p m

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52.67 p m

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335.15 p m

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154.15 p m

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Solution

The correct option is D $154.15 p m$
In a body centred cubic lattice, oppositely charged ions touch each other along the cross-diagonal of the cube.
So,
$2 r^{+} + 2 r^{-} = \sqrt{3} a$

$r^{+} + r^{-} = \frac{\sqrt{3} a}{2}$

$r^{+} = (\frac{\sqrt{3} \times 387}{2} - 181) p m r^{+} = 154.15 p m$

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Similar questions

Q. Ammonium chloride crystallises in a body-centred cubic lattice with a unit distance of 387 pm. Calculate (a) the distance between oppositely charged ions in the lattice and (b) the radius of the

N H_{4}^{+}

ion if the radius of

C l^{-}

ions is 181 pm.

$N H_{4} C l$ 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?

Q. Ammonium chloride crystallises in a body centered cubic lattice with a unit distance of

386 p m

. Calculate the radius (in pm) of

N H_{4}^{+}

ion. If the radius of

C l^{-}

ion is

181 p m

(T a k e \sqrt{3} = 1.73)

Q. Ammonium chloride crystallizes in a body-centered cubic lattice with a unit distance of

387

pm. If the size of

C l^{-}

ion is

181

pm, the size of

N H_{4}^{+}

ion would be :

$N H_{4} C l$ 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

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NH4Cl crystallizes in a body centred cubic lattice , with a unit cell distance of 387 pm. Calculate the radius of the NH+4 ion if the radius of the Cl− ion is 181 pm.

The correct option is D 154.15 pmIn a body centred cubic lattice, oppositely charged ions touch each other along the cross-diagonal of the cube. So, 2r++2r−=√3a r++r−=√3a2 r+=(√3×3872 −181) pmr+=154.15 pm

$N H_{4} C l$ crystallizes in a body centred cubic lattice , with a unit cell distance of $387 p m$ . Calculate the radius of the $N H_{4}^{+}$ ion if the radius of the $C l^{-}$ ion is $181 p m$ .

The correct option is D $154.15 p m$
In a body centred cubic lattice, oppositely charged ions touch each other along the cross-diagonal of the cube.
So,
$2 r^{+} + 2 r^{-} = \sqrt{3} a$

$r^{+} + r^{-} = \frac{\sqrt{3} a}{2}$

$r^{+} = (\frac{\sqrt{3} \times 387}{2} - 181) p m r^{+} = 154.15 p m$