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 :

206

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116

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174

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154

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Solution

The correct option is C $154$ pm
In BCC lattice structure, we have $\sqrt{3} a = 2 (r_{+} {+ r}_{-})$

$\sqrt{3} \times 387 = 2 (r_{+} + 181)$

$r_{+} = 154$ pm

Therefore, the size of $N H_{4}^{+}$ ion would be $154$ pm.

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Similar questions

$N H_{4} C l$ 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

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)

$N H_{4} C l$ 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)

$N H_{4} C l$ 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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Ammonium chloride crystallizes in a body-centered cubic lattice with a unit distance of 387 pm. If the size of Cl− ion is 181 pm, the size of NH+4 ion would be :

The correct option is C 154 pmIn BCC lattice structure, we have √3a=2(r++r−)√3×387=2(r++181)r+=154 pmTherefore, the size of NH+4 ion would be 154 pm.

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 :

The correct option is C $154$ pm
In BCC lattice structure, we have $\sqrt{3} a = 2 (r_{+} {+ r}_{-})$

$\sqrt{3} \times 387 = 2 (r_{+} + 181)$

$r_{+} = 154$ pm

Therefore, the size of $N H_{4}^{+}$ ion would be $154$ pm.