You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Types of Crystalline Solids

Cesium chlori...

Question

Cesium chloride forms a body-centered cubic lattice. Cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is $412$ pm and $C l^{-}$ ion has a radius of $181$ pm. Calculate the radius (in pm) of $C s^{+}$ ion.

176

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

350

No worries! We‘ve got your back. Try BYJU‘S free classes today!

88

No worries! We‘ve got your back. Try BYJU‘S free classes today!

201

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $176$
As we know, for body-centered cubic lattice,

$r = \frac{\sqrt{3} a}{4}$

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

$181 + r_{+} = \frac{\sqrt{3} \times 412}{2}$

$r^{+} = 176$ pm

Hence, the correct option is $A$

Suggest Corrections

Similar questions

Q. In

C s C l

structure, cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is

412

pm and chloride ion has a radius of

181

pm. Calculate the radius (in pm) of caesium ion.

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 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-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.

Join BYJU'S Learning Program

Cesium chloride forms a body-centered cubic lattice. Cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is 412 pm and Cl− ion has a radius of 181 pm. Calculate the radius (in pm) of Cs+ ion.

The correct option is A 176As we know, for body-centered cubic lattice,r=√3a4r−+r+=2r=√3a2181+r+=√3×4122r+=176 pmHence, the correct option is A

Cesium chloride forms a body-centered cubic lattice. Cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is $412$ pm and $C l^{-}$ ion has a radius of $181$ pm. Calculate the radius (in pm) of $C s^{+}$ ion.

The correct option is A $176$
As we know, for body-centered cubic lattice,

$r = \frac{\sqrt{3} a}{4}$

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

$181 + r_{+} = \frac{\sqrt{3} \times 412}{2}$

$r^{+} = 176$ pm

Hence, the correct option is $A$