Single Choice

$$CsCl$$ crystallises in body centred cubic lattice. If $$ a$$ is its edge length, then which of the following expressions is correct?

ArCs++rCl−=√32a
Correct Answer
BrCs++rCl−=√3a
CrCs++rCl−=3a
DrCs++rCl−=3a2

Solution

In $$ CsCl$$ structure, $$ Cs^{+}$$ ion is in contact with $$ Cl^{-}$$ ion at the nearest distance which is equal to $$\dfrac{\sqrt3a}{2}$$.

SIMILAR QUESTIONS

Solid State

Co-ordination number of the cation is maximum in :

Solid State

The position of $$Cs^+$$ ion in $$CsCl$$ structure is:

Solid State

The number of oppositely charged nearest neighbours to a Caesium ion in Caesium Chloride lattice are:

Solid State

CsCl has bcc structure with $$Cs^{+}$$ at the centre and $$Cl^{-}$$ ion at each corner. If $$r_{Cs^+}$$ is $$1.69 \,\mathring {A}$$ and $$r_{Cl^-}$$ is $$1.81 \,\mathring {A},$$ what is the edge length of the cube ?

Solid State

What is the co-ordination number of $$Cl^{-}$$ in $$CsCl$$ structure ?

Solid State

In a crystal of $$CsCl$$, the nearest neighbours of each $$Cs$$ ion are:

Solid State

If the length of the body diagonal for CsCl which crystallises into a cubic structure with $$Cl^-$$ ions at the corners and $$Cs^+$$ ions at the centre of the unit cells is $$7\ \mathring A$$ and the radius of the $$Cs^+$$ ion is $$1.69\ \mathring A$$. What is the radii (in $$\mathring A$$) for $$Cl^-$$ ion?

Solid State

For cesium chloride structure, the interionic distance (in terms of edge length, a) is equal to $$\displaystyle \frac{\sqrt{3a}}{2}$$.

Solid State

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.

Solid State

The $$8\,\colon\,8$$ type of packing is present in___________.

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