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Standard XII

Chemistry

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If the length...

Question

If the length of the body diagonal for CsCl which crystallises into a cubic structure with $C l^{-}$ ions at the corners and $C s^{+}$ ions at the centre of the unit cells is $7 o A$ and the radius of the $C s^{+}$ ion is $1.69 o A$ . What is the radii for $C l^{-}$ ion?

1.81 o A

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5.31 o A

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3.62 o A

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None of these

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Solution

The correct option is C $1.81 o A$
$C s^{+ 2}$ at body centre, $C l^{-}$ at corners.
So, $2 (r_{C s^{+}} + r_{C l^{-}}) =$ length of body diagonal $= 7 ˚ A$
$r_{C s^{+}} = 1.69 ˚ A$
$\Rightarrow r_{C l^{+}} = 3.5 - 1.69 = 1.81 ˚ A$

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Similar questions

Q. If the length of body diagonal for

CsCl

, which crystallises into a cubic structure with

{Cl}^{-}

ions at the corners and

{Cs}^{+}

ions at the centre of unit cell is

7 ˚ A

and the radius of

{Cs}^{+}

ion is

1.69 ˚ A

. The radius of

{Cl}^{-}

ion?

Q. If the length of the body diagonal for CsCl which crystallizes into a cubic structure with

C l^{-}

ions at the corners and

C s^{+}

ions at the centre of the unit cells is

7 ˚ A

and the radius of the

C s^{+}

ion is

1.69 ˚ A

, what is the radius of

C l^{-}

ion in

˚ A

C s C l

has cubic structure of ions in which

C s^{+}

ions is present in the body-centre of the cube. If density is

3.99 g c m^{- 3}

.
(a) Calculate the length of the edge of a unit cell.
(b) What is the distance between

C s^{+}

and

C l^{-}

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(c) What is the radius of

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ion if the radius of

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Q. In a cubic crystal of CsCl (density

= 3.97

g cm

^{- 3}

), the eight corners are occupied by

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ions with

C s^{+}

ions at the body centre. Calculate the distance (in

˚ A

) between the neighbouring

C s^{+}

and

C l^{-}

ions.

C s C l

crystallizes in a cubic lattice that has a

C l^{-}

at each corner and

C s^{+}

at the centre of the unit cell. If

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, what is the value of edge length a of the cube?

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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 7oA and the radius of the Cs+ ion is 1.69oA. What is the radii for Cl− ion?

The correct option is C 1.81oACs+2 at body centre, Cl− at corners.So, 2(rCs++rCl−)= length of body diagonal =7˚ArCs+=1.69˚A⇒rCl+=3.5−1.69=1.81˚A

If the length of the body diagonal for CsCl which crystallises into a cubic structure with $C l^{-}$ ions at the corners and $C s^{+}$ ions at the centre of the unit cells is $7 o A$ and the radius of the $C s^{+}$ ion is $1.69 o A$ . What is the radii for $C l^{-}$ ion?

The correct option is C $1.81 o A$
$C s^{+ 2}$ at body centre, $C l^{-}$ at corners.
So, $2 (r_{C s^{+}} + r_{C l^{-}}) =$ length of body diagonal $= 7 ˚ A$
$r_{C s^{+}} = 1.69 ˚ A$
$\Rightarrow r_{C l^{+}} = 3.5 - 1.69 = 1.81 ˚ A$