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JEE Advance - Chemistry (2008 - Paper 2 Offline - No. 19)

In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.
In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.
In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.
The volume of this HCP unit cell is :
$$24\sqrt 2 \,{r^3}$$
$$16\sqrt 2 \,{r^3}$$
$$12\sqrt 2 \,{r^3}$$
$${{64} \over {3\sqrt 3 }}\,{r^3}$$

Explanation

Volume of unit cell = Area of base $$\times$$ Height of the unit cell

The (HCP) is a hexagonal prism structure. The base of (HCP) is a hexagon. We know that the hexagon is made of the 6 equilateral triangles.

Area of equilateral triangle $$ = {{\sqrt 3 } \over 4}\,{a^2}$$

Therefore, the area of the base is given as,

Area of base $$ = 6\times{{\sqrt 3 } \over 4}\,{a^2}$$

Let's the 'h' be the height of the hexagonal unit

Therefore, we have

$${h \over a} = \sqrt {{8 \over 3}} $$

Now, substitute these values in the volumes of (HCP) we have,

Volume of HCP = Area of base $$\times$$ height of the hexagon

$$\therefore$$ Volume of HCP $$ = 6 \times {{\sqrt 3 } \over 4}\,{a^2} \times \sqrt {{8 \over 3}} a$$

Since, spheres touch each other, the edge length 'a' is equal to twice the radius of the atom. That is a = 2r

Substitute these values in the volume we get,

Volume of HCP $$ = 6 \times {{\sqrt 3 } \over 4}\,{a^2} \times \sqrt {{8 \over 3}} a$$

$$ = 6 \times {{\sqrt 3 } \over 4}\,{(2r)^2} \times {{\sqrt 8 } \over {\sqrt 2 }}2r = 24\sqrt 2 {r^3}$$

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