Chemistry Class 12 NCERT Solutions: Chapter 1 The Solid State Part 7

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Q: 11. Silver crystallises in fcc lattice. If edge length of the cell is and density is , calculate the atomic mass of silver.

Ans:

It is given that the edge length,

Density,

As the lattice is fcc type, the number of atoms per unit cell,

We also know that,

Using the relation

Therefore, atomic mass of silver

Q: 12. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Ans:

It is given that the atoms of Q are present at the corners of the cube.

Therefore, number of atoms of Q in one unit cell

It is also given that the atoms of P are present at the body-centre.

Therefore, number of atoms of P in one unit cell = 1

This means that the ratio of the number of P atoms to the number of Q atoms,

Hence, the formula of the compound is PQ.

The coordination number of both P and Q is 8.

Q: 13. Niobium crystallises in body-centred cubic structure. If density is , calculate atomic radius of niobium using its atomic mass .

Ans:

It is given that the density of niobium,

Atomic mass,

As the lattice is bcc type, the number of atoms per unit cell,

We also know that,

Applying the relation:

So,

For body-centred cubic unit cell:

Q: 14. If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Ans:

Q 10 Ans Image of Octahedral Void

Q 10 Ans Image of Octahedral Void

Q 10 Ans Image of Octahedral Void

A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that is right-angled

Now, applying Pythagoras theorem, we can write: