Class 12 Chemistry

Solid State

NCERT Solution (part 3)

Question – 1.11 – Silver crystallizes in fcc lattice. If edge length of the cell is 4.07 `xx` 10– 8 cm and density is 10.5 g cm– 3. Calculate the atomic mass of silver.

Answer: Given, edge length (a) `=4.07xx10^(-8)`

Density (d) = 10.5 g cm-3

For fcc number of atoms per unit cell (z) = 4

We know that Atomic Mass (M) `=(da^3N_A)/z`

Thus, M = `(10.5gcm^3xx4.07xx10^(-8)xx6.033xx10^(23)text(mol)^(-1))/4`

Or, M = `(10.5gxx67.419xx10^(-24)xx6.022xx10^(23)text(mol)^(-1))/4`

Or, M = `(10.5gxx67.419xx10^(-1)xx6.022text(mol)^(-1))/4`

Or, M `=(426.2970789)/4g\text(mol)^(-1)`

Or, `M=106.574g\text(mol)^(-1)=107 g\text(mol)^(-1)`


Question – 1.12 - 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?

Answer: Given, Atoms of Q are at the corners of the cube and P at the body-centre.

So, number of atoms Q in one unit cell `=8xx1/8=1`

Number of atoms of P in one unit cell = 1

So, ratio of P and Q atoms = P : Q = 1 : 1

So, formula of given compound = PQ

Since it is bcc

Hence, coordination number of P and Q = 9


Question – 1.13 - Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

Answer: Given , density (d) = 8.55 g cm-3

Atomic Mass (M) = 93 u = 93 g mol-1

Atomic radius (r) = ?

We know, Avogadro Number NA = 6.022 `xx` 1023mol-1

Since, given lattice is bcc

Hence, number of atoms per unit cell (z) = 2

We know that, `d=(zM)/(a^3N_A)`

Or, 8.55 g cm-3 = `(2xx93g\text(mol)^(-1))/(a^3xx6.022xx10^(23)text(mol)^(-1))`

Or, `a^3=(2xx93g)/(8.55g\cm^(-3)xx6.022xx10^(23))`

Or, `a^3=(186)/(51.4881xx10^(23)\cm^3`

Or, `a^3=3.6124xx10^(-23)cm^3`

Or, `a^3=36.124xx10^(-24)cm^3`

Or, `a=3.3057xx10^(-8)cm`

Thus atomic radius of niobium = 14.31 nm


Question – 1.14 - 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.

Answer:

class 12 solid state ncert exercise solution18

In the given figure, let an sphere having centre ‘O’ is fitted in the octahedral void.

As given, radius of the sphere fitted in the octahedral void = r

And radius of the atoms in close packing = R

Here, ∠AOD = 90°

In ΔAOD, `DA^2=OA^2+OD^2`

Or, `(R+R)^2=(R+r)^2+(R+r)^2`

Or, `4R^2=2(R+r)^2`

Or, `sqrt2R=R+r`

Or, `r=sqrt2R-R`

Or, `r=R(sqrt2-1)`

Or, `r=R(1.414-1)`

Or, `r=0.414R`

This is the required relation between r and R.




Introduction

NCERT Exercise Solution (Part 2)

NCERT Exercise Solution (Part 3)

NCERT Exercise Solution (Part 4)

NCERT Exercise Solution (Part 5)