Solid State Test - 7

1. HCl is an example of polar molecular solids which are bonded together by dipole-dipole interaction.

2. CCl₄ is an example of non polar solids which are bonded together by dispersion or London forces.

3. CaF₂ is an example of ionic solids which are bonded together by coulombic or electrostatic force of attraction.

4. Al-N is an example of covalent solids which are bonded together by covalent bonding.

Hence, the correct option is (C).

Crystalline solids are solids that have their atoms, molecules, and ions highly repetitive in nature and arranged in a specific pattern. 

Potassium Chloride, Caesium Chloride, and rhombic sulphur are crystalline solids whereas glass is an amorphous solid.

Hence, the correct option is (C).

Given that:

Isotopic mass \(=55.92066\)

Mass number \(=56\)

We know that:

Packing fraction \(=\frac{\text { Isotopic mass - Mass number }}{\text { Mass number }} \times 10^{4}\)

\(=\frac{55.92066-56}{56} \times 10^{4}\)

\(=-14.167\)

Hence, the correct option is (A).

Given:

\(a=564 \mathrm{pm} = 564 \times 10^{-10}cm\)

We know that:

The density is given by the formula:

\(\rho=\frac{z M}{a^{3} N_{A}}\)

As we know that,

Each unit cell of \(\mathrm{NaCl}\) has \(4 \mathrm{Na}^{-}\)and \(4 \mathrm{Cl}^{-}\)ions.

\(\Rightarrow z=4\)

Avogadro's number, \(N_{A}=6.022 \times 10^{23} / \mathrm{mol}\)

The total mass of \(\mathrm{NaCl}\):

\(M=22.99+34.34=58.5 \mathrm{~g} / \mathrm{mol}\) 

Therefore,

\( \text { Density}\), \(\rho=\frac{z M}{a^{3} N_{A}}\)

\(=\frac{4 \times 58.5}{\left(564 \times 10^{-10}\right)^{3} \times 6.022 \times 10^{23}} \mathrm{~g~cm}^{-3} \)

\(= 2.165 \mathrm{~g~cm}^{-3}\)

Hence, the correct option is (B).

The total number of octahedral voids per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. 

It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. 

Only \(\frac{1}{4}\)th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.

12 octahedral voids located at each edge and shared between four unit cells

\(=12\times\frac{1}{4}=3\)

Total number of octahedral voids = 4

We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids \(=\frac{4}{4}=1\).

Hence, the correct option is (A).

In sugar, the constituent particles are arranged in a specific manner. Therefore, it is a crystalline solid. 

The solids in which the constituent particles of matter are arranged and organized in a specific manner are called crystalline solids.

In coal, wax, and clay the constituent particles are arranged in a random manner. Therefore, these are an amorphous solid.

Hence, the correct option is (D).

The three dimensional arrangement of constituent particles in a crystal is represented in such a way each particle is taken as a point, the arrangement is called as crystal lattice.

Thus, a regular arrangement of the points in space is the correct definition of crystal lattice.

Hence, the correct option is (C).

All the mentioned compounds are examples of molecular solids: 

• \(I_{2}\) is an example of non polar molecular solid, 
• Solid \(\mathrm{NH}_{3}\) is an example of polar molecular solid and 
• Ice is an example of hydrogen bonded molecular solid.

Hence, the correct option is (D).

Common salt, sugar and iron are crystalline solids whereas rubber is an amorphous solid as it does not have well developed perfectly ordered crystalline structure. Other amorphous solids include tar, glass, plastic, butter etc.

Hence, the correct option is (D).

Crystalline solids are anisotropic in nature.  

• It is because the arrangement of constituent particles is regular and ordered along all the directions. 
• Therefore, the value of any physical property (electrical resistance or refractive index) would be different along each direction.
  
• That is, some of their physical index show different values when measured along different directions in the same crystal.
  

Hence, the correct option is (B).

Crystalline solid is a solid material whose constituents, such as atoms, molecules or ions are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

Calcite has its all molecules arranged in regular fashion. Structure of this solid is shown in the figure.

Graphite is a crystalline form of the element, carbon with its atoms arranged in a hexagonal structure. It occurs naturally in this form and is the most stable form of carbon under standard conditions. The structure of graphite is shown below.

Hence, the correct option is (D).

In ionic solids the vacancies are produced due to absence of cations and anions in stoichiometric proportions resulting in defect called Schottky defect.

This defect arises when cation and anion are of similar sizes.

CsCl shows Schottky defect.

Hence, the correct option is (D).

The appearance of colour in solid alkali metal halides is generally due to F - centres.

• F - centres (Farbe - centre) are generated when an electron occupies a vacancy generated by anion. This electron excites and de-excites losing photons which is of visible region in spectrum.
• Irradiations of alkali halides with gamma radiations creates anion vacancies where an electron is trapped and is responsible for the absorption of light.
• On heating the alkali halide in excess of alkali metal vapours creates vacancies in which an electron is trapped. So, there appears colour in solid alkali metal halides.

Hence, the correct option is (D).

Solids have a regular repeated molecular pattern in three dimensional space and this is due to their property of high inter-molecular forces. But in liquids and gases, the arrangement of molecular pattern is irregular.

Hence, the correct option is (C).

NaCl has a face centred cubic unit cell. Chloride ions (being larger in size) form cubic close packing. They are present at the corners and at the centre of each face of the cube. Smaller sodium ions are present in all the octahedral holes.

Hence, the correct option is (A).

In ionic solids the vacancies are produced due to absence of cations and anions in stoichiometric proportions resulting in defect called Schottky defect.

The density is less than expected due to missing spaces.

Hence, the correct option is (B).

Since \({A}\) is present at the 8 corners, number of \({A}\) ions per unit cell \(=\frac{1}{8} \times 8=1\)

Since \(B\) is present at the centre of 6 faces, number of \(B\) per unit cell \(=\frac{1}{2} \times 6=3\)

\(\therefore\) The ratio of \(A\) and \(B\) is \(1: 3\), so the formula of the compound is \(A B_{3}\).

Hence, the correct option is (C).

If \(a\) is the edge length of the cube and \(r\) is the radius of each atom. 

Distance of nearest approach between two atoms \(=2 r=1.73 \) Å.

Therefore, 

\(r=\frac{1.73}{2}=0.865 \) Å

For BCC lattice, the relation of edge length and radius of the atom is,

\(\sqrt{3} a=4 r\)

\(a=\frac{4 r}{\sqrt{3}}\)

\(a=\frac{4 \times 0.865}{\sqrt{3}}\)

\(a=1.99=2\)Å \(=200 \mathrm{pm}\)

The edge length of the BCC lattice is \(200 \mathrm{pm}\).

Hence, the correct option is (C).

Seven of the eight corners are occupied by atom \({A}=\frac{1}{8}\times 7=\frac{7}{8}\)

Faces are occupied by \(B=\frac{1}{2}\times 6=3\)

\(A: B=\frac{7}{8}: 3=7: 24\)

Thus, the general formula will be \({A}_{7} {B}_{24}\).

Hence, the correct option is (C).

Dimagnetic materials have a weak, negative susceptibility to magnetic fields. Diamagnetic substances are slightly repelled by a magnetic field and it does not retain its magnetic property when external field is removed.

Hence, the correct option is (B).

Graphite is one of the allotropes of carbon. Unlike diamond, it is an electrical conductor and a good lubricant. Graphite is a good conductor of electricity due to the presence of free valence electrons.

This property of graphite is due to the carbon atoms arranged in different layers and each atom is covalently bonded on three of its neighbouring atoms in the same layer. The fourth valence electrons of each atom are present between different layers and are free to move about. These free electrons in graphite make it a good conductor of electricity.

Hence, the correct option is (D).

Effective number of atoms of \(X, Y\) and \(Z\) are:

According to question:

For X:

Number of atoms \(=\frac{1}{8} \times 7\) (out of 8 corners, \(X\) is occupying 7)

For Y:

Number of atoms \(=\frac{1}{2} \times 6=3 (\mathrm{Y}\) is occupying all face centers)

For Z:

Number of atoms \(=\frac{1}{8}({Z}\) is occupying one corner out of 8 )

Therefore, the formula of the compound will be:

\(X_{\frac{7}{8}} Y_{3} Z_{\frac{1}{8}}\)

\(=X_{7} Y_{24} Z\)

Hence, the correct option is (B).

Let the edge length or side of the cube be 'a', and the radius of each particle be 'r'.

Then, we know that:

For simple cubic lattice:

\({a}=2 {r}\)

Volume of cube \(=a^{3}\)

\( =8 {r}^{3}\)

In simple cube, each unit cell has only one sphere,

Volume of sphere \(=\frac{4}{3} \pi {r}^{3}\)

Packing efficiency \(=\frac{\text { Volume of a sphere }}{\text { Volume of the cube }} \times 100 \%\) 

\(=\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100\)

\(=52.4 \%\)

Hence, the correct option is (A).

FeO has metal deficiency defect. Metal deficiency defect is supposed to arise when there are lesser number of positive ions than negative ions. In case of FeO, the positive ions are missing from their lattice sites. 

The additional negative charge is balanced by some nearby metal ion by acquiring one more positive charge. It happens in FeO because Fe has capacity of showing variable oxidation states.

Hence, the correct option is (B).

Given:

\(a=390~pm\)

For bcc structure:

Interionic distance \(=r^{+}+r^{-}=\frac{\sqrt{3}}{2} a\)

where, \(a:\) edge length 

\(r_{N H_{4}^{+}}^{+}+r_{Cl^{-}}^{-}=\frac{\sqrt{3}}{2} a\)

\(r_{N H_{4}^{+}}^{+}+180=\frac{\sqrt{3}}{2} \times 390\)

\(r_{N H_{4}^{+}}^{+}+180 = 338\)

\(r_{N H_{4}^{+}}^{+}=338 {~pm}-180 {~pm}\)

\(r_{N H_{4}^{+}}^{+}=158 {~pm}\)

Hence, the correct option is (B).

Given:

Distance between a Pb²⁺ ion and S^2– ion is 297 pm.

\(\Rightarrow\left(r+r^{\prime}\right)=297pm=297 \times 10^{-10} cm\)

\({NaCl}\) structure - fcc structure

\(\Rightarrow {Pb}^{2+}\) occupies \({O}^{-}\)voids at edge centers.

We know that:

\(a=2\left(r+r^{\prime}\right)\)

\( \frac{a}{2}=297 \times 10^{-10} {~cm}\)

\(a=297 \times 2 \times 10^{-10} {~cm}\)

Volume \(=a^{3}=209584584 \times 10^{-30} {~cm}^{3}\)

\(=209.58 \times 10^{-24} {~cm}^{3}\)

Hence, the correct option is (A).

We know that:

Density, \(\rho=\frac{Z \times M}{a^{3} \times N_{0}}\)

where,

\(Z:\) no. of atoms in the bcc unit cell \(=2\) 

\({M}:\) molar mass of \({CsBr}=133+80=213\)

\({a}:\) edge length of unit cell \(=436.6 {pm}\)

\(=436.6 \times 10^{-10} {~cm}\)

Therefore,

Density, \(\rho=\frac{2 \times 213}{\left(436.6 \times 10^{-10}\right)^{3} \times 6.02 \times 10^{23}}\)

\(=8.50 {~g} / {cm}^{3}\)

For a unit cell \(=\frac{8.50}{2}=4.25 {~g} / {cm}^{3}\)

Hence, the correct option is (D).

In the hexagonal closest packed (hcp) each ion has 12 neighboring ions, therefore, it has a coordination number of 12 and contains 6 atoms per unit cell.

The centre atom in layer B of HCP structure is touched with 12 other atoms of the same cell. 

Hence, the correct option is (A).

The amorphous solids are the solids which have short range order and are isotropic in nature which means that the physical properties are same along the crystal length.

Thus the option (D) is incorrect as amorphous solids are isotropic nature.

Hence, the correct option is (D).

Hexagonal close packing (hcp) and cubic close packing (ccp) have same packing efficiency. Let us take a unit cell of edge length “a”. Length of face diagonal, b can be calculated with the help of Pythagoras theorem,

\(b^{2}=a^{2}+a^{2}\)

\(\Rightarrow b^{2}=2 a^{2}\)

\(\Rightarrow b=\sqrt{2} a\)

From the figure, radius of the sphere, r 

= \(\frac{1}{4}\) × length of face diagonal, b

\(\mathrm{r}=\frac{1}{4} b=\frac{\sqrt{2}}{4} a\)

\(\Rightarrow \mathrm{a}=2 \sqrt{2} r\)

In ccp structures, each unit cell has four atoms,

Packing efficiency \(=\frac{\text { volume occupied by four spheres in unit cell }}{\text { Total volume of unitcell }} \times 100\)

\(=\frac{4 \times\left(\frac{4}{3}\right) \pi r^{3} \times 100}{(2 \sqrt{2} r)^{3}}\)

\(=74 \%\)

Hence, the correct option is (C).