Chemistry Class 12 NCERT Solutions: Chapter 1 The Solid State Part 2

Q: 13. Explain how much portion of an atom located at (I) corner and (II) body-centre of a cubic unit cell is part of its neighbouring unit cell.


(I) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells.

Therefore, Equation portion of the atom is shared by one unit cell.

(II) An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e., its contribution to the unit cell is 1.

Q: 14. What is the two dimensional coordination number of a molecule in square close packed layer?


In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4.

Q: 15. A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?


Number of close-packed particles Equation

Therefore, number of octahedral voids Equation

And, number of tetrahedral voids Equation

Therefore, total number of voids Equation

Q: 16. A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy Equation of tetrahedral voids. What is the formula of the compound?


The ccp lattice is formed by the atoms of the element N.

Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.

According to the question, the atoms of element M occupy Equation of the tetrahedral voids.

Therefore, the number of atoms of M is equal to Equation of the number of atoms of N.

Therefore, ratio of the number of atoms of M to that of N is M: N Equation


Thus, the formula of the compound is Equation .

Q: 17. Which of the following lattices has the highest packing efficiency

(I) simple cubic

(II) body- centred cubic and

(III) hexagonal close- packed lattice


Hexagonal close-packed lattice has the highest packing efficiency of Equation . The packing efficiencies of simple cubic and body-centred cubic lattices are Equation and Equation respectively.

Q: 18. An element with molar mass Equation Equation forms a cubic unit cell with edge length 405 pm. If its density is Equation , what is the nature of the cubic unit cell?


It is given that density of the element, Equation

Molar mass, Equation

Edge length, Equation


It is known that, Avogadro's number, Equation

Applying the relation






This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).

Q: 19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?


When a solid is heated, vacancy defect can arise. A solid crystal is said to have vacancy defect when some of the lattice sites are vacant.

Vacancy defect leads to a decrease in the density of the solid.

Q: 20. What type of stoichiometric defect is shown by:

(I) Equation

(II) Equation


(I) Equation shows Frenkel defect.

(II) Equation shows Frenkel defect as well as Schottky defect

Q: 21.


When a cation of higher valence is added to an ionic solid as an impurity to it, the cation of higher valence replaces more than one cation of lower valence so as to keep the crystal electrically neutral. As a result, some sites become vacant. For example, when Equation is added to Equation , each Equation ion replaces two Equation . However, one Equation ion occupies the site of one Equation and the other site remains vacant. Hence, vacancies are introduced.

Q: 22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.


The colour develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and get excited.

For example, when crystals of Equation are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the chloride ions from the crystal diffuse to the surface to form Equation with the deposited Equation atoms. During this process, the Na atoms on the surface lose electrons to form Equation and the released electrons diffuse into the crystal to occupy the vacant anionic sites. These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.

Q: 23. A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?


Ann-type semiconductor conducts because of the presence of extra electrons. Therefore, a group 14 element can be converted to n-type semiconductor by doping it with a group 15 element.

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