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NEET-XII-Chemistry
01: The Solid State
- Qstn #12-iiFace-centred and end-centred unit cells.Ans : Face-centred unit cell
In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face.
End-centred unit cell
An end-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.
- Qstn #13Explain 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.
Ans : (i) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells.
Therefore,
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.
- Qstn #14What is the two dimensional coordination number of a molecule in square close packed layer?
Ans : 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.
- Qstn #15A 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?
Ans : Number of close-packed particles = 0.5 × 6.022 × 1023 = 3.011 × 1023
Therefore, number of octahedral voids = 3.011 × 1023
And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023
Therefore, total number of voids = 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023
- Qstn #16A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
Ans : 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
of the tetrahedral voids.
Therefore, the number of atoms of M is equal to
of the number
of atoms of N.
Therefore, ratio of the number of atoms of M to that of N is M: N_20-11-08_Utpal_12_Chemistry_1_24_html_m3affbdf1.gif)
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Thus, the formula of the compound is M2 N3.
- Qstn #17Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?
Ans : Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of simple cubic and body-centred cubic lattices are 52.4% and 68% respectively.
- Qstn #18An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m-3, what is the nature of the cubic unit cell?
Ans : It is given that density of the element, d = 2.7 × 103 kg m-3
Molar mass, M = 2.7 × 10-2 kg mol-1
Edge length, a = 405 pm = 405 × 10-12 m
= 4.05 × 10-10 m
It is known that, Avogadro’s number, NA = 6.022 × 1023 mol-1
Applying the relation,
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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).
- Qstn #19What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Ans : 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.
- Qstn #21Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ans : 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 Sr2+ is added to NaCl, each Sr2+ ion replaces two Na+ ions. However, one Sr2+ ion occupies the site of one Na+ ion and the other site remains vacant. Hence, vacancies are introduced.
- Qstn #22Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Ans : 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 NaCl 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 NaCl with the deposited Na atoms. During this process, the Na atoms on the surface lose electrons to form Na+ ions 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.
- Qstn #23A 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?
Ans : An n-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.
- Qstn #24What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ans : Ferromagnetic substances would make better permanent magnets.
In solid state, the metal ions of ferromagnetic substances are grouped together into small regions. These regions are called domains and each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance, the domains are randomly oriented. As a result, the magnetic moments of the domains get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced.
The ordering of the domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.