Bihar Board - Class 12 Chemistry - Chapter 1: The Solid State Long Answer Question
Long Question Answer
Q. 1. How can you determine the atomic mass of an unknown metal if you know its density and The dimension of its unit cell ? Explain.
Answer ⇒ Let ‘a’ be the edge length of a cubic unit cell and it contains ‘Z’ atoms of a substance of atomic mass ‘M’.
Mass of unit cell = No. of atoms in unit cell x mass of each atom= Z M where M is mass of one atom
=MNA=ZMNA ( NA is Avogrado’s number)
Volume of unit cell =a
Density of unit cell =Mass of unit cellVolume of unit cell
= ZMNA 1a3=ZMa3 NA
Density d=ZMa3 NA g/cm3
From the above expression for density, d we can calculate the atomic mass, M of metal as other parameters are known.
Q. 2. Classify each of the following as being either a p-type or an n-type semiconductor :
(i) Ge doped with In
(ii) B doped with Si.
Answer ⇒ (i) Ge is an element of 14th group (like Si) and has configuration 4s2-4p2. It has been doped with In, a 13th group element having 5s2 5p1 configuration i.e., element of the 14th group has been doped with the 13th group element. All three valence electrons of impurity atom (In) get bonded with three out of four electrons of Ge and one electron of Ge remains unbonded. Conductivity is due to the unbonded electron of the insulator, Ge. Therefore, it is a p-type semiconductor.
(ii) Boron, B is an element of the 13th group and has a 2s22pl configuration. It is doped with Si, an element of the 14th group having 352 3p2 configuration. All three electrons of boron gets bonded with 3 out of 4 electrons of Si and 4th electron of impurity atom (i.e. Si) is responsible for conductivity. Thus it is a n-type semiconductor
Q. 3. How will you distinguish between the following pairs of terms:
(i) Hexagonal close packing and cubic close packing.
(ii) Crystal lattice and unit cell.
(iii) Tetrahedral void and octahedral void
Answer ⇒ (i) Hexagonal close packing follows AB AB AB type arrangement i.e., third layer duplicated first one and fourth layer duplicates second one. In cubic close packing ABC ABC …. type arrangement is present i.e., third layer lies in the depression of the second layer that does not directly over the atom of the first layer.
(ii) Regular three dimensional arrangement points in space is known as space lattice when these points are replaced with actual atoms or ions is known as crystal lattice. On the other hand, the smaller portion of the space lattice which can generate the complete lattice by repeating its own dimensions in various directions is known as a unit cel
(a) Octahedral void (b) Tetrahedral void
(iii) The void created by six spheres in contact is called an octahedral void Fig. (a) and the void created by the four spheres in contact is called a tetrahedral void Fig. (b).
Q. 4. Explain the following with suitable examples :
(a) Paramagnetism
(b) Diamagnetism
(c) Ferromagnetism
(d) Antiferromagnetism
(e) Ferrimagnetism
(f) Piezoelectric effect.
Answer ⇒ (a) Paramagnetism : The substances which are weakly attracted by a magnetic field are called paramagnetic substances and the property of the substances is called paramagnetism. Paramagnetism is due to the presence of one or more unpaired electrons which are attracted by magnetic fields.
Example : O2 , Cu2+ ,Fe3+ ,Cr3+ etc.
(b) Diamagnetism : Diamagnetic substances are weakly repelled by a magnetic field. H2O , NaCl and C6H6 are some examples of such substances. They are weakly magnetized in a magnetic field in the opposite direction. Diamagnetism is shown by those substances in which all the electrons are paired and there are no unpaired electrons. Pairing of electrons cancels their magnetic moments and they lose their magnetic character.
(c) Ferromagnetism : A few substances like iron,cobalt, nickel, gadolinium’ and CrO2 are attracted very strongly by a magnetic field. Such substances are called ferromagnetic substances. Besides strong attractions, these substances can be permanently magnetized. In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet.
Example :Iron, Cobalt, Nickel etc.
(d) Antiferromagnetism : Substances like MnO Anti ferromagnetism has a domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each other’s magnetic moment.
(e) Ferrimagnetism : This magnetic property is caused by particles on inter-penetratiaties lattices with unequal number of electrons with parallel and antiparallel spins so that there dipole moment. FeOFe2O3= (Fe3O4) is an example of ferrimagnetic materials.
(f) Piezoelectric effect (or pressure electricity) : Insulators do not conduct electricity because the electrons present in them are held tightly to the individual atoms or ions and do not move. However, when an electric field is applied polarization takes place and newly formed dipoles may align themselves in an ordered manner so that such crystals have a net moment.
When mechanical stress is applied on polar crystals so as to deform them, electricity is produced due to displacement of ions. This is known as piezoelectric effect and electric so this product is known as Piezo electricity or pressure electricity.
Examples are : titanium, barium and led, lead zirconate (PbZrO3) , ammonium dihydrogen phosphate and quartz.
Q. 5. Calculate the atomic radius in Sc, bcc and fcc crystal.
Answer ⇒ Atomic radius (r) in sc :
Let edge length =a
From geometry, r+ r=a or, 2r= a
r=a2
Atomic radius (r) in sc :
From geometry,
In BDH
(BH)2=(BD)2+(HD)2
(BH)2=(BD)2+a2
(4 r)2=(BD)2+a2
16 r2=(BD)2+a2 …………..(i)
Now from BAD
(BD)2=(AD)2+(AB)2
(BD)2=a2+a2=2a2 …….(ii)
Putting the value of (BD)2 in eq. (i)
16 r2=2 a2+a2
r2=3 a216
r =34a
Atomic radius in fcc :
From geometry
In ABC
(AC)2=(AB)2+(BC)2
(4r)2=a2+a2 or , 16 r2 =2 a2
r2 =2 a216
r =24a
Q. 6. Explain : (a) The basis of similarities and differences between metallic and ionic crystals
(b) Ionic solids are hard and brittle.
Answer ⇒ (a) Basis of Similarities : (i) Both ionic and metallic crystals have electrostatic forces of attraction In ionic crystals these are between the oppositely charged ions. In metals, these are among the valence electrons which are mobile or delocalised and the fixed kernels which are positively charged. That is why both have high melting points.
(ii) In both cases the bond is non-directional.
Basis of differences : (i) In ionic crystals, the ions are not free to move. Hence they cannot conduct electricity in the solid state. They can do so only in the molten or aqueous state when they become free to move. In metals, the valence electrons being MOBILE/ DELOCALIZED are free to move/flow. Hence they can conduct electricity in the solid state. Ions carry electricity in ionic crystals when present in the molten or aqueous state and mobile electrons carry electricity in metallic crystals.
(ii) Ionic bond is a strong bond, whereas metallic bond is a weak bond.
(b) Ionic crystals are hard, because there are strong electrostatic froces of attraction among the positively charged ions. They are brittle because ionic bonds are non-directional.
Q. 7. Classify the following solids in different categories based on the nature of intermolecular forces operating is them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Ans ⇒ Potassium sulphate : Ionic solid; coulombic forces of attraction.
Tin : Metallic solid; metallic bonding
Benzene (solid) : Molecular solid; dispersion or London forces.
Urea : Molecular solid, dipole-dipole attractions.
Ammonia : Molecular solid (polar) Hydrogen bonding.
Water (ice) : Molecular solid; hydrogen-bonding.
Zinc sulphide : Ionic solid, coulombic forces of attraction.
Graphite : Net-work or covalent, covalent bonds.
Rubidium: Metallic solid, metallic bonds (attraction between positive metal ions called kernels and mobile electrons)
Argon : Molecular solid, van der Waals forces.
Silicon Carbide (Sic) : Network or covalent solid. covalent bonds.
Q. 8. Analysis shows that nickel oxide has formula Ni0.98 O1.00 What fractions of Nickel exist as Ni2+ and Ni3+ ions ?
Answer ⇒ Ni0.98 O1.00 is a non-stoichiometric compound and is a mixture of Ni2+ and Ni3+ ions. Let x atoms of Ni3+ are present in the compound. This means that a Ni2+have been replaced by Ni3+ ions.
Therefore, no. of Ni2+ ions =(0.98 -x) .
For electrical neutrality, positive charge on compound = negative charge on compound.
Therefore, 2 (0.98 -x)+3x=2
1.96-2x+3x=2
or, x=2-1.96 =0.4
Hence, % of Ni ions =0.040.98100=4.08 %
Thus, Ni2+ is 96 % and Ni3+ is 4 % in given sample of nickel oxide having formula, Ni0.98 O1.00 .
Q 9. Which substance would be a better choice to make a permanent magnet, ferromagnetic or ferrimagnetic. Explain your answer.
Answer ⇒ Substances like iron, cobalt, nickel, gadolinium and CrO2, are known as ferromagnetic substances and make better permanent magnets as they are attracted very strongly by the magnetic field. These substances can also be permanently magnetized. The metal ions of ferromagnetic substances, in solid-state, are grouped into small regions, which are known as domains. Each domain acts as a tiny magnet. Meanwhile, in a magnetized piece of a ferromagnetic substance, the domains are randomly oriented and so the magnetic moments of the domains get canceled. Nevertheless, when the substance is placed in a magnetic field, all the domains are oriented in the direction of the magnetic field and produce a strong magnetic effect. The ordering of the domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.
Q 10. What is semiconductor ? Describe the two main types of semiconductor and contrast their conduction mechanism.
Answer ⇒ Substances whose conductance lies between that of metals (conductors and insulators are called semiconductors.
Two main types of semiconductors are n-type and p-type.
n-type semiconductors : Group 14 elements like Ge, Si doped with group 15 elements such as P and As are called n-type semiconductors. The symbol ‘n’ indicates that negative charge flows in them. The group 15 elements have one excess valence electron (compared to Si or Ge) after forming the four covalent bonds normally formed by group 14 elements. The excess electrons give rise to electrical conduction.
p-type semiconductors : A group 13 element which has only three valence electrons forms an electron deficient bond or a hole. Such holes can move across the crystal giving rise to electrical conductivity. Thus impurity doped Si and Ge are semiconductors whose conductivity increases with increasing temperature unlike metals whose conductivity decreases with increase in temperature. Group 14 elements doped with group 13 elements are called p-type seni-conductors. The symbol ‘p’ is used because in an electric field, the holes move through the crystal like positive charge, i.e., in a direction opposite to the flow of electrons.
Q 11. ‘Stability of a crystal is reflected in the magnitude of its melting points’.Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules ?
Answer ⇒ Melting point of solid water =273 K
M.Pt. of ethyl alcohol =155.7 K
M.Pt, of diethyl ether =156.8 K
M.Pt. of methane =90.5 K
Higher the melting point of a crystal, greater are the forces holding the constituent particles (ions/atoms) together and hence greater is its stability.
The intermolecular forces in solid water (ice) and ethyl alcohol are mainly due to hydrogen bonding. Higher melting point of ice than ethyl alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in solid water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in diethyl ether actually are dipole-dipole attractions. Methane, on the other hand, is a nonpolar molecule. The only forces present in them are the weak van der Waals forces and so the melting point of methane is least out of all of them and is a gas at room temperature.
Q 12. Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer ⇒ Potassium sulphate → Ionic solid
Benzene → Molecular (non-polar) solidTin → Metallic solid
Urea → Polar molecular solid
Water → Hydrogen bonded molecular solid
Ammonia → Polar molecular solid
Zinc sulphide → Ionic solid
Rubidium → Metallic solid
Graphite → Covalent or network solid
Silicon carbide → Covalent or network solid
Argon → Non-polar molecular solid
Q 13 . Explain using an appropriate example of how ionic solids, with anionic vacancies caused due to the metal excess defect, start developing a color.
Answer ⇒ Let us take an example of NaCl to explain this when NaCl crystals are heated in a sodium vapor atmosphere, sodium atoms are deposited on the crystal’s surface. This causes the Cl- ions to leave their lattice sites to form NaCl /with the deposited sodium atoms. In this process, the sodium atoms on the surface lose their electrons to form Na+ ions and the released electrons move into the crystal to fill in the vacant anionic sites. These electrons absorb energy from the incoming visible light and get excited to a higher energy level. Thereby, imparting a yellow color to the crystals.
Q 14. Distinguish between :
(i) Hexagonal and monoclinic unit cells.
(ii) Face-centered and end-centered unit cells.
Answer ⇒ (i) Difference between Hexagonal and monoclinic unit cells are as follows :
| Hexagonal Unit cell | Monoclinic unit cell |
(1) Axial distance or edge lengths | a=bc | abc |
(2) Axial angles | ==90 | ==90, 90 |
(3) Examples | =120 Graphite, ZnO , CdS | Monoclinic Sulphur, PbCrO2, Na2SO4 .10H2O |
(ii) Difference between Face-centered and end centered unit cells are as follows :
Face centered unit cell | End centered unit cell |
It contains one particle present at the center of its each face, besides the one that is at its corners. | In this unit cell, one Constituent particle is present at the center of any two opposite faces besides the one present at its corners. |
Q 15. (a) What is meant by the term 'coordination number’ ?
(b) What is the coordination number of atoms :
(i) in a cubic close packed structure,
(ii) in a body centered cubic structure.
Answer ⇒ (a) Co-ordina- tion number : The total number of nearest
Immediate neighbor atoms of a tion number : The total number of nearest neighbor atoms of a particular atom in a crystal lattice is known as its coordination
Example : In hexagonal close-packing (hcp) arrange-ment, each particle is surrounded by 12 equidistant spheres, i.e., 6 spheres in contact with in the same plane and 3 each in adjacent layers. One Representation of coordination immediately above and the number 12 in hcp packing another immediately below. This coordination number in help packing is 12.
(b) (i) 6 : 6, (ii) 8 : 8.
Q 16. Define the term amorphous and crystalline solids with examples.
Answer ⇒Amorphous Solids : A solid is said to be amorphous if the constituent particles are not arranged in any regular manner. They may have only short range orders. Some examples of amorphous solids are glass, plastics, rubber etc.
Crystalline Solids : The substances whose constituents are arranged in a definite orderly arrangement are called crystalline solids. For example, NaCl, S, diamond, sugar etc. The crystalline substances have sharp melting points and have physical properties different in different directions, i.ecrystalline substances are Anisotropic. They have long range and short range order.
Q 17. Name the parameters that characterize a unit cell.
Answer ⇒A unit cell is characterized by the following parameters :
(i) Its dimensions along the three edges a, band as shown. These edges may or may not be mutually perpendicular.
. Illustration of parameters of a unit cell
(ii) Angles between the edges (between b and c ), (between a and c) and (between a and b ).
Thus a unit cell is characterized by six parameters a, b, c and , , .
Q 18. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Answer ⇒ If molten NaCl containing a little amount of SrCl2 is crystallized, some of the sites of Na+ ions are occupied by Sr2+ . Each Sr2+ replace two sodium ions to maintain electrical neutrality. It occupies the site of one Nation and the other site remains vacant. The cationic valances thus produced are equal in number to that of Sr2+ atoms.
Introduction of cation vacancy in Na+ Cl- by substitution of Na+ by Sr2+
Q. 19 . Ionic solids do not conduct electricity in the solid state but only in the molten state. Why?
Answer ⇒ The ions of ionic solids are responsible for conducting electricity. In the solid state, however, these ions are not free to move around inside the solid because of the strong electrostatic forces. Hence, ionic solids do not conduct electricity in a solid state. Whereas, in a molten state, the ions are free to move and thus they can conduct electricity.
Q. 20. Find the number of voids in 0.2 moles of a compound forming hexagonal close-packed structure. What number of these are tetrahedral voids?
Answer ⇒ Given:
No. of close-packed particles =0.26.022 1023 =1.2044 1023
Thus, no. of octahedral voids =1.2044 1023
And, no. of tetrahedral voids =2 1.2044 1023 =2.4088 1023
Thus, total number of voids =1.2044 1023 +2.4088 1023 =3.6132 1023
Q.21 . Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(a) Tetra phosphorus decaoxide (P4O10)
(b) Ammonium phosphate (NH4)3PO4
(c) SiC
(d) I2
(e) P4
(f) Plastic
(g) Graphite
(h) Brass
(i) Rb
(j) LiBr
(k) Si
Answer ⇒
Metallic : ( h ) Brass, ( i ) Rb
Molecular : (a) Tetra phosphorus decaoxide (P4O10), (d) I2, (e) P4
Ionic : ( b ) Ammonium phosphate (NH4)3PO4 , ( j ) LiBr
Amorphous : ( f ) Plastic
Covalent : ( c ) SiC, ( g ) Graphite, ( k ) Si
Q.22 . List four distinctions between crystalline and amorphous solids with one example of each.
Answer ⇒ The difference between crystalline and amorphous solids are:
Crystalline solids | Amorphous solids |
They have a symmetric arrangement. | They have an unsymmetric arrangement. |
Melting points of these are sharp | A wide range of melting points is observed |
They are anisotropic | They are isotropic |
They go through a clean cleavage procedure. | They do not go through a clean cleavage procedure. |
Example- Quartz | Example- Glass |
Q. 23 . Explain: (a) List two differences between metallic and ionic crystals.
Answer ⇒(a) The differences between metallic and ionic crystals are:
Ionic solids | Metallic solids |
They have coulombic or electrostatic forces between the constituent particles. | They have metallic bonds between the constituent particles. |
In solid form they are insulators but in the molten form, they are conductors. | They are conductors in both solid as well as molten form. |
(b) Sodium chloride is hard but sodium metal is soft.
Ans: Due to the fact that sodium has only one valence electron, there is a weak metallic link. As a result, sodium metal is a brittle substance. Because the chlorine atom has received an electron from the sodium atom, sodium chloride exists in crystal form.
Q.24 . Explain how many portions of an atom is located at
(a) corner
(b) body center
(c) face-centered
(d) edge center of a cubic unit cell.
Answer ⇒ (a) In the corner of a cubic unit cell, eight neighboring unit cells share the same atom. One unit cell, therefore, shares 18 th of the atom.
(b) Each unit cell in a cubic unit cell has its own atom in the body center, which is not shared with its neighbor unit cell. Due to its presence in a single unit cell, an atom's contribution to the unit cell is one.
(c) On the face of a cubic unit cell, six neighboring unit cells share the same atom. One unit cell, therefore, shares 12 th of the atom.
(d) A cubic unit cell's edge is always shared by four unit cells at a time. Each unit cell will get 14 th of an atom located at the edge center.
Q .25 . An element crystallizes in a cubic close packed structure having a fcc unit cell of an edge 200 pm . Calculate the density if 200 g of this element contain 24 1023 atoms.
Answer ⇒ We know the formula of density is:
d= Z Ma3 N0
We are given that the unit cell is FCC, and for FCC the number of atoms is 4 (Z=4).
We are given the edge length as 200 pm . This can be written as:
a= 200 10-8 cm
NO=24 1023
M=200 g
Putting the values, we get
d = 4 2008 10-24 24 1023
d= 41.67 g/cm3
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