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Published on: 28/05/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
Lithium borohydride,LiBH4,crystallize in an orthorhombic system with 4 molecules per unit cell.The unit cell dimensions are: a=6.81\(\overset { 0 }{ A } \) ,b=4.43\(\overset{0}{A}\) and c=7.17\(\overset{0}{A}\).Calculate the density of the crystal.Take atomic mass of Li=7,B=11 and H=1 a.m.u.
2.
What is the arrangement of atoms in the lattice structure of diamond and give contribution of each C atom?
3.
In a crystal,Frenkel defect is not shown by alkali metal halides but silver halides show.Why?
4.
Out of NaCl and CsCl,which is more stable and why?
5.
If the atoms of an element have the radius r,then in a primitive cubic unit cell,calculate
(A) the length of the face diagonal
(B) the length of the body diagonal
1.
0.676g cm-3
2.
Diamond is a covalent crystal in which each C-atom is sp3 hybridized.Thus, each C-atom is covalently bonded to four other C-atom tetrahedrally.These tetrahedra are linked together into a three dimensional giant molecule.
The structure of diamond is similar to that o Zns.
In diamond, we have C-atoms in place of Zn2+ and S2- ions.Thus, diamond has face-centred cubic structure in which C-atoms are present at the corners as well as face-centres and alternate tetrahedral voids.
Contribution of C-atoms at the corners = \(8\times{1\over 8}=1\)
Contribution of C-atoms at face-centres = \(6\times{1\over 2}=3\)
As four C-atoms are present on the body diagonals i.e., in the alternate voids, their contribution = 4
Total no. of C-atoms in the unit cell = 1 + 3 + 4 = 8
3.
Frenkel defect is found in silver halides due to small size of Ag+ ion.It is not found in alkali metal halides,as the alkali metal ions are too big to fit into the interstitial sites.
4.
CsCl is more stable than NaCl.This is higher the coordination number,greater are the forces of attraction between the cation and the anions in the close-packed arrangement.As CsCl has co-ordination number of 8:8 while NaCl has a coordination number of 6:6,therefore,CsCl is more stable.
5.
\(2\sqrt{2}r\)
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