## Solid State Quiz-6

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Q1. A solid is made of two elements XandZ. The atoms Z are in ccp arrangement while the atom
X occupy all the tetrahedral sites. What is the formula of the compound?
•  XZ
•
•
•
Solution
Given, A solid has two elements=X and Z
Z are in ccp arrangement and X occupy all tetrahedral sites.
Let the number of atoms of Z in ccp arrangement = 100
∴ Number of atoms of tetrahedral sites = 200
∴ Number of atoms of X=200 (∵ They occupy all tetrahedral sites)
∴Ratio of X :Z=200 :100 =2 :1
∴ The formula of compound is

Q2.The number of octahedral sites in a cubical close pack array of N sphere is :
•  N/2
•  2N
•  4N
•  N
Solution
Each sphere has one octahedral hole and two tetrahedral holes.

Q3.  Maximum ferromagnetism is found in :
•   Fe
•  Ni
•  Co
•  None of these
Solution
More is the number of unpaired electron, more is magnetic nature.

Q4. The lattice points of a crystal of hydrogen iodide are occupied by
•  HI molecules
•  H atoms and I atoms
•
•
Solution
HI molecules

Q5.In a cubic close packing of spheres in three dimensions, the co-ordination number of each sphere
is :
•  6
•  9
•  3
•  12
Solution
In hexagonal close packing and cubic close packing, the co-ordination number is 12.

Q6. The crystal system of a compound with unit cell dimensions a=0.387, b=0.387, and c=0.504 nm and Î±= Î²=90° and Î³=120° is :
•  Cubic
•  Hexagonal
• Orthorhombic
•  Rhombohedral
Solution
For hexagonal a=b ≠c and Î±=Î²=90° and Î³=120°.

Q7.The radii of ions are 95 pm and 181 pm respectively. The edge length of NaCl
unit cell is
•  276 pm
•  138 pm
•  552 pm
•  415 pm
Solution
NaCl has fcc structure. In fcc lattice
Where, a= edge length
Edge length =
=(2×95+2×181) pm
=190+362=552 pm

Q8.A solid XY has NaCl structure. If radius of X+ is 100 pm. What is the radius of Y− ion?
•  120 pm
•  136.6 to 241.6 pm
•  136.6 pm
•  241.6 pm
Solution
The =0.414 to 0.732 (due to fcc structure)
∴  =241.54 to 136.6 pm

Q9. The number of hexagonal faces that are present in a truncated octahedron is
•  2
•  4
•  6
•  8
Solution
The truncated octahedron is the 14-faced Archimedean solid, with 14 total faces : 6 squares and 8 regular hexagons. The truncated octahedron is formed by removing the six right square pyramids one from each point of a regular octahedron as :

Q10. The limiting radius ratio for tetrahedral shape is
•  0 to 0.155
•  0.255 to 0.414
•  0.155 to 0.225
• 0.414 to 0.732
Solution
For tetrahedral shape, limiting radius ratio is 0.225 − 0.414.

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