SOLID-STATE-QUIZ-6

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 α= β=90° and γ=120° is :

•  Cubic
•  Hexagonal
• Orthorhombic
•  Rhombohedral

Solution
For hexagonal a=b ≠c and α=β=90° and γ=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.