The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry..

**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?

Solution

Given, A solid has two elements=X and Z

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

**Q2.**The number of octahedral sites in a cubical close pack array of N sphere is :

Solution

Each sphere has one octahedral hole and two tetrahedral holes.

Each sphere has one octahedral hole and two tetrahedral holes.

**Q3.**Maximum ferromagnetism is found in :

Solution

More is the number of unpaired electron, more is magnetic nature.

More is the number of unpaired electron, more is magnetic nature.

**Q4.**The lattice points of a crystal of hydrogen iodide are occupied by

Solution

HI molecules

HI molecules

**Q5.**In a cubic close packing of spheres in three dimensions, the co-ordination number of each sphere

is :

Solution

In hexagonal close packing and cubic close packing, the co-ordination number is 12.

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 :

Solution

For hexagonal a=b ≠c and Î±=Î²=90° and Î³=120°.

For hexagonal a=b ≠c and Î±=Î²=90° and Î³=120°.

unit cell is

=(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?

**Q9.**The number of hexagonal faces that are present in a truncated octahedron is

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 :

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

Solution

For tetrahedral shape, limiting radius ratio is 0.225 − 0.414.

For tetrahedral shape, limiting radius ratio is 0.225 − 0.414.