IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..

**Q1.**Na and Mg crystallize in bcc- and fcc-type crystals, respectively, then the number of atoms of Na and Mgpresent in the unit cell of their respective crystal is

Solution

(d) Number of unit cell for bcc is 2 and fcc is 4

(d) Number of unit cell for bcc is 2 and fcc is 4

**Q2.**Potassium crystallizes with a :

Solution

(b) All factual statements

(b) All factual statements

**Q3.**The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have co-ordination number of 8. The crystal class is

Solution

b) CN of (8:8) of Li⨁and Ag⨁ suggest bcc structure

b) CN of (8:8) of Li⨁and Ag⨁ suggest bcc structure

**Q4.**The atomic fraction (d) of tin in bronze (fcc) with a density of 7717 kg m-3 and a lattice parameter of 3.903 Å is (Aw Cu=63.54, Sn=118.7, 1 amu=1.66×10^-27kg)

Solution

(b) ρ=Number of atom of each kind (Mw of each kind)×1.66×10-27kg a3 7717 kg m-3=Number of Sn atoms×118.7×1.66×10^-27) +(Number of Cu atoms)×(63.54×1.66×10^-27) 3.903×10-103m3 276.4=nSn118.7+nCu(63.54) 4.35=1.86nSn+nCu nCu=4⇒nSn=0.188 Atomic fraction =nSn/nSn+nCu=0.05

(b) ρ=Number of atom of each kind (Mw of each kind)×1.66×10-27kg a3 7717 kg m-3=Number of Sn atoms×118.7×1.66×10^-27) +(Number of Cu atoms)×(63.54×1.66×10^-27) 3.903×10-103m3 276.4=nSn118.7+nCu(63.54) 4.35=1.86nSn+nCu nCu=4⇒nSn=0.188 Atomic fraction =nSn/nSn+nCu=0.05

**Q5.**The ratio of packing density in fcc, bcc, and cubic structure is, respectively ,

Solution

(a) Packing fraction ion fcc, bcc, and sc are 0.74, 0.68 and 0.52, respectively ∴Ratio=0.74/0.74:0.68/0.74:0.52/0.74 =1:0.92:0.70

(a) Packing fraction ion fcc, bcc, and sc are 0.74, 0.68 and 0.52, respectively ∴Ratio=0.74/0.74:0.68/0.74:0.52/0.74 =1:0.92:0.70

**Q6.**A metallic crystal crystallizes into a lattice containing a sequence of layers ABABAB... …any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space?

Solution

(b) ABAB type of packing means ccp packing in which 74% space is occupied and 26% is empty

(b) ABAB type of packing means ccp packing in which 74% space is occupied and 26% is empty

**Q7.**In NaCl, the chloride ions occupy the space in a fashion of

Solution

(a) Factual statement

(a) Factual statement

**Q8.**If the lattice parameter of Si = 5.43 Å and the mass of Siatom is 28.08 × 1.66 ×10^-27 kg, the density of silicon in kg m-3 is (Given: Silicon has diamond cubic structure)

Solution

(a) Sihas fcc structure (Zeff=4 and Si is also present in alternate TVs (=4). So the total number of atom =4+4=8/unit cell Note: Si has diamond cubic structure, refer section 1.17 ∴ =Zeff×Aw/NAxa^3 Mass of Si atom=AwNA =8×Mass of Si atom /a^3 =8×28.08×1.66×10-27kg / 5.43×10-103m3 [1 Å=10-10m] =2330 kg m-3

(a) Sihas fcc structure (Zeff=4 and Si is also present in alternate TVs (=4). So the total number of atom =4+4=8/unit cell Note: Si has diamond cubic structure, refer section 1.17 ∴ =Zeff×Aw/NAxa^3 Mass of Si atom=AwNA =8×Mass of Si atom /a^3 =8×28.08×1.66×10-27kg / 5.43×10-103m3 [1 Å=10-10m] =2330 kg m-3

**Q9.**Due to Frenkel defect, the density of the ionic solids

Solution

(c) Due to Frenkel defects, density does not change

(c) Due to Frenkel defects, density does not change

**Q10.**In the structure of diamond, carbon atoms appears at