What is the formula of the compound in which element N forms CCP lattice and atoms M occupy 1/3rd of tetrahedral voids?

1. Solid State:

(i) Solid is the form of matter which possesses a definite shape and a definite volume.

(ii) If intermolecular forces > thermal energy, substance exists as solid

(iii) Classification of Solid: Crystalline (Regular arrangement of particles, anisotropic), Amorphous (No regular arrangement of particles, isotropic).

(iv) Classification of crystalline solids: Ionic, Molecular, Covalent/Network, Metallic

2. Crystal systems (types and characteristics):

(i) Types: 

(a) Cubic (a=b=c, α=β=γ=90°)

(b) Tetragonal (a=b≠c, α=β=γ= 90°)

(c) Orthorhombic/Rhombic (a≠b≠c, α=β=γ=90°)

(d) Monoclinic (a≠b≠c, α=γ=90°≠β)

(e) Triclinic (a≠b≠c, α≠β≠γ≠90°)

(f) Rhombohedral/Trigonal (a=b=c, α=β=γ≠90°)

(g) Hexagonal (a=b≠c, α=β=90°, γ=120°).

(ii) Contribution by particles present at different positions

Corner =18, Face-centre =12, Body-centre =1, Edge-centre =14

(iii) Number of particles per unit cell of a cubic crystal

Simple =8×18=1, Body-centred =8×18+1=2, Face-centre =8×18+6×12=4

(iv) If the number of close packed spheres be N, then:

The number of octahedral voids generated =N and the number of tetrahedral voids generated =2N

(v) Relationship between radius (r) of an atom and edge length (a)

Simple cubic r=a2, Face centred cubic r=a22, Body centred cubic r=34a

Density of unit cell, d=Z×Ma3×NAg cm-3

Where Z=No. of atoms in unit cell; M=Atomic mass; a=Edge length (in cm)

NA=Avogadro number (6.022×1023)

(vi) Packing efficiency:

ccp and hcp: Packing efficiency =Volume occupied by 4 sphere sin the unit cellTotal volume of the unit cell×100%=74%

(vii) BCC:

Packing efficiency =Volume occupied by 2 sphere sin the unit cellTotal volume of the unit cell×100%= 68% 

(viii) Simple cubic:

Packing efficiency =Volume of one atomVolume of the cubic unit cell×100%=52.4%

3. Imperfections/defects in solids:

(i) Any departure from perfectly ordered arrangement of constituent particles is called defect or imperfection.

(ii) Stoichiometric defects: When ratio between cations and anions remains the same. Two types are Schottky defect and Frenkel defect.

(iii) Non-stoichiometric defects: When ratio of cations and anions changes as a result of the defect. Metal excess and Metal deficiency are the two types of this defect.

(iv) Impurity defects. Adding impurities to crystalline solids to change their properties is called doping.

5.

Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:

Ag+ (aq) + e-  →  Ag(s)          E° = +0.80 V

 H+ (aq) + e-   → 1/2 H2 (g)     E° = 0.00 V

 On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?

We have given:

 Ag+ (aq) + e- → Ag(s)          E° = +0.80 V

 H+ (aq) + e-→ 1/2 H2 (g)      E° = 0.00 V

The relationship between the standard free energy change and emf of a cell reaction is given by 
∆ G = – nFE(cell)

Thus, the more positive the standard reduction potential of a reaction, the more negative is the standard free energy change associated with the process and, consequently, the higher is the feasibility of the reaction.
Since E 0 Ag+/Ag has a greater positive value than E0 H+ /H, the reaction which is feasible at the cathode is given by

 Ag+ (aq) + e- → Ag(s)         

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Text Solution

Answer : `M_(2) N_(3)`

Solution : ccp packing is close packing <br> Let the number of `OV = 1//atom` <br> `:.` Number of `TV = 2//atom` <br> Number of `M` atom `= (1)/(3) xx 2 = (2)/(3)` <br> Thus formula is `M_((2)/(3)) N_(1)`. <br> Simplifying, formula is `M_(2) N_(3)`.