ORGANIC NAMED REACTIONS

Solid State ( MLL)

UNIT-1-SOLID STATE Minimum Level Learning · Solids are of two types: (1) Amorphous and Crystalline solids

CRYSTALLINE SOLIDS  1) Have a well-defined geometrical shape 2) Have long range order

3) Have sharp melting point 

 4) Anisotropic in nature 

5)True solids

 6) Show cleavage property 

 AMORPHOUS SOLIDS 1 Do not have a well-defined geometrical shape                                              2) Have short range order 

3 )Donot have sharp melting point

4)Isotropic in nature 

5) Pseudo solids or super cooled liquids

6)Do not show cleavage property ·

ANISOTROPY: It is the property due to which crystals show different values of their physical properties like refractive index when measured along different directions in the same crystal. It arises from different arrangement of particles in different directions. · 

ISOTROPY: is the property due to which amorphous substances show same values of their physical properties like refractive index in all directions in the same substance. It arises due to irregular arrangement of particles in all the directions. · 

CRYSTAL LATTICE: A regular three dimensional arrangement of points (constituting particles) in space is called a crystal lattice. · CHARACTERISTICS OF A CRYSTAL LATTICE: (i) Each point in a lattice is called lattice point or lattice site. (ii) Each point in a lattice represents one constituent particle. (iii) Lattice points are joined by straight lines to bring out the geometry of the lattice. · 

UNIT CELL: The smallest portion of a crystal lattice which when repeated in different directions, generates the whole crystal lattice. · CHARACTRISTIC FEATURES OF A UNIT CELL: (i) Its dimensions along the three edges, a,b and c. (ii) Angles between the edges, α, β and γ. ·

TYPE OF UNIT CELLS: These are of two types: (i) Primitive Unit Cell: In such cells, constituting particles are present only on the corner positions of unit cell. These are also known as simple cubic unit cell. 

(ii) Centred Unit Cell: In such cells, constituting particles are present at the corner positions as well as at positions other than corners of unit cell. These are also known as Non-primitive unit cell. Centred unit cells are of three types: 

(i) Body-Centred Unit Cells (BCC): In such a cell one constituting particle is present at each corner and one at the centre of the body. 

(ii) Face-Centred Unit Cells (FCC): In such a cell one constituting particle is present at each corner and one at the centre of each face. 

(iii) End-Centred Unit Cells: In such a cell one constituting particle is present at each corner and one at the centre of any two opposite faces. · NUMBER OF ATOMS IN A UNIT CELL: 

(i) Primitive Unit Cell: Each cubic unit cell contains 8 atoms on its corners and each corner is shared by 8 unit cells, so contribution of each corner to its unit cell = 1/8. Total no. of atoms in a primitive unit cell = 8 × 1/8 = 1 

(ii) Body-Centred Unit Cells (BCC): Contribution of points at corners = 8 × 1/8 = 1 Contribution of point at body-centre = 1 (as atom at the body-centre wholly belongs to the unit cell in which it is present) So total no. of atoms per unit cell = 1 + 1 = 2 (iii) Face-Centred Unit Cells (FCC): Contribution of points at corners = 8 × 1/8 = 1 Contribution of 6 points at each face = 6 × ½ = 3 (as each face is shared by two unit cells) t corners = of atoms per unit cell = 1 + 3 = 4 · 

COORDINATION NUMBER: The number of nearest neighbors of a particle is known as its coordination number. (i) In one dimensional closed pack arrangement, coordination number is 1. (ii) In square close packing in two dimensions, coordination number is 4. (iii) In hexagonal close packing in two dimensions, coordination number is 6. (iv) In hexagonal close packing and cubic close packing (fcc) in three dimensions, coordination number is 12.

TETRAHEDRAL VOID

A void which is surrounded by 4 spheres which lie at the vertices of a regular tetrahedron.

OCTAHEDRAL VOID

A void which is surrounded by 6 spheres which lie at the vertices of a regular octahedron.

NOTE: The number of these two types of voids depend upon the number of close packed spheres. Let the number of close packed spheres = N The no. of octahedral voids = N The no. of tetrahedral voids = 2N · 

PACKING EFFICIENCY: It is the percentage of the total space filled by the particles. 

(i) Packing efficieny in simple cubic lattice: Let the side of the cube (edge length) = a

 Let the radius of the sphere (constituting particles are assumed to be spherical in shape) = r 

a = 2r 

Volume of the cubic unit cell = a3 = (2r)3 = 8r3

 No. of atoms per unit cell in simple cubic = 1 Percentage of packing efficiency = volume of one atom × 100 % 

 Volume of cubic unit cell = (4/3)πr3 × 100 8r3 = 52.4% · 

IMPERFECTIONS(DEFECTS) IN SOLIDS: These are the irregularities in the arrangement of the constituting particles. These are also known as crystal defects. These defects are of two types:

 (i) POINT DEFECT: these are the deviations (irregularity) from ideal arrangement around a point(atom) in a crystalline substance.

 (ii) LINE DEFECT: these are the deviations (irregularity) from ideal arrangement in entire rows of lattice points in a crystalline substance. · TYPES OF POINT DEFECTS: These are of three types : (i) Stoichiometric defects (ii) Non-stoichiometric defects (iii) Impurity defects STOICHIOMETRIC DEFECT 1) Do not disturb the stoichiometry (composition) of the solid.

2) Also known as intrinsic or thermodynamic defects.

It is of two types : (i) vacancy defect (ii) interstitial defect

NON-STOICHIOMETRIC DEFECT

1) change the stoichiometry of the substance. 2) It is also of two types: (i) Metal excess defect (ii) Metal deficiency defect · 

Differences b/w vacancy and interstitial defects: These defects are shown by non-ionic solids. VACANCY DEFECT 1) When some of the lattice sites are vacant then the crystal is said to have vacancy defect.

2) This results in decrease in the density of the substance

INTERSTITIAL DEFECT 1) When some constituting particles are present at the interstitial sites also then the crystal is said to have interstitial defect. .2) This results in increase in the density of the substance. ·

Ionic solids must always maintain electrical neutrality, so instead of vacancy or interstitial defect, they show these defects as Frenkel and Schottky defects.

Differences b/w Frenkel and Schottky defects: FRENKEL DEFECT 1) Smaller ion is dislocated from its normal site to interstitial site. 2) Density remains unchanged. Ex : AgBr, AgCl 

SCHOTTKY DEFECT

1) Equal number of cations and anions are missed from their positions 2) Overall density of the substance deceases. Ex: AgBr, NaCl ·

METAL EXCESS DEFECT: is due to (a) anion vacancies (b) presence of extra cations in interstitial sites · 

F-Centres: The anion vacancy occupied by an unpaired electron is known as F-centre. · ELECTRICAL PROPERTIES: 

Band Theory: 

(i) METALS: Valence band is either partially filled or it overlaps with a higher energy unoccupied conduction band so that electron s can easily flow under an applied electric field, so these are good conductor of electricity. 

(ii) INSULATORS: the gap between filled valence band and unoccupied conduction band is large; therefore electrons cannot jump to it and do not show any conductivity. 

(iii) SEMICONDUCTORS: the gap between filled valence band and unoccupied conduction band is small; therefore some electrons may jump to conduction band and show some conductivity. · DOPING: The process of introducing small amount of impurities in semiconductors to increase their conductivity is known as doping. · D

b/w n-type and p-type semiconductors:

n-TYPE SEMICONDUCTOR Semiconductor is doped with an electron rich impurity. Extra e- becomes delocalized which is responsible for conductivity. So conductivity is due to negatively charged e-

p-TYPE SEMICONDUCTOR: Semiconductor is doped with an electron deficit impurity.The lack of electrons creates holes which is responsible for conductivity.

· Differences b/w 13-15 compounds and 12-16 compounds: 

13-15 COMPOUNDS Formed by combination of elements of group 13 and 15 to simulate average valence of 4. Ex: AlP, GaAs 

12-16 COMPOUNDS Formed by combination of elements of group 12 and 16 to simulate average valence of 4. Ex: ZnS, CdS · 

MAGNETIC PROPERTIES:

(a) Paramagnetic Substances: are weakly attracted by magnetic field and lose their magnetism in the absence of magnetic field. Paramagnetism is due to unpaired electrons. Ex: O2, Fe3+ 

(b) Diamagnetic Substances: are weakly repelled by magnetic field .Diamagnetism is due to paired electrons. Ex: NaCl, H2O 

(c) Ferromagnetic Substances: are attracted very strongly by magnetic field and can be permanently magnetized. All the magnetic moments are aligned in the same direction so having a strong net magnetic moment. Ex: Fe, Co ↑↑↑↑↑↑↑↑↑↑↑

(d) Antiferromagnetic Substances: Magnetic domains are aligned oppositely and cancel out each other’s magnetic moment so net magnetic moment is zero. Ex: MnO ↑↓↑↓↑↓↑↓↑↓↑↓ 

(e) Ferrimagnetic Substances: Magnetic domains are aligned in parallel and antiparallel directions in unequal numbers resulting in small net magnetic moment. They are weakly attracted by magnetic field. Ex: Fe3O4 ↑↑↓↑↑↓↑↑↑↓

P- Block Elements ( Group 14)

REASONING QUESTIONS OF P-BLOCK ELEMENTS CLASS-XII Q1) Nitrogen exists as a diatomic molecule and Phosphorous as P4 .Why? OR Nitrogen forms a larger number of oxides than phosphorous. Explain. OR Oxygen exists as a diatomic molecule and Sulphur as S8 .Why?

Ans) As nitrogen has a tendency to form pπ-pπ multiple bond due to small size and high electronegativity while Phosphorous can’t form due to large size. 

Q2) Nitrogen is inert(less reactive) at room temperature. Why?

Ans) Because it has triple bond, so its bond dissociation enthalpy is very high.

Q3)Why does nitrogen show catenation property less than phosphorous? OR Sulphur exhibits tendency for catenation but oxygen doesn’t. Why?

Ans) As single N-N bond is weaker than the single P-P bond because of high interelectronic repulsion of the non-bonding electrons, due to small bond length in N2. Q4) Q4) N2is a gas while P4 is a solid at room temperature. Why? OR Explain why Oxygen is a gas while sulphur is a solid.

Ans) N2 molecules are attracted to each other through weak vanderwaal forces which can be easily overcome by the energy available at room temperature while 4 P (8 S) atoms are linked through covalent bonds which can’t be broken at room temperature.

Q5) The stability of +5 oxidation state decreases on moving down the group. Why? OR Stability of higher oxidation state decreases on moving down the group. Explain. OR Bi is a strong oxidizing agent in pentavalent state. OR The +5 oxidation state of Bi is less stable than its +3 state. Why OR SbF5 is much more stable than BiF5. Why? Ans) Due to inert pair effect, the ns electrons can’t participate in the bond formation.

Q6) Nitrogen shows maximum covalence of 4. Why? OR Though nitrogen exhibits +5 oxidation state, it doesn’t form pentahalides.Why? OR Why phosphorous form PF5 while nitrogen does not form NF5.OR Oxygen shows maximum covalence of 4. Why? OR ClF3 exists but FCl3 doesn’t. Why? OR Why does oxygen not show an oxidation state of +4 and +6 like sulphur? 

Ans) Due to absence of d orbitals in nitrogen.(oxygen or fluorine)

Q7) Why does nitrogen show anomalous behavior? OR Why does oxygen show anomalous behavior?

Ans) Because of smaller size, absence of d-orbitals, high electronegativity and high ionization enthalpy.

Q8) NH3 is liquid at room temperature while PH3 is a gas. Why? OR NH3 has higher boiling point than PH3. Why? H2O is liquid at room temperature while H2S is a gas. Why?

Ans) Because NH3 molecules are associated to each other through intermolecular hydrogen bonds while PH3 molecules can’t form hydrogen bonds.

Q9) PH3 is a weaker base than NH3. Why? OR NH3 is basic while BiH3 is only feebly basic. Why?

Ans) P is having larger size and less electronegativity than N, therefore tendency to donate the lone pair is less in PH3.

Q10) Why is NH3 a good complexing agent? OR Why does NH3 act as a Lewis base? OR NH3 act as a ligand.

Ans) Due to lone pair on N in NH3.

ORGANIC CHEMISTRY: REASONING QUESTIONS , CLASS: XII

POLYMERS

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

CHEMISTRY IN EVERYDAY LIFE