Wednesday, October 19, 2022

Dalton’s Atomic Theory

Limitations of Dalton’s Theory

• It fails to explain why atoms of different kinds should differ in mass and valency etc.

• The discovery of isotopes and isobars showed that atoms of same elements may have different atomic masses (isotopes) and atoms of different kinds may have same atomic masses (isobars).

• The discovery of various sub-aomic particles like X-rays, electrons, protons etc. during late 19th century lead to the idea that the atom was no longer an indivisible and smallest particle of the matter.

Saturday, October 15, 2022

Magnetic Property,Colours and Formation of complexes,Catalytic properties,Formation of Interstitial Compounds of Transition Metals

Magnetic Property of Transition Metals

Diamagnetic substances contain electron pairs with opposite spins and are repelled by applied magnetics filled.

· Ex.:

Ti⁺⁴, ------ [Ar] 3d⁰4s⁰ ----- no of unpaired electron ----->Zero---> Diamagnetic

V⁺⁵ -----[Ar] 3d⁰4s⁰ -----no of unpaired electron ----->Zero---> Diamagnetic

· Sc³⁺ ------[Ar] 3d⁰4s⁰ ------no of unpaired electron ----->Zero---> Diamagnetic

· Zn --------[Ar] 3d¹⁰4s² ------no of unpaired electron ----->Zero---> Diamagnetic

· Hg---------[Ar] 3d¹⁰4s² ------no of unpaired electron ----->Zero---> Diamagnetic

· Cd ------- [Ar] 3d¹⁰4s^{2 ------} no of unpaired electron ----->Zero---> Diamagnetic

Paramagnetic substances contain unpaired electron spins or unpaired electrons and are attracted strongly in applied magnetic field.

Sc^2+, ---------[Ar] 3d¹4s⁰ ------no of unpaired electron ----->1e^- ---> paramagnetic

Cr³⁺ ---------[Ar] 3d³4s⁰ ------no of unpaired electron ----->3e^- ---> paramagnetic

· Paramagnetic character increase with increase in no of unpaired electron.

· Each unpaired electron having magnetic moment associated with its spin angular momentum and orbital angular momentum.so magnetic momentum can be calculated based on spin only formula

μ = √(n(n+2)) BM (Bohr Magnetons

PROBLEM: Calculate the 'spin only' magnetic moment of M2+(aq) ion (Z = 27).

SOLUTION: Z = 27 = [Ar] 3d7 4s2

M2+ = [Ar] 3d7

This means that it has 3 unpaired electrons.

n = 3

Colours of Transition Metal Ions

· Colour in transition metal ions depends upon presence of unpaired electron which show d-d transition of unpaired electron from t_2g to e_g set of energies when electron absorbs energy to jump from t_2g to e_g and come back to t_2g from e_g by emission of energy that appear with colour

Ex.:

Ti⁺⁴, ------ [Ar] 3d⁰4s⁰ ----- no of unpaired electron ----->Zero---> colourless

V⁺⁵ -----[Ar] 3d⁰4s⁰ -----no of unpaired electron ----->Zero---> colourless

Ans - Because of absence due to presence of paired electrons which do not show d-d transition

Ex-

Sc^2+, ---------[Ar] 3d¹4s⁰ ------no of unpaired electron ----->1e^- ---> coloured

Cr³⁺ ---------[Ar] 3d³4s⁰ ------no of unpaired electron ----->3e^- ---> coloured

Ans- due to presence of unpaired electrons which show d-d transition.

The great tendency of transition metal ions to form complexes is due to

a) small size of the atoms and ions

b) high nuclear charge and

c) availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons donated by ligands.

[Fe (CN)6]3–

[Fe(CN)6]4–

[Cu(NH3)4]2+

Catalytic properties of transition metals

· Good catalysts due to the presence of free valencies and also variable oxidation states

some Examples are as

· Iron (III) catalyses the reaction between iodide and persuphate ions-

2I^- + S₂O₈^2- -------> I₂+ 2SO₄^2-

Explanation of catalytic action of Fe³⁺in above reaction

Step-1

2Fe³⁺ + 2I^--------> 2Fe²⁺ + 2I₂

Step-2

2 Fe²⁺+ S₂O₈^2--------> 2Fe³⁺ + 2SO₄^2-

in this catalytic action of Fe, Fe shows variable oxidation state i.e. Fe³⁺changes into Fe²⁺ Again into Fe³⁺

Pt-used as a catalyst in the manufacture of H₂S0₄.

· Fe-used as a catalyst in the manufacture of NH₃ by Haber process. A small amount of molybdenum is added as a promoter.

· Ni.-used as a catalyst in the hydrogenation of oils.

· V₂0₅-used as a catalyst for the oxidation of S0₂ into S0₃ for the manufacture of H₂S0₄ in the contact process.

· Mn02-used as a catalyst in the decomposition of KCI0₃ for preparation of oxygen.

Formation of Interstitial Compounds-

· interstitial compounds are those which are formed when small atoms like H, C, N, B etc are trapped inside the crystal lattice of metals.

· The general characteristic physical and chemical properties of these compounds are:

a). High melting points which are higher than those of pure metals.

b) Retain metallic conductivity i.e. of pure metals.

c). Very hard and some borides have hardness as that of diamond.

d). Chemically inert.

Alloy Formation –

· These form alloy because of similar in atomic size and other characteristics of transition metal

· Alloys are hard and having high melting point. e.g., Brass (Cu + Zn) Bronze (Cu + Sn) etc. Hg when mix with other metals form semisolid amalgam except Fe, Co, Ni, Li.