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] 3d104s2 ------ no of unpaired electron ----->Zero---> Diamagnetic
Sc2+, ---------[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 t2g to eg set of energies when electron absorbs
energy to jump from t2g to eg and come back to t2g
from eg 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-
Sc2+,
---------[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-
+ S2O82-
-------> I2 + 2SO42-
Explanation
of catalytic action of Fe3+ in above reaction
Step-1
2Fe3+ + 2I- ------->
2Fe2+ + 2I2
Step-2
2 Fe2+ + S2O82- ------->
2Fe3+ + 2SO42-
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.
