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Wednesday, August 28, 2019

Why [Mn(H2O)6]+2 is colourless although in which Mn+2 ion had five unpaired electrons ?

There are 5 unpaired electrons in Mn+2 ion and we know that.
                                 Spin multiplicity is = (2S+1) 
                     Where S is some of spin thus  S = 5×1/2=5/2
                                            Hence multiplicity = (2S+1)= 6 

we know multiplicity of excited state cannot be six, thus electronic transition in Mn+2  are 
spin forbidden, hence [Mn(H2O)6]+2 salt appear colourless.

Also [Mn(H2O)6]+2 has centre of symmetry and in such cases electronic transition are expected to be laporate forbidden.

Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?

The intense colour of the latter complex is strongly suggestive of a charge transfer transition and since the metal is already fully reduced (zero oxidation state), it is highly likely that this involves a MLCT transition. The π* levels of the bipy or CO ligands are possible acceptors but the fact that Fe(CO)5 doesn’t show this colour suggests that it is the bipy π* levels that are involved in Fe(bipy)(CO)3. Since there should be MLCT transitions to the CO π* levels as well, we assume that the lack of colour for Fe(CO)5 means that these transitions fall in the UV rather than the visible.

Why all the tetrahedral Complexes are high spin Complexes ?

The magnitude of crystal field splitting energy (CFSE) in tetrahedral Complexes is quite small and it is always less than the pairing energy. Due to this reason pairing of electron is energetically unfavorable. Thus all the tetrahedral Complexes are high spin Complexes. In fact no tetrahedral Complex with low spin has been found to exist.


Tuesday, August 27, 2019

FACTOR'S AFFECTING STABILITY OF COMPLEXES:

 (1) Charge on central metal cation:
In general stability of Complexes is directly proportional to the magnitude of charge on central metal atom. Thus Complexes of Fe +3 are more stable than Fe+2
(2) Size of central metal cation:
As size of the central atom ion decreases the stability of the Complex increases. This is applicable when Oxidation state of central metal ion is same in all the cases. For example Zn+2 < Cu+2 < Ni+2 < Co+2 < Fe+2 < Mn+2
(3) Nature of the ligands:
The size and charge of ligands is also an important factor of in deciding the stability of complexes.
(i) If the Ligand is smaller, it can approach the central metal ion more closely forming a stable bond.
(ii) High charge ligand will form a strong bond. Thus high charge and small size of ligands leads to Formation of stable Complexes.
(iii) Charge density of the ligand= charge/size. More the charge density more is the stability of the Complex. For example fluoride (F- ) will form more stable Complexes and Iodide (I-) forms least stable Complexes.
(4) Chelating effects:
Formation of five and six membered chelate from polydentate ligands enhance the stability of Complexes in comparison to monodentate ligands. This is called chelate effect of chelation. In general, the more number of chelate rings in the Complex, the more will be stability of the Complex.
(5) Formation constant of Complexes:
Stronger is the metal- ligand bond, less is the dissociation of Complex ion in the Solution and hence greater is the stability of complex. Thus the larger the numerical value of formation constant, the thermodynamically more stable is the complex.

Sunday, August 25, 2019

What is d-d transition in complexes and explain colour of complex by d-d transition ?

Most of the transition metal compounds are coloured both in the solid state and in aqueous solution. This is because of the presence of incompletely filled d-orbitals.  When a transition metal compound is formed the degenerate d-orbitals of the metal  split into two sets, one having three orbitals dxy, dyz and dxz called t2g orbitals with lower energy and the other having two orbitals dx2 –y2 and dz2 called eg orbitals with slightly higher energy in an octahedral field. This is called crystal field splitting When white light falls on these compounds, some wavelength is absorbed for promotion of electrons from one set of lower energy orbitals to another set of 
slightly higher energy within the same d-subshell. This is called d-d transition. The remainder light is reflected which has a particular colour.


The colours of some 3d metal ions:

SN
d-configuration
Examples with colour
1
d0 (No d-d transition)

2
d1
Ti3+ (3d1) Purple,  V+4(3d1) Blue
3
d2
V+3(3d1) Green
4
d3
Cr3+ (3d3) Violet green
5
d4
Mn+3(3d4) Violet , Cr2+ (3d4) Blue
6
d5
Mn+2(3d5) Pink, Fe+3(3d5) Yellow
7
d6
Fe+2(3d6) Brown , Co+2(3d6) Green ,
8
d7

9
d8
Ni+2(3d8) Green
10
d9
Cu+2(3d9) Blue
10
d10
Sc+3(3d0)  colourless

ILLUSTRATION (1): The mechanism of light absorption in coordination compounds is that photons of appropriate energy can excite the coordination entity from its ground state to an excited state. Consider [Ti(H2O)6]3+  In which  Ti(+3) ion has one electron in d sub shell ( in lower energy t2g d- orbital) . In aqueous solution, [Ti(H2O)6]3+. Appear as purple due to the absorption of light from visible range ( green and yellow portion) resulting  d-d transition ( electron jump from t2g level to eg level) as result  complex has complementary ie purple .
The variety of color among transition metal complexes has long fascinated the chemists.

ILLUSTRATION (2) : Aqueous solutions of [Fe(H2O)6]3+ are red, [Co(H2O)6]2+ are pink, [Ni(H2O)6]2+ are green, [Cu(H2O)6]2+ are blue and [Zn(H2O)6]2+ are colorless. Although the octahedral [Co(H2O)6]2+ are pink, those of tetrahedral [CoCl4]2- are blue. The green color of [Ni(H2O)6]2+ turns blue when ammonia is added to give [Ni(NH3)6]2+. Many of these facts can be rationalized from CFT.

Related Question:


(1) Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?