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Sunday, February 2, 2020

What is Laporte Selection rule?

Laporte Selection Rule is given by Otto Laporte a German American Physicist

According to Laporte selection rule only allowed transitions are those occurring with a change in parity (flip in the sign of one spatial coordinate.) OR During an electronic transition the azimuthal quantum number can change only by ± 1 (Δ l = ±1) The Laporte selection rule reflects the fact that for light to interact with a molecule and be absorbed, there should be a change in dipole moment.

Practical meaning of the Laporte rule:

Laporte allowed transitions: are those which occur between gerade to ungerade or ungerade to gerade orbitals.

Laporte forbidden transitions: are those which occur between gerade to gerade or ungerade to ungerade orbitals.

Gerade = symmetric with respect to centre of inversion i.e. atomic or molecular orbital with center of symmetry or number of nodal plane = 0, 2, 4 (even number)

Ungerade = anti symmetric with respect to centre of inversion i.e. atomic or molecular orbital without center of symmetry or number of nodal plane = 1, 3, 5, (odd numbers)

Important Note:

This rule affects Octahedral and Square planar complexes as they have center of symmetry. Tetrahedral complexes do not have center of symmetry therefore this rule does not apply

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?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ?
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ?
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?

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?

Both [Mn(H₂O)₆]²⁺and [FeF₆]^3-have a d⁵ configuration and high-spin complexes, but electronic transitions are not only Laporte-forbidden, but also spin-forbidden. thus the dilute solutions of Mn²⁺ and Fe ⁺³ complexes are colorless.

Important Note:

For first transition series d⁵ system, weak ligand field , and coordination number six (6) Complexes are found to be colourless due to violation of selection rule.

What is the Selection rule for colour in complexes?

Spin selection rule states that transitions that involve a change in spin multiplicity as compare to ground state are forbidden.

(1) According to this rule, any transition for which Δ S = 0 (it means no change in spin multiplicity after d-d transition) is allowed.

(2) If Δ S ≠ 0 (change in spin multiplicity after transition) then it is forbidden (transition not allowed)

Intensity of colour due to d-d transition:

(1) Intensity of colour due to d-d transition will found to be high if transition follow laporte selection rule.

(2) Intensity of colour due to d-d transition will found to be poor due violation of laporte selection rule.

(3) Intensity of colour in tetrahedral Complexes for (non centre of symmetry) is found to be higher than octahedral (centre of symmetry).

Important Note:

For first transition series d⁵system, weak ligand field, and coordination number six (6) Complexes are found to be colourless due to violation of selection rule.

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?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ?
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ?
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?

Which of the Complex of the following pairs has the largest value of CFSE? (1) [Co(CN)6]3- and [Co(NH3)6]3+ (2) [Co(NH3)6]3+ and [CoF6]3- (3) [Co(H2O)6]3+ and [Rh(H2O)6]3+ (4) [Co(H2O)6]2+ and [Co(H2O)6]3+

(1) CN is the stronger ligand than NH₃ therefore CFSE of [Co(CN)₆]³^- will be more than [Co(NH₃)₆]³⁺

(2) NH₃ is stronger ligand than F therefore CFSE of [Co(NH₃)₆]³⁺ will be more than [CoF₆]^3-.
(3) Co belongs to 3d series whereas The Rh belong to 4d series. More the value of n more is CFSE therefore CFSE of [Rh(H₂O)₆]³⁺ is more than [Co(H₂O)₆]3+ .
(4) Oxidation number of Co in [Co(H₂O)₆]³⁺ is more than the Oxidation number of [Co(H₂O)₆]²⁺ therefore, CFSE of [Co(H₂O)6]³⁺ is more than [Co(H₂O)₆]²⁺.

Saturday, February 1, 2020

What is oil of vitriol ?

Concentrated sulphuric acid is a dense and oily liquid which is also known as oil of vitriol. Concentrated sulphuric acid has a specific gravity of 1.84 and a boiling point of 611 K. The high boiling point and high viscosity indicate that sulphuric acid has associated structure due to hydrogen bonding as shown below:

The concentrated acid is soluble in water and the dissolution process is highly exothermic. So acid is always diluted by adding acid to the water slowly and not by adding water to acid. This is done because in the later case, lot of heat is produced which causes the acid to spurt out of the container.

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