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

Related Question:

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⁺.

Related Question:

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 ?

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.

Related Questions:

(1) Arrange in increasing order of extent of hydrolysis [ CCI4, MgCI2, AICI3, PCl5, SiCI4].

(2) Although Sulphur contain vacant d-orbital but SF6 does not under go hydrolysis. Why ?

(3) CCl4 can not be hydrolysed but SiCl4 can be. Why?

(4) Silianol (SiH3OH) is more acidic than methanol (CH3OH) why?

(5) What happened when Graphite treated concentrated acids like HNO3/HF or KMnO4 ?.

(6) Why Boric acid become strong acid in the presence of cis 1,2-diol or 1,3-diol ?

(7) What is the molecular formula of Borax ?

(8) What is oil of vitriol ?

What are the groups that LiAlH4 can and cannot reduce?

Lithium aluminium hydride (LiAIH4) is a nucleophilic reducing agent, and used to reduce polar multiple bonds like >C=O, -CN bond because it is hydride (H-) donor. And it is simply abbreviated as LAH.

LIAIH₄ can reduce:
(1) Aldehydes to primary alcohols,
(2) Ketones to secondary alcohols,
(3) Carboxylic acids and esters to primary alcohols.
(4) Acid chlorides to primary alcohols.
(5) Ester to primary alcohols.
(6) Acid anhydrides to primary alcohols.
(7) Amides and nitriles to primary amines.
(8) Isonitriles to secondary amines.
(9) Epoxides (Cyclic ethers) to alcohols.
(10) lactones (Cyclic esters) to diols. 

Important Notes:
LiAH4 can reduce Aldehydes, ketones, carboxylic acids, ester, acid chlorides, acids anhydrides, acid amides,  Nitro Compounds and nitriles and isonitriles  without attacking (Isolated non polar bonds , like -C=C- ) double bonds present in the Compounds. The only exception in this case alpha- Beta unsaturated compounds containing phenyl group in the Beta position, in this case LiAH₄ also attack on double bond.
We can say that the double or triple bonds in conjugation with the polar multiple bonds can be reduced.

Oxymercuration demercuration hydration procedure is superior to acid- catalysed hydration of most alkenes. Why ?

Oxymercuration demercuration hydration procedure is superior to acid- catalysed hydration of most alkenes. Because (1) the two stage process of oxymercuration demercuration is fast and takes place under mild condilions and gives more than 90 % yield of alcohol and (2) rearrangement doesn't take place hence most desire product easily obtained.

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