Since Al3- ion is highly charged and quite small the bond uniting the hydrogen and oxygen atoms with coordinated H2O molecules is considerable weakened and other solvent water molecules are able to act as bases and abstract protons - Al H2O 6 - aq- 3-H2O - - - Al H2O 5 OH - aq- 2-H3O aq- - - Al H2O 5OH - aq- 2-H2O - - - Al H2O 4 OH 2 - aq- -H3O aq- -If the solution also contains a strongly basic anion such as CO32- or S2- further ionization of hydrated aluminium ions can occur - Al H2O 4 OH 2 - aq- -CO3 aq- 2- - Al H2O 3 OH 3 - solid -HCO3 aq- -Behaviour of the cation in CO32- solution in the above passage can be considered as-

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Byju's Answer

Standard XII

Chemistry

Ammonia

Since Al3+ ...

Question

Since $A l^{3 +}$ ion is highly charged and quite small the bond uniting the hydrogen and oxygen atoms with coordinated $H_{2} O$ molecules is considerable weakened and other solvent water molecules are able to act as bases and abstract protons
${[A l {(H_{2} O)}_{6}]}_{6}^{3 +} + H_{2} O_{(ι)} \to {[A l {(H_{2} O)}_{5} (O H)]}_{5}^{2 +} + H_{3} O_{(a q .)}^{+}$
${[A l {(H_{2} O)}_{5} O H]}_{5}^{2 +} + H_{2} O_{(ι)} \to {[A l {(H_{2} O)}_{4} {(O H)}_{2}]}_{4}^{+} + H_{3} O_{(a q .)}^{+}$
If the solution also contains a strongly basic anion such as $C O_{3}^{2 -} o r S^{2 -}$ further ionization of hydrated aluminium ions can occur
${[A l {(H_{2} O)}_{4} {(O H)}_{2}]}_{4}^{+} + C O_{3 (a q .)}^{2 -} ⇌ {[A l {(H_{2} O)}_{3} {(O H)}_{3}]}_{(s o l i d)} + H C O_{3 (a q .)}^{-}$
Behaviour of the cation in $C O_{3}^{2 -}$ solution in the above passage can be considered as:

Bronsted acid

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Arrhenius base

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Lewis acid

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Lewis base

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Solution

The correct option is D Lewis base
$C O_{3}^{2 -}$ behaves as a Lewis base by donating electron and accepting hydrogen hence reducing itself.

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Similar questions

Consider following reactions:

[C o (N H_{3})_{6}]^{3 +} (a q) + 6 H_{3} O^{+} (a q) ⇌ [C o (H_{2} O)_{6}]^{3 +} (a q) + 6 N H_{4}^{+} (a q)

The products are favoured in the above reaction:

II.

[C o (N H_{3})_{6}]^{2 +} (a q) + 6 H_{3} O^{+} (a q) \to [C o (H_{2} O)_{6}]^{2 +} (a q) + 6 N H_{4}^{+} (a q)

The reaction (II) takes place with

H_{2} O

in a matter of seconds.

Which is thermodynamically unstable and also kinetically labile?

Q. The initial rate of oxidation of iodide ion

(I^{-})

by the peroxodisulphate ion

[S_{2} O_{8}^{2 -}]

is studied by the following given reactions.

2 I^{-} (a q) + S_{2} O_{8}^{2 -} (a q) \to I_{2} (a q) + 2 S_{2} O_{4}^{- 2} (a q) (s l o w)

I_{2} (a q) + 2 S_{2} O_{3}^{2 -} (a q) \to 2 I^{-} (a q) + S_{4} O_{6}^{2 -} (a q) (f a s t)

A little of these sulphate ions along with starch are added in the first reaction. The iodine formed in the first reaction reacts with sulphate ions and the blue colour of iodine starts developing as soon as the whole

S_{2} O_{3}^{2 -}

are used up. The sulphur containing atoms in the course of reactions are:

Consider following reactions:

[C o (N H_{3})_{6}]^{3 +} (a q) + 6 H_{3} O^{+} (a q) ⇌ [C o (H_{2} O)_{6}]^{3 +} (a q) + 6 N H_{4}^{+} (a q)

The products are favoured in the above reaction.

II.

[C o (N H_{3})_{6}]^{2 +} (a q) + 6 H_{3} O^{+} (a q) \to [C o (H_{2} O)_{6}]^{2 +} (a q) + 6 N H_{4}^{+} (a q)

The reactions (II) takes place with

H_{2} O

in a matter of seconds.

Select correct statement:

Balanced ionic equation for the reaction of Potassium Dichromate (VI) K₂Cr₂O₇ with Sodium Sulphite Na₂SO₃ in acid solution to give Chromium (III) ion and sulphate ion is

Q. Peroxodisulphate salts (e.g.,

N a_{2} S_{2} O_{8}

) are strong oxidising agents used as bleaching agents for fats, oils and fabrics. Can oxygen gas oxidise sulphate ion to peroxodisulphate ion

(S_{2} O_{8}^{2 -})

in acidic solution, with the

O_{2} (g)

being reduced to water? Given,

O_{2} (g) + 4 H^{+} (a q) + 4 e \to 2 H_{2} O E^{o} = 1.23 V

S_{2} O_{8}^{2 -} (a q) + 2 e \to 2 S O_{4}^{2 -} (a q) E^{o} = 2.01 V

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The correct option is D Lewis baseCO2−3 behaves as a Lewis base by donating electron and accepting hydrogen hence reducing itself.

The correct option is D Lewis base
$C O_{3}^{2 -}$ behaves as a Lewis base by donating electron and accepting hydrogen hence reducing itself.