Answer:
Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺
Explanation:
Data Given:
Reactants:
Ni(s) + H₂O(l)
Product = ?
Solution:
Names of the Reactants
Ni = Nickel
H₂O = water
Reaction:
Normally nickel directly does not react with water under normal condition. But indirectly in acidic or neutral condition it form complex ion.
For this it first dissolve slowly in dilute acid, in this reaction it liberate Ni²⁺ ions these nickel ions form light green complex ions in aqueous solution.
Ni(s) + H₂SO₄(aq) -------> Ni²⁺(aq) + SO₄²⁻(aq) + H₂
This is a type of complex formation in which Nickel react with water and produced a light green color new complex ion or product.
Complete reaction is as under
Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺
Balance Reaction:
Ni(s) + 6 H₂O(l) -------------> [Ni(H₂O)₆]²⁺
So.
by this reaction one product is formed that is [Ni(H₂O)₆]²⁺ named as hexaqua nickel ion
