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2 years ago

How do you predict a reaction outcome for Ni(s) + H2O(L)?

Chemistry

1 answer:

ElenaW [278]2 years ago

7 0

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

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Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

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Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

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