Draw the kinetic and thermodynamic enolates formed when 3-methylbutan-2-one (methyl isopropyl ketone) is treated with base. Incl
1 answer:
You might be interested in
<u>Answer: </u>The correct statement is X is the effective nuclear charge, and it increases across a period.
<u>Explanation:</u>
We are given that:
X = number of protons − number of core electrons
Effective nuclear charge is defined as the actual nuclear charge (Z = number of protons) minus the screening effect caused by the electrons present between nucleus and valence electrons. These electrons are the core electrons.
The formula used for the calculation of effective nuclear charge given by Slater is:
where,
= effective nuclear charge
Z = atomic number or actual nuclear charge or number of protons
= Screening constant
The effective nuclear charge increases as we go from left to right in a period because nuclear charge increases with no effective increase in screening constant.
Hence, the correct answer is X is the effective nuclear charge, and it increases across a period.
The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
- Oxidation reaction
Li⁰(s) → Li⁺(aq) + e⁻ (2)
- Reduction reaction
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
Learn more here:
I hope it helps you!
