In the following balanced reaction, what volume of a 3.0 M H2SO4 is required to completely neutralize 200.0 ml of a 2.5 M NaOH?
2 answers:
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For 7A(17) :
Electronic configuration
So, there are 5 unpaired electrons present in group 7A(17).
<h3>What are Unpaired Electrons?</h3>- An unpaired electron is an electron that doesn't form part of an electron pair when it occupies an atom's orbital in chemistry.
- Each of an atom's three atomic orbitals, designated by the quantum numbers n, l, and m, has the capacity to hold a pair of two electrons with opposing spins.
- Unpaired electrons are extremely uncommon in chemistry because an object carrying an unpaired electron is typically quite reactive. This is because the production of electron pairs, whether in the form of a chemical bond or as a lone pair, is frequently energetically advantageous.
- They play a crucial role in describing reaction pathways even though they normally only appear momentarily during a reaction on a thing called a radical in organic chemistry.
To learn more about unpaired electrons with the given link
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<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction
<u>Explanation:</u>
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.
For the given chemical reaction:
<u>On the reactant side:</u>
Oxidation number of H = +1
Oxidation number of N = +5
Oxidation number of O = -2
Oxidation number of S = 0
<u>On the product side:</u>
Oxidation number of H = +1
Oxidation number of N = +2
Oxidation number of O = -2
Oxidation number of S = +6
As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.
The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.
Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction