Answer:
Explanation:
In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.
We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.
The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.
To ascertain the oxidation state, we have to comply with some rules:
- The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
- The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion.
For example, let us find the oxidation state of Cr in Cr₂O₇²⁻
This would be: 2x + 7(-2) = -2
x = +6
We see that the oxidation number of Cr, a transition metal in the given ion is +6.
The aim is to use less space while demonstrating the distribution of electrons in shells
If you want to depict how an atom's electrons are scattered across its subshells, an orbital notation is more suited.
This is due to the fact that some atoms have unique electronic configurations that are not readily apparent from textual configurations.
<h3>How does electron configuration work?</h3>
The placement of electrons in orbitals surrounding an atomic nucleus is known as electronic configuration, also known as electronic structure or electron configuration.
<h3>What sort of electron arrangement would that look like?</h3>
- For instance: You can see that oxygen contains 8 electrons on the periodic table.
- These 8 electrons would fill in the following order: 1s, 2s, and finally 2p, according to the aforementioned fill order. O 1s22s22p4 would be oxygen's electron configuration.
learn more about electronic configuration here
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The answer is D.timber good luck
Answer:
D
Explanation:
31 / 2.8 = 11.0714286 L per mole of helium
3.5 / 4 = 0.875 moles
2.8 + 0.875 = 3.675 moles
11.0714286 x 3.675 = 40.6875 L
