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Electrons are subatomic particles with the symbol e−. They have a negative electric charge (-1 elementary charge. )
1
vote
Accepted
What kind of electronic configurations are possible?
This misconception arrises because most of the textbooks and websites starts pairing electrons from left to right in a series. …
1
vote
Finding excited stage of electron from its potential energy
The potential energy ($U$) of an electron in nth orbit of hydrogen atom is given below : $$ U = \frac{-27.2}{n^2} \space (in \space eV)$$ $$ So, \space -6.8 = \frac{-27.2}{n^2} $$ $$ \implies n=2 …
2
votes
Why is orbital energy not the mean between ionization energy and electron affinity when the ...
The orbital energy simply refers to the ionisation energy. So the orbital energy of the hydride ion should be numerically equal to the electron affinity value of the hydrogen atom.
And there is no vio …
1
vote
Accepted
How to calculate ionization energy of an atom provided energy of the electron at ground state?
Ionisation energy : It is the minimum amount of energy required to remove an electron from the neutral gaseous atom.
If the electron is bound to the nucleus then its total energy will be negative. T …
4
votes
How do we write spin multiplicity for Mn(2+), Mn(7+)?
Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$.
Now coming to your question,
$\ce{Mn}$ has 5 unpaired electrons in it. … $\ce{Mn^{7+}}$ has zero number of unpaired electrons. Therefore, its spin multiplicity $(S) = 0+1=1$. …