To determine the number of unpaired electrons in the $Mn^{2+}$ ion, let's first consider the electron configuration of a neutral manganese (Mn) atom.
Manganese (Mn) has an atomic number of 25, so the electron configuration of neutral Mn is:
$[Ar] 4s^2 3d^5$
When manganese forms a $Mn^{2+}$ ion, it loses two electrons. These electrons are removed first from the 4s orbital because it is higher in energy than the 3d orbital.
So, the electron configuration of $Mn^{2+}$ is:
$[Ar] 3d^5$
The 3d subshell has 5 electrons in the case of $Mn^{2+}$, and since the 3d orbitals can hold a maximum of 10 electrons, the 5 electrons will occupy the 5 available d orbitals with each orbital having 1 electron (this is the maximum number of unpaired electrons).
Therefore, the number of unpaired electrons in $Mn^{2+}$ is:
Answer: 5
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: