Answer
a) Neutral Manganese atom has 25 proptns & 25 electrons.
b) No. of electrons in the cation $Mn^{2+}$ = 23.
No. of protons in the cation $Mn^{2+}$ = 25.
c) Electronic configuration of Mn atom = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}$
Two electrons from 4s orbital are lost to form the $Mn^{2+}$cation.
Work Step by Step
Atomic number of Mn = 25.
So, no. of protons = 25.
no. of electrons = 25 for neutral Mn atom.
Now, in $Mn^{2+}$ there is two positive charge. So, no. of electrons that are lost is 2. hence, the no. of electrons will be 23. No. of protons will be the same i.e. 25.
4s orbital is the outermost orbital and we know that electrons are emitted from the valence shell i.e. the outermost orbital. So, two electrons from 4s orbital are lost to form the $Mn^{2+}$cation.
