4 and 2 electrons are present.
The 2nd ionization energy is removing a 2nd electron from that resulting cation:
<span>Li+ --> Li2+ + 1e- </span>
There are no states in the picture, but Na should have a "(s)" after it, and Cl2 should have a "(g)" after it. NaCl should have an "(s)". Chlorine is a diatomic element so it has a "2" subscript on it.
Hope this helped! :)
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An atom of vanadium
(V) has 23 electrons.
Given :
A vanadium (V) atom with 23 protons and has a net charge of 0.
To find:
The number of electrons in a vanadium atom
Solution:
Number of protons in vanadium atom = 23
The 0 net charge on the atom indicates that the atom has an equal number of protons and electrons which makes it a neutral atom.
Number of electrons in vanadium atom = Number of protons in vanadium atom = 23
An atom of vanadium
(V) has 23 electrons.
Learn more about subatomic particles here:
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