# How would you draw NiCl2 lewis dot diagram?

Ni has 28 electrons, and my teacher told me that all elements have 2,8,8,2…. maximum number of electrons in shells. If I follow this method and try to draw Nickel, it has 8 outer valence electrons. Then how am I supposed to connect 2 Chlorine to the Nickel?

Best to consider it as ionic that is Ni^2+ plus two Cl^- ions. Ni^2+ as you have correctly deduced has 8 valence e⁻s but they are d e⁻s (Ni: [Ar] 3d^8) so you could just put the eight e⁻s around the Ni^2+ ion like a regular octet.

If you are doing Lewis diagram of an ion such as the sulfide ion, S= knowing the sulfur atom has 6 electrons in outer energy level, and of course the ion will have 8 electrons having gained 2 from another atom, so the Lewis structure would be just the syboll S surrounded by four pairs of electrons. Now if you are doing Lewis diagram for say H2S, each hydrogen has one electron as you know, and the sulfur has 6 electrons in outer energy level. And you probably know that the overall charge on any molecule is zero. so just do the diagram and don’t worry about formal charge, in this case it would be set up H=S=H with a lone pair above and below the sulfur atom, and the shape of the molecule would be bent at about 90 degrees .

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Other than the elements that are exceptions to the octet rule, all elements want to have a full shell of 8 electrons. Formal charge calculatom is for judging which structure is more correct. To answer your question, you would draw the resonance structure of each Lewis Dot Structure and then apply Formal Charge calculation to see which structure is more correct. Any element that reaches the 3d shell will have an expanded octet. So in a sense, even Phosphorus can go beyond 8 electrons.