✔ 最佳答案
Both structures are incorrect.
The main point is that : Before bonding, O atom (1s2 2s2 2p4) has 6 electrons in its outermost shell. After bonding, O atom should have 8 electrons in its outermost shell, and this is the electronic configuration of a noble gas. Besides, since O atom does not have low-energy empty d orbitals, it cannot have more than 8 electrons in its outermost shell.
====
In the first structure, the central O atom forms 1 double bond and 1 single bond, and thus has 3 non-bonding electrons in the outermost shell. It has 9 electrons in the outermost shell, and this is impossible.
To draw the correct structure, the O-O single bond should be a dative bond instead. It can be represented in two ways : O→O or +O-O-, in which the first O atom is the central atom, and the last O atom is the terminal O atom.
Now, the central O atom form a double bond and supply a pair of electrons to form a dative bond, and thus it has 2 non-bonding electrons in the outermost shell. It has now 8 electrons in the outermost shell.
In the terminal →O atom, it does not give any electrons in bond formation. It has 8 electrons in the outermost shell.
In the teminal =O atom, it forms a double bond, and thus has 4 non-bonding electrons in the outermost shell. It has also 8 electrons in the outermost shell.
=====
In the second structure, each O atom form two double bonds and thus has 2 non-bonding electrons in the outermost shell. Each O atom has 10 electrons in its outermost shell, and this is impossible.
