H2SeO4

[Ian_Lee_1E]

hello,

I was wondering how you know when drawing the lewis structure that the 2 Hydrogens go to the oxygen rather than making 6 single bonds, 4 oxygen and 2 hydrogens attached to Se.

[Lea Chamoun 2J]

If the Se in H2SeO4 formed 6 single bonds, one with each H and O atom, then the formal charges would not add up to the total charge of the molecule. When oxygen is in a single bond and has 6 lone pairs, its formal charge is -1. Since there are 4 O's in this molecule, the formal charges of the O would add up to -4 if Se formed single bonds with O and H. The formal charge on Se would be 6-(12/2+0)=0, making the overall charge of the molecule -4, which is not the case. When Se is double bonded to 2 O atoms, its formal charge is still 0. The formal charge of all the other atoms in the molecule would also be 0 because 2 of the O's are double bonded to Se and the other 2 O's have a single bond with Se and a single bond with H. Since this structure minimizes the formal charge on the molecule, it is more favorable than the structure in which Se forms a single bond with each atom in the molecule.

[Jasraj Parmar 3H]

It is because if you have single bonds, it would not be the most stable lewis structure.If you double bond the SE and O together, you are able to get the most stable lewis structure. I was confused at first also.