Question

draw the best lewis/electron - dot structure for h₂so₄. how many single, double, and/or triple bonds does the molecule possess? hint: the best structure will have zero formal charge on all atoms.
___ single bonds ; _ double bonds ; ___ triple bonds
formatting example: h₂o
___ 2 _ single bonds ; _ 0 _ double bonds ; _ 0 ___ triple bonds should be entered as 2;0;0
short answer
6;0;0

Explanation:

Step1: Determine the central atom

The central atom in \(H_2SO_4\) is \(S\) (sulfur).

Step2: Calculate the total number of valence electrons

• \(H\) has \(1\) valence electron, and there are \(2\) \(H\) atoms (\(2\times1 = 2\) valence electrons).
• \(S\) has \(6\) valence electrons.
• \(O\) has \(6\) valence electrons, and there are \(4\) \(O\) atoms (\(4\times6=24\) valence electrons).
• Total valence electrons \(=2 + 6+24=32\)

Step3: Arrange the atoms and form bonds

• Connect \(H\) to \(O\) and \(O\) to \(S\).
• The structure that gives zero formal charge on all atoms has \(2\) \(O\) atoms double - bonded to \(S\) (\(S = O\)) and \(2\) \(O\) atoms single - bonded to \(S\) (\(S - O\)), and each of those single - bonded \(O\) atoms is also single - bonded to a \(H\) atom (\(O - H\)).
• Count the single bonds: There are \(2\) \(O - H\) bonds and \(2\) \(S - O\) bonds, so \(4\) single bonds from the non - double - bonded \(O\) atoms.
• Count the double bonds: There are \(2\) \(S = O\) double bonds.
• There are no triple bonds in \(H_2SO_4\) as sulfur does not form triple bonds in this context and the valence electron count and formal charge requirements are met without triple bonds.

Answer:

\(4;2;0\)