Draw Lewis structures for the following: (a) \(\operatorname{SiH}_{4},\) (b) \(\mathrm{CO}\) \((\mathbf{c}) \mathrm{SF}_{2},(\mathbf{d}) \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{H}\) is bonded to \(\mathrm{O})\) , \((\mathbf{e}) \mathrm{ClO}_{2}^{-},(\mathbf{f}) \mathrm{NH}_{2} \mathrm{OH}\)

Count the total number of valence electrons for each molecule/ion based on the valence electron configuration of the individual elements. a) SiH4: Si has 4 valence electrons and each H has 1 valence electron, so for SiH4 we have: 4 (for Si) + 4x1 (for 4 H atoms) = 8 valence electrons b) CO: C has 4 valence electrons and O has 6 valence electrons, so for CO we have: 4 + 6 = 10 valence electrons c) SF2: S has 6 valence electrons and each F has 7 valence electrons, so for SF2 we have: 6 + 2x7 = 20 valence electrons d) H2SO4: S has 6 valence electrons, each O has 6 valence electrons, and each H has 1 valence electron, so for H2SO4 we have: 6 + 4x6 (for 4 O atoms) + 2x1 (for 2 H atoms) = 32 valence electrons e) ClO2-: Cl has 7 valence electrons, each O has 6 valence electrons, and there is an extra electron gained by the ion, so for ClO2- we have: 7 + 2x6 + 1 = 20 valence electrons f) NH2OH: N has 5 valence electrons, each H has 1 valence electron, and O has 6 valence electrons, so for NH2OH we have: 5 + 3x1 + 6 = 14 valence electrons

Form single bonds between the central atom and the surrounding atoms. Then add electrons to complete the octet for each atom. a) SiH4 Lewis structure: Si is the central atom, forming single bond with hydrogen atoms. Si-H | H-Si-H | H b) CO Lewis structure: C is the central atom, forming a triple bond with oxygen to complete their octet. C≡O c) SF2 Lewis structure: S is the central atom, forming single bond with fluorine atoms. F | S - F d) H2SO4 Lewis structure: S is the central atom, forming single bonds with four oxygen atoms, and two oxygen atoms have H atoms attached to them. O | O-S-O | O | H e) ClO2- Lewis structure: Cl is the central atom, forming single bonds with two oxygen atoms. O | Cl-O f) NH2OH Lewis structure: N is the central atom, forming single bonds with two hydrogen atoms and the oxygen atom. Oxygen is then attached to a hydrogen atom. H | H-N-O-H

For any remaining electrons, distribute them to the outer atoms (except hydrogen) to fulfill the octet rule. In some cases, refine the structure by forming double or triple bonds, if needed. a) SiH4 - no need to modify; all atoms have a full outer shell. b) CO - triple bond already formed; no modification needed. c) SF2: To complete the octet rule for sulfur, we need to add two non-bonding electron pairs: F | S - F | / \ LP LP d) H2SO4: Add two non-bonding electron pairs to the oxygen atoms that are not bonded to hydrogen atoms, forming double bonds: O | O=S-O | O | H e) ClO2-: Add three non-bonding electron pairs to the outer oxygen atoms, and complete the octet rule for chlorine by adding two non-bonding electron pairs. O | Cl=O | / \ LP LP f) NH2OH: Add a non-bonding electron pair to nitrogen to complete the octet rule. Hydrogen and oxygen are already complete. H | H-N-O-H | LP