Be sure to answer all parts. the sulfate ion can be represented with four s―o bonds or with two s―o and two s═o bonds. (a) which representation is better from the standpoint of formal charges? four s―o bonds two s―o bonds and two s═o bonds (b) what shape is the sulfate ion, and what hybrid orbitals of s are postulated for the σ bonding? bent tetrahedral trigonal planar trigonal pyramidal spmdn where m = and n = (c) in view of the answer to part (b), what orbitals of s must be used for the π bonds? what orbitals of o? sulfur: 3d 3p 2p 3s oxygen: 3p 2p 2s 3s

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) hybrid orbitals invovle sigma bonding.

orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

a)

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

Therefore, structure (b) is the better representation of sulfate ion.

b)

In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.

There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is

c)

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves bonding.

Therefore, bonds are formed by the overlapping of d-orbitals of sulfur with  p-orbitals of oxygen.

pure water does not conduct electricity very well. however, when certain substances are dissolved in water, the solution does conduct electricity. you can make a simple device that shows how well a solution conducts electricity. this device uses a flashlight bulb to indicate how well the solution conducts electricity. the better the solution conducts electricity, the brighter the bulb will glow.

explanation: