Question

procedure part 2 : expanded octet lewis structures
several species containing central atoms from third period or below in the periodic table possess expanded octets in order to lower the formal charges on individual atoms.
complete the following table; take into account the expanded octet for each species.
this is not the report sheet
formula name skeleton # of valence e⁻ e⁻ used in single bonds remaining e⁻ complete lewis structure
po₄³⁻ phosphate anion o - p - o
so₄²⁻ sulfate anion
pcl₅ phosphorous pentachloride
sf₄ sulfur tetrafluoride f - s - f
sf₆ sulfur hexafluoride f - f - s - f - f
brf₃ bromine trifluoride f - br - f

Explanation:

Step1: Determine valence electrons for $[PO_4]^{3 -}$

P has 5 valence e⁻, each O has 6 valence e⁻ and add 3 for the - 3 charge. So total valence e⁻ = 5+(4×6)+3 = 32. There are 4 single - bonds in the skeleton, using 8 e⁻. Remaining e⁻ = 32 - 8=24.

Step2: Determine valence electrons for $[SO_4]^{2 -}$

S has 6 valence e⁻, each O has 6 valence e⁻ and add 2 for the - 2 charge. So total valence e⁻ = 6+(4×6)+2 = 32. There are 4 single - bonds in the skeleton (assumed similar to phosphate), using 8 e⁻. Remaining e⁻ = 32 - 8 = 24.

Step3: Determine valence electrons for $PCl_5$

P has 5 valence e⁻, each Cl has 7 valence e⁻. So total valence e⁻ = 5+(5×7)=40. There are 5 single - bonds in the skeleton, using 10 e⁻. Remaining e⁻ = 40 - 10 = 30.

Step4: Determine valence electrons for $SF_4$

S has 6 valence e⁻, each F has 7 valence e⁻. So total valence e⁻ = 6+(4×7)=34. There are 4 single - bonds in the skeleton, using 8 e⁻. Remaining e⁻ = 34 - 8 = 26.

Step5: Determine valence electrons for $SF_6$

S has 6 valence e⁻, each F has 7 valence e⁻. So total valence e⁻ = 6+(6×7)=48. There are 6 single - bonds in the skeleton, using 12 e⁻. Remaining e⁻ = 48 - 12 = 36.

Step6: Determine valence electrons for $BrF_3$

Br has 7 valence e⁻, each F has 7 valence e⁻. So total valence e⁻ = 7+(3×7)=28. There are 3 single - bonds in the skeleton, using 6 e⁻. Remaining e⁻ = 28 - 6 = 22.

Formula	Name	Skeleton	# of valence e⁻	e⁻ used in single bonds	Remaining e⁻	Complete Lewis structure
$[SO_4]^{2 -}$	sulfate anion	O - S - O (with 4 O around S)	32	8	24	S is the central atom with 4 single - bonds to O. Each O has 3 lone pairs and S has no lone pairs. The overall charge is - 2.
$PCl_5$	phosphorous pentachloride	Cl - P - Cl (with 5 Cl around P)	40	10	30	P is the central atom with 5 single - bonds to Cl. Each Cl has 3 lone pairs.
$SF_4$	Sulfur tetrafluoride	F - S - F (with 4 F around S)	34	8	26	S is the central atom with 4 single - bonds to F and 1 lone pair on S. Each F has 3 lone pairs.
$SF_6$	Sulfur hexafluoride	F - S - F (with 6 F around S)	48	12	36	S is the central atom with 6 single - bonds to F. Each F has 3 lone pairs.
$BrF_3$	Bromine trifluoride	F - Br - F (with 3 F around Br)	28	6	22	Br is the central atom with 3 single - bonds to F and 2 lone pairs on Br. Each F has 3 lone pairs.

Answer:

Formula	Name	# of valence e⁻	e⁻ used in single bonds	Remaining e⁻
$[SO_4]^{2 -}$	sulfate anion	32	8	24
$PCl_5$	phosphorous pentachloride	40	10	30
$SF_4$	Sulfur tetrafluoride	34	8	26
$SF_6$	Sulfur hexafluoride	48	12	36
$BrF_3$	Bromine trifluoride	28	6	22