We can draw three inequivalent Lewis structures for the selenite ion. SeO3^2-. T

Question

We can draw three inequivalent Lewis structures for the selenite ion. SeO3^2-. The concepts of formal charge and electronegativity can help to choose the structure that is the best representation 1. Assign formal charges to the elements in each of the structures below. Note: Count oxygen atoms starting from the left for each structure. 2. The best Lewis structure for SeO3^2- is A B C Use the References button above to access important values if needed for this question. The length of a covalent bond depends upon the size of the atoms and the bond order. For each pair of covalently bonded atoms, choose the one expected to have the shortest bond length (A) H-Br (B) C-Br (A,B) (C) P-Br (D) P-Cl (C,D) 9 mole group attempts remaining Use the References button above to access important values if needed for this question The length of a covalent bond depends upon the size of the atoms and the bond order. For each pair of covalently bonded atoms, choose the one expected to have the shortest bond length. (A) N-N (B) N triple bond N (A,B) (C) C-O (D) C=O (C,D) 9 more group attempts remaining

Explanation / Answer

Picture 1

For A

Formal charge for Se = +1

for O1 =-1

for O2 = -1

for O3 = -1

For B

Formal charge on Se =-1

for O1 = 0

for O2 = 0

for O3 = -1

ForC

Formal charge on Se = 0

For O1 =0

for O2 = -1

for O3 = -1

Picture 2

Out of A,B the one with shorter bondlength is B (H-Br)

Out of C,D the one with shorter bond length is D (P-Cl)

Picture 3

Out of A,B the one with shorter bond length is B

Out of C,D the one with shorter bond length is D (C=O)

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