2- Indicate the oxidation number of phosphorus- iodine- nitrogen- tell.docx

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2. Indicate the oxidation number of phosphorus, iodine, nitrogen, tellurium, and sli- con in the following ions. b, PO33- c. HP042- e. I03- f. 102- Io2i. N n. Teo o. Si03 O3 NO, Solution 2. a. PO4^3- x + (-2*4) = -3 x-8+3 = 0 x= +5 The oxidation number of phosphorus is +5 b. PO3^3- x+ (-2*3) = -3 x-6+3 = 0 x = +3 The oxidation number of phosphorus is +3 c. HPO4^2- 1+ x + (-2*4) = -2 1+x-8+2 = 0 x-5 = 0 x = +5 The oxidation number of phosphorus is +5 d. P3O10^5- 3x+ (-2*10) = -5 3x-20+5 = 0 3x-15 = 0 3x = 15 x = +5 The oxidation number of phosphorus is +5 e. IO3^- x+ (-2*3) = -1 x-6+1 = 0 x = +5 The oxidation number of iodine is +5 f. IO2^- x + (-2*2) = -1 x-4+1 = 0 x = +3 The oxidation number of iodine is +3 g. IO^- x+(-2*1) = -1 x-2+1 = 0 x -1 = 0 x = +1 The oxidation number of iodine is +1 h. NH4^+ x + 1*4 = +1 xÂ Â = +1-4 x = -3 The oxidation number of nitrogen is -3 i. NO3^- x + (-2*3) = -1 x-6+1 = 0 x-5 = 0 x = +5 The oxidation number of nitrogen is +5 J. NO2^- x+(-2*2) = -1 x = -1+4 x = +3 The oxidation number of nitrogen is +3 K. NO^+ x + (-2*1) = +1 x-2-1 = 0 x-3 = 0 x = +3 The oxidation number of nitrogen is +3 l. NO2^+ x + (-2*2) = +1 x-4-1 = 0 x = +5 The oxidation number of nitrogen is +5 m. N2O2^2- 2x+ (-2*2) = -2 2x-4+2 = 0 2x-2 = 0 2x = 2 x + 1 The oxidation number ofnitrogen is +1 n. TeO4^2- x + (-2*4) = -2 x-8+2 = 0 x-6 = 0 x = +6 The oxidation number of tellurium is +6 o. SiO3^2- x + (-2*3) = -2 x-6+2 = 0 x = +4 The oxidation number of silicon is +4 .

Educação

x-4+1 = 0
x = +3
The oxidation number of iodine is +3
g. IO^-
x+(-2*1) = -1
x-2+1 = 0
x -1 = 0
x = +1
The oxidation number of iodine is +1
h. NH4^+
x + 1*4 = +1
xÂ Â = +1-4
x = -3
The oxidation number of nitrogen is -3
i. NO3^-
x + (-2*3) = -1
x-6+1 = 0
x-5 = 0
x = +5
The oxidation number of nitrogen is +5
J. NO2^-
x+(-2*2) = -1
x = -1+4
x = +3
The oxidation number of nitrogen is +3
K. NO^+
x + (-2*1) = +1
x-2-1 = 0
x-3 = 0
x = +3
The oxidation number of nitrogen is +3
l. NO2^+
x + (-2*2) = +1
x-4-1 = 0
x = +5
The oxidation number of nitrogen is +5
m. N2O2^2-
2x+ (-2*2) = -2
2x-4+2 = 0
2x-2 = 0
2x = 2
x + 1
The oxidation number ofnitrogen is +1
n. TeO4^2-
x + (-2*4) = -2
x-8+2 = 0

x-6 = 0
x = +6
The oxidation number of tellurium is +6
o. SiO3^2-
x + (-2*3) = -2
x-6+2 = 0
x = +4
The oxidation number of silicon is +4

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2- Indicate the oxidation number of phosphorus- iodine- nitrogen- tell.docx

1. 2. Indicate the oxidation number of phosphorus, iodine, nitrogen, tellurium, and sli- con in the following ions. b, PO33- c. HP042- e. I03- f. 102- Io2i. N n. Teo o. Si03 O3 NO, Solution 2. a. PO4^3- x + (-2*4) = -3 x-8+3 = 0 x= +5 The oxidation number of phosphorus is +5 b. PO3^3- x+ (-2*3) = -3 x-6+3 = 0 x = +3 The oxidation number of phosphorus is +3 c. HPO4^2- 1+ x + (-2*4) = -2 1+x-8+2 = 0 x-5 = 0 x = +5 The oxidation number of phosphorus is +5 d. P3O10^5- 3x+ (-2*10) = -5 3x-20+5 = 0 3x-15 = 0 3x = 15 x = +5 The oxidation number of phosphorus is +5 e. IO3^- x+ (-2*3) = -1 x-6+1 = 0 x = +5 The oxidation number of iodine is +5 f. IO2^- x + (-2*2) = -1

2. x-4+1 = 0 x = +3 The oxidation number of iodine is +3 g. IO^- x+(-2*1) = -1 x-2+1 = 0 x -1 = 0 x = +1 The oxidation number of iodine is +1 h. NH4^+ x + 1*4 = +1 xÂ Â = +1-4 x = -3 The oxidation number of nitrogen is -3 i. NO3^- x + (-2*3) = -1 x-6+1 = 0 x-5 = 0 x = +5 The oxidation number of nitrogen is +5 J. NO2^- x+(-2*2) = -1 x = -1+4 x = +3 The oxidation number of nitrogen is +3 K. NO^+ x + (-2*1) = +1 x-2-1 = 0 x-3 = 0 x = +3 The oxidation number of nitrogen is +3 l. NO2^+ x + (-2*2) = +1 x-4-1 = 0 x = +5 The oxidation number of nitrogen is +5 m. N2O2^2- 2x+ (-2*2) = -2 2x-4+2 = 0 2x-2 = 0 2x = 2 x + 1 The oxidation number ofnitrogen is +1 n. TeO4^2- x + (-2*4) = -2 x-8+2 = 0

3. x-6 = 0 x = +6 The oxidation number of tellurium is +6 o. SiO3^2- x + (-2*3) = -2 x-6+2 = 0 x = +4 The oxidation number of silicon is +4

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