Assign oxidation states for all atoms in each of the following compounds a. \({KMnO}_{4} \quad\quad\quad f. {Fe}_{3} {O}_{4}\) b. \({NiO}_{2} \quad\quad\quad\quad g. {XeOF}_{4}\) c. \({Na}_{4} {Fe}({OH})_{6} \quad h. {SF}_{4}\) d. \({NH}_{4} {h}_{2} {HPO}_{4} \quad i. {CO}\) e. \({P}_{4} {O}_{6} \quad\quad\quad\quad\quad j. {C}_{6} {H}_{12} {O}_{6}\)

Short Answer

Expert verified
The oxidation states for the given compounds are: a. K(+1), Mn(+7), O(-2) b. Ni(+4), O(-2) c. Na(+1), Fe(+2), O(-2), H(+1) d. N(-3), H(+1), P(+5) e. P(+3), O(-2) f. Fe(+2, +3), O(-2) g. Xe(+6), O(-2), F(-1) h. S(+4), F(-1) i. C(+2), O(-2) j. C(0), H(+1), O(-2)

Step by step solution

01

Assign oxidation states for K, Mn, and O

Assign +1 for K, -2 for O. The sum of the oxidation states must be zero since it's a neutral compound. \(+1 + x + 4(-2) = 0\) \(x = +7\) So, the oxidation states are K(+1), Mn(+7), and O(-2). b. NiO2
02

Assign oxidation states for Ni and O

Assign -2 for O. The sum of oxidation states must be zero since it's a neutral compound. \(x + 2(-2) = 0\) \(x = +4\) So, the oxidation states are Ni(+4) and O(-2). c. Na4Fe(OH)6
03

Assign oxidation states for Na, Fe, O, and H

Assign +1 for Na, -2 for O, and +1 for H. The sum of oxidation states must be zero since it's a neutral compound. \(4(+1) + x + 6(-2) + 6(+1) = 0\) \(x = +2\) So, the oxidation states are Na(+1), Fe(+2), O(-2), and H(+1). d. NH4H2HPO4
04

Assign oxidation states for N, H, and P

Assign +1 for H (both in NH4 and H2). Assign -2 for O. The sum of oxidation states must be zero since it's a neutral compound. \(+1 + 4(+1) + 2(+1) + x + 4(-2) = 0\) \(x = +5\) So, the oxidation states are N(-3), H(+1), and P(+5). e. P4O6
05

Assign oxidation states for P and O

Assign -2 for O. The sum of oxidation states must be zero. \(4x + 6(-2) = 0\) \(x = +3\) So, the oxidation states are P(+3) and O(-2). f. Fe3O4
06

Assign oxidation states for Fe and O

Assign -2 for O. We know iron can have oxidation states of +2 or +3, and the sum of oxidation states must be zero. \(3x + 4(-2) = 0\) It becomes clear that there are two Fe(+3) and one Fe(+2). So, the oxidation states are Fe(+2, +3) and O(-2). g. XeOF4
07

Assign oxidation states for Xe, O, and F

Assign -2 for O and -1 for F. The sum of oxidation states must be zero. \(x - 2 + 4(-1) = 0\) \(x = +6\) So, the oxidation states are Xe(+6), O(-2), and F(-1). h. SF4
08

Assign oxidation states for S and F

Assign -1 for F. The sum of oxidation states must be zero. \(x + 4(-1) = 0\) \(x = +4\) So, the oxidation states are S(+4) and F(-1). i. CO
09

Assign oxidation states for C and O

Assign -2 for O. The sum of oxidation states must be zero. \(x + (-2) = 0\) \(x = +2\) So, the oxidation states are C(+2) and O(-2). j. C6H12O6
10

Assign oxidation states for C, H, and O

Assign -2 for O and +1 for H. In most organic compounds, carbon has an oxidation state of -4, 0, or +4. Each carbon atom \(x = \frac{12(+1) + 6(-2)}{6}\); thus, the oxidation state of carbon is 0. So, the oxidation states are C(0), H(+1), and O(-2).

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