Question

drop each atom/ion in the correct category. an atom/ion with the oxidation number +7 an atom/ion with the oxidation number +4 an atom/ion with the oxidation number -2 an atom/ion with the oxidation number +1 an atom/ion with the oxidation number -3 mn in kmno4 au in aubr o in h2o n in nh3 pb in pbo2 h in h2o cl in cl2o7 v in vcl4

Explanation:

Step1: Recall oxidation - number rules

Oxidation number of oxygen in most compounds is - 2, hydrogen is + 1. In a neutral compound, the sum of oxidation numbers of all atoms is zero. In a poly - atomic ion, the sum of oxidation numbers equals the charge of the ion.

Step2: Calculate oxidation number of Mn in \(KMnO_4\)

Let the oxidation number of Mn be \(x\). In \(KMnO_4\), K has an oxidation number of + 1 and O has an oxidation number of - 2. The compound is neutral, so \(+1+x + 4\times(-2)=0\). Solving for \(x\): \(1 + x-8 = 0\), \(x=+7\).

Step3: Calculate oxidation number of Au in AuBr

Let the oxidation number of Au be \(y\). Br has an oxidation number of - 1. The compound is neutral, so \(y+( - 1)=0\), \(y = + 1\).

Step4: Oxidation number of O in \(H_2O\)

Oxygen has an oxidation number of - 2 in \(H_2O\) (by the general rule for oxygen in most compounds).

Step5: Calculate oxidation number of N in \(NH_3\)

Let the oxidation number of N be \(z\). H has an oxidation number of + 1. The compound is neutral, so \(z + 3\times(+1)=0\), \(z=-3\).

Step6: Calculate oxidation number of Pb in \(PbO_2\)

Let the oxidation number of Pb be \(w\). O has an oxidation number of - 2. The compound is neutral, so \(w+2\times(-2)=0\), \(w = + 4\).

Step7: Oxidation number of H in \(H_2O\)

Hydrogen has an oxidation number of + 1 in \(H_2O\) (by the general rule for hydrogen in most compounds).

Step8: Calculate oxidation number of Cl in \(Cl_2O_7\)

Let the oxidation number of Cl be \(a\). O has an oxidation number of - 2. The compound is neutral, so \(2a+7\times(-2)=0\), \(2a = 14\), \(a=+7\).

Step9: Calculate oxidation number of V in \(VCl_4\)

Let the oxidation number of V be \(b\). Cl has an oxidation number of - 1. The compound is neutral, so \(b+4\times(-1)=0\), \(b = + 4\).

Oxidation Number	Atoms/Ions
+ 4	Pb in \(PbO_2\), V in \(VCl_4\)
- 2	O in \(H_2O\)
+ 1	Au in AuBr, H in \(H_2O\)
- 3	N in \(NH_3\)

Answer:

An atom/ion with the oxidation number + 7: Mn in \(KMnO_4\), Cl in \(Cl_2O_7\)
An atom/ion with the oxidation number + 4: Pb in \(PbO_2\), V in \(VCl_4\)
An atom/ion with the oxidation number - 2: O in \(H_2O\)
An atom/ion with the oxidation number + 1: Au in AuBr, H in \(H_2O\)
An atom/ion with the oxidation number - 3: N in \(NH_3\)