Order the following molecules from lowest to highest oxidation state of the nitrogen atom: $\mathrm{HHO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}$ \(\mathrm{NaNO}_{2}\)

Each atom in a molecule has an oxidation state which helps in keeping track of electrons in a chemical reaction. The following rules should be followed to assign oxidation states: 1. The oxidation state of any pure element (in its elemental form) is 0. 2. The oxidation state of a monatomic ion is equal to its charge. 3. Oxygen usually has an oxidation state of -2, with some exceptions. 4. Hydrogen has an oxidation state of +1 when bonded to non-metals and -1 when bonded to metals. 5. The sum of the oxidation states of all atoms in a compound is equal to the overall charge on the compound. Now, we can apply these rules to find the oxidation states of nitrogen in the given molecules.

1. HHO3: In this molecule, H (hydrogen) has an oxidation state of +1 and O (oxygen) has an oxidation state of -2. Since there are 3 oxygen atoms, the total contribution from O atoms is 3*(-2) = -6. The contribution from H atoms is 1*1 = 1. Hence, the oxidation state of N (nitrogen) is x + 1 - 6 = 0; x = 5. 2. NH4Cl: The compound can be written as (NH4)+ (Cl)- to show that it's an ionic compound, where (NH4)+ is the ammonium ion and (Cl)- is the chloride ion. In this compound, Cl has an oxidation state of -1. In the ammonium ion (NH4)+, the contribution from four H atoms is 4*1 = 4. Hence, the oxidation state of N is x + 4 = 1; x = -3. 3. N2O: In this molecule, O has an oxidation state of -2. So, the oxidation state of N is x + x - 2 = 0; x = 1. 4. NO2: In this molecule, O has an oxidation state of -2. Since there are 2 oxygen atoms, the total contribution from O atoms is 2*(-2) = -4. Hence, the oxidation state of N is x - 4 = 0; x = 4. 5. NaNO2: Here, we have an ionic compound Na+ (NO2)-, where Na has an oxidation state of +1. The NO2- ion contribution is -1, since it's a nitrite ion. In the NO2- ion, O has an oxidation state of 2*(-2) = -4. Therefore, the oxidation state of N is x - 4 = -1; x = 3.

Now that we have the oxidation states of nitrogen in each molecule, we can arrange them from lowest to highest: 1. Lowest oxidation state: NH4Cl (-3) 2. N2O (1) 3. NaNO2 (3) 4. NO2 (4) 5. Highest oxidation state: HHO3 (5) So, the correct order is: \(NH_4Cl\), \(N_2O\), \(NaNO_2\), \(NO_2\), \(HHO_3\).