Question

Arrange the following in order of decreasing oxidation number of the nitrogen atom: \({N}_{2} {NH}_{3}\) \({N}_{2} {O}_{4}, {N}_{2} {O}, {N}_{2} {H}_{4}\) and \({NO}_{3}^{-}\)

Short answer

\(NO_3^- > N_2O_4 > N_2O > N_2 > N_2H_4 > NH_3\)

Step-by-step solution

Determine Oxidation Number Rules

Recall the general rules for oxidation numbers: (1) The oxidation number of an atom in a free element is always 0. (2) The oxidation number of a monatomic ion is equal to the charge of that ion. (3) The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it is equal to the charge of the ion.

Oxidation Number for \(N_2\)

For \(N_2\) (dinitrogen), since it is a diatomic molecule of the same element, the oxidation number of nitrogen is 0.

Oxidation Number for \(NH_3\)

In \(NH_3\) (ammonia), hydrogen has an oxidation number of +1. Let the oxidation number of nitrogen be \(x\). Then the equation becomes: \([x + 3(+1) = 0]\), solving for \(x\) gives \([-3]\).

Oxidation Number for \(N_2O_4\)

In \(N_2O_4\) (dinitrogen tetroxide), oxygen has an oxidation number of -2. Let the oxidation number of nitrogen be \(x\). The equation is: \([2x + 4(-2) = 0]\), solving for \(x\) gives \([+4/2 = +2]\).

Oxidation Number for \(N_2O\)

In \(N_2O\) (dinitrogen monoxide), oxygen has an oxidation number of -2. Let the oxidation number of nitrogen be \(x\). The equation becomes: \([2x + (-2) = 0]\), solving for \(x\) gives \([+1]\).

Oxidation Number for \(N_2H_4\)

In \(N_2H_4\) (hydrazine), hydrogen has an oxidation number of +1. Let the oxidation number of nitrogen be \(x\). The equation is: \([2x + 4(+1) = 0]\), solving for \(x\) gives \([-4/2 = -2]\).

Oxidation Number for \(NO_3^-\)

In \(NO_3^-\) (nitrate ion), oxygen has an oxidation number of -2. Let the oxidation number of nitrogen be \(x\). The equation becomes: \([x + 3(-2) = -1]\), solving for \(x\) gives \([+5]\).

Arrange in Decreasing Order

Based on the oxidation numbers found: \(NO_3^- = +5\), \(N_2O_4 = +2\), \(N_2O = +1\), \(N_2 = 0\), \(N_2H_4 = -2\), \(NH_3 = -3\). So, the order is: \(NO_3^- > N_2O_4 > N_2O > N_2 > N_2H_4 > NH_3\).

Key concepts

oxidation states

Oxidation states, also known as oxidation numbers, are a key concept in understanding chemical reactions, particularly redox (reduction-oxidation) reactions. They represent the charge an atom would have if all bonds to atoms of different elements were 100% ionic.
Oxidation states help in determining the electron transfer in reactions, indicating how many electrons an atom gains, loses, or shares.
In the step-by-step solution above, oxidation numbers were assigned to each nitrogen compound by following specific rules.

• In \(N_2\), the oxidation number of nitrogen is 0 because it is a diatomic molecule of the same element.
• In \(NH_3\), the nitrogen has an oxidation state of -3, calculated by balancing the sum of the oxidation numbers in the compound to zero.
• In \(NO_3^-\), nitrogen has an oxidation state of +5, determined by ensuring the overall sum matches the charge on the ion.

This systematic approach allows us to rank the compounds by their oxidation states.

chemical formulas

Chemical formulas represent compounds using symbols for their constituent elements and subscripts indicating the number of each atom.
These formulas help in understanding the composition and structure of compounds.
For example, in the exercise:

• \(NH_3\) represents ammonia, composed of one nitrogen and three hydrogen atoms.
• \(N_2\) represents dinitrogen, a diatomic molecule consisting of two nitrogen atoms.
• \(N_2O_4\) is dinitrogen tetroxide, with two nitrogen and four oxygen atoms.

The subscripts provide critical information needed to balance oxidation numbers and understand chemical behavior.
Chemical formulas are essential in writing and balancing chemical equations, predicting reaction outcomes, and calculating molecular masses.

balancing equations

Balancing equations is an essential skill in chemistry. It ensures that the same number of each type of atom appears on both sides of the equation, maintaining the law of conservation of mass.
During the process of assigning oxidation numbers, the equations inside each step must be balanced to correctly determine the oxidation state of atoms.
For instance, in \(N_2O_4\), the equation \[2x + 4(-2) = 0\] balances the atoms and allows the calculation of nitrogen's oxidation number.
Without balancing equations, the calculations of oxidation states would be inaccurate, leading to incorrect predictions about the chemical reactions.
Balancing also plays a crucial role in understanding stoichiometry, determining reactant-product ratios, and conserving elemental mass.

oxidation rules

Oxidation rules are guidelines for assigning oxidation numbers to elements within compounds and ions.
These rules help with accurate calculations and consistency in determining oxidation states.

• The oxidation number of a free element is always 0. For example, \(N_2\) has an oxidation state of 0.
• For monatomic ions, the oxidation number is equal to the charge. Thus, a \((NH_4)^+\) ion sees nitrogen as -3 to balance the positives from hydrogen.
• The sum of oxidation numbers in a neutral compound must be 0, and in a polyatomic ion, it must equal the charge of the ion. This helps in calculating unknown oxidation states as in \(NO_3^-\).

Applying these rules systematically ensures the correct determination of oxidation states across different chemical compounds.

redox reactions

Redox reactions involve the transfer of electrons between species, leading to changes in their oxidation states.
Understanding oxidation states is crucial in identifying which atoms are oxidized (lose electrons) and which are reduced (gain electrons).

• In the example given, nitrogen in \(NH_3\) (oxidation state -3) can be oxidized to \(NO_3^-\) (oxidation state +5), showing it loses electrons.
• Conversely, a reduction reaction occurs if nitrogen moves from \(NO_3^-\) to \(NH_3\).

Balancing redox reactions requires balancing both the mass and charge, ensuring that electron transfer is accounted for on both sides.
Mastery of redox reactions is essential for understanding many biological, industrial, and environmental processes.