Li3N(s) + 2 H2(g) <-> LiNH2(s) + 2 LiH(s) dH° = -192 kJ/mol_rxn Because pure H2 is a hazardous subst — Stoichiometry Chemistry Question

STEPS:

1. Identify the reactant containing hydrogen and determine its initial oxidation state:
In the reactants, hydrogen exists as elemental hydrogen gas, $\text{H}_2(g)$. By definition, the oxidation state of any atom in its pure, elemental form is exactly $0$.

2. Identify the products containing hydrogen:
In the products, hydrogen is partitioned into two different solid compounds: lithium amide ($\text{LiNH}_2(s)$) and lithium hydride ($\text{LiH}(s)$).

3. Determine the oxidation state of hydrogen in $\text{LiNH}_2$:
* Lithium ($\text{Li}$) is an alkali metal (Group 1) and always has an oxidation state of $+1$ in its compounds.
* The amide ion is $\text{NH}_2^-$. Since nitrogen ($\text{N}$) is significantly more electronegative than hydrogen ($\text{H}$), nitrogen takes its standard oxidation state of $-3$.
* Set up the charge balance equation for the neutral $\text{LiNH}_2$ compound:
$$(+1)_{\text{Li}} + (-3)_{\text{N}} + 2(x)_{\text{H}} = 0 \implies 2x = +2 \implies x = \mathbf{+1}$$
* Thus, the hydrogen atoms in $\text{LiNH}_2$ have an oxidation state of $+1$.

4. Determine the oxidation state of hydrogen in $\text{LiH}$:
* Lithium hydride is an ionic compound where metal cations are bonded to hydride anions.
* Lithium ($\text{Li}$) maintains its stable oxidation state of $+1$.
* Since $\text{LiH}$ is a neutral compound, the hydride ion ($\text{H}^-$) must balance this charge:
$$(+1)_{\text{Li}} + (y)_{\text{H}} = 0 \implies y = \mathbf{-1}$$
* Thus, the hydrogen atoms in $\text{LiH}$ have an oxidation state of $-1$.

5. Compare the initial and final oxidation states of hydrogen:
* Oxidation: Some hydrogen atoms transition from an oxidation state of $0$ in $\text{H}_2$ to $+1$ in $\text{LiNH}_2$. Because their oxidation number increased, these hydrogen atoms are oxidized.
* Reduction: Other hydrogen atoms transition from an oxidation state of $0$ in $\text{H}_2$ to $-1$ in $\text{LiH}$. Because their oxidation number decreased, these hydrogen atoms are reduced.

6. Conclude:
Because hydrogen atoms in this single chemical process undergo both an increase and a decrease in oxidation state, the H atoms are both oxidized and reduced, which makes Option C the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: While it is true that some hydrogen atoms are oxidized to $+1$ in $\text{LiNH}_2$, this statement is incomplete because it ignores the hydrogen atoms that are simultaneously reduced to $-1$ in $\text{LiH}$.
• Option B is incorrect: While it is true that some hydrogen atoms are reduced to $-1$ in $\text{LiH}$, this statement is incomplete because it ignores the hydrogen atoms that are simultaneously oxidized to $+1$ in $\text{LiNH}_2$.
• Option D is incorrect: This option falsely claims that no redox chemistry occurs. Because the oxidation states of hydrogen shift from $0$ to both $+1$ and $-1$, a clear oxidation-reduction (disproportionation-like) reaction has occurred.