Reactivity with Hydrogen: The Hydrides
Master the critical trends of $H_2E$ compounds: Acidity, Stability, and the Water Anomaly.
All elements of Group 16 (Oxygen, Sulfur, Selenium, Tellurium, and Polonium) react with hydrogen to form hydrides of the general formula $H_2E$. While $H_2O$ is an odorless liquid essential for life, the other hydrides ($H_2S, H_2Se, H_2Te$) are highly toxic, foul-smelling gases. The trends in their bond angles, acidic character, and thermal stability are high-yield topics for competitive exams.
1. Structure and Bond Angle
The central atom in these hydrides possesses 6 valence electrons. Two are used to form covalent bonds with hydrogen, leaving two lone pairs. This gives 4 electron domains, corresponding to $sp^3$ hybridization.
Due to strong lone pair - lone pair repulsion, the perfect tetrahedral angle of $109.5^\circ$ is compressed, resulting in a Bent or V-shaped geometry.
The Bond Angle Trend & Drago's Rule
As we move down the group, the bond angle drops dramatically from $104.5^\circ$ in water to nearly $90^\circ$ for the heavier hydrides.
Down the group, the electronegativity of the central atom decreases. The shared bonding electrons are pulled closer to the Hydrogen atoms. As the bond pairs move further away from the central atom, the repulsion between them decreases. This allows the two lone pairs on the central atom to push the bonds much closer together.
Advanced Explanation (Drago's Rule):
For elements in Period 3 and below (S, Se, Te), the energy difference between the $s$ and $p$ orbitals is large, and their electronegativity is low ($\leq 2.5$). Therefore, hybridization does not occur. The central atom uses pure, unhybridized $p$-orbitals to bond with hydrogen. Since $p$-orbitals are exactly $90^\circ$ apart, the bond angles of $H_2S, H_2Se,$ and $H_2Te$ are very close to $90^\circ$.
Figure 1: The bond angle drops from the tetrahedral ideal in H₂O to pure orthogonal p-orbitals in H₂Te.
2. Thermal Stability & Acidic Character
These two properties are inversely related. They both depend entirely on the strength of the $E-H$ bond. Because atomic size increases from Oxygen to Polonium, the $E-H$ bond length increases, and its Bond Dissociation Enthalpy drastically decreases.
Thermal stability decreases down the group.
Reason: The long $Te-H$ and $Po-H$ bonds are very weak. They break easily upon heating, meaning these heavier hydrides are highly unstable.
Acidic character increases down the group.
Reason: An acid releases $H^+$ ions in solution. Because the bond gets weaker and easier to break down the group, $H_2Te$ releases its hydrogen ion most readily, making it the strongest acid.
3. Reducing Character
A good reducing agent readily supplies hydrogen to other substances. The ability to supply hydrogen depends directly on how easily the $E-H$ bond breaks.
- Water ($H_2O$) is extremely stable and does not break easily to provide hydrogen. Thus, it is not a reducing agent.
- Hydrogen Telluride ($H_2Te$) has a very weak bond, breaks easily, and is therefore a strong reducing agent.
4. Physical State & Boiling Point Anomaly
Hydrides are covalent compounds. Their boiling points should depend on van der Waals forces, which increase with molecular mass. Therefore, boiling point should increase steadily down the group. However, Water breaks the rule spectacularly.
Due to the small size and extraordinarily high electronegativity of Oxygen, water molecules form extensive and strong intermolecular Hydrogen Bonds ($\cdots H-O-H \cdots OH_2 \cdots$).
Sulfur and the heavier elements have low electronegativity and large sizes; they cannot form hydrogen bonds. They rely purely on weak van der Waals forces. Therefore, $H_2S$, $H_2Se$, and $H_2Te$ are foul-smelling, toxic gases at room temperature, while $H_2O$ is a liquid with an anomalously high boiling point ($100^\circ\text{C}$).
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