Among the hydrides given below which is reasonably electron precise hydride?

Options

A. `NH_(3)`
B. `SiH_(4)`
C. `NaH`
D. `H_(2)S`

Answer

To determine which among the given hydrides is a reasonably electron precise hydride, we need to analyze each hydride based on the definition of electron precise hydrides. Electron precise hydrides are those that have an exact number of electrons used in forming covalent bonds, without any excess electrons.

### Step-by-Step Solution:

1. **Understand the Definition of Electron Precise Hydrides**:
- Electron precise hydrides have an exact number of electrons that are used to form covalent bonds without any excess electrons.

2. **Analyze NH3 (Ammonia)**:
- Structure: NH3 has one nitrogen atom bonded to three hydrogen atoms.
- Nitrogen has 5 valence electrons (1s² 2s² 2p³).
- In NH3, nitrogen uses 3 electrons to bond with 3 hydrogen atoms, leaving 2 electrons as a lone pair.
- Conclusion: NH3 is an electron-rich hydride because it has excess electrons.

3. **Analyze SiH4 (Silane)**:
- Structure: SiH4 has one silicon atom bonded to four hydrogen atoms.
- Silicon has 4 valence electrons (1s² 2s² 2p⁶ 3s² 3p²).
- In SiH4, silicon uses all 4 valence electrons to bond with 4 hydrogen atoms, with no lone pairs remaining.
- Conclusion: SiH4 is an electron precise hydride because all electrons are used in bonding.

4. **Analyze NaH (Sodium Hydride)**:
- Structure: NaH is an ionic hydride.
- Sodium donates one electron to hydrogen, forming Na⁺ and H⁻.
- Conclusion: NaH does not form covalent bonds and is not an electron precise hydride.

5. **Analyze H2S (Hydrogen Sulfide)**:
- Structure: H2S has one sulfur atom bonded to two hydrogen atoms.
- Sulfur has 6 valence electrons (1s² 2s² 2p⁶ 3s² 3p⁴).
- In H2S, sulfur uses 2 electrons to bond with 2 hydrogen atoms, leaving 4 electrons as lone pairs.
- Conclusion: H2S is an electron-rich hydride because it has excess electrons.

6. **Final Conclusion**:
- Among the hydrides analyzed, SiH4 is the only hydride that is reasonably electron precise, as it uses all of its valence electrons in covalent bonding without any excess.

### Answer:
The reasonably electron precise hydride among the given options is **SiH4**.

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