Which hydride among the following is less stable?
- HF
- BeH2
- NH3
- LiH
The Correct Option is B
Solution and Explanation
The stability of hydrides is related to the bond strength between the element and hydrogen. Among the given hydrides, \( \text{BeH}_2 \) is less stable due to its electron deficiency or hypovalency.
Explanation:
- \( \text{BeH}_2 \):
- Beryllium (Be) has 2 valence electrons, each forming a covalent bond with hydrogen to give \( \text{BeH}_2 \).
- According to the octet rule, Be needs 4 electrons in its valence shell to gain stability, but it only has two bonds with hydrogen.
- This electron deficiency makes \( \text{BeH}_2 \) less stable and prone to polymerization, forming polymeric \( (\text{BeH}_2)_n \).
- \( \text{HF}, \text{NH}_3, \) and \( \text{LiH} \):
- \( \text{HF} \): Has a strong polar covalent bond, making it stable.
- \( \text{NH}_3 \): Satisfies the octet rule and forms stable covalent bonds.
- \( \text{LiH} \): Is an ionic hydride formed between \( \text{Li}^+ \) and \( \text{H}^- \), making it stable.
Conclusion:
Final Answer: \( \text{BeH}_2 \) is less stable among the given hydrides.
Top Questions on Hydrides
- Consider the following statements:
Statement-I: H$_2$Se is more acidic than H$_2$Te.
Statement-II: H$_2$Se has higher bond dissociation enthalpy than H$_2$Te.
In light of the above statements, choose the correct option: A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is …….
- Identify the hydride which is not correctly matched with the example given in brackets.
- Which of the following reactions give $\mathrm{H_2}$ as one of the products? (Reactions are not balanced.)
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- $\mathrm{B_2H_6 + N(CH_3)_3}$
- $\mathrm{Al + NaOH + H_2O}$
- $\mathrm{BF_3 + NaH}$
- Among the following, which is an ionic hydride?
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