Question:
Identify the electron rich hydrides from the following:
Identify the electron rich hydrides from the following:
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Electron-rich hydrides contain lone pairs on the central atom (e.g., H\(_2\)O, HF, NH\(_3\)).
Updated On: Jun 20, 2026
- II & IV only
- III & IV only
- I & III only
- I & II only
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The Correct Option is C
Solution and Explanation
Step 1: Understand electron rich hydrides.
Electron rich hydrides are those hydrides in which the central atom has lone pairs or excess electron density, typically Group 15–17 hydrides or molecules with nonbonding electrons.
Step 2: Analyze H\(_2\)O (I).
Oxygen has two lone pairs in H\(_2\)O. These nonbonding electrons make it electron rich in nature. Hence H\(_2\)O is electron rich hydride.
Step 3: Analyze B\(_2\)H\(_6\) (II).
Diborane is electron deficient due to 3-center 2-electron bonds. Boron lacks octet completion, so it is electron deficient, not electron rich.
Step 4: Analyze HF (III).
Fluorine has three lone pairs and high electron density. Hence HF is electron rich hydride due to strong electronegativity and lone pairs.
Step 5: Analyze CH\(_4\) and ZrH (IV, V).
CH\(_4\) is neutral covalent hydride with no lone pairs on carbon. ZrH is metallic/interstitial hydride, not electron rich in this classification.
Step 6: Final conclusion.
Only H\(_2\)O and HF satisfy electron-rich hydride criteria. Therefore correct option is I & III.
Final Answer: \[ \boxed{\text{I \& III only}} \]
Electron rich hydrides are those hydrides in which the central atom has lone pairs or excess electron density, typically Group 15–17 hydrides or molecules with nonbonding electrons.
Step 2: Analyze H\(_2\)O (I).
Oxygen has two lone pairs in H\(_2\)O. These nonbonding electrons make it electron rich in nature. Hence H\(_2\)O is electron rich hydride.
Step 3: Analyze B\(_2\)H\(_6\) (II).
Diborane is electron deficient due to 3-center 2-electron bonds. Boron lacks octet completion, so it is electron deficient, not electron rich.
Step 4: Analyze HF (III).
Fluorine has three lone pairs and high electron density. Hence HF is electron rich hydride due to strong electronegativity and lone pairs.
Step 5: Analyze CH\(_4\) and ZrH (IV, V).
CH\(_4\) is neutral covalent hydride with no lone pairs on carbon. ZrH is metallic/interstitial hydride, not electron rich in this classification.
Step 6: Final conclusion.
Only H\(_2\)O and HF satisfy electron-rich hydride criteria. Therefore correct option is I & III.
Final Answer: \[ \boxed{\text{I \& III only}} \]
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