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प्रश्न
How will you account for 104.5° bond angle in water?
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उत्तर
In water hybridisation of oxygen is sp3. Angle should be 109° (approx.) but due to LP – LP repulsion bond angle reduces to 104.5°.
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संबंधित प्रश्न
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| \[\ce{H, O}\] | \[\ce{D2O}\] | |
| Melting point / K | 373.0 | 374.4 |
| Enthalpy of vapourisation at (373 K)/kJ mol–1 | 40.66 | 41.61 |
| Viscosity/centipoise | 0.8903 | 1.107 |
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