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प्रश्न
Write the formula to calculate the bond order of molecule.
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उत्तर
Bond order of a molecule = `("N"_"b"-"N"_"a")/2`
where Nb is the number of electrons present in bonding MOs and Na is the number of electrons present in antibonding MOs.
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संबंधित प्रश्न
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Using data from the Table, answer the following:
| Examples | C2H6 Ethane | C2H4 Ethene | C2H2 Ethyne |
| Structure |
\[\begin{array}{cc} \backslash \phantom{......}/\phantom{.}\\ \ce{—C – C —}\\ /\phantom{......}\backslash\phantom{.}\end{array}\] |
\[\begin{array}{cc} \backslash \phantom{......}/\\ \ce{C \text{=} C}\\ /\phantom{......}\backslash\end{array}\] |
\[\ce{- C ≡ C -}\] |
| Type of bond between carbons | single | double | triple |
| Bond length (nm) | 0.154 | 0.134 | 0.120 |
| Bond Enthalpy kJ mol-1 | 348 | 612 | 837 |
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| Molecular Formula | Structural Formula | Shape/ Geometry | Bond angle |
| BeCl2 | 180° | ||
| O=C=O | Linear | ||
| C2H2 |
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(Nb = bonding electrons, Na = antibonding electrons)
