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प्रश्न
Arrange the following in the order of property indicated for the given set:
F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
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उत्तर १
- Bond dissociation enthalpy decreases as the bond distance increases from F2 to I2 due to the increase in the size of the atom on moving from F to I.
- The F-F bond dissociation enthalpy is smaller than the Cl-Cl and even smaller than the Br-Br.
- This is because the F atom is very small and has large electron-electron repulsion among the lone pairs of electrons in F2 molecule where they are much closer to each other than in the case of Cl2.
- The increasing order of bond dissociation enthalpy is:
I, < F2 < Br2 < Cl2
उत्तर २
I2 < F2 < Br2 < Cl2
Due to an increase in atom size, bond dissociation enthalpy reduces as the X-X bond distance in the X2 molecule increases. But F2 is an anomaly since its bond dissociation enthalpy is lower than Br2 and Cl2. This results from the F atom’s extremely small size, as the F-F bond pair in the F2 molecule is repelled by the three lone pairs of electrons on it.
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(iii) \[\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}\]
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