The bond dissociation energy of `F_(2)` ,is very low due to

Define bond dissociation energy.

The bond dissociation energy of N_(2)^(+) is less than that of N_(2) . Show why this is the case.

The average energy required to break a P-P bond in P_(4)(s) into gaseous atoms is 53.2 kcal mol^(-1) . The bond dissociation energy of H_(2)(g) is 104.2kcal mol^(-1) , Delta H_(f)^(0) of PH_(3)(g) from P_(4)(s) is 5.5 kcal mol^(-1) . The P-H bond energy in kcal mol^(-1) is [ Neglect presence of Van der Waals force in P_(4)(s) ]

If the bond dissociation energies of XY , X_(2) and Y_(2) are in the ratio of 1:1:0.5 and DeltaH_(f) for the formation of Xy is -200 KJ//mol . The bond dissociation energy of X_(2) will be :-

Statement-1 : Bond dissociation energy of F_(2) is more than Cl_(2) . Statement-2 : Due to smaller size of fluorine there are greater electron repulsions between the F atoms than Cl atoms.

The bond dissociation energy of B-F in BF_(3) is 646 kJ mol^(-1) whereas that of C-F in CF_(4) is 515 kJ mol^(-1) . The correct reason for higher B-F bond dissociation energy as compared to that of C-F in CF_(4) is

Account for the following : (i) NH_(3) is a stronger base than PH_(3) (ii) Sulphur has a greater tendency for catenation than oxygen. (iii) Bond dissociation energy of F_(2) is less than that of Cl_(2) .

Bond dissociation energy of F_(2) is less that of Cl_(2) give reason.

Assertion: Bond dissociation energy of F_(2) molecule is less than that of Cl_(2) molecule. Reason: Due to inter-electronic repulsion between F atom, F-F bond length in F_(2) molecule is higher than Cl-Cl bond lengthh in Cl_(2) molecule.

The enthalpy of formation of UF(g) is 22kcal mol^(-1) and that of U(g) is 128kcal mol^(-1) . The bond energy of the F-F bond is 37kcal mol^(-1) . The bond dissociation energy of UF(g) is (are):