The heat of atomisation of PH_3(g) is 228 K.Cal mol^–1 and that of P₂H_4(g) is 355 K.Cal mol^–1 The energy of the P–P bond is (in K.Cal);

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answer is B.

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Detailed Solution

PH_3(g) → P_(g) + 3H_(g) ΔH=228 Kcal
3B.E(P-H) = 228
BE_P-H = 228/3
$\large B{E_{{P_2}{H_4}}} = 4B{E_{P - H}} + B{E_{P - P}}\$
$\large 355 = 4 \times \frac{{228}}{3} + B{E_{P - P}}\$
355=304+BE_p-p
BE_P-P= 335-304
= 51 Kcal

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