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Q.

The heat of atomisation of PH3(g) is 228 K.Cal mol–1 and that of P2H4(g) is 355 K.Cal mol–1 The energy of the P–P bond is (in K.Cal);

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a

26

b

204

c

51

d

102

answer is B.

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Detailed Solution

PH3(g) → P(g) + 3H(g) ΔH=228 Kcal
3B.E(P-H) = 228
BEP-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+BEp-p
BEP-P = 335-304
           = 51 Kcal

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The heat of atomisation of PH3(g) is 228 K.Cal mol–1 and that of P2H4(g) is 355 K.Cal mol–1 The energy of the P–P bond is (in K.Cal);