Question

$\frac{1}{2}{H}_2\left(g\right)\to H\left(g\right)\Delta H_f=50\text{kcal/mol}$
$P{H}_3\left(g\right)\to P\left(g\right)+3H\left(g\right)\Delta H_{\text{dissociation}}=243\text{kcal/mol}$
P - P bond dissociation energy is $x\times \text{10 kcal/mol}$. The value of $x$ in nearest possible integer is

Answer 1

$D_{H-H}=50\times 2kcal/mol{D}_{P-H}=\frac{243}{3}=81kcal/mol{X}_p-X_H=0.208\sqrt{\Delta };X-electronegativity,$
Since P and H have half filled electronic configuration,
$X_P=X_H,Hence\Delta =0$
$\Delta =D_{P-H}-\sqrt{\left(D_{P-P}\right)\left(D_{H-H}\right)}=0$
$D_{P-H}^2=\left(D_{P-P}\right)\left(D_{H-H}\right)$
$D_{P-P}=\frac{D_{P-H}^2}{D_{H-H}}$
$=\frac{81\times 81}{100}$
$=65.61kcal/mol{D}_{P-P}=6.5\times 10kcal/molx=7$