White phosphorus is a tetra-atomic solid $(P_{4})$ at room temperature.

Given: $Δ H_{s u b l ima t i o n}$ of P₄(s) = 59 kJ/mole
$Δ H_{a t o mi s a t i o n}$ of P₄(s) = 1506.5 kJ/mole.
Use 1 eV/atom = 96.5 kJ mol⁻¹ and $h c = 12400 e V \overset{˚}{A}$ .

Question

White phosphorus is a tetra-atomic solid $(P_{4})$ at room temperature.

Given: $Δ H_{s u b l ima t i o n}$ of P₄(s) = 59 kJ/mole

$Δ H_{a t o mi s a t i o n}$ of P₄(s) = 1506.5 kJ/mole.

Use 1 eV/atom = 96.5 kJ mol⁻¹ and $h c = 12400 e V \overset{˚}{A}$ .

TestHub · Accepted Answer

1506.5 = 59 + 6 BE P − P ​ or, BE P − P ​ = 241.25 kJ mol − 1 = 2.5 eV per bond. E = λ 12400 ​ ⇒ λ = 2.5 12400 ​ = 4960 A ˚

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