1. Tardigrade
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  3. Chemistry
  4. The boiling point of Kr and Rn are -152°C and -62°C respectively. Find the approximate boiling point of Xe

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Q. The boiling point of $Kr$ and $Rn$ are $-152^{\circ}C$ and $-62^{\circ}C$ respectively. Find the approximate boiling point of $Xe$

Classification of Elements and Periodicity in Properties

Solution:

According to the law of triad, the properties of middle element is average of rest two
$\therefore $ Boiling point of $Xe$ in a triad of $Kr, Xe, Rn$ =
(boiling point of $Kr$ + boiling point of Rn) / 2
$=(-152-62)/2=-107 \,{}^{\circ}C$