A xenon compound 'A' upon partial hydrolysis gives XeO2F2. The number of lone pair of electrons present in compound A is _________. (Round off to the Nearest Integer)
Answer (integer)
19
Solution
XeF<sub>6</sub> on partial hydrolysis form XeO<sub>2</sub>F<sub>2</sub>.<br><br>XeF<sub>6</sub> + H<sub>2</sub>O $\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{Hydrolysis}^{Partial}}$ XeO<sub>2</sub>F<sub>2</sub> + 2HF<br><br>In XeF<sub>6</sub>, central atom Xe has one lone pair all 6 fluorine have 3 lone pairs each.<br><br>So, total number of lone pair on XeF<sub>6</sub> = 1 + (6 $\times$ 3) = 19
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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