JEE Advance - Chemistry (2019 - Paper 1 Offline - No. 17)
At 143 K, the reaction of XeF4 with O2F2 produces a xenon compound Y. The total number of lone pair(s) of electrons present on the whole molecule of Y is .................
Answer
19
Explanation
XeF4 reacts with O2F2 to form XeF6.O2F2 is fluoronating reagent.
$$Xe{F_4} + {O_2}{F_2}\buildrel {143\,K} \over \longrightarrow \mathop {Xe{F_6}}\limits_{(Y)} + {O_2}$$
The structure of XeF6 is
Y compound (XeF6) has 3 lone pair in each fluorine and one lone pair in xenon.
Hence, total number of lone pairs electrons is 19.
$$Xe{F_4} + {O_2}{F_2}\buildrel {143\,K} \over \longrightarrow \mathop {Xe{F_6}}\limits_{(Y)} + {O_2}$$
The structure of XeF6 is
Y compound (XeF6) has 3 lone pair in each fluorine and one lone pair in xenon.
Hence, total number of lone pairs electrons is 19.
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