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JEE Advanced 2019Paper 1

Exam: JEE Advanced
Year: 2019
Paper: Paper 1
Subject: Chemistry
Chapter: p Block Elements (Group 15, 16, 17 & 18)

JEE Advanced 2019 · Paper 1 · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

At 143K, the reaction of Xe F _ 4 with O _ 2 F _ 2 produces a xenon...

At

143K,

the reaction of

Xe{F}_{4}

with

{O}_{2}{F}_{2}

produces a xenon compound

Y .

The total number of lone pair(s) of electrons present on the whole molecule of

Y

is __________ At 143K, the reaction of Xe F _ 4 with O _ 2 F _ 2 produces a xenon compound Y . The total number of lone pair(s) of

At 143K, the reaction of Xe F _ 4 with O _ 2 F _ 2 produces a xenon compound Y . The total number of lone pair(s) of

Correct answer

0

Step-by-step solution

${XeF}_{4} + {O}_{2}{F}_{2} \to {XeF}_{6} + {O}_{2}$

So $Y$ is ${XeF}_{6}$

Each $F$ atom contains three lone pairs and $Xe$ atom contain one lone pair.

so total lone pair $= 3 \times 6 + 1 = 19$

Total $L . P = 19$ .
At 143K, the reaction of Xe F _ 4 with O _ 2 F _ 2 produces a xenon compound Y . The total number of lone pair(s) of

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