Question

The ratio of number of lone pairs on central atom in ammonia, Water and XeF2 are
3.21
2:13
1:23
23.1

Answer 1

Answer:

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Answer 2

1:2:3

The ratio of number of lone pairs on central atom in $NH_{3}$, $H_{2}O$ and $XeF_{2}$ are 1:2:3

• $NH_{3}$ : Because only 2 of the nitrogen's 5 valence electrons are still available to form a lone pair, there are 3 bond pairs and 1 lone pair.

• $H_{2}O$ : Due to the fact that only 4 of the oxygen's 6 valence electrons remain after forming 2 lone pairs, there are two bond pairs and two lone pairs.

• $XeF_{2}$ : Because just six of the xenon's eight valence electrons are left, there are two bond pairs and three lone pairs.

The ratio of  number of lone pairs on central atom in $NH_{3}$, $H_{2}O$ and $XeF_{2}$ are therefore 1:2:3.

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