For each of the following, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybridization of the central atom. a. \(\mathrm{KrF}_{2}\) b. \(\mathrm{KrF}_{4}\) c. \(\mathrm{XeO}_{2} \mathrm{~F}_{2}\) d. \(\mathrm{XeO}_{2} \mathrm{~F}_{4}\)

Short Answer

Expert verified
a. \(\mathrm{KrF}_{2}\): Lewis structure: Kr[F]-2; Molecular structure: linear; Bond angle: 180°; Hybridization of Kr: \(\mathrm{sp}^3\) b. \(\mathrm{KrF}_{4}\): Lewis structure: \({{Kr\!\left(\underset{\bullet}{\phantom{F}}_{}^{F}\right)}}\!\!{}^{-}_{F}^{-}_{F}\); Molecular structure: square planar; Bond angle: 90°; Hybridization of Kr: \(\mathrm{sp}^3\mathrm{d}\) c. \(\mathrm{XeO}_{2}\mathrm{F}_{2}\): Lewis structure: \({{Xe\underset{\bullet}{\phantom{O}}_{}^{O}\Big(\!\!{}^{F}\backprime\:\!}_{F}\Big)}\); Molecular structure: square planar; Bond angle: 90°; Hybridization of Xe: \(\mathrm{sp}^3\) d. \(\mathrm{XeO}_{2}\mathrm{F}_{4}\): Lewis structure: \({{Xe\underset{\bullet}{\phantom{O}}_{}^{O}\Big(\!\!{}^{F}\backprime\:\!}_{F}\Big)}\!\left\{\!\!\!\!\!\!{}_{F}^{F}\right\}\); Molecular structure: octahedral; Bond angle: 90°; Hybridization of Xe: \(\mathrm{sp}^3\mathrm{d}^2\)

Step by step solution

01

Valence electrons

Kr has 8 valence electrons and F has 7. Since there are two F atoms, the total number of valence electrons in KrF2 is 8 + 2(7) = 22.
02

Lewis structure

In the Lewis structure, Kr is surrounded by two F atoms. The remaining 4 electrons are two lone pairs on Kr: Kr[F]-2
03

Steric number (SN)

There are two bonding pairs and two lone pairs, which gives a steric number of 4.
04

Molecular structure

Based on the steric number and the presence of two lone pairs, the molecular structure is linear.
05

Bond angles

The bond angle in a linear molecular structure is 180° between the two F atoms.
06

Hybridization

With a steric number of 4, the hybridization of Kr is \(\mathrm{sp}^3\) b. \(\mathrm{KrF}_{4}\)
07

Valence electrons

Kr has 8 valence electrons, and F has 7. Since there are four F atoms, the total number of valence electrons in KrF4 is 8 + 4(7) = 36.
08

Lewis structure

In the Lewis structure, Kr is surrounded by four F atoms: \({{Kr\!\left(\underset{\bullet}{\phantom{F}}_{}^{F}\right)}}\!\!{}^{-}_{F}^{-}_{F}\)
09

Steric number (SN)

There are four bonding pairs and one lone pair, which gives a steric number of 5.
10

Molecular structure

Based on the steric number and the presence of one lone pair, the molecular structure is square planar.
11

Bond angles

The bond angle in a square planar molecular structure is 90° between adjacent F atoms.
12

Hybridization

With a steric number of 5, the hybridization of Kr is \(\mathrm{sp}^3\mathrm{d}\) c. \(\mathrm{XeO}_{2}\mathrm{F}_{2}\)
13

Valence electrons

Xe has 8 valence electrons, O has 6, and F has 7. There are two O atoms and two F atoms, so the total number of valence electrons is 8 + 2(6) + 2(7) = 34.
14

Lewis structure

In the Lewis structure, Xe is surrounded by two O atoms and two F atoms: \({{Xe\underset{\bullet}{\phantom{O}}_{}^{O}\Big(\!\!{}^{F}\backprime\:\!}_{F}\Big)}\)
15

Steric number (SN)

There are four bonding pairs and no lone pairs, which gives a steric number of 4.
16

Molecular structure

Based on the steric number and no lone pairs, the molecular structure is square planar.
17

Bond angles

The bond angle in a square planar molecular structure is 90° between adjacent O and F atoms.
18

Hybridization

With a steric number of 4, the hybridization of Xe is \(\mathrm{sp}^3\) d. \(\mathrm{XeO}_{2}\mathrm{F}_{4}\)
19

Valence electrons

Xe has 8 valence electrons, O has 6, and F has 7. There are two O atoms and four F atoms, so the total number of valence electrons is 8 + 2(6) + 4(7) = 46.
20

Lewis structure

In the Lewis structure, Xe is surrounded by two O atoms and four F atoms: \({{Xe\underset{\bullet}{\phantom{O}}_{}^{O}\Big(\!\!{}^{F}\backprime\:\!}_{F}\Big)}\!\left\{\!\!\!\!\!\!{}_{F}^{F}\right\}\)
21

Steric number (SN)

There are six bonding pairs and no lone pairs, which gives a steric number of 6.
22

Molecular structure

Based on the steric number and no lone pairs, the molecular structure is octahedral.
23

Bond angles

The bond angle in an octahedral molecular structure is 90° between adjacent O and F atoms.
24

Hybridization

With a steric number of 6, the hybridization of Xe is \(\mathrm{sp}^3\mathrm{d}^2\)

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