/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 Selenium tetrafluoride reacts wi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Selenium tetrafluoride reacts with fluorine to form selenium hexafluoride. (a) Draw Lewis structures for both selenium fluorides, and predict any deviations from ideal bond angles. (b) Describe the change in orbital hybridization of the central Se during the reaction.

Short Answer

Expert verified
SeF4 is a see-saw shape with bond angles deviating from 90° and 120°. SeF6 is octahedral with 90° bond angles. Se changes hybridization from sp^3d in SeF4 to sp^3d^2 in SeF6.

Step by step solution

01

- Draw Lewis Structure for Selenium Tetrafluoride (SeF4)

To draw the Lewis structure for SeF4:1. Determine the total number of valence electrons: Se (6) + 4F (7 each) = 34 electrons.2. Place selenium (Se) in the center and arrange the four fluorine (F) atoms around it.3. Connect each F atom to Se with a single bond.4. Distribute the remaining electrons to complete each F atom's octet.5. Place any remaining electrons on the Se atom.6. The final structure should show Se with two lone pairs and four Se-F bonds, creating a see-saw shape, with bond angles deviating from the ideal 90° and 120° due to the lone pairs.
02

- Draw Lewis Structure for Selenium Hexafluoride (SeF6)

To draw the Lewis structure for SeF6:1. Determine the total number of valence electrons: Se (6) + 6F (7 each) = 48 electrons.2. Place selenium (Se) in the center and arrange the six fluorine (F) atoms around it.3. Connect each F atom to Se with a single bond.4. Distribute the remaining electrons to complete each F atom's octet.5. There are no remaining electrons to place on Se.6. The final structure forms an octahedral shape, with 90° bond angles between all Se-F bonds.
03

- Predict Bond Angle Deviations in SeF4

In SeF4, the ideal bond angles for a see-saw shape derived from a trigonal bipyramidal electron-domain geometry would be 90°, 120°, and 180°. However, due to the lone pairs on Se, the bond angles will be slightly less in actual structure as lone pairs occupy more space and repel adjacent bonds, reducing bond angles.
04

- Predict Bond Angles in SeF6

In SeF6, the molecule forms an ideal octahedral shape. All bond angles are 90° or 180°, as lone pairs do not affect the structure. Therefore, no deviations from the ideal bond angles are expected.
05

- Describe Orbital Hybridization Change

For Se in SeF4, the central atom is sp^3d hybridized as there are five electron pairs around Se (four bonding pairs and one lone pair). In SeF6, the central atom is sp^3d^2 hybridized as there are six bonding electron pairs. The change from sp^3d to sp^3d^2 hybridization occurs during the reaction of SeF4 with additional F2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structures
Lewis structures help us understand molecule structure by showing the arrangement of electrons around atoms in a compound. To draw the Lewis structure for SeF4, we start by calculating the total number of valence electrons: Selenium (Se) contributes 6 electrons, and each of the four fluorine (F) atoms contributes 7, totaling 34 electrons. We place Se in the center and arrange the fluorine atoms around it. Connecting each F to Se with a single bond uses up 8 electrons, leaving 26 to complete the octets of the F atoms and Se itself. Se in SeF4 has two lone pairs, resulting in a 'see-saw' shape due to electron repulsion effects.
The Lewis structure for SeF6 is calculated similarly: Se (6) + 6F (7) total 48 electrons. We again place Se centrally and surround it with six F atoms, connecting each with a single bond. The remaining electrons complete the octets of the F atoms. In this structure, all pairs of electrons are bonding pairs, forming an octahedral shape.
Orbital Hybridization
Orbital hybridization is a concept where atomic orbitals mix to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. In SeF4, selenium is surrounded by five electron pairs: four bonding pairs and one lone pair. These electron pairs require an sp^3d hybridization on the Se atom, forming one s orbital, three p orbitals, and one d orbital to generate five hybrid orbitals.
When SeF4 reacts with fluorine to form SeF6, the selenium needs to accommodate six bonding pairs. This requires sp^3d^2 hybridization, involving one s orbital, three p orbitals, and two d orbitals to produce six hybrid orbitals. This shift from sp^3d to sp^3d^2 hybridization accounts for the structural changes that occur during the reaction.
Bond Angles
Bond angles are the angles between adjacent bonds within a molecule, indicative of its shape and electron pair repulsion. In SeF4, the molecule adopts a 'see-saw' structure due to the lone pairs on Se. Ideally, a 'see-saw' derived from trigonal bipyramidal geometry would feature 90°, 120°, and 180° angles. However, the actual bond angles deviate due to lone pair repulsions compressing the F-Se-F angles.
On the other hand, SeF6 has an ideal octahedral shape. All bond angles are either 90° or 180° because no lone pairs are present to distort the geometry. The regular octahedral structure of SeF6 ensures that there are no deviations from the ideal bond angles, maintaining symmetry and uniformity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which group(s) of the periodic table is (are) described by each of the following general statements? (a) The elements form compounds of VSEPR class \(A X_{3} E\). (b) The free elements are strong oxidizing agents and form monatomic ions and oxoanions. (c) The atoms form compounds by combining with two other atoms that donate one electron each. (d) The free elements are strong reducing agents, show only one nonzero oxidation state, and form mainly ionic compounds. (c) The elements can form stable compounds with only three bonds, but, as a central atom, they can accept a pair of electrons from a fourth atom without expanding their valence shell. (f) Only larger members of the group are chemically active.

Copper(II) hydrogen arsenite (CuHAsO \(_{3}\) ) is a green pigment once used in wallpaper. In damp conditions, mold metabolizes this compound to trimethylarsine \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{As}\right],\) a highly toxic gas. (a) Calculate the mass percent of As in each compound. (b) How much CuHAsO must react to reach a toxic level in a room that measures \(12.35 \mathrm{~m} \times 7.52 \mathrm{~m} \times 2.98 \mathrm{~m}\) (arsenic is toxic at \(\left.0.50 \mathrm{mg} / \mathrm{m}^{3}\right) ?\)

Complete and balance the following equations. If no reaction occurs, write NR: (a) \(\mathrm{H}_{3} \mathrm{PO}_{4}(l)+\operatorname{NaI}(s) \longrightarrow\) (b) \(\mathrm{Cl}_{2}(g)+\mathrm{I}^{-}(a q) \longrightarrow\) (c) \(\mathrm{Br}_{2}(l)+\mathrm{Cl}^{-}(a q) \longrightarrow\) (d) \(\mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow\)

(a) What is the range of oxidation states shown by the elements of Group \(5 \mathrm{~A}(15)\) as you move down the group? (b) How does this range illustrate the general rule for the range of oxidation states in groups on the right side of the periodic table?

Complete and balance the following equations. If no reaction occurs, write NR: (a) \(\mathrm{Rb}(s)+\mathrm{Br}_{2}(l) \longrightarrow\) (b) \(\mathrm{I}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (c) \(\mathrm{Br}_{2}(l)+\mathrm{I}^{-}(a q) \longrightarrow\) (d) \(\mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.