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(a) \(\mathrm{PF}_{5}\) has \(D_{3 \mathrm{h}}\) symmetry. What is its structure? (b) Suggest an appropriate bonding scheme for \(\mathrm{PF}_{5}\) within VB theory, giving appropriate resonance structures.

Short Answer

Expert verified
PF₅ has a trigonal bipyramidal structure. VB theory suggests sp³d hybridization, forming resonance structures with partial double bonds using d orbitals.

Step by step solution

01

Understanding the Symmetry

The molecule \(\mathrm{PF}_5\) is given as having \(D_{3h}\) symmetry. This means it belongs to a group of molecules that have a trigonal bipyramidal shape. The \(D_{3h}\) point group is characterized by having a horizontal mirror plane and a threefold rotation axis along with two vertical mirror planes.
02

Describing the Geometrical Structure

Given the \(D_{3h}\) symmetry, \(\mathrm{PF}_5\) should have a trigonal bipyramidal structure. This structure consists of three equatorial fluorine atoms forming a triangle around the phosphorus atom, and two axial fluorine atoms perpendicular to the triangle.
03

Visualizing the Bonding in VB Theory

Within Valence Bond (VB) theory, the phosphorus atom in \(\mathrm{PF}_5\) may utilize \(d\) orbitals in addition to \(s\) and \(p\) orbitals for hybridization. This leads to \( ext{sp}^3d\) hybridization, wherein one \(s\), three \(p\), and one \(d\) orbitals mix to form five equivalent hybrid orbitals for bonding with fluorine atoms.
04

Formulating Resonance Structures

In VB theory, resonance structures can be proposed for \(\mathrm{PF}_5\) to demonstrate expanded octet bonding. A possible approach is showing that the \(d\) orbitals can help form \(\pi\) bonds by forming partial double bonds with axial fluorine atoms, involving resonance between single and double bond character between different P-F bonds.

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Key Concepts

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

D3h Symmetry
Understanding the concept of symmetry in molecules is key to grasping their physical properties and behavior during chemical reactions. In particular, the PD__{3h} symmetry of PF5 tells us a lot about its spatial configuration. PD__{3h} symmetry is defined by specific symmetry elements that hold a molecule in an organized form. It involves the following characteristics:
  • A horizontal mirror plane (Psigma_h) which divides the molecule into two symmetrical halves.
  • A threefold axis of rotation (PC_3) passing vertically through the phosphorus atom, which means the molecule looks the same after a 120-degree rotation around this axis.
  • Additionally, it includes vertical mirror planes (Psigma_v) that parallel the rotational axis.
Understanding this symmetry helps predict physical and chemical properties, like how the molecule interacts with light through spectroscopy. Having such symmetry simplifies the understanding of its spectroscopic signatures, providing insights into its character and interactions with other molecules.
Trigonal Bipyramidal Structure
The PPF_5 molecule adopts a trigonal bipyramidal structure due to its PD__{3h} symmetry. Structurally, this means:
  • There is a central phosphorus atom surrounded by five fluorine atoms in a distinctive configuration.
  • Three of these fluorine atoms, known as equatorial atoms, form a plane with angles of 120 degrees between them.
  • The other two fluorine atoms lie perpendicular to this plane and are referred to as the axial atoms.
This configuration minimizes repulsion according to VSEPR (Valence Shell Electron Pair Repulsion) theory. By placing three atoms in an equatorial plane and two on opposite poles, the structure achieves stability against electron pair repulsion, a key concept that dictates molecular geometry in chemistry.
Valence Bond Theory
The Valence Bond (VB) theory provides a way to visualize how atoms come together to form molecules by overlapping their atomic orbitals. In the case of PPF_5, phosphorus must form five bonds with fluorine. Normally, this would seem impossible within the octet rule, as phosphorus typically has only three s and p orbitals available.
To form five bonds, phosphorus takes part in resonance and utilizes additional orbitals:
  • The d orbitals are invoked, allowing the formation of more bonds by accommodating more than eight electrons.
  • Projection of resonance structures shows partial double bond character by mixing with d orbitals despite bonds supposedly being single.
This use of expanded octets illustrated by VB theory reflects the flexibility of phosphorus in bonding beyond its conventional octet, by involving higher energy d orbitals to form a stable molecule through a blend of single and partial double bonds.
sp3d Hybridization
Hybridization is a concept key to understanding molecular structure and bonding. In PPF_5, sp3d hybridization occurs due to mixing of one s, three p, and one d orbitals from phosphorus to form five hybrid orbitals.
This allows:
  • The phosphorus atom to form five equivalent sigma bonds with five fluorine atoms.
  • These orbitals are oriented in space to support the trigonal bipyramidal geometry.
sp3d hybridization is crucial in explaining the accommodation of five electron pairs around the phosphorus, permitting the formation of stable structures like PPF_5. This configuration underscores why hybrid orbitals are more advantageous than atomic orbitals alone, offering an energetically feasible way to maintain equilibrium within molecules through optimized bonding.

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Most popular questions from this chapter

(a) Draw the structure of \(\left[\mathrm{CO}_{3}\right]^{2-}\). (b) If all the \(\mathrm{C}-\mathrm{O}\) bond distances are equal, write a set of resonance structures to describe the bonding in \(\left[\mathrm{CO}_{3}\right]^{2-}\). (c) Describe the bonding in \(\left[\mathrm{CO}_{3}\right]^{2-}\) in terms of a hybridization scheme and compare the result with that obtained in part (b).

\(\mathrm{VB}\) and \(\mathrm{M} \mathrm{O}\) approaches to the bonding in linear \(\mathrm{XH}_{2}\) (X has \(2 s\) and \(2 p\) valence atomic orbitals) give pictures in which the \(\mathrm{X}-\mathrm{H}\) bonding is localized and delocalized respectively. Explain how this difference arises.

(a) \(\mathrm{BCl}_{3}\) has \(D_{3 \mathrm{h}}\) symmetry. Draw the structure of \(\mathrm{BCl}_{3}\) and give values for the bond angles. \(\mathrm{NCl}_{3}\) has \(C_{3 \mathrm{v}}\) symmetry. Is it possible to state the bond angles from this information? (b) Derive the symmetry labels for the atomic orbitals on \(\mathrm{B}\) in \(\mathrm{BCl}_{3}\) and on \(\mathrm{N}\) in \(\mathrm{NCl}_{3}\).

Suggest an appropriate hybridization scheme for the central atom in each of the following species: (a) \(\mathrm{SiF}_{4}\); (b) \(\left[\mathrm{PdCl}_{4}\right]^{2-} ;(\mathrm{c}) \mathrm{NF}_{3} ;(\mathrm{d}) \mathrm{F}_{2} \mathrm{O} ;(\mathrm{e})\left[\mathrm{CoH}_{5}\right]^{4-}\); \((\mathrm{f})\left[\mathrm{FeH}_{6}\right]^{4-} ;(\mathrm{g}) \mathrm{CS}_{2} ;(\mathrm{h}) \mathrm{BF}_{3}\).

\(\operatorname{In}\left[\mathrm{B}_{2} \mathrm{H}_{7}\right]^{-}(5.11),\) each \(\mathrm{B}\) atom is approximately tetrahedral. (a) How many valence electrons are present in the anion? (b) Assume that each \(\mathrm{B}\) atom is \(s p^{3}\) hybridized. After localization of the three terminal \(\mathrm{B}-\mathrm{H}\) bonds per \(\mathrm{B}\), what B-centred orbital remains for use in the bridging interaction? (c) Following from your answer to part (b), construct an approximate orbital diagram to show the formation of \(\left[\mathrm{B}_{2} \mathrm{H}_{7}\right]^{-}\) from two \(\mathrm{BH}_{3}\) units and \(\mathrm{H}^{-}\). What does this approach tell you about the nature of the \(\mathrm{B}-\mathrm{H}-\mathrm{B}\) bridge?

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