/*! 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 45 For each statement, indicate whe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each statement, indicate whether it is true or false. (a) In order to make a covalent bond, the orbitals on each atom in the bond must overlap. (b) A p orbital on one atom cannot make a bond to an s orbital on another atom. (c) Lone pairs of electrons on an atom in a molecule influence the shape of a molecule. (d) The 1 s orbital has a nodal plane. (e) The \(2p\) orbital has a nodal plane.

Short Answer

Expert verified
(a) True - Covalent bonds require orbital overlap. (b) False - p orbitals can bond with s orbitals. (c) True - Lone pairs affect the molecular shape. (d) False - 1s orbitals do not have nodal planes. (e) True - 2p orbitals have a nodal plane.

Step by step solution

01

Statement (a) - Covalent Bond Requirement

Statement (a) is true. In order to form a covalent bond, the orbitals on each atom in the bond must overlap. This overlap allows for the sharing of electrons between the two atoms, resulting in a stable covalent bond.
02

Statement (b) - Bonding Between p and s Orbitals

Statement (b) is false. A p orbital on one atom can indeed form a bond with an s orbital on another atom. This can occur, for example, in the formation of a sigma bond, where the probability of finding the shared electrons is highest along the axis that connects the two nuclei.
03

Statement (c) - Lone Pairs and Molecular Shape

Statement (c) is true. Lone pairs of electrons on an atom in a molecule have a significant influence on the shape of the molecule. This is because lone pairs of electrons still occupy space and create repulsion, pushing other atoms in the molecule away from them, ultimately affecting the overall geometry of the molecule.
04

Statement (d) - Nodal Plane in 1s Orbital

Statement (d) is false. The 1s orbital does not have a nodal plane. A nodal plane is a region in an orbital where the probability of finding an electron is zero. The 1s orbital is a spherical shape with uniform electron density, and there is no such region with zero electron probability.
05

Statement (e) - Nodal Plane in 2p Orbital

Statement (e) is true. The 2p orbital has a nodal plane. The 2p orbital has a dumbbell shape, with two lobes separated by a nodal plane. In this nodal plane, the probability of finding an electron is zero.

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.

Orbital Overlap
When atoms come together to form covalent bonds, something interesting happens at the atomic level: orbital overlap. This is a key requirement for covalent bonding. Orbital overlap involves the sharing of electrons between two atoms' orbitals, which are the spaces where electrons are most likely to be found. The overlapping allows atoms to achieve a more stable electronic configuration by sharing electrons.
Some common types of overlaps include:
  • s-s overlap: This occurs when two s orbitals from different atoms overlap. It's often seen in molecules like hydrogen gas ( H_2 ).
  • s-p overlap: An example is seen in hydrogen fluoride ( HF ), where a hydrogen's 1s orbital overlaps with a fluorine's 2p orbital.
  • p-p overlap: When two p orbitals overlap, electrons can be shared along the axis connecting the two nuclei, like in fluorine gas ( F_2 ).
Each type of orbital overlap contributes to the molecule's stability and dictates the type of bond—such as sigma or pi bonds—that forms.
Molecular Geometry
The shape of a molecule, also known as molecular geometry, is not just about how atoms are connected. It's influenced heavily by the presence of lone electron pairs. These lone pairs are pairs of valence electrons that are not shared with other atoms and occupy space around an atom.
These lonely pairs exert a repulsive force on nearby electrons, impacting the arrangement of atoms in a molecule. This concept is crucial when using the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict molecular shapes. For example:
  • Tetrahedral: A molecule like methane ( CH_4 ) where all four positions around the central carbon atom are occupied by hydrogen atoms.
  • Trigonal Pyramidal: Ammonia ( NH_3 ) has one lone pair, which pushes the three hydrogen atoms into a trigonal pyramidal shape.
  • Bent: Water ( H_2O ), with two lone pairs on oxygen, results in a bent molecular shape.
Overall, the distribution of lone pairs and bonding pairs determines the geometry, influencing properties like polarity and reactivity.
Nodal Planes
In atomic orbitals, a nodal plane is a fascinating concept. It refers to an area within an orbital where the chance of finding an electron is zero. The nature and number of nodal planes depend on the shape and type of the orbital.
The simplest orbital, the 1s, is spherical and has no nodal planes. Electrons are evenly distributed throughout, with no area of zero electron probability. However, higher energy levels introduce nodal planes.
In a 2p orbital, which is shaped like a dumbbell, there is a nodal plane between the two lobes. This plane means that an electron can't exist right in the middle of the 2p orbital. Such structures play a critical role in bonding and molecular formation by influencing how atoms interact.
Understanding nodal planes helps chemists appreciate and predict how orbitals might overlap and influence molecular geometry. Especially in higher orbitals, the number of nodal planes increases, becoming essential in complex molecular structures.

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

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

(a) An AB \(_{6}\) molecule has no lone pairs of electrons on the A atom. What is its molecular geometry? (b) An AB \(_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the AB \(_{4}\) molecule in part (b), predict the molecular geometry.

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each.(c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right),\) ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule?

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.