The bonding in the nitrate ion is to be explained in terms of delocalized molecular orbitals. Concept introduction: According to the valence bond approach, a bond is formed by the overlapping of singly occupied orbitals present on two atoms. A hybrid orbital is formed by the mixing of two or more atomic orbitals. Atomic or hybrid orbitals overlap along the internuclear axis to form a sigma bond, while a sideways overlap of parallel p orbitals forms a pi bond. Two atomic orbitals combine to form a bonding and an antibonding molecular orbital. Orbitals that lie on the internuclear axis combine to form the sigma σ molecular orbital and orbitals parallel to each other combine to form the π molecular orbitals.

Question

Want to see the full answer?

Answer 1

Question

Chapter 9, Problem 66QP

Interpretation Introduction

Interpretation:

The bonding in the nitrate ion is to be explained in terms of delocalized molecular orbitals.

Concept introduction:

According to the valence bond approach, a bond is formed by the overlapping of singly occupied orbitals present on two atoms.

A hybrid orbital is formed by the mixing of two or more atomic orbitals.

Atomic or hybrid orbitals overlap along the internuclear axis to form a sigma bond, while a sideways overlap of parallel p

orbitals forms a pi bond.

Two atomic orbitals combine to form a bonding and an antibonding molecular orbital.

Orbitals that lie on the internuclear axis combine to form the sigma σ

molecular orbital and orbitals parallel to each other combine to form the π molecular orbitals.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

I

Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option command

AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, N