The bond orders for N 2 2 + , N 2 − and N 2 2 − are to be determined. Also the species are to be ranked in the order of decreasing bond energy and decreasing bond length. Concept introduction: The molecular orbital diagram is a tool that is used to describe the chemical bonding formed between different molecules. It is used to predict the bond strength and the electronic transitions that a molecule can undergo. Bond order is the number of bonds that an atom can form with other atoms. Higher the bond order, more will be the stability of the molecule and vice-versa. Bond energy is the amount of heat that is needed to break the bonds between various atoms. It depends on bond order. Higher the bond order, more will be its bond energy and vice-versa. Bond length is the average distance between the centres of the atomic nuclei of the combining atoms. Its value is inversely proportional to bond order.

Question

Want to see the full answer?

Answer 1

Question

Chapter 11.3, Problem 11.4AFP

Interpretation Introduction

Interpretation:

The bond orders for N22+, N2− and N22− are to be determined. Also the species are to be ranked in the order of decreasing bond energy and decreasing bond length.

Concept introduction:

The molecular orbital diagram is a tool that is used to describe the chemical bonding formed between different molecules. It is used to predict the bond strength and the electronic transitions that a molecule can undergo.

Bond order is the number of bonds that an atom can form with other atoms. Higher the bond order, more will be the stability of the molecule and vice-versa.

Bond energy is the amount of heat that is needed to break the bonds between various atoms. It depends on bond order. Higher the bond order, more will be its bond energy and vice-versa. Bond length is the average distance between the centres of the atomic nuclei of the combining atoms. Its value is inversely proportional to bond order.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Complete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. More... No reaction. my ㄖˋ + 1. Na O Me Click and drag to start drawing a structure. 2. H +

Predict the intermediate 1 and final product 2 of this organic reaction: NaOMe H+ + 1 2 H H work up You can draw 1 and 2 in any arrangement you like. Note: if either 1 or 2 consists of a pair of enantiomers, just draw one structure using line bonds instead of 3D (dash and wedge) bonds at the chiral center. Click and drag to start drawing a structure. X $ dm

Predict the major products of this organic reaction: 1. NaH (20°C) 2. CH3Br ? Some notes: • Draw only the major product, or products. You can draw them in any arrangement you like. • Be sure to use wedge and dash bonds where necessary, for example to distinguish between major products that are enantiomers. • If there are no products, just check the box under the drawing area. No reaction. Click and drag to start drawing a structure. G Cr