The reason for the formation oflow-spin and high spin octahedral complexes by Mn 3+ and not by Mn 4 + should be determined. Concept introduction: A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration

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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... ☐ ☐ : ☐ + G 1. NaOMe Click and drag to start drawing a structure. 2. H +

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Chapter 19, Problem 38QAP

Interpretation Introduction

Interpretation:

The reason for the formation oflow-spin and high spin octahedral complexes by Mn³⁺ and not by

Mn⁴+ should be determined.

Concept introduction: A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration

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