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Chapter 19: Q40E (page 1096)

Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.

Short Answer

Expert verified

No, it is not possible for a complex of a metal in the transition series to have six unpaired electrons

Step by step solution

01

of 1: Explanation

It is not possible for a complex of a metal in the transition series to have six unpaired electrons because the maximum number of electrons in $d$ - orbital is $10$ .
According to the Hund's rule of maximum multiplicity: the electrons will be filled singly first. Which indicates that the maximum of unpaired electrons of any element in the d-block can’t be greater than $5.$

02

Result:-

It is not possible as the maximum number of unpaired electrons in the $d$ -block is $5.$

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Most popular questions from this chapter

How many unpaired electrons are present in each of the following?

$\begin{array}{r} (a) {(C o F_{6})}^{3 -} (h i g h s p i n) \\ (b) {(M n {(C N)}_{6})}^{3 -} (l o w s p i n) \\ (c) {(M n {(C N)}_{6})}^{4 -} (l o w s p i n) \\ (d) {(M n C l_{6})}^{4 -} (h i g h s p i n) \\ (e) {(R h C l_{6})}^{3 -} (l o w s p i n) \end{array}$

Determine the number of unpaired electrons expected for ${(F e {(N O_{2})}_{6})}^{3 -}$ and for ${(F e F_{6})}^{3 -}$ in terms of crystal field theory.

Name each of the compounds or ions given in $E x e r c i s e 19.30$ .

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M₃O₄ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO∙M₂O₃, to permitestimation of the metal’s two oxidation states.)

(a) Sc₂O₃

(b) TiO₂

(c) V₂O₅

(d) CrO₃

(e) MnO₂

(f) Fe₃O₄

(g) Co₃O₄

(h) NiO

(i) Cu₂O

Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:

(a) tetrahydroxozincate(II) ion (tetrahedral)

(b) hexacyanopalladate(IV) ion

(c) dichloroaurate(I) ion (note that aurum is Latin for “gold”)

(d) diamminedichloroplatinum(II)

(e) potassium diamminetetrachlorochromate(III)

(f) hexaamminecobalt(III) hexacyanochromate(III)

(g) dibromobis(ethylenediamine) cobalt(III) nitrate

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