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The sum of coordination number and oxidation number of the metal M in the complex \left [ M(en)_{2} (C_{2}O_{4})\right ]Cl (where en is ethyenediamine) is:

  • Option 1)

    9

  • Option 2)

    6

  • Option 3)

    7

  • Option 4)

    8

 

Answers (1)

best_answer

As we learned in concept

Types of Ligands on the basis of Connectivity -

(i) Unidentate - ligand is bound to metal ion through a single donor atom.

eg: Cl^{-}, H_{2}O and NH_{3}

(ii) Bidentate : when ligand can bind through two donor atoms

eg: CH_{2}NH_{2}-CH_{2}NH_{2}

(iii) Polydentate - when ligand bind to two or more donor atoms

(iv) Hexadentate - type of polydentate having six donor atoms

(v) Ambidentate- which can ligate through two different atoms at a time

eg. M\leftarrow SCN & M\leftarrow NCS

- wherein

All ambidentate ligands are monodentate but all monodentate ligands are not ambidentate

   

 

[M(en)_2(C_2O_4)]Cl has 3 bidentate ligands which can lead with 2 sites simultaneously. So, the coordination number is 2\times+2 = 6.

C_2O_4^{2-} has negative charge of 2. Add that the negative charge on Cl and total negative charge lucomes. The actral atom must have +3 charge to netralize the complex.

therefore, O.N = +3

therfore, ON + CN = 3+6 = 9

therefore, solution is 1


Option 1)

9

correct

Option 2)

6

incorrect

Option 3)

7

incorrect

Option 4)

8

incorrect

Posted by

Plabita

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