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# Tell me? - Coordination compounds: - NEET

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

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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

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