Test: Introduction to Coordination Compounds - NEET MCQ

In [Pt(en)₂CI₂] the charge on the complex ion is +2 and coordination number is 6. Hence, the primary and secondary valency are +2 and 6 respectively.

Let the oxidation state of pt be x.
Oxidation state of cl is -1.
So x + 0 – (1*3) must be equal to -1 since the charge of the whole compound is -1.
x + 0 – (1*3) = -1
x -3 = -1
x = -1 + 3
x = +2
So the oxidation state of platinum is +2.
Secondary is due to legend there are mono deadened legend 
then 1×4= 4
 

Pt(NH₃)₂CI₄] is neutral complex and no ions are formed in aqueous solution.

K,(PtCl₆] → 2K⁺ + [PtCI₆]^2-
It does not have Cl^- ion toffive precipitate of AgCI.

Total chlorine molecules are 3 and one third means one chlorine molecule precipitates. So, one chlorine has to be outside the coordination sphere and that is in option C.

Mohr's salt is a doub le salt and it will give test for Fe²⁺ ion. It dissociates into simple substances or ions completely when dissolved in water.

Cr(24) = [Ar]3d54s1
Since (CO) is strong ligand, in Cr(CO)6 none of the electron is unpaired. Therefore ‘spin only’ magnetic moment is zero.

The acetylacetonate (acac) forms six membered ring as follows:

Correct Answer : a

Explanation : NO in brown ring complex has +1 oxidation state.

Let oxidation state of Fe be x.

Since the overall charge on the coordination sphere, [Fe(H2O)5NO+]  is +2, the sum of oxidation states of all elements in it should be equal to +2.

Therefore, x+1=+2

or, x=+1 

Hence, the oxidation state of iron in the brown ring complex is +1.

In [Ni(CO)₄]
Let the oxidation state of Ni is x.
x + 4 x (0) = 0 (Carbonyl (CO) is a neutral ligand and have zero charge.)
x = 0
Hence, Ni have zero oxidation state.

CH₃COCH₂COCH₃

It can ligate through two oxygen atoms.

Common alum is K₂SO₄.AI₂(SO₄)₃.24H₂O and carnallite is KCl.MgCl₂.6H₂O.

The complex K₄[Fe(CN)₆] whose formula can be written like that of double salt. Fe(CN)₂ . 4KCN, dissociates to give K⁺ and [Fe(CN)₆]^4- ions in the aqueous solution.

The answer is 36 because in coordination compound we use EAN method due to which inert group atomic number or noble gas atomic number is more stable.

[Co(en)₃]CI₃ produces 4 ions in solution as follows :
[Co(en)₃] CI₃→ [Co(en)₃]³⁺ + 3CI^-

1. is correct as charge on the complex ion will be +1

2.  is incorrect as the complex will form 4 ions in solution

3. is correct as there is no charge on the complex

4. is also correct as cu+2 has blue color in solution

Hence A, C and D are correct.

In the aqueous solution of Pd(NH₃)₂CI₂, the atoms of chlorine I are in coordination sphere and the van't Hoff factor of the ] compound are unite.

Since, the oxidation number of Co is + 3 hence, the complex compound would be [CoCI . 5H₂O]CI₂.

Hence, 1 mole of [CoCI . 5H₂O]CI₂ gives 2 moles of Cl^- ion.
Given, moles of [CoCI . 5H₂O]CI₂ = 200 mL x 0.01 M = 2 millimol
2 millimol of [CoCI . 5H₂O]CI₂ will give 4 millimol of Cl^-.

Now, to neutralise 4 millimol of Cl^-, 4 millimol of AgNO₃ is required.
Let the volume o! AgNO₃ = VmL
Millimoles of AgNO₃ = 0.1 x V
0.1 x V = 4 millimol

[Co(NH₃)₅CI]CI₂→[Co(NH₃)5CI]²⁺+ 2Cl^- 
Hence, 3 ions are present for one mole of the complex in the solution.

(i) → (r), (ii) → (q), (iii) → (p), (iv) → (s)

Formula of Mohr’s salt is FeSO₄ . (NH₄)₂SO₄ . 6H₂O . Here, x is 6

AgNO₃ solution is added in excess of 1 M solution of CoCI₃ . xNH₃.
CoCl₃​.xNH_3​+AgNO₃​→AgCl (1mole)

This precipitation of 1 mol of AgCl by this reaction shows that there is only one Cl outside the coordination sphere, which is not as a ligand ( as ligands are not ionisable).

Hence, the compound must be as follows:(showing the coordination sphere) [Co(NH₃​)₄​Cl₂​]Cl, as this is the octahedral complex, where it is clear that there are only 2 Cl as ligand and other ligands are NH₃​. 

So, 6−2 = 4 NH₃​ ligands.

The correct answer is Option D
Number of ionisable Cl− in [Cr(H₂O)₅Cl]Cl₂ is 2.
∴ Millimoles of Cl− =30×0.01×2=0.6
∴ Millimoles of Ag+ required =0.6
Now, 0.6/V(in ml)   = 0.1
or, V=6 ml.

ln Co(CO)₆
Let the oxidation state of Co is x.
x + 6 x 0 = 0
[Carbonyl (CO) is a neutral ligand hence, zero charg]
x = 0
Hence, Co has zero oxidation state.

Coordination number of platinum is 6.
∴ Number of chloride ions precipitated by adding AgCI are 3.

Statement I is true because CO is a neutral ligand, Statement II is true because
EAN of Fe
= Atomic number of Fe + Electrons gained by coordination
= 26 + 2 x 5 = 36
But Statement II is not the correct explanation. Correct explanation is that oxidation state of Fe in Fe(CO)₅ is zero because CO is a neutral ligand.

Statement I The oxidation number of Pt in Zeis e’s salt is +2.
Statement II Zeise’s salt is ionic complex with the formula K[PtCl₃(C ₂H₄)].
Hence, statement II is correct but statment I is incorrect.