## Question

### Solution

Correct option is

I < II < III < IV

Complex                Unpaired electrons

I                                      0

II                                     2

III                                    3

IV                                    4

For N unpaired electrons, magnetic moments is:

#### SIMILAR QUESTIONS

Q1

Select the complex in which secondary valency is satisfied before the primary valency.

Q2

We get a copper complex with ammonia having the formula [Cu(NH3)4]2+only in alkaline solution and not in acidic solution. It can be explained on the basis of the fact that :

Q3

The molecular formula of both ‘A’ and ‘B’ is same.

A’ can be converted to B be boiling in dil. HCl.

A’ on reaction with oxalic acid yields a complex having the formula Ni(NH3)2(C2O4) but ‘B’ does not.

From the above information we can say that

Q4

Select the complex that can be reduced most easily.

Q5

0.001 mol of cobalt complex having molecular formula represented by Co(NH3)5(NO3)(SO4) was passed through a cation exchange (RSO3H) and the acid coming out of it was titrated with 0.1 M NaOH solution. For complete neutralization of acid coming out of cation exchanger the volume of NaOH required was 20.00 ml. From the above data we can say that the complex can be represented as

Q6

Three different solutions were prepared by dissolving same amounts of the following complexes I, II, III in water. The freezing points of these solutions were then determined. The correct orders of freezing points will be :

I. [Co(NH3)3(NO2)3

II. [Co(NH3)5(NO2)]2+ [Co(NH3)2(NO2)4]2

III. [Co(NH3)4(NO2)2][Co(NH3)2(NO2)4

[Given : Kf(H2O) = 1.86oC/m]

Q7

The correct order of ligands in  the trans-directing series is

Q8

Effective atomic number (EAN) of Fe in brown ring complex [Fe(H2O)5NO]2+

Q9

If FeCl3 . 6H2O would exist as a complex, one mol of it in aqueous solution on reaction with excess of AgNO3 forms:

Q10

[Fe(H2O)5NO2+] is brown-ring complex in NO3. In this complex: