1

### JEE Main 2017 (Online) 9th April Morning Slot

[Co2(CO)8] displays :
A
one Co−Co bond, six terminal CO and two bridging CO
B
one Co−Co bond, four terminal CO and four bridging CO
C
no Co−Co bond, six terminal CO and two bridging CO
D
no Co−Co bond, four terminal CO and four bridging CO
2

### JEE Main 2018 (Offline)

The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are
A
+3, 0 and +4
B
+3, +4 and +6
C
+3, +2 and +4
D
+3, 0 and +6

## Explanation

Assume oxidation state of Cr in all the compounds = x

(i)$\,\,\,$ In [cr(H2O)6] Cl3 oxidation state of Cr is

x + 0 $\times$ 6 + ($-$1 $\times$ ) = O

$\Rightarrow \,\,\,\,\,$ x + 0 $-$ 3 = O

$\Rightarrow \,\,\,\,$ x = +

(ii)$\,\,\,$ [Cr (C6 H6)2] oxidation state of Cr is

x + 0 $\times$ 2 = 0

$\Rightarrow \,\,\,\,$ x = 0

(iii) $\,\,\,$ In K2 [ Cr(CN)2 (O)2 (O2) (NH3)] oxidation state of Cr is

1 $\times$ 2 + x + ($-$ 1 $\times$ 2) + ($-$2 $\times$ 2) + ($-$2) + 0 = 0

$\Rightarrow \,\,\,\,$ x = + 6

$\therefore\,\,\,$ + 3, 0 and + 6 is the correct answer.

Note :

O2 molecule can have 0, $-$ 1, $-$ 2 oxidation state but in K2 [ Cr (CN)2 (O)2 (O2) NH3 ] if we choose zero as the oxidation state of O2 then for Cr oxidation state will be $+$ 4. But + 4 oxidation state of Cr is unstable and $+$ 6 is most stable that is why we choose $-$ 2 oxidation state of O2.
3

### JEE Main 2018 (Offline)

Consider the following reaction and statements:
[Co(NH3)4Br2]+ + Br- $\to$ [Co(NH3)3Br3] + NH3
(I) Two isomers are produced if the reactant complex ion is a cis-isomer
(II) Two isomers are produced if the reactant complex ion is a trans-isomer
(III) Only one isomer is produced if the reactant complex ion is a trans-isomer
(IV) Only one isomer is produced if the reactant complex ion is a cis – isomer
The correct statements are
A
(II) and (IV)
B
(I) and (II)
C
(I) and (III)
D
(III) and (IV)

## Explanation

When reactant is cis isomer then following reaction takes place.

So, if reactant is cis - isomer then two isomers are produced.

When reactant is trans isomer then following reaction takes place.

So, if the reactant is trans isomer then only one isomer is produced.
4

### JEE Main 2018 (Online) 15th April Morning Slot

The correct combination is :
A
[Ni(CN)4]2- $-$ tetrahedral ;

[Ni(CO)4] $-$ paramagnetic
B
[Ni(Cl)4]2- $-$ paramagnetic ;

[Ni(CO)4] $-$ tetrahedral
C
[Ni(Cl)4]2- $-$ square-planar ;

[Ni(CN)4] 2-$-$ paramagnetic
D
[Ni(Cl)4]2- $-$ diamagnetic ;

[Ni(CO)4] $-$square-planar

## Explanation

[NiCl4]2– : Oxidation state of Ni in [NiCl4]2– = + 2

Cl is a weak field ligand and cannot take part in pairing of electrons.
Hence, the complex is tetrahedral and paramagnetic with two unpaired electrons.

[Ni(CO)4] : Oxidation state of Ni in [Ni(CO)4] is zero. CO is a strong field ligand thus pairing of electrons takes place in d-orbitals.
Hence, the complex is tetrahedral and diamagnetic.