The oxidation states of Cr in $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3},\left[\mathrm{Cr}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)_{2}\right]$, and $\mathrm{K}_{2}\left[\mathrm{Cr}(\mathrm{CN})_{2}(\mathrm{O})_{2}\left(\mathrm{O}_{2}\right)\left(\mathrm{NH}_{3}\right)\right]$ respectively are :

JEE Main · 2018 · 08 Apr 2018 · Offline · d and f Block Elements

The oxidation states of Cr in

[Cr (H_{2} O)_{6}] Cl_{3}, [Cr (C_{6} H_{6})_{2}]

, and

K_{2} [Cr (CN)_{2} (O)_{2} (O_{2}) (NH_{3})]

respectively are :

Answer: (D)

[Cr (H X_{2} O) X_{6}] Cl X_{3}

the water ligands are neutral and three chlorides balance the charge, leaving chromium at

+ 3

. In

[Cr (C X_{6} H X_{6}) X_{2}]

both benzene rings are neutral, so chromium is

0

. In

K X_{2} [Cr (CN) X_{2} (O) X_{2} (O X_{2}) (NH X_{3})]

the two cyanides contribute

- 2

, the two oxo groups

- 4

, the peroxo group

- 2

and ammonia nothing, while the two potassiums make the complex ion

- 2

, so

x - 2 - 4 - 2 = - 2

gives

x = + 6

. The oxidation states are

+ 3

0

and

+ 6

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option