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The oxidation number of Cr in [Cr(NH3)4C...

The oxidation number of `Cr` in `[Cr(NH_3)_4Cl_2]^(+)` is:

A

`+3`

B

`+2`

C

`+1`

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
A

`a + (4 xx 0) + 2 xx -1 = 1 therefore a = +3`
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Knowledge Check

  • The oxidation number of Cr in [Cr(NH_3)_4CI_2]^(+) is:

    A
    `+3`
    B
    `+2`
    C
    `+1`
    D
    zero
  • The ionization isomer of [Cr(H_2O)_4Cl(NO_2)]Cl is

    A
    `[Cr(H_2O)_4 (O_2N)]Cl_2`
    B
    `[Cr(H_2O)_4 Cl_2]NO_2`
    C
    `[Cr(H_2O)_4Cl(ONO)]Cl`
    D
    `[Cr(H_2O)_4Cl_2(NO_2)]H_2O`
  • The ionization isomer of [Cr(H_2O)_4 Cl(NO_2)]Cl is

    A
    `[Cr(H_2O)_4 (O_2N)]Cl_2`
    B
    `[Cr(H_2O)_4Cl_2] NO_2`
    C
    `[Cr(H_2O)_4 Cl(ONO)]Cl`
    D
    `[Cr(H_2O)_4 Cl_2 (NO_2)]H_2O`
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