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Na(2)S+Na(2)[Fe(CN)(5)NO]rarrNa(4)[Fe(NH...

`Na_(2)S+Na_(2)[Fe(CN)_(5)NO]rarrNa_(4)[Fe(NH)_(5)nos],` oxidation number of Fe in reactant ( complex ) and product ( complex) are `:`

A

`2,1`

B

`2,2`

C

`2,3`

D

`3,3`

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To determine the oxidation numbers of iron (Fe) in the reactant and product complexes, we can follow these steps: ### Step 1: Identify the reactant complex and its components The reactant complex is \( \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \). ### Step 2: Assign oxidation states in the reactant complex - Let the oxidation state of Fe be \( x \). - The cyanide ion (CN) has a charge of -1, and since there are 5 CN groups, their total contribution is \( 5 \times (-1) = -5 \). - The nitrosyl ion (NO) has a charge of +1. The overall charge of the complex is -2 (since it is part of the sodium salt, which balances the +2 charge from 2 sodium ions). Thus, we can set up the equation: \[ x + (-5) + (+1) = -2 \] ### Step 3: Solve for \( x \) Rearranging the equation gives: \[ x - 4 = -2 \implies x = +2 \] So, the oxidation state of Fe in the reactant \( \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \) is +2. ### Step 4: Identify the product complex and its components The product complex is \( \text{Na}_4[\text{Fe(NH}_5\text{NO})] \). ### Step 5: Assign oxidation states in the product complex - Let the oxidation state of Fe in the product be \( y \). - The ammine ion (NH) has a charge of -1, and since there are 5 NH groups, their total contribution is \( 5 \times (-1) = -5 \). - The nitrosyl ion (NO) again has a charge of +1. The overall charge of the complex is -1 (since it is part of the sodium salt, which balances the +4 charge from 4 sodium ions). Thus, we can set up the equation: \[ y + (-5) + (+1) = -1 \] ### Step 6: Solve for \( y \) Rearranging the equation gives: \[ y - 4 = -1 \implies y = +2 \] So, the oxidation state of Fe in the product \( \text{Na}_4[\text{Fe(NH}_5\text{NO})] \) is also +2. ### Conclusion The oxidation number of Fe in both the reactant and product complexes is +2. ### Final Answer The oxidation numbers of Fe in the reactant and product complexes are both +2. ---

To determine the oxidation numbers of iron (Fe) in the reactant and product complexes, we can follow these steps: ### Step 1: Identify the reactant complex and its components The reactant complex is \( \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \). ### Step 2: Assign oxidation states in the reactant complex - Let the oxidation state of Fe be \( x \). - The cyanide ion (CN) has a charge of -1, and since there are 5 CN groups, their total contribution is \( 5 \times (-1) = -5 \). ...
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Na_(2)S +Na_(2) [Fe (CN)_(5)NO] to Na_(4) [Fe (CN)_(5)NOS] Find sum of oxidation number of Fe in reactant (complex) and product (complex) are.

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Knowledge Check

  • Na_(2)S +Na_(2)[Fe(CN)_(5)NO] to Na_(4) [Fe(CN)_(5)NOS] . Oxidation number of Fe in reactant (complex) and product (complex) are :-

    A
    2, 1
    B
    2, 2
    C
    2, 3
    D
    3, 3
  • Na_(2)S + Na_(2)[Fe(CN)_(5)NO]to Na_(4)[Fe(CN)_(5)NOS], oxidation number of Fe in rectant (complex) and product (complex) are:

    A
    2, 1
    B
    2, 2
    C
    2, 3
    D
    3, 3
  • The oxidation number of Fe in K_(4) [Fe(CN_(6))] is -

    A
    `+6`
    B
    `+4`
    C
    `+3`
    D
    `+2`
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