Question

If trans-[Cr(en)2(NCS)2]SCN is heated, it forms gaseous ethylenediamine and solid [Cr(en)2(NCS)2][Cr(en)(NCS)4]. Write a balanced chemical equation for this reaction. What are the oxidation states of the Cr ions in the reactant and in the two complex ions in the product?

Answer 1

Balanced equation: 2 [Cr(en)₂(NCS)₂](SCN)(s) → en(g) + [Cr(en)₂(NSC)₂][Cr(en)(NCS)₄](s)

Oxidation states on Cr in the reactant and the products: +3

Step-by-step explanation:

The balanced chemical equation is the following:

2 [Cr(en)₂(NCS)₂](SCN)(s) → en(g) + [Cr(en)₂(NSC)₂][Cr(en)(NCS)₄](s) (1)

In the balanced reaction (1) we have that 2 moles of [Cr(en)₂(NCS)₂](SCN) decompose in 1 mol of ethylenediamine and 1 mol of Cr(en)₂(NSC)₂][Cr(en)(NCS)₄.

Determination of oxidation state of the Cr:

In the reactant of the balanced equation (1), we know that the charge on SCN is -1, so for the complex [Cr(en)₂(NCS)₂](SCN) to be neutral, the charge on the Cr(en)₂(NCS)₂ must be +1, hence the state oxidation on the Cr is:

`2 (x + 0 + 2 \cdot (-1)) = 2 (+1)`

where x: is the oxidation state on Cr, 0: is the oxidation state on ethylenediamine since is neutral, and (+1): is the oxidation state on [Cr(en)₂(NCS)₂].

`2x - 4 = 2 \rightarrow x = +3`

So, the oxidation state of Cr in the reactant is +3.

In the products of the balanced equation (1), the charge on the two complex ions must be (+1) for [Cr(en)₂(NSC)₂] and (-1) for [Cr(en)(NCS)₄] since total charge on the complex [Cr(en)₂(NSC)₂][Cr(en)(NCS)₄] is zero (is neutral), so we have:

`[x + 2(0) + 2(-1)] + [x + 0 + 4(-1)] = 0`

`2x - 2 - 4 = 0`

`x = +3`

Therefore, the oxidation state on Cr is +3 in the two complex ions in the products.

I hope it helps you!

Answer 2

Balanced chemical equation:

2[Cr(en)₂(NCS)₂]SCN(s) → [Cr(en)₂(NCS)₂][Cr(en)(NCS)₄](s) + en(g)

Oxidation states: +3 for both reactant and product.

Step-by-step explanation:

The ethylenediamine has molecular formula C₂H₈N₂, which is called "en" in the compound. The reaction is:

[Cr(en)₂(NCS)₂]SCN(s) → [Cr(en)₂(NCS)₂][Cr(en)(NCS)₄](s) + en(g)

To balance the reaction, the number of elements must be equal on both sides of it. So, we must multiply the reactant by 2:

2[Cr(en)₂(NCS)₂]SCN(s) → [Cr(en)₂(NCS)₂][Cr(en)(NCS)₄](s) + en(g)

The ions NCS⁻ have oxidation number -1, the ethylenediamine has oxydation number 0, and the complex has neutral charge. So, calling the oxidation number of Cr as x:

Reactant

x + 2*0 + 2*(-1) + (-1) = 0

x -3 = 0

x = +3

Product

x + 2*0 + 2*(-1) + x + 0 + 4*(-1) = 0

2x -2 -4 = 0

2x = +6

x = +3

So, the Cr ions in both reactant and product have oxidation number +3.