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zheka24 [161]
3 years ago
8

Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -

Chemistry
2 answers:
Ann [662]3 years ago
6 0

The oxidation state of the Ru metal in the complex {\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-} is \boxed{{\text{zero}}} .

Further explanation:

Oxidation number:

The oxidation number is used to represent the formal charge on an atom. It also shows that gain or loss of electrons by the atom. Oxidation number can be a positive or negative number but cannot be fractional.

The rules to identify the oxidation state:

(1) The oxidation state of an atom in the elemental form is zero.

(2) The total charge on the species is equal to the sum of the oxidation state of individual atoms.

(3) The oxidation state of halides is -1. For example, fluorine, chlorine, bromine, iodine have -1 oxidation state.

(4)Hydrogen has a +1 oxidation state.

(5) Oxygen has an oxidation state -2.

(6) In a coordination compound, neutral ligands have zero oxidation state and negative ligands such as CN have -1 oxidation.

The given compound is {\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}

Here, CN is a negative ligand thus oxidation state is -1 and CO is a neutral ligand thus it has 0 oxidation state. Also, the complex has -1 negative charge.

The expression to calculate the oxidation state in {\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-} is,

\left[{\left({{\text{oxidation state of Ru}}}\right)+\left({{\text{oxidation state of CN}}}\right)+4\left({{\text{oxidation state of CO}}}\right)}\right]=-1

…… (1)

Rearrange equation (1) for the oxidation state of Ru.

{\text{Oxidation state of Ru}}=\left[{-\left({{\text{oxidation state of CN}}}\right)-4\left({{\text{oxidation state of CO}}}\right)-1}\right]

…… (2)  

Substitute -1for the oxidation {\text{state}}  of CN and 0 for the oxidation state of CO in equation (2).

\begin{aligned}{\text{Oxidation state of Ru}}&=\left[{-\left({-{\text{1}}}\right)-4\left({\text{0}}\right)-1}\right]\\&=0\\\end{aligned}

The oxidation state of Ru is zero.

Learn more:

1. General statement that is not applied on metals: <u>brainly.com/question/2474874 </u>

2. The neutral element represented by the excited state electronic configuration: <u>brainly.com/question/9616334 </u>

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Coordination complex

Keywords: Oxidation state, metal complex, ru(cn)(co)4-, formal charge, cynide, carbonyl, zero, hydrogen, oxygen, 0 and -1.

kogti [31]3 years ago
3 0
The given complex ion is as follow,

                                              [Ru (CN) (CO)₄]⁻

Where;
            [ ]  =  Coordination Sphere

            Ru  =  Central Metal Atom  =  <span>Ruthenium

            CN  =  Cyanide Ligand

            CO  =  Carbonyl Ligand

The charge on Ru is calculated as follow,

                               Ru + (CN) + (CO)</span>₄  =  -1
Where;
            -1  =  overall charge on sphere

             0  =  Charge on neutral CO

            -1  =  Charge on CN

So, Putting values,


                               Ru + (-1) + (0)₄  =  -1

                               Ru - 1 + 0  =  -1

                               Ru - 1  =  -1

                               Ru  =  -1 + 1

                               Ru  =  0
Result:
          <span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
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Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

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3 years ago
The density of iron is about ___ times greater than water.<br><br> 6, 2, 8, 10, or 4
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Explanation:

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3 years ago
How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
aksik [14]

Complete Question:

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

Answer:

2.23x10⁶ g

Explanation:

The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:

0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³

0.7976 g/m³

The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:

A = πR², where R is the radius = 1.01x10² m (half of the diameter)

A = π*(1.01x10²)²

A = 32047 m²

The volume is then:

V = 32047 * 87.32

V = 2.7983x10⁶ m³

The mass of the F⁻ is the concentration multiplied by the volume:

m = 0.7976 * 2.7983x10⁶

m = 2.23x10⁶ g

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