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maks197457 [2]
3 years ago
12

Predict the ground-state electron configuration of the following ions. Write the answer in abbreviated form beginning with noble

gas brackets for Ru2+ and W3+

Chemistry
2 answers:
AlexFokin [52]3 years ago
7 0

Ru ²⁺ :[Kr] 4d⁶

W ³⁺ : [Xe] 4f¹⁴5d³

<h3>Further explanation </h3>

The Atomic Number (Z) indicates the number of protons in an atom of an element. If the atom is neutral then the number of protons will be equal to the number of electrons. So the atomic number can also indicate the number of electrons.

This energy level is expressed in the form of electron configurations.

Charging electrons in the sub shell use the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Writing the electron configuration can be abbreviated using the electron configuration of the noble gas in brackets because it has the same configuration

  • a. Ru element-Ruthenium

It is in group 8B and period 5 with atomic number 44

So the electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s¹, 4d⁷

This configuration includes an exception to the Aufbau rule which is that Electrons occupy orbitals of the lowest energy level, which should be 5s² 4d⁶ (exceptions occur mainly in the transition group), so that with the configuration of the noble gases from Kr [1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶], the Ru element can be written:

Ruthenium-Ru: [Cr] 5s¹4d⁷

Because it releases 2 electrons to form Ru²⁺ ions, the configuration becomes

Ru ²⁺ :[Kr] 4d⁶

  • W element - Wolfram / Tungsten

It is in group 6B and period 6 with atomic number 74

So the electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s²4f¹⁴5d⁴

Because according to Xe configuration [1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶] it can be written:

W: [Xe] 6s²4f¹⁴5d⁴

Because it releases 3 electrons to form a W³⁺ ion, the configuration becomes (electrons with the largest n , 6s and 5d, the electrons are removed first):

W ³⁺ : [Xe] 4f¹⁴5d³

<h3>Learn more  </h3>

element X  

brainly.com/question/2572495  

electrons and atomic orbitals  

brainly.com/question/1832385  

about subatomic particles statement  

brainly.com/question/3176193  

Tanzania [10]3 years ago
5 0

Answer: The ground-state electronic configuration of

1) Ru^{2+}=[Kr]4d^6

2) W^{3+]=4f^{14}5d^3

Explanation: The electronic configuration of elements is defined as the distribution of electrons around the nucleus of that element. It depends on the atomic number of the element.

Atomic number of the element = Number of electrons.

  • For Ru-element

Atomic number = 44

Number of electrons = 44

Electronic configuration of Ru-element = [Kr]4d^75s^1

To form Ru^{2+}, 2 electrons are released from the neutral Ru-element.

So, the electronic configuration of Ru^{2+}=[Kr]4d^6

  • For W-element

Atomic number = 74

Number of electrons = 74

Electronic configuration of W-element = [Xe]4f^{14}5d^46s^2

To form W^{3+}, 3 electrons are released from the neutral W-element.

So, the electronic configuration of W^{3+}=[Xe]4f^{14}5d^3

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For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed
Cloud [144]

Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅]  =  0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅]  = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

6 0
3 years ago
4
matrenka [14]

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

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2 years ago
A container of oxygen with a volume of 60 L is heated from 300 K to 400 k, What is the new volume?
Lunna [17]

Answer:

80L

Explanation:

V1/T1 = V2/T2

V2 = V1 T2/T1

T1 = 300K

V1 = 60L

T2 = 400K

V2 = ?

V2 = V1 T2/T1

V2 = (60L)(400K) / (300K)

V2 = 80L

7 0
2 years ago
Why are alkalis sometimes<br> described as the opposite of<br> acids?
zlopas [31]

Answer: Because Alkalis are bases that can dissolve in water, which can also neutralize acid.

7 0
3 years ago
Acetic acid has a Kb of 2.93 °C/m and a normal boiling point of 118.1 °C. What would be the boiling point of a solution made by
alexdok [17]

Boiling point elevation is given as:

ΔTb=iKbm

Where,

ΔTb=elevation in the boiling point

that is given by expression:

ΔTb=Tb (solution) - Tb (pure solvent)

Here Tb (pure solvent)=118.1 °C

i for CaCO3= 2

Kb=2.93 °C/m

m=Molality of CaCO₃:

Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)

=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)

=(100.0÷100 g/mol)/0.4

= 2.5 m

So now putting value of m, i and Kb in the boiling point elevation equation we get:

ΔTb=iKbm

=2×2.93×2.5

=14.65 °C

boiling point of a solution can be calculated:

ΔTb=Tb (solution) - Tb (pure solvent)

14.65=Tb (solution)-118.1

Tb (solution)=118.1+14.65

=132.75

3 0
3 years ago
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