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maks197457 [2]
4 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]4 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]4 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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alisha [4.7K]

<u>Answer:</u> The Gibbs free energy of the given reaction is 1.379\times 10^3kJ

<u>Explanation:</u>

The equation used to calculate Gibbs free energy change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

For the given chemical reaction:

2CH_3OH(g)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(CO_2(g))})+(4\times \Delta G^o_f_{(H_2O(g))})]-[(2\times \Delta G^o_f_{(CH_3OH(g))})+(3\times \Delta G^o_f_{(O_2(g))})]

We are given:

\Delta G^o_f_{(H_2O(g))}=-228.57kJ/mol\\\Delta G^o_f_{(CO_2(g))}=-394.36kJ/mol\\\Delta G^o_f_{(CH_3OH(g))}=-161.96kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-394.36))+(4\times (-228.57))]-[(2\times (-161.96))+(3\times (0))]\\\\\Delta G^o_{rxn}=-1379.08kJ/mol

The equation used to Gibbs free energy of the reaction follows:

\Delta G=\Delta G^o+RT\ln Q_{p}

where,

\Delta G = free energy of the reaction

\Delta G^o = standard Gibbs free energy = -1379.08 kJ/mol = -1379080 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/K mol

T = Temperature = 25^oC=[273+25]K=298K

Q_{p} = Ratio of concentration of products and reactants = \frac{(p_{CO_2})^2(p_{H_2O})^4}{(p_{CH_3OH})^2(p_{O_2})^3}

p_{CO_2}=5.29atm\\p_{H_2O}=3.89atm\\p_{CH_3OH}=3.82atm\\p_{O_2}=7.56atm

Putting values in above expression, we get:

\Delta G=-1379080J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(5.29)^2\times (3.89)^4}{(3.82)^2\times (7.56)^3}))\\\\\Delta G=-1379039J=1379.039kJ=1.379\times 10^3kJ

Hence, the Gibbs free energy of the given reaction is 1.379\times 10^3kJ

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