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shutvik [7]
2 years ago
10

Og is the noble gas after rn. to go from [rn] to [og], you must fill four subshells (s, p, d, and f) with a total of 32 electron

s. thus, the atomic numbers of 6th and 7th period elements of the same group differ by 32. to go from [og] to the next noble gas, however, you would theoretically fill five subshells (s, p, d, f, and g). how many electrons are needed to fill all five subshells?

Chemistry
2 answers:
Alenkinab [10]2 years ago
7 0

The total number of electrons that are required to fill all five subshells namely s, p, d, f, g is \boxed{50{\text{ electrons}}}.

Further Explanation:

The electronic configuration is the distribution of electrons of an atom in the atomic orbitals. There are two states for an electron: ground as well as the excited state. The configuration of the atom in the lowest possible energy levels is called the ground-state electronic configuration. When an electron jumps from the stable ground state to some higher level, that state is called the excited state and the electronic configuration corresponding to this state is known as the excited-state half-filled configuration.

The filling of electrons in different energy levels or orbitals is done in accordance with the following three rules.

1. Aufbau principle: The principle states that the electrons are filled in various orbitals in the increasing order of their energies as follows:

1s

2. Hund’s rule: Electron pairing will not start until each orbital is singly occupied.

3. Pauli’s exclusion principle: According to this principle, all the four quantum numbers \left({n,\;l,\;{m_l},\;{m_s}}\right)  for any two electrons can never be the same. In an orbital, the spin of two electrons has to be different. If one electron has the clockwise spin, the other would have the anticlockwise spin and vice-versa.

The elements that fully filled valence shells are known as noble gas elements or inert gases. They occupy the group 18 of the periodic table.

The electronic configuration of known noble gas elements according to the Aufbau rule is shown in the attached image.

The element Og having atomic number 118 is the noble gas element expected to be filled after Rn. In accordance with the Aufbau principle, the orbitals to be filled should be 5f, 6d,7s,7p in addition to the electrons in the radon core.

The g subshell has nine orbitals each of which would contain 18 electrons. Hence, the total number of electrons required to be filled is calculated as follows:

\begin{aligned}{\text{Total electrons}}&=s+p+d+f+g\\&=2+6+10+14+18\\&=50\\\end{aligned}

Learn more:

1. Determine the ion with configuration \left[{{\text{Ar}}}\right]\;3{d^2} : brainly.com/question/7599542

2. Oxidation and reduction reaction: brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Electronic configuration of the elements

Keywords: Aufbau principle, Pauli’s exclusion principle, Hund’s rule of maximum multiplicity,g subshell, Noble gas configuration, energy levels, orbital, Og and radon.

Juliette [100K]2 years ago
5 0

Answer : 50 electrons are needed to fill all five subshells

Explanation :

To move from Rn to Og, we need to fill 4 subshells which are s, p d & f.

The maximum number of electrons that can be added to each orbital are given below.

s = 2 e⁻

p = 6 e⁻

d = 10 e⁻

f = 14 e⁻

If we add all these electrons, we get 2 + 6 + 10 + 14 = 32 electrons.

We need to add 32 electrons to the atomic number of Rn to get the atomic number of Og.

Hence we say that the atomic numbers of 6th and 7th period elements of the same group differ by 32.

To go from Og to the next noble gas, we have to theoretically fill 5 subshells.

The maximum electrons that can be added to g orbital are 18.

Therefore in order to go from Og to next element, we would add

2 + 6 + 10 + 14 + 18 = 50 electrons.

50 electrons are needed to fill all five subshells.


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<h3>Answer:</h3>

28.96 kJ/°C

<h3>Explanation:</h3>

We are given;

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We are required to find the heat capacity of benzoic acid;

<h3>Step 1: Moles of benzoic acid </h3>

Moles = Mass ÷ molar mass

Molar mass of benzoic = 122.12 g/mol

Therefore;

Moles = 3.1007 g ÷ 122.12 g/mol

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<h3>Step 2: Determine the specific heat capacity </h3>

Heat change for 1 mole = 3226.7 kJ

Moles of Benzoic acid = 0.0254 moles

But;

Specific heat capacity × ΔT = Moles × Heat change

  cΔT = nΔH

Therefore;

Specific heat capacity,c = nΔH ÷ ΔT

                                     = (3226.7 kJ × 0.0254 moles) ÷ 2.83°C

                                    = 28.96 kJ/°C

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Calculate the density of a sample of 1.00 mole of NH3 at 793mmHg and -9.00 C
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A mixture of 2.00 moles of H., 3.00 moles of NH3, 4.00 moles of Co, and 5.00 moles.

Explanation:

A mixture of 2.00 moles of H., 3.00 moles of NH3, 4.00 moles of Co, and 5.00 moles.

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If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen
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This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

2NOBr(g)\rightarrow 2NO(g)+Br_2(g)

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

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k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken  = 142 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.0440 M

Putting values in above equation, we get:

0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)

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2 years ago
certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 40.00 gram sample of the alcohol pro
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Answer:

The answer to your question is:    C₂H₆O₁   = C₂H₆O

Explanation:

Data

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mass = 40 g  produced 76.40 g of CO2

                                       46.96 g of H2O

Empirical formula = ?

Process

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                         44g of CO2 --------------------  12 g of Carbon

                         76.40 g of CO2 --------------- x

                          x = 20.84 g of Carbon

                         12 g of Carbon ---------------  1 mol

                         20.84 g of C    ---------------   x

                         x = (20.84 x 1) / 12

                         x = 1.74 mol of Carbon

                        18 g of H2O --------------------  2 g of H

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                        x = (46.96 x 2) / 18

                        x = 5.22 g of H

                        1 g of H ------------------------  1 mol of H

                        5.22 g of H -------------------   x

                        x = 5.22 mol of H

Mass of Oxygen = 40 - 20.84 - 5.22g

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                         16 g of Oxygen ----------------  1 mol

                         13.94 g of O --------------------   x

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Divide by the lowest number of moles

Carbon = 1.74 / 0.87 = 2

Hydrogen = 5.22 / 0.87 = 6

Oxygen = 0.87 / 0.87 = 1

                           

Empirical formula

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