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Elodia [21]
3 years ago
8

Determine if the electron configurations are valid or invalid

Chemistry
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer:

1) A (valid)

2) B (Invalid)

3) B (Invalid)

4) A (valid)

5) B (Invalid)

6) B (Invalid)

Explanation:

Electronic configuration:

The electronic configuration is the arrangnment of electrons in orbitals (s, p, d, f, g, h so on) according to quantum  number and Aufbau rule.

An atom of element have orbitals, and electrons are revolving in it.

There are quantum numbers that describe the number of electron in specific shell.

The Principle quantum number is denoted by "n" and can not be "0".

These "n" is K, L, M, N and electrons are distribution in it.

K =2

L= 8

M= 18

According to Aufbau electron occupied in the lowest energy level before occupying the highest energy level.

According to this the electronic configuration is

1s², 2s², 2p⁶ and so on....

So,

"s" orbital accommodate maximum of 2 electrons

"p" orbital accommodate maximum of 6 electrons

"d" orbital accommodate maximum of 10 electrons

"f" orbital accommodate maximum of 14 electrons

and the coefficient 1, 2, 3, 4 are energy level of orbitals.

The 1st orbital is 1s and it only can occupy 2 electrons the next coming electron goes into 2s that also can occupy 2 electrons and the further electrons will go to 2p orbital.

So keeping in mind all above details we investigate all option one by one

1.  A (valid)

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵

This is valid option. number of electrons are rightly accommodated in orbitals. Also 3d is higher energy level than 4p.

______________

2.  B (Invalid)

1s², 2s², 2p⁶, 3s², 3p⁷

This is invalid option. number of electrons are not rightly accommodated in orbitals as "p" orbital can only accommodate 6 electrons and not more than 6.

________________

3.  B (Invalid)

[Xe], 7s², 6d¹, 5f⁸

This is Invalid option. Number of electrons are not rightly accommodated in orbitals. There should 5f orbital before 6d.

______________

4.  A (valid)

[Kr] 5s², 4d¹⁰, 5p⁵

This is valid option. number of electrons are rightly accommodated in orbitals.

______________

5.  B (Invalid)

[He]--

This is invalid option. There is no exact explanation of electron distribution in orbitals.

______________

6.  B (Invalid)

1s², 2s², 2p⁶, 3s³, 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s--

This is invalid option. number of electrons are not rightly accommodated in orbitals. As "s" can accommodate only 2 electrons not 3 electrons. also there is no details of 5s

______________

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g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc
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\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

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Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

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Question 34 (1 point)
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Answer:

– 1

Explanation:

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