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3 years ago

Write a complete electron configuration for borin

Chemistry

1 answer:

olga2289 [7]3 years ago

6 0

The electron configuration for Boron (B) is given as $1s^2 2s^2 2p^1$ or $[He] 2s^2 2p^1$

Answer:

Boron (B) has an atomic number equal to 5 and is an element from Group 13 of the periodic table. It has a total of 5 electrons of which 3 are valence electrons.

The electron configuration for Boron can be written with the first two electrons in the 1s orbital. Since 1s orbital can hold only two electrons, out of the remaining three electrons; two electrons will go in the 2s orbital. The remaining one electron in the 2p orbital. Therefore the Boron electron configuration will be $1s^2 2s^2 2p^1$

Since $1s^2$ is the electron configuration for Helium, Electron configuration for Boron can also be written as $[He] 2s^2 2p^1$

<em>(Refer attached figure)</em>

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3 years ago

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For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

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Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

$KNO_{3}$ act as an oxidizing agent.

The oxidation number of N in $KNO_{3}$ is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in $KNO_{2}$ is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number reduced this means $KNO_{3}$ itself get reduced to $KNO_{2}$ and it can act as an oxidizing agent.

$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).

Now, we conclude that the oxidation number increases this means $KNO_{3}$ itself get oxidized to $O_{2}$ and it can act as reducing agent.

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