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Varvara68 [4.7K]
2 years ago
7

How many electrons are typically involved in bonding for group 1, group 2 and group 3 elements?

Chemistry
1 answer:
olga55 [171]2 years ago
8 0

Here, we are required to determine how many electrons are typically involved in bonding for group 1, group 2 and group 3 elements and How many electrons are typically involved in bonding for group 6, 7 and 8 elements.

  • For group 1: Only one electron.
  • For group 2: Only two electrons
  • For group 3: Only 3 electrons
  • For group 6: Only 2 electrons
  • Fir group 7: Only 1 electron
  • For group 8: No electron

<em>The Valence electrons are the electrons which are available for bonding and chemical reactions in elements.</em>

<em>The Valence electrons are the electrons which are available for bonding and chemical reactions in elements.However, not all Valence electrons are actively involved in bonding. This is evident in the group 7(halogens) elements, although they have 7 Valence electrons, only one is actively involved in bonding.</em>

  • For group 1 elements: They have only one Valence electron and it is actively involved in bonding (ionic bonding)

  • For group 2 elements: They have only two Valence electron and both are actively involved in bonding (ionic bonding)

  • For group 3 elements: They have only three Valence electrons and all three are typically used for bonding.

  • For group 6 elements: They have 6 Valence electrons and are electronegative. As such, they only need to receive 2 electrons from an electropositive element and consequently, they only contribute two electrons into the bonding. As such, only 2 electrons are typically used for bonding.

  • For group 7: They have 7 Valence electrons and are highly electronegative. As such, they only need to receive 1 electrons from an electropositive element and consequently, they only contribute 1 electrons into the bonding. As such, only 1 electrons are typically used for bonding.

  • For group 8: They have a full octet configuration and are unreactive. Therefore, no electron is typically used for bonding.

Read more:

brainly.com/question/18258856

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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