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The electronic configuration ofhte eleme...

The electronic configuration ofhte elements. A, B and C are given below. Answer the question from 14 to 17 on the basis of these configuration.
`{:(A,1s^(2),2s^(2),2p^(6),,),(B,1s^(2),2s^(2),2p^(6),3s^(2),3p^(3)),(C,1s^(2),2s^(2),2p^(6),3s^(2),3p^(5)):}``
The bond between B and C will be

A

ionic

B

covalent

C

hydrogen

D

coordinate

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To determine the type of bond between elements B and C based on their electronic configurations, we will follow these steps: ### Step 1: Write the Electronic Configurations - For element B: \(1s^2, 2s^2, 2p^6, 3s^2, 3p^3\) - For element C: \(1s^2, 2s^2, 2p^6, 3s^2, 3p^5\) ### Step 2: Determine the Stability of Each Element - Element B has 5 electrons in its outer shell (3s and 3p orbitals). To achieve a stable electronic configuration (like a noble gas), it can lose 3 electrons from the 3p orbital. Thus, it can achieve a +3 oxidation state (B³⁺). - Element C has 7 electrons in its outer shell (3s and 3p orbitals). To achieve stability, it needs 1 more electron to complete its p orbital. Therefore, it can gain 1 electron to achieve a -1 oxidation state (C⁻). ### Step 3: Identify the Type of Bond Formed - Since B loses electrons and C gains electrons, there is a transfer of electrons from B to C. This results in the formation of ions: B³⁺ and C⁻. - The bond formed due to the electrostatic attraction between these oppositely charged ions is known as an **ionic bond**. ### Step 4: Write the Chemical Formula - The resulting compound from the bonding of B and C can be represented as \(BC_3\), indicating that three C⁻ ions bond with one B³⁺ ion. ### Conclusion The bond between B and C is an **ionic bond**. ---
To determine the type of bond between elements B and C based on their electronic configurations, we will follow these steps: ### Step 1: Write the Electronic Configurations - For element B: \(1s^2, 2s^2, 2p^6, 3s^2, 3p^3\) - For element C: \(1s^2, 2s^2, 2p^6, 3s^2, 3p^5\) ### Step 2: Determine the Stability of Each Element - Element B has 5 electrons in its outer shell (3s and 3p orbitals). To achieve a stable electronic configuration (like a noble gas), it can lose 3 electrons from the 3p orbital. Thus, it can achieve a +3 oxidation state (B³⁺). ...
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