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Draw the Lewis dot structure for the cyanide ion (CN-) and identify the hybridization state of the carbon and nitrogen atoms. Also identify the sigma and pi bonds present in the structure and any non-bonding electrons in the molecule.

Sagot :

1) The Lewis dot structure of the cyanide ion (CN-) is as follows:

Cyanide ion (CN-):

C: N:

2) Electron configuration:

C: [He] 2s2 2p2

N: [He] 2s2 2p3

The hybridization state of the carbon and nitrogen atoms in the cyanide ion is sp.

3) In the cyanide ion, the sigma bonds are the single bonds between the carbon and nitrogen atoms, and the pi bonds are the double bonds between the carbon and nitrogen atoms. There are no non-bonding electrons in the molecule.

The cyanide ion has a total of 10 valence electrons, which are used to form bonds between the carbon and nitrogen atoms. The carbon atom forms a single bond with the nitrogen atom by sharing one of its 2p electrons, and the nitrogen atom forms a triple bond with the carbon atom by sharing three of its 2p electrons. This results in a linear molecule with a carbon atom bonded to a nitrogen atom, with both atoms having a single bond and a double bond.

The hybridization state of the carbon and nitrogen atoms in the cyanide ion is sp. This is because the carbon atom has two unpaired electrons in its 2p orbitals, and the nitrogen atom has three unpaired electrons in its 2p orbitals. These unpaired electrons form sigma bonds with each other, resulting in an sp hybridization state.

Learn more about Lewis dot structure, here https://brainly.com/question/20300458

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