Question

How many \(\sigma\) and \(\pi\) bonds are there in the molecule of tetracyanoethylene \((\mathrm{CN})_{2} \mathrm{C}=\mathrm{C}(\mathrm{CN})_{2} ?\) (1) Nine \(\sigma\) and nine \(\pi\) (2) Five \(\sigma\) and nine \(\pi\) (3) Nine \(\sigma\) and seven \(\pi\) (4) Five \(\sigma\) and eight \(\pi\)

Short answer

(3) Nine \( \text{σ} \) and seven \( \text{π} \)

Step-by-step solution

Draw the tetracyanoethylene molecule

The molecular formula for tetracyanoethylene is \(\text{(CN)}_{2}C=C(\text{CN})_{2}\). Begin by drawing this structure to visualize the bonding. Each \(CN\) group consists of a carbon triple-bonded to a nitrogen.

Identify all the \( \text{C} \) and \( \text{N} \) atoms

In the molecule, there are four \( \text{CN} \) groups and one \( \text{C=C} \) double bond between the central carbon atoms. In total, there are four carbons in the \( \text{CN} \) groups and two central carbons.

Count the \( \text{C–N} \) single bonds

Each of the four \( \text{CN} \) groups has a \( \text{C–N} \) single bond. So, there are four \( \text{C–N} \) single bonds.

Count the \( \text{N≡C} \) triple bonds

Each of the four \( \text{CN} \) groups has a \( \text{N≡C} \) triple bond containing one \( \text{σ} \) and two \( \text{π} \) bonds, resulting in four \( \text{σ} \) bonds and eight \( \text{π} \) bonds.

Count the \( \text{C=C} \) double bond

The \( \text{C=C} \) double bond between the two central carbon atoms consists of one \( \text{σ} \) bond and one \( \text{π} \) bond.

Sum up the \( \text{σ} \) and \( \text{π} \) bonds

Adding all the bonds together, there are 4 \( \text{C–N} \) single bonds (\(\text{σ}\text{-bonds}\)), 4 \( \text{N≡C} \) triple bonds (\(\text{σ}\text{-bonds}\) and \(\text{π}\text{-bonds}\)), and 1 \( \text{C=C} \) double bond (\(\text{σ}\text{-bonds}\) and \(\text{π}\text{-bonds}\)). So, in total, there are 9 \( \text{σ} \) bonds and 7 \( \text{π} \) bonds.

Key concepts

sigma bonds

In tetracyanoethylene, we need to understand the concept of \( \sigma \) bonds. A \( \sigma \) bond is the strongest type of covalent bond, formed by the direct overlap of atomic orbitals.
These bonds are typically single bonds, but every double and triple bond also contains one \( \sigma \) bond.
In tetracyanoethylene (\text{(CN)}_{2}C=C(\text{CN})_{2}), the four carbon-nitrogen single bonds (\( \text{C-N} \)) are \( \sigma \) bonds.
Additionally, the carbon-carbon double bond (\text{C=C}) between the central carbons contributes another \( \sigma \) bond.
Each of the carbon-nitrogen triple bonds (\text{N\text{≡}C}) includes one \( \sigma \) bond.
This results in the molecule having a total of 9 \( \sigma \) bonds.

pi bonds

Now, let's look at \( \pi \) bonds in tetracyanoethylene. A \( \pi \) bond is formed by the sideways overlap of adjacent p-orbitals. These bonds are typically found in double and triple bonds.
In a triple bond, there would be 2 \( \pi \) bonds along with 1 \( \sigma \) bond.
In tetracyanoethylene, each of the \text{N\text{≡}C} triple bonds contains 2 \( \pi \) bonds.
Moreover, the \text{C=C} double bond between the two central carbons also contains 1 \( \pi \) bond.
Therefore, adding these up, there are a total of 7 \( \pi \) bonds in the molecule.

molecular structure analysis

Analyzing the molecular structure of tetracyanoethylene (\text{(CN)}_{2}C=C(\text{CN})_{2}) helps us understand the distribution of \( \sigma \) and \( \pi \) bonds.
Start by drawing the structure, identifying the central carbon-carbon double bond (\text{C=C}) and the carbon-nitrogen bonds.
Notice that each \text{CN} group has a carbon atom triple-bonded to a nitrogen atom and single-bonded to the central carbon.
Examining these bonds allows us to identify 4 \( \sigma \) bonds from the \text{C-N} single bonds, 4 \( \sigma \) and 8 \( \pi \) bonds from the \text{N\text{≡}C} triple bonds, plus 1 \( \sigma \) and 1 \( \pi \) bond from the \text{C=C} bond.
This comprehensive analysis confirms that tetracyanoethylene has 9 \( \sigma \) bonds and 7 \( \pi \) bonds.

tetracyanoethylene

Tetracyanoethylene, (\text{(CN)}_{2}C=C(\text{CN})_{2}), is an organic molecule containing two cyanide groups attached on either side of a central ethylene (C=C) structure.
The four cyanide groups are crucial in determining the bonding nature within this molecule.
Cyanide groups are represented as \text{N\text{≡}C}, where the carbon and nitrogen atoms are connected by a triple bond.
This gives the molecule unique properties and bond distribution, resulting in a total of 9 \( \sigma \) bonds and 7 \( \pi \) bonds.

chemical bonding

Chemical bonding is essential to understand how atoms connect and form molecules. There are several types of bonds, but within tetracyanoethylene, we observe two main types: \( \sigma \) and \( \pi \) bonds.
\( \sigma \) bonds form the framework of the molecule, while \( \pi \) bonds provide additional stability and influence the molecule's properties.
Understanding the distribution of these bonds within tetracyanoethylene (\text{(CN)}_{2}C=C(\text{CN})_{2}) reveals 4 \( \sigma \) and 8 \( \pi \) bonds derived from the \text{N\text{≡}C} triple bonds, and additional contributions from the \text{C-N} and \text{C=C} bonds.
This results in 9 \( \sigma \) bonds and 7 \( \pi \) bonds in total.