Question

Write Lewis structures for these four isoelectronic species: (a) CO, (b) \(\mathrm{NO}^{+},\) (c) \(\mathrm{CN}^{-},\) (d) \(\mathrm{N}_{2}\). Show formal charges.

Short answer

The Lewis structures would depict an arrangement where \(CO\) has a single bond with Oxygen carrying a negative formal charge and Carbon a positive one. \(\mathrm{NO}^{+}\) features a single bond with Nitrogen carrying a negative charge and Oxygen a positive one. For \(\mathrm{CN}^{-}\), a single bond is present with Nitrogen showing a negative formal charge. Lastly, \(\mathrm{N}_{2}\) is observed to have a triple bond with no formal charges present on Nitrogen atoms.

Step-by-step solution

Drawing Lewis Structure for \(CO\)

First, calculate the total number of valence electrons available. Carbon has 4 and oxygen has 6. Therefore, there are 10 electrons in total. Draw C and O side by side, and connect them with a straight line (which represents two shared electrons or one covalent bond). Distribute the remaining 8 electrons amongst the atoms in pairs. The oxygen atom will end up with 6 electrons around it, which gives it a negative formal charge, while the carbon atom will hold a positive formal charge.

Drawing Lewis Structure for \(\mathrm{NO}^{+}\)

In \(\mathrm{NO}^{+}\), the nitrogen atom provides 5 valence electrons and oxygen provides 6, but since the molecule carries a +1-positive charge, you have to subtract one electron from the total count, leaving us with 10 electrons. After forming a single bond between the N and O atom, which uses up two electrons, distribute the remaining 8 electrons in pairs to each atom. The Nitrogen atom will own five electrons (a negative formal charge) while Oxygen will own a positive charge.

Drawing Lewis Structure for \(\mathrm{CN}^{-}\)

In \(\mathrm{CN}^{-}\), carbon contributes 4 valence electrons while nitrogen provides 5. The molecule carries a negative charge, so add an extra electron to the total. There are 10 electrons in total. After single bonding C and N, distribute the remaining 8 electrons giving nitrogen a formal charge of -1.

Drawing Lewis Structure for \(\mathrm{N}_{2}\)

For \(\mathrm{N}_{2}\), each Nitrogen atom has 5 valence electrons for a total of 10 electrons. After forming a triple bond between the Nitrogen atoms (using 6 electrons), distribute the remaining 4 electrons in pair to each atom. The Nitrogens atoms remain neutral with no formal charges due to all the electrons being evenly distributed.

Key concepts

Isoelectronic Species

Isoelectronic species are fascinating entities in the world of chemistry. They refer to different molecules or ions that have the same number of electrons. While their compositions differ, their electron count matches, leading to similar electronic structures. In the original exercise, the species CO, NO⁺, CN^-, and N₂ are all isoelectronic as they each have a total of 10 electrons.
For students, recognizing isoelectronic species is crucial, as it helps understand concepts related to molecular stability and reactivity. These species are prime examples of how different atoms or molecules can be linked by their electronic similarities.

• CO: In the molecule of carbon monoxide, a total of 10 valence electrons are distributed between carbon and oxygen.
• NO⁺: The addition or removal of a charge in species like NO⁺ can maintain the isoelectronic status by adjusting the electron count.
• CN^-: The negative charge in cyanide adjusts total electrons to 10, keeping it isoelectronic with the others.
• N₂: Molecular nitrogen is naturally stable with 10 electrons, fitting it into this group effortlessly.

Recognizing these events aids in predicting molecular similarities and property comparisons.

Formal Charge

Understanding formal charge is pivotal when drawing Lewis Structures, as it allows us to evaluate the stability and distribution of electrons in a molecule or ion. The formal charge is essentially the theoretical charge on an atom in a molecule, calculated by:
\[\text{Formal Charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{\text{Bonding electrons}}{2}\]Applying this formula helps us assess whether atoms in a molecule are distributing electrons optimally.
The Lewis structure of CO demonstrates the importance of formal charges as it shows that the distribution of electrons leaves oxygen with a negative charge and carbon with a positive one.

• This imbalance occurs because oxygen, being more electronegative, tends to retain more electrons.
• The structure of NO⁺ showcases a similar scenario, where nitrogen ends up with a negative charge due to its electron arrangement.
• On the other hand, CN^- displays a negative charge primarily on nitrogen, aligning with standard electronegativity trends.

For N₂, the formal charges balance out perfectly, indicating a stable, neutral molecule. Calculating formal charges ensures the most probable arrangement of electrons, aiding in hypothesizing molecular behaviors.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a central role in chemical bonding and reactions. These electrons determine how an atom will interact and bond with others, making them fundamental in the formation of molecules.
In the context of Lewis structures, counting valence electrons accurately is the first step:

• Carbon (C): Has 4 valence electrons.
• Oxygen (O): Possesses 6 valence electrons.
• Nitrogen (N): Has 5 valence electrons.

To construct a valid Lewis structure, tally the valence electrons and allocate them to fulfill the molecule's or ion's bonding requirements. For instance, in CO: - Carbon's 4 and Oxygen's 6 valence electrons sum up to 10, forming bonds and fulfilling each atom's octet needs.
This theme repeats with all other isoelectronic species like NO⁺, CN^-, and N₂. Understanding valence electrons not only helps in creating accurate Lewis structures but also provides insight into molecular interactions and stability.