Draw the Lewis structure for a dinitride(2-) ion N22-
Select the reagent(s) required for the following transformation.
a. NaNO2, HCl; (2) H2
b. NaNO2, HCl; (2) H2O
c. NaNO2, HCl; (2) H3PO4
d. NaNO2, HCl; (2) H3PO2
The Correct Answer and Explanation is :
Lewis Structure of the Dinitride Ion (N₂²⁻):
The dinitride ion, N₂²⁻, consists of two nitrogen atoms and a total of two extra electrons (due to the 2⁻ charge). Here’s how to draw its Lewis structure:
- Count the total number of valence electrons:
- Each nitrogen atom has 5 valence electrons (since nitrogen is in group 15 of the periodic table).
- The 2⁻ charge adds 2 more electrons.
- Total electrons = (5 × 2) + 2 = 12 valence electrons.
- Structure Setup:
- Place the two nitrogen atoms in the center, with a single bond between them (which will consume 2 electrons).
- We have 10 remaining electrons to be distributed.
- Distribute the remaining electrons:
- Assign 6 electrons (3 pairs) as lone pairs around each nitrogen atom to complete their octets (8 electrons around each nitrogen atom).
- Examine the bonding:
- The single bond accounts for 2 electrons, but to complete the octet and give the molecule the correct charge, we need to make a double or triple bond between the two nitrogen atoms.
- Given that the total valence electrons are 12, and each nitrogen atom needs 8 electrons, the most likely bonding configuration is a triple bond between the two nitrogen atoms, with a lone pair remaining on each nitrogen atom.
Thus, the Lewis structure of N₂²⁻ is:
N≡NWith two lone pairs on each nitrogen atom, and the 2⁻ charge is shared equally.
Select the Reagent(s) for the Following Transformation:
This question involves a series of reactions. Let’s break down each option:
Reagents Explanation:
The reactions are likely related to the conversion of one compound to another. The key here is identifying the correct reagents to facilitate a particular transformation, typically involving nitrosonium (NO⁺) chemistry, redox reactions, or the influence of different acids. Here’s the breakdown of each option:
- a. NaNO₂, HCl; (2) H₂:
- NaNO₂ (Sodium nitrite) and HCl (Hydrochloric acid) are commonly used for diazotization reactions, where an amine group is converted into a diazonium ion. The subsequent treatment with H₂ (Hydrogen) would likely reduce a nitroso or a diazonium species to an amine or other reduced form.
- b. NaNO₂, HCl; (2) H₂O:
- This option likely involves the diazotization reaction, where the nitrite (NaNO₂) reacts with the amine in acidic conditions (HCl), forming a diazonium salt. Instead of using hydrogen gas (H₂), water (H₂O) is likely used to stabilize or facilitate the diazonium ion’s further reactions, possibly leading to substitution or coupling reactions.
- c. NaNO₂, HCl; (2) H₃PO₄:
- Here, H₃PO₄ (Phosphoric acid) is used instead of water or hydrogen. Phosphoric acid can act as a dehydrating agent and may help stabilize intermediates, potentially favoring nitration or nitration-like reactions rather than simply diazotization.
- d. NaNO₂, HCl; (2) H₃PO₂:
- H₃PO₂ (Phosphorous acid) is a reducing agent. When used in this context, it can reduce diazonium ions to the corresponding amines or other reduced nitrogen compounds. This is likely the correct choice if the aim is to reduce a diazonium intermediate or related species.
Correct Answer: d. NaNO₂, HCl; (2) H₃PO₂
Explanation:
The combination of NaNO₂ and HCl is used to form a diazonium salt (as part of a diazotization reaction), and the presence of H₃PO₂ (phosphorous acid) serves as a reducing agent. This is commonly used to reduce diazonium ions to the corresponding amines or other nitrogen-containing compounds. Therefore, the correct reagents to complete the transformation in this case are NaNO₂, HCl, and H₃PO₂.