Problem 25
Question
Amongst the trihalides of nitrogen, which one is the least basic? (a) \(\mathrm{NF}_{3}\) (b) \(\mathrm{NCl}_{3}\) (c) \(\mathrm{NBr}_{3}\) (d) \(\mathrm{NI}_{3}\)
Step-by-Step Solution
Verified Answer
The least basic trihalide of nitrogen is \(\mathrm{NF}_{3}\).
1Step 1: Understanding Trihalides of Nitrogen
Trihalides of nitrogen are compounds where nitrogen forms a single bond with three halogen atoms. These compounds include \(\mathrm{NF}_{3}\), \(\mathrm{NCl}_{3}\), \(\mathrm{NBr}_{3}\), and \(\mathrm{NI}_{3}\). The basicity of these compounds depends on the ability of the nitrogen atom to donate its lone pair of electrons.
2Step 2: Basicity and Electronegativity
The basicity of a nitrogen halide is inversely related to the electronegativity of the halogen attached to the nitrogen. The more electronegative the halogen, the more it attracts the shared electrons towards itself, making it less available for donation, which reduces the basicity. Among the halogens, fluorine is the most electronegative, followed by chlorine, bromine, and iodine.
3Step 3: Analyzing Each Trihalide
\(\mathrm{NF}_{3}\) has nitrogen attached to the most electronegative halogen, fluorine, which makes the lone pair of electrons on nitrogen the least available for donation, making it the least basic. In contrast, \(\mathrm{NI}_{3}\) has iodine, which is the least electronegative, making \(\mathrm{NI}_{3}\) potentially more basic than \(\mathrm{NF}_{3}\).
4Step 4: Conclusion on Basicity
Since fluorine is the most electronegative among the halogens mentioned, it results in \(\mathrm{NF}_{3}\) being the least basic among the nitrogen trihalides due to the decreased availability of nitrogen's lone pair for other reactions or binding due to the strength of fluorine's electronegativity.
Key Concepts
BasicityElectronegativityLone Pair Donation
Basicity
Basicity refers to the ability of a molecule to accept protons or donate a pair of electrons. More simply, in the context of nitrogen trihalides, it is about how "willing" the nitrogen atom is to share its electrons with another atom or molecule.
In nitrogen trihalides like \( \mathrm{NF}_{3} \), \( \mathrm{NCl}_{3} \), \( \mathrm{NBr}_{3} \), and \( \mathrm{NI}_{3} \), the nitrogen atom has a lone pair of electrons. These electrons can potentially be donated to other atoms or compounds, which would influence the basicity of the trihalide. However, the presence of different halogens influences this ability significantly.
The more basic a compound is, the stronger its ability to donate this lone pair. This becomes key for distinguishing between different nitrogen trihalides.
In nitrogen trihalides like \( \mathrm{NF}_{3} \), \( \mathrm{NCl}_{3} \), \( \mathrm{NBr}_{3} \), and \( \mathrm{NI}_{3} \), the nitrogen atom has a lone pair of electrons. These electrons can potentially be donated to other atoms or compounds, which would influence the basicity of the trihalide. However, the presence of different halogens influences this ability significantly.
The more basic a compound is, the stronger its ability to donate this lone pair. This becomes key for distinguishing between different nitrogen trihalides.
Electronegativity
Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself. This property is crucial in determining the behavior of a nitrogen trihalide compound.
This is why \( \mathrm{NF}_{3} \) is the least basic among the presented trihalides, as the nitrogen is "hogged" by fluorine, preventing it from sharing its electrons effectively.
- Fluorine, with the highest electronegativity, pulls the shared electrons closer to itself compared to other halogens.
- As a result, in \( \mathrm{NF}_{3} \), the nitrogen's lone pair is less available for donation, thereby decreasing its basicity compared to other trihalides.
- Chlorine, bromine, and iodine follow fluorine in decreasing order of electronegativity, affecting the availability of nitrogen's lone pair accordingly.
This is why \( \mathrm{NF}_{3} \) is the least basic among the presented trihalides, as the nitrogen is "hogged" by fluorine, preventing it from sharing its electrons effectively.
Lone Pair Donation
Lone pair donation is a fundamental concept for understanding the basicity of compounds. When we talk about lone pair donation in nitrogen trihalides, it involves the nitrogen atom sharing its pair of non-bonded electrons with another atom or ion.
The ability of the nitrogen in these trihalides, like \( \mathrm{NF}_{3} \), to donate its lone pair is heavily influenced by the electronegativity of the halogens it is bonded to.
In compounds where nitrogen is bonded to more electronegative atoms like fluorine, the lone pair is less readily available for donation. This reduced availability of the lone pair makes \( \mathrm{NF}_{3} \) the least basic option. Conversely, in \( \mathrm{NI}_{3} \), where iodine is less electronegative, nitrogen's lone pair is more freely available, enhancing its basicity compared to \( \mathrm{NF}_{3} \).
The ability of the nitrogen in these trihalides, like \( \mathrm{NF}_{3} \), to donate its lone pair is heavily influenced by the electronegativity of the halogens it is bonded to.
In compounds where nitrogen is bonded to more electronegative atoms like fluorine, the lone pair is less readily available for donation. This reduced availability of the lone pair makes \( \mathrm{NF}_{3} \) the least basic option. Conversely, in \( \mathrm{NI}_{3} \), where iodine is less electronegative, nitrogen's lone pair is more freely available, enhancing its basicity compared to \( \mathrm{NF}_{3} \).
Other exercises in this chapter
Problem 23
Lithium is the strongest reducing agent among alkali metals due to which of the following factors? (a) hydration energy (b) electron affinity (c) lattice energy
View solution Problem 24
The alkali metal that reacts with nitrogen directly to form nitride is (a) \(\mathrm{K}\) (b) \(\mathrm{Na}\) (c) Li (d) \(\mathrm{Rb}\)
View solution Problem 26
Which one of the following arrangement is in the order of increasing density? (a) \(\mathrm{Na}
View solution Problem 27
The decreasing order of solubility of alkaline earth metal hydroxides \(\left[\mathrm{M}(\mathrm{OH})_{2}\right]\) is given as (a) \(\mathrm{Ba}>\mathrm{Sr}>\ma
View solution