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The molecular nitrogen is not reactive due to (a) small inter-nuclear distance of \(\mathrm{N}\) atom (b) triple bond in \(\mathrm{N}\) atom (c) double bond in \(\mathrm{N}\) atom (d) high electronegativity of \(\mathrm{N}\)

Short Answer

Expert verified
The triple bond in N_2 (option b) makes it unreactive.

Step by step solution

01

Understand the Nature of Nitrogen

Molecular nitrogen ( N_2) is a diatomic molecule composed of two nitrogen atoms. It is known for its stability and inertness.
02

Analyze Bond Characteristics

Nitrogen atoms in N_2 are connected by a triple bond. The triple bond consists of one sigma bond and two pi bonds, making it one of the strongest types of covalent bonds. This strong bond provides significant stability to the molecule.
03

Effect of High Electronegativity

While nitrogen does have high electronegativity, it is not the primary reason for its lack of reactivity. Electronegativity tends to impact polarity and hydrogen bonding, more so than the stability due to covalent bonds in N_2.
04

Consider Inter-Nuclear Distance

Small inter-nuclear distances can contribute to bond strength, but the key factor here is the type and number of bonds (not just the distance) that contribute to the stability of the molecule.
05

Assess Bond Type Importance

The correct factor by which N_2 is particularly unreactive is due to the presence of the triple bond. This bond is much stronger than a single or double bond, requiring significant energy to break.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triple Bond
The molecular nitrogen ( N_2) molecule is held together by one of the strongest covalent linkages known as a triple bond. This bond formation involves three shared pairs of electrons between the two nitrogen atoms. The triple bond is composed of one sigma bond and two pi bonds.
  • **Sigma Bond**: The sigma bond is the first of the bonds to form and involves end-to-end overlap of orbitals.
  • **Pi Bonds**: The two pi bonds provide additional strength and are formed above and below the sigma bond when orbitals overlap side-to-side.
This powerful connection results in an exceptionally stable molecule that is not easily broken apart, rendering nitrogen largely inert under most conditions. It is this stability that is primarily responsible for nitrogen’s lack of reactivity.
Electronegativity
Electronegativity is a measure of how strongly an atom attracts electrons in a bond. Nitrogen is known for its high electronegativity value. However, in the context of molecular nitrogen (N_2), this property is not the main reason for the molecule's reactivity.
  • In a covalent bond such as in N_2, two identical atoms (both nitrogen) have equal electronegativity.
  • This means they share electrons equally, maintaining a nonpolar bond and contributing to the molecule's overall stability.
Overall, while electronegativity influences factors like bond polarity and the potential for hydrogen bonding in other compounds, it has minimal impact on the bond strength within the N_2 molecule.
Inter-Nuclear Distance
The inter-nuclear distance refers to the physical space between the nuclei of two bonded atoms. In N_2, the small inter-nuclear distance is a reflection of the strong triple bond that holds the nitrogen atoms together.
  • A smaller inter-nuclear distance generally means stronger bonds due to increased overlap between atomic orbitals.
  • In N_2, the triple bond results in a very short inter-nuclear distance, further strengthening the molecule.
While this short distance contributes to the stability of the N_2 molecule, the nature and number of the bonds (especially the triple bond) are more significant in determining why N_2 is not reactive.
Covalent Bond Stability
Covalent bond stability is crucial in understanding the lack of reactivity in molecular nitrogen. A covalent bond involves the sharing of electrons between atoms, and its strength depends on the number of shared electron pairs.
  • In N_2, the triple bond involves three electron pairs being shared, forming a very stable molecular structure.
  • The more electron pairs that are shared, the stronger the bond becomes, preventing the molecule from breaking apart easily.
Thus, the stability of the covalent bond in nitrogen, due to the presence of a triple bond, ensures that the molecule remains unreactive under standard conditions. This bond strength is a key reason why breaking apart N_2 to allow reactions to occur requires significant energy.

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Most popular questions from this chapter

Consider the following statements: I. Rate of transfer of \(\mathrm{D}^{+}\)from \(\mathrm{D}_{2} \mathrm{O}\) is slower than that of \(\mathrm{H}^{+}\)from \(\mathrm{H}_{2} \mathrm{O} .\) II. \(\mathrm{K}_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}\)is nearly similar to that of \(\mathrm{K}_{\mathrm{a}}\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}^{+} \mathrm{H}_{3}\) III. \(_{1} \mathrm{H}^{3}\) is a radioactive isotope. Here, correct statements are: (a) I, II, III (b) II, III (c) I, II (d) I, III

An orange solid (A) on heating gives a green residue (B), a colourless gas (C) and water vapours. The dry gas (C) on passing over heated magnesium gave a white solid (D). (D) on reaction with water gives a gas (E) which gives dense white fumes with HCl. Here, (E) will be (a) \(\mathrm{N}_{2}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{N}_{2} \mathrm{O}\)

The industrial preparation of nitric acid by Ostwald's process involves (a) hydrolysis of \(\mathrm{NH}_{3}\) (b) reduction of \(\mathrm{NH}_{3}\) (c) hydrogenation of \(\mathrm{NH}_{3}\) (d) oxidation of \(\mathrm{NH}_{3}\)

Ammonia can be dried by (a) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{P}_{4} \mathrm{O}_{10}\) (c) \(\mathrm{CaO}\) (d) anhydrous \(\mathrm{CaCl}_{2}\)

Which of the following reactions shows the correct sequence of the Ostwald process in the manufacture of nitric acid? (a) \(4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \stackrel{750^{\circ} \mathrm{C}-900^{\circ} \mathrm{C}, \text { catalyst }}{\longrightarrow} 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{NO} \stackrel{\text { heat } \mathrm{O}_{2}}{\longrightarrow} \mathrm{NO}_{2} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\mathrm{O}} \mathrm{HNO}_{3}\) (b) \(\mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2} \stackrel{\mathrm{O}_{2}}{\longrightarrow} \mathrm{SO}_{3} \frac{3}{+\mathrm{HNO}_{3}}\) \(\longrightarrow \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NaHSO}_{4}\) (d) both (a) and (b)

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