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What is the order of basic nature of hydrides of group \(\mathrm{V}\) elements? (a) \(\mathrm{AsH}_{3}>\mathrm{SbH}_{3}>\mathrm{PH}_{3}>\mathrm{NH}_{3}\) (b) \(\mathrm{NH}_{3}>\mathrm{SbH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}\) (c) \(\mathrm{NH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3}\) (d) \(\mathrm{PH}_{3}>\mathrm{NH}_{3}>\mathrm{SbH}_{3}>\mathrm{AsH}_{3}\)

Short Answer

Expert verified
The correct order is (c)  ext{NH}_{3}> ext{PH}_{3}> ext{AsH}_{3}> ext{SbH}_{3}.

Step by step solution

01

Understand Basic Nature

The basic nature of a molecule is determined by its ability to donate a lone pair of electrons. In hydrides of Group V elements, this ability is influenced by the electronegativity of the central atom.
02

Analyze Electronegativity Trend

In Group V, the electronegativity decreases down the group: nitrogen is more electronegative than phosphorus, arsenic, and antimony. Therefore, the ability to donate lone pairs decreases as you move down the group.
03

Relate Electronegativity to Basicity

The higher the electronegativity, the stronger the ability to hold onto electrons, making the compound a better base. Thus, the order of basic nature should follow the order of electronegativity.
04

Determine Order of Basicity

Based on the electronegativity trend, the most basic hydride will be ammonia ( ext{NH}_{3}), followed by phosphine ( ext{PH}_{3}), arsine ( ext{AsH}_{3}), and stibine ( ext{SbH}_{3}).
05

Check Against Given Options

Compare the calculated order ( ext{NH}_{3} > ext{PH}_{3} > ext{AsH}_{3} > ext{SbH}_{3}) with the options provided. Option (c) matches this order.

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Key Concepts

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

Group V Elements
Group V elements in the periodic table include nitrogen, phosphorus, arsenic, and antimony. These elements are part of the p-block and belong to the group known as pnictogens. A key characteristic of these elements is that they have five electrons in their outermost electron shell. This makes them quite unique because they need three more electrons to reach a full octet, which is a stable electronic configuration.

Because they are part of the same group, these elements share some chemical properties. For example, all of them can form hydrides, which are compounds with hydrogen. However, as you move down the group, these elements show variations in their properties due to the increasing atomic size and decreasing electronegativity. Understanding these variations is essential for predicting the behavior of their hydrides.
Electronegativity Trend
Electronegativity is the ability of an atom to attract electron pairs towards itself. In Group V, the electronegativity decreases as you move down the group from nitrogen to antimony. This happens because the atomic size increases, resulting in a weaker pull on electrons.

To visualize this trend:
  • Nitrogen ( ext{N}) is the most electronegative.
  • Followed by phosphorus ( ext{P}).
  • Then arsenic ( ext{As}).
  • And finally, antimony ( ext{Sb}), which is the least electronegative.
This trend impacts the basicity of hydrides formed by these elements. A more electronegative element tends to be a better base because it can hold electron pairs more tightly, making nitrogen's hydride ( ext{NH}_{3}) the most basic in the group.
Lone Pair Donation
Lone pair donation is a concept in chemistry where an atom donates its unshared electron pairs to form bonds. Hydrides of Group V elements, like ammonia (\(\text{NH}_{3}\)), have lone pairs that they can donate, influencing their basic nature.

In hydrides:
  • Ammonia has a readily available lone pair and, due to its high electronegativity, it can easily donate this pair, making it a strong base.
  • Phosphine ( ext{PH}_{3}) has a lone pair, but it's less readily donated compared to ammonia because of lower electronegativity.
  • Arsine ( ext{AsH}_{3}) and stibine ( ext{SbH}_{3}) have even less tendency to donate their lone pairs because of the further decrease in electronegativity.
Hence, the order of basic nature of these hydrides follows their ability to donate lone pairs, influenced substantially by electronegativity.
Amphoteric Behavior
Amphoteric behavior refers to the ability of a substance to act as both an acid and a base. In the context of hydrides of Group V elements, understanding amphoteric behavior is important, even though these hydrides are primarily basic.

Most Group V hydrides act as bases, like ammonia, which is well-known for its basicity. However, compounds like metals of arsenic and antimony can exhibit amphoteric behavior when combined with certain elements, although this is more common in oxides and hydroxides.

This dual behavior arises because amphoteric substances can react both with acids, donating electrons, and with bases, accepting electrons. While Group V hydrides themselves are not traditionally amphoteric, their related compounds can exhibit this interesting behavior, emphasizing the nuanced chemistry of these elements.

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Most popular questions from this chapter

Both \(\mathrm{NF}_{3}\) and \(\mathrm{NCl}_{3}\) are covalent but they differ in the extent of hydrolysis because (a) Cl can expand its octet by using d-orbitals (b) electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{Cl}\) (c) dipole moment of \(\mathrm{NF}_{3}\) is greater than that of \(\mathrm{NCl}_{3}\) (d) \(\mathrm{NF}_{3}\) is more stable than \(\mathrm{NCl}_{3}\) and hydrolysis product of \(\mathrm{NF}_{3}\), HFO does not exist.

Maximum number of covalent bonds formed by \(\mathrm{N}\) and \(\mathrm{P}\) respectively are (a) 3,6 (b) 4,6 (c) 3,5 (d) 4,5

Which of the following represents the correct order of increasing first ionization enthalpy for \(\mathrm{Ca}, \mathrm{Ba}, \mathrm{S}, \mathrm{Se}\) and Ar? (a) \(\mathrm{Ba}<\mathrm{Ca}<\mathrm{Se}<\mathrm{S}<\mathrm{Ar}\) (b) \(\mathrm{Ca}<\mathrm{Ba}<\mathrm{S}<\mathrm{Sr}<\mathrm{Ar}\) (c) \(\mathrm{Ca}<\mathrm{S}<\mathrm{Ba}<\mathrm{Se}<\mathrm{Ar}\) (d) \(\mathrm{S}<\mathrm{Se}<\mathrm{Ca}<\mathrm{Ba}<\mathrm{Ar}\)

The oxidizing power of \(\mathrm{HXO}_{4}\) type compounds of halogens is in the order (a) \(\mathrm{HIO}_{4}>\mathrm{HBrO}_{4}>\mathrm{HClO}_{4}\) (b) \(\mathrm{HIO}_{4}>\mathrm{HClO}_{4}>\mathrm{HBrO}_{4}\) (c) \(\mathrm{HBrO}_{4}>\mathrm{HIO}_{4}>\mathrm{HClO}_{4}\) (d) \(\mathrm{HClO}_{4}>\mathrm{HBrO}_{4}>\mathrm{HIO}_{4}\)

White phosphorus on reaction with \(\mathrm{NaOH}\) gives \(\mathrm{PH}_{3}\) as one of the products. This is a (a) Dimerization reaction (b) Disproportional reaction (c) Condensation reaction (d) Precipitation reaction

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