
Which hydride among the following strong reducing agents?
(A) $As{{H}_{3}}$
(B) $Bi{{H}_{3}}$
(C) $P{{H}_{3}}$
(D) $Sb{{H}_{3}}$
Answer
445.5k+ views
Hint: The term hydride is commonly named after binary compounds that hydrogen forms other elements of the periodic table. Except for noble gases, all the elements in the periodic table will form hydrides with H elements. In hydrides, negative charge with hydride ions is reasonable for reducing properties of hydrides. Write the answer based on this fact.
Complete step by step solution:
In the previous classes of chemistry, we have studied the basic concepts of the general part of chemistry that says about the oxidizing agents, reducing agents and other related definitions.
Let us see which hydride acts as the reducing agent.
In group 15 elements, the general formula of hydrides $M{{H}_{3}}$, where M is the elements of group-15 elements and in these hydrides negative charge ion ${{H}^{-}}$ is called hydride ion, except for ammonia.
The hydrides of group 15 elements are, $N{{H}_{3}},P{{H}_{3}},As{{H}_{3}},Sb{{H}_{3}},Bi{{H}_{3}}$
The stability of these hydrides decreases from ammonia to bismuthine, because with increasing the size then the overlapping between the orbitals of these elements and hydrogen will not be effective.
The group 15 elements of hydrides or strong reducing agents. This reducing property increases from going down to the group with increasing the size of the atom due to increases the distance between metal and hydrogen. So the strength of the M-H bond decreases.
So, hydrogen forms these hydrides easily release from $P{{H}_{3}}$ to $Bi{{H}_{3}}$, and the increasing order of reducing property as follows:
\[P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}>Bi{{H}_{3}}\]
Therefore, $Bi{{H}_{3}}$ is the strongest reducing agent among all hydrides in the group 15 elements.
So, the correct answer is B.
Note: These hydrides of these group 15 elements are basic in nature and act as Lewis base due to the availability of lone pairs of electrons at the central metal atom. The basic character will increase in this group by increasing the size of the atom. There is an abnormal behaviour of boiling point and melting points of these hydrides in this group as follows
\[P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}>Bi{{H}_{3}}\]
Complete step by step solution:
In the previous classes of chemistry, we have studied the basic concepts of the general part of chemistry that says about the oxidizing agents, reducing agents and other related definitions.
Let us see which hydride acts as the reducing agent.
In group 15 elements, the general formula of hydrides $M{{H}_{3}}$, where M is the elements of group-15 elements and in these hydrides negative charge ion ${{H}^{-}}$ is called hydride ion, except for ammonia.
The hydrides of group 15 elements are, $N{{H}_{3}},P{{H}_{3}},As{{H}_{3}},Sb{{H}_{3}},Bi{{H}_{3}}$
The stability of these hydrides decreases from ammonia to bismuthine, because with increasing the size then the overlapping between the orbitals of these elements and hydrogen will not be effective.
The group 15 elements of hydrides or strong reducing agents. This reducing property increases from going down to the group with increasing the size of the atom due to increases the distance between metal and hydrogen. So the strength of the M-H bond decreases.
So, hydrogen forms these hydrides easily release from $P{{H}_{3}}$ to $Bi{{H}_{3}}$, and the increasing order of reducing property as follows:
\[P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}>Bi{{H}_{3}}\]
Therefore, $Bi{{H}_{3}}$ is the strongest reducing agent among all hydrides in the group 15 elements.
So, the correct answer is B.
Note: These hydrides of these group 15 elements are basic in nature and act as Lewis base due to the availability of lone pairs of electrons at the central metal atom. The basic character will increase in this group by increasing the size of the atom. There is an abnormal behaviour of boiling point and melting points of these hydrides in this group as follows
\[P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}>Bi{{H}_{3}}\]
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