Answer
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Hint: The values of bond dissociation enthalpy have been given which means we can compare the molecules and decide which molecules will get dissociated first. The strongest reducing agent will be the one which will require less energy to give away or donate proton molecules from its structure.
Complete step by step answer:
1) First of all, we will learn the concept of bond dissociation enthalpy where it can be defined as the energy required to break the bonds present in its structure.
2) The values of bond dissociation enthalpy are given which are in decreasing order. The more value of Bond dissociation enthalpy means there will be more energy required to break the bonds present in that molecular structure.
3) Hence, the strongest structure is of $N{H_3}$ which has the highest value of bond dissociation enthalpy and the weakest structure is of $Sb{H_3}$ which has the lowest value of bond dissociation enthalpy.
4) Now the reducing agent is the chemical species that reduces other chemicals by donating the proton atom present in its structure. Hence, to be the strongest reducing agent the structure must be broken its bonds at lower energy.
5) As the compound $Sb{H_3}$ has the lowest bond dissociation enthalpy which means it needs less energy to break its bonds and after breaking of bonds it can easily donate the hydrogen atoms to other atoms.
6) Hence, we can say that the compound $Sb{H_3}$ is the strongest reducing agent
which shows the option D is the correct choice.
Note:
The reduction means the gain of electrons and the loss of electrons. But the reducing agent means which reduces other chemical species and gets oxidized itself which means a reducing agent will donate electrons and lose its electrons. This is very important to remember the difference between these two concepts.
Complete step by step answer:
1) First of all, we will learn the concept of bond dissociation enthalpy where it can be defined as the energy required to break the bonds present in its structure.
2) The values of bond dissociation enthalpy are given which are in decreasing order. The more value of Bond dissociation enthalpy means there will be more energy required to break the bonds present in that molecular structure.
3) Hence, the strongest structure is of $N{H_3}$ which has the highest value of bond dissociation enthalpy and the weakest structure is of $Sb{H_3}$ which has the lowest value of bond dissociation enthalpy.
4) Now the reducing agent is the chemical species that reduces other chemicals by donating the proton atom present in its structure. Hence, to be the strongest reducing agent the structure must be broken its bonds at lower energy.
5) As the compound $Sb{H_3}$ has the lowest bond dissociation enthalpy which means it needs less energy to break its bonds and after breaking of bonds it can easily donate the hydrogen atoms to other atoms.
6) Hence, we can say that the compound $Sb{H_3}$ is the strongest reducing agent
which shows the option D is the correct choice.
Note:
The reduction means the gain of electrons and the loss of electrons. But the reducing agent means which reduces other chemical species and gets oxidized itself which means a reducing agent will donate electrons and lose its electrons. This is very important to remember the difference between these two concepts.
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