The correct answer is option 4. \(SbH_3\).
The reducing ability of a compound is often related to its ability to donate electrons. In the given compounds \(NH_3\), \(PH_3\), \(AsH_3\), and \(SbH_3\), the central atoms are nitrogen (N), phosphorus (P), arsenic (As), and antimony (Sb), respectively.
The bond dissociation enthalpy (\(\Delta_{diss}(E-H)\)) is a measure of the energy required to break the E-H bond. A lower bond dissociation enthalpy implies a weaker bond and a greater tendency to release electrons, making the corresponding compound a stronger reducing agent.
Let's analyze the given data:
- \(NH_3\) (\(\Delta_{diss}(N-H) = 389 \, kJ/mol\))
- \(PH_3\) (\(\Delta_{diss}(P-H) = 322 \, kJ/mol\))
- \(AsH_3\) (\(\Delta_{diss}(As-H) = 297 \, kJ/mol\))
- \(SbH_3\) (\(\Delta_{diss}(Sb-H) = 255 \, kJ/mol\))
As we move down the group from nitrogen (N) to antimony (Sb), the size of the atom increases. This increase in size leads to weaker E-H bonds because the outermost electrons are farther from the nucleus and experience weaker attractive forces.
Therefore, \(SbH_3\) has the lowest bond dissociation enthalpy (255 kJ/mol), indicating a weaker Sb-H bond. Consequently, \(SbH_3\) is the strongest reducing agent among the given compounds.
In summary, the trend in reducing ability is as follows: \(NH_3\) < \(PH_3\) < \(AsH_3\) < \(SbH_3\), and the correct answer is \(SbH_3\) (option 4). |
