The decreasing order of the bond moment of `E-H` bond in `NH_(3), PH_(3), AsH_(3) and SbH_(3)` is given by

Options

A. `NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3)`
B. `SbH_(3) gt AsH_(3) gt PH_(3) gt NH_(3)`
C. `NH_(3) gt AsH_(3) gt PH_(3) gt SbH_(3)`
D. `NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3)`

Answer

To determine the decreasing order of the bond moment of the E-H bond in NH₃, PH₃, AsH₃, and SbH₃, we need to analyze the polarity of the E-H bonds in each of these compounds. The bond moment is directly related to the polarity of the bond, which in turn is influenced by the difference in electronegativity between the elements involved in the bond.

### Step-by-Step Solution:

1. **Identify the Electronegativity Values**:
- For Nitrogen (N): 3.0
- For Phosphorus (P): 2.1
- For Arsenic (As): 2.0
- For Antimony (Sb): 1.9
- For Hydrogen (H): 2.1

2. **Calculate the Electronegativity Differences**:
- For NH₃ (N-H bond):
\[ \text{Difference} = |3.0 - 2.1| = 0.9 \]
- For PH₃ (P-H bond):
\[ \text{Difference} = |2.1 - 2.1| = 0.0 \]
- For AsH₃ (As-H bond):
\[ \text{Difference} = |2.0 - 2.1| = 0.1 \]
- For SbH₃ (Sb-H bond):
\[ \text{Difference} = |1.9 - 2.1| = 0.2 \]

3. **Determine the Polarity**:
- The greater the electronegativity difference, the more polar the bond will be, and thus the higher the bond moment.
- From the calculations:
- NH₃ has the highest electronegativity difference (0.9), hence the highest bond moment.
- SbH₃ has the next highest difference (0.2), followed by AsH₃ (0.1), and finally PH₃ (0.0).

4. **Rank the Compounds Based on Bond Moment**:
- NH₃ > SbH₃ > AsH₃ > PH₃

5. **Write the Final Order**:
- The decreasing order of the bond moment of E-H bonds is:
\[ \text{NH}_3 > \text{SbH}_3 > \text{AsH}_3 > \text{PH}_3 \]

### Final Answer:
The decreasing order of the bond moment of E-H bonds in NH₃, PH₃, AsH₃, and SbH₃ is:
**NH₃ > SbH₃ > AsH₃ > PH₃**