The reducing strength of hydrides of Group 15 elements increases as we move down the group. This trend can be explained based on the following factors: - As we move down the group, the size of the central atom increases, which weakens the E − H bond (where E is the central atom.) The bond dissociation energy decreases. This makes it easier for the hydride to donate hydrogen (as H$^-$), thereby increasing its reducing power. Among the given options:
NH$_3$ (ammonia) has the strongest N--H bond and, therefore, is the weakest reducing agent.
PH$_3$, SbH$_3$, and BiH$_3$ follow the trend of increasing reducing strength due to weaker bonds as we move down the group.
BiH$_3$ (bismuthine) has the weakest Bi--H bond, making it the strongest reducing agent among the given hydrides.
Conclusion: The strongest reducing agent is BiH$_3$.
Concentration of KCl solution (mol/L) | Conductivity at 298.15 K (S cm-1) | Molar Conductivity at 298.15 K (S cm2 mol-1) |
---|---|---|
1.000 | 0.1113 | 111.3 |
0.100 | 0.0129 | 129.0 |
0.010 | 0.00141 | 141.0 |
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: