Analysis of Statement A: The stability of hydrides does indeed decrease as we go down the group due to increasing atomic size and decreasing bond strength. Thus, the order is NH3 > PH3 > AsH3 > SbH3 > BiH3 is correct.
Analysis of Statement B: The reducing ability of the hydrides increases down the group due to the increase in bond length and decrease in bond dissociation energy, making NH3 < PH3 < AsH3 < SbH3 < BiH3 correct.
Analysis of Statement C: NH3 does exhibit some reducing properties but is generally considered a weak reducing agent. On the other hand, BiH3 can act as a mild reducing agent. Therefore, the statement is correct.
Analysis of Statement D: Basicity generally increases down the group, so NH3 < PH3 < AsH3 < SbH3 < BiH3 is true.
Conclusion: Statements A and D are true, while statements B and C are indeed correct. Therefore, the best answer that reflects this is: B and C only
Given below are two statements.
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In the light of the above statements, choose the correct answer from the options given below:
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: