Among $\mathrm{H}_2 \mathrm{~S}, \mathrm{H}_2 \mathrm{O}, \mathrm{NF}_3, \mathrm{NH}_3$ and $\mathrm{CHCl}_3$, identify the molecule $(\mathrm{X})$ with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule $(X)$ is :

$$ \text { The final product }[\mathrm{B}] \text { is : } $$

Given below are two statements :

Statement I : $\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$ is the correct order in terms of first ionization enthalpy values.

Statement II : $\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{Po}>\mathrm{O}$ is the correct order in terms of the magnitude of electron gain enthalpy values.

In the light of the above statements, choose the correct answer from the options given below :

Consider the following reduction processes :

$$ \begin{aligned} & \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{~s}), \mathrm{E}^0=-1.66 \mathrm{~V} \\ & \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^0=+0.77 \mathrm{~V} \\ & \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^0=+1.81 \mathrm{~V} \\ & \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{~s}), \mathrm{E}^0=-0.74 \mathrm{~V} \end{aligned} $$

The tendency to act as reducing agent decreases in the order :