Question

# Consider the elements $$N, P, O, S$$ and arrange them in order of(a) increasing first ionization enthalpy,(b) increasing negative electron gain enthalpy,(c) increasing non-metallic character.

Solution

## Arranging all the elements into different groups and periods in order of their increasing atomic numbers, we have,(a) Since $$\Delta_{i}H_1$$ decreases down a group, therefore, $$\Delta_{i}H_1$$ of $$N$$ and $$O$$ are higher than those of $$P$$ and $$S$$. Further since $$N$$ has more stable exactly half-filled electronic configuration in the 2p-subshell, therefore, it is more difficult to remove out an electron from $$N$$ than from $$O$$ even though $$O$$ has higher nuclear charge. Similarly, $$P$$ has exactly half-filled electronic configuration in the 3p-subshell. Therefore, $$\Delta_{i}H_1$$ of $$P$$ is higher than that of $$S$$. Thus, the overall increasing order of first ionization enthalpy of these elements follows the order: $$S < P < 0 < N$$.(b) Since adding an electron to smaller size 2p-orbital causes greater repulsion than adding an electron to larger 3p-orbital, therefore, electron gain enthalpies of $$P$$ and $$S$$ are more negative than those of $$N$$ and $$O$$ respectively. Further since $$S$$ has higher nuclear charge but $$P$$ has more stable exactly half-filled electronic configuration in the 3p-subshell, therefore, it is easier to add an electron to $$S$$ than to $$P$$. In other words, electron gain enthalpy of $$S$$ is more negative than that of $$P$$. Similarly, $$N$$ has exactly half-filled electronic configuration in the 2p-subshell but $$O$$ has higher nuclear charge.But the addition of an electron to $$N$$ causes repulsions to such an extent that electron gain enthalpy of $$N$$ is actually positive while that of $$O$$ as expected in negative. Combining all the above results, the increasing order of negative electron gain enthalpy of these elements follows the order: $$N < P < 0 < S$$.(c) Since non-metallic character decreases down a group but increases along a period, therefore, $$O$$ is the most non-metallic element while $$P$$ is the least non-metallic element. The actual order of increasing non-metallic character is: $$P < S < N < O$$.Chemistry

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