The increasing order of the first ionization enthalpies enthalpies of the elements B, P, S and F (lowest first) is:

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$F < S < P < B$

$B < P < S < F$

$P < S < B < F$

$B < S < P < F$

answer is D.

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Detailed Solution

The increasing order of the first ionization enthalpies enthalpies of the elements B, P, S and F (lowest first) is $B < S < P < F$ .

In general, the ionization enthalpy rises with increasing atomic number as we move from left to right in a period. As we move down a group, the ionization enthalpy decreases.

Because P $(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3})$ possesses a stable half-filled electronic state, it has a larger ionization enthalpy than S. Ionization energies are increased in sequence.