What is the increasing order of the first ion | vedclass

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(D) The general trend for ionization enthalpy across a period from left to right is that it increases due to an increase in effective nuclear charge.T

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$B < S < P < F$

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(D) The general trend for ionization enthalpy across a period from left to right is that it increases due to an increase in effective nuclear charge.The elements $B, P, S, F$ belong to the second and third periods.The electronic configurations are: $B (Z=5): 1s^2 2s^2 2p^1$,$P (Z=15): [Ne] 3s^2 3p^3$,$S (Z=16): [Ne] 3s^2 3p^4$,$F (Z=9): 1s^2 2s^2 2p^5$.Phosphorus $(P)$ has a stable half-filled $p$-orbital $(3p^3)$,which makes its ionization enthalpy higher than that of Sulfur $(S)$.Comparing the values: $B$ has the lowest,followed by $S$,then $P$,and $F$ has the highest.Therefore,the increasing order is $B < S < P < F$.

$(IP)_1$	$7.1 \ eV$
$(IP)_2$	$10.3 \ eV$
$(IP)_3$	$36.2 \ eV$
$(IP)_4$	$39.1 \ eV$

What is the increasing order of the first ionization enthalpy of $B, P, S$ and $F$ (lowest first)?

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The ionization potential $(IP)$ values for an element are given below. Identify the element:
$(IP)_1$ $7.1 \ eV$
$(IP)_2$ $10.3 \ eV$
$(IP)_3$ $36.2 \ eV$
$(IP)_4$ $39.1 \ eV$

The correct order in which the first ionisation potential increases is

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