Which of the following order of given propert | vedclass

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(C) Let us analyze each option:$1$. $PF_3 < PCl_3$ (Basic character): $F$ is more electronegative than $Cl$,so $PF_3$ has lower electron density on $P

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$HCCl_3 > HCF_3$ (Acidic character)

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(C) Let us analyze each option:$1$. $PF_3 $2$. $N(CH_3)_3 > N(SiH_3)_3$ (Basic character): In $N(SiH_3)_3$,the lone pair on $N$ is involved in $ppi-dpi$ back bonding with $Si$,making it less available for donation. Thus,$N(CH_3)_3$ is more basic. This order is correct.$3$. $HCCl_3 > HCF_3$ (Acidic character): The acidity of haloforms depends on the stability of the conjugate base (carbanion). The $CCl_3^-$ ion is stabilized by the $-I$ effect of $Cl$ and $d$-orbital resonance,whereas $CF_3^-$ is less stable due to the lack of $d$-orbitals on $F$. Thus,$HCCl_3$ is more acidic. This order is correct.$4$. $SiH_3-OH > CH_3-OH$ (Acidic character): In $SiH_3-OH$,the $O$ lone pair participates in $ppi-dpi$ back bonding with the empty $d$-orbital of $Si$. This weakens the $O-H$ bond,making it more acidic than $CH_3-OH$. This order is correct.Wait,re-evaluating option $C$: The acidity of $HCX_3$ is determined by the stability of the carbanion $CX_3^-$. $CF_3^-$ is actually more stable than $CCl_3^-$ due to the strong $-I$ effect of $F$ atoms,despite the lack of $d$-orbital participation. Therefore,$HCF_3$ is more acidic than $HCCl_3$. Thus,the order $HCCl_3 > HCF_3$ is incorrect.

Which of the following order of given properties is incorrect?

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