Answer
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Hint :In order to this question, to choose the incorrect option, we will check each given option with respect to their given hints, and then compare whether the given statement is right or wrong with respect to their hints.
Complete Step By Step Answer:
In option(A)- $P{F_3} > PC{l_3}$ (Basic character)
The basic character is inversely proportional to covalent character. $C{l^ - }$ is greater in size than ${F^ - }$ . So \[\;PC{l_3}\] is more covalent than $P{F_3}$ .
Hence, $P{F_3}$ is more basic, which is correct.
In option(B)- $N{(CH)_3} > N{(Si{H_3})_3}$ (Basic character)
$N{(Si{H_3})_3}$ contains a lone pair of $N$ delocalised into the empty d-orbital of silicon via backbonding. Thus the backbonding makes $N{(Si{H_3})_3}$ less basic than $N{(CH)_3}$ , which is correct.
In option(C)- $HCC{l_3} > HC{F_3}$ (Acidic character)
$HCC{l_3}$ has effective backbonding and hence the negative charge partially gets established by back donation to the vacant 3d-orbital of $C{l^ - }$ .
Thus, $HCC{l_3}$ is stronger acid than $HC{F_3}$ , which is correct.
In option(D)- $Si{H_3} - OH < C{H_3} - OH$ (Acidic character)
Acid strength of an acid also depends upon stability of its conjugate base. So, silikon $Si{H_3}OH$ is more acidic than methanol ( $C{H_3}OH$ ) because conjugate base silianol stabilised by dispersal of negative charge.
Hence, the incorrect option is (D).
Note :
Some acids show acidic character as they dissociate in the aqueous solution which results in the production of hydrogen ions (acids like \[HCl,{\text{ }}HN{O_3}\] ). Compounds similar to glucose or alcohol also do contain hydrogen elements but they do not show signs of acidic nature.
Complete Step By Step Answer:
In option(A)- $P{F_3} > PC{l_3}$ (Basic character)
The basic character is inversely proportional to covalent character. $C{l^ - }$ is greater in size than ${F^ - }$ . So \[\;PC{l_3}\] is more covalent than $P{F_3}$ .
Hence, $P{F_3}$ is more basic, which is correct.
In option(B)- $N{(CH)_3} > N{(Si{H_3})_3}$ (Basic character)
$N{(Si{H_3})_3}$ contains a lone pair of $N$ delocalised into the empty d-orbital of silicon via backbonding. Thus the backbonding makes $N{(Si{H_3})_3}$ less basic than $N{(CH)_3}$ , which is correct.
In option(C)- $HCC{l_3} > HC{F_3}$ (Acidic character)
$HCC{l_3}$ has effective backbonding and hence the negative charge partially gets established by back donation to the vacant 3d-orbital of $C{l^ - }$ .
Thus, $HCC{l_3}$ is stronger acid than $HC{F_3}$ , which is correct.
In option(D)- $Si{H_3} - OH < C{H_3} - OH$ (Acidic character)
Acid strength of an acid also depends upon stability of its conjugate base. So, silikon $Si{H_3}OH$ is more acidic than methanol ( $C{H_3}OH$ ) because conjugate base silianol stabilised by dispersal of negative charge.
Hence, the incorrect option is (D).
Note :
Some acids show acidic character as they dissociate in the aqueous solution which results in the production of hydrogen ions (acids like \[HCl,{\text{ }}HN{O_3}\] ). Compounds similar to glucose or alcohol also do contain hydrogen elements but they do not show signs of acidic nature.
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