
The correct increasing order of trans effect of following species is:
A. $C{N^ - } > B{r^ - } > {C_6}H_5^ - > N{H_3}$
B. $B{r^ - } > C{N^ - } > N{H_3} > {C_6}H_5^ - $
C. $N{H_3} > B{r^ - } > {C_6}H_5^ - > C{N^ - }$
D. $C{N^ - } > {C_6}H_5^ - > B{r^ - } > N{H_3}$
Answer
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Hint: We need to remember that in simple words trans effect is the position that an incoming ligand will attach to a square planar complex or it is the effect of ligand towards the substitution on another ligand which is trans to it. It was first explained in $1926$ by Chernyaev in platinum complexes. It is a process of labialization which means formation of more reactive ligands. It is important to remember that incoming ligands will add trans to the ligand far to the left in trans series.
Complete step by step answer:
Complete step by step answer:
We must have to remember that the Trans effect can be a nucleophilic substitution where two identical ligands undergo replacement and forms two different products and it depends on the side where the ligand left and it forms two products that is trans and cis.
We will use the following sequence which shows the intensity of trans effect ( i.e. the rate of substitution of trans ligand).
$\begin{array}{*{20}{c}}
{{F^ - },{H_2}O,OH < N{H_3} < py < C{l^ - } < B{r^ - } < {I^ - },SC{N^ - },N{O_2}^ - ,SC{{\left( {N{H_2}} \right)}_2},Ph} { < SO_3^{2 - } < P{R_3},As{R_3},S{R_2},CH_3^ - < NO,CO,C{N^ - },{C_2}{H_4}}
\end{array}$
In the given question, it is $C{N^ - }$ which is lying towards the end of series and shows higher trans effect because it has vacant π orbital and can easily form metal ligand π bond. So we can say that ligands which are able to form this type of bond will show more trans effect. Now we have two options A and D with $C{N^ - }$ . Then look at the series we will find ${C_6}H_5^ - $ , then $B{r^ - }$ and lastly $N{H_3}$
So, the correct answer is Option D.
Note: As we know that some authors use the term trans effect as structural trans effect or thermodynamic trans effect. Structural trans effect can be found experimentally by X-ray crystallography technique. It should be noted that trans effect is applicable to square planar complexes but cis effect is found in octahedral complexes. Although chances of cis effect are very less as compared to trans effect. In biology ligands.
Complete step by step answer:
Complete step by step answer:
We must have to remember that the Trans effect can be a nucleophilic substitution where two identical ligands undergo replacement and forms two different products and it depends on the side where the ligand left and it forms two products that is trans and cis.
We will use the following sequence which shows the intensity of trans effect ( i.e. the rate of substitution of trans ligand).
$\begin{array}{*{20}{c}}
{{F^ - },{H_2}O,OH < N{H_3} < py < C{l^ - } < B{r^ - } < {I^ - },SC{N^ - },N{O_2}^ - ,SC{{\left( {N{H_2}} \right)}_2},Ph} { < SO_3^{2 - } < P{R_3},As{R_3},S{R_2},CH_3^ - < NO,CO,C{N^ - },{C_2}{H_4}}
\end{array}$
In the given question, it is $C{N^ - }$ which is lying towards the end of series and shows higher trans effect because it has vacant π orbital and can easily form metal ligand π bond. So we can say that ligands which are able to form this type of bond will show more trans effect. Now we have two options A and D with $C{N^ - }$ . Then look at the series we will find ${C_6}H_5^ - $ , then $B{r^ - }$ and lastly $N{H_3}$
So, the correct answer is Option D.
Note: As we know that some authors use the term trans effect as structural trans effect or thermodynamic trans effect. Structural trans effect can be found experimentally by X-ray crystallography technique. It should be noted that trans effect is applicable to square planar complexes but cis effect is found in octahedral complexes. Although chances of cis effect are very less as compared to trans effect. In biology ligands.
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