Compare the carbon-carbon sigma bond length i | vedclass

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(C) The carbon-carbon bond length depends on the hybridization of the carbon atoms involved. In $Ethane$ $(CH_3-CH_3)$, the carbon atoms are $sp^3$ hy

Vedclass · Accepted Answer

$Ethane > Ethylene > Acetylene$

Vedclass · Answer

(C) The carbon-carbon bond length depends on the hybridization of the carbon atoms involved. In $Ethane$ $(CH_3-CH_3)$, the carbon atoms are $sp^3$ hybridized, resulting in a bond length of approximately $154 \ pm$. In $Ethylene$ $(CH_2=CH_2)$, the carbon atoms are $sp^2$ hybridized, resulting in a bond length of approximately $134 \ pm$. In $Acetylene$ $(CH \equiv CH)$, the carbon atoms are $sp$ hybridized, resulting in a bond length of approximately $120 \ pm$. As the percentage of $s$-character increases ($sp^3 = 25\%$, $sp^2 = 33.3\%$, $sp = 50\%$), the bond length decreases. Therefore, the order of bond length is $Ethane > Ethylene > Acetylene$.

$LIST$-$I$ (Hybridization)	$LIST$-$II$ (Orientation in Space)
$A$. $sp^3$	$I$. Trigonal bipyramidal
$B$. $dsp^2$	$II$. Octahedral
$C$. $sp^3 d$	$III$. Tetrahedral
$D$. $sp^3 d^2$	$IV$. Square planar

Compare the carbon-carbon $\sigma$ bond length in ethane, ethylene and acetylene.

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$(ii)$ $SF_4 + 3H_2O \to H_2SO_3 + 4HF$
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Then according to given information the incorrect statement is

Suggest reasons why the $B-F$ bond lengths in $BF_3\ (130\ pm)$ and $BF_4^-\ (143\ pm)$ differ.

Match List-$I$ with List-$II$ :

$LIST$-$I$ (Hybridization) $LIST$-$II$ (Orientation in Space)

$A$. $sp^3$ $I$. Trigonal bipyramidal

$B$. $dsp^2$ $II$. Octahedral

$C$. $sp^3 d$ $III$. Tetrahedral

$D$. $sp^3 d^2$ $IV$. Square planar

Choose the correct answer from the options given below :

$sp^3$ hybridisation is found in

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