Question

The hybridisation state of carbon atoms in the product formed by the reaction of ethyl chloride with aqueous KOH is:
A. sp
B. \[s{{p}^{2}}\]
C. \[s{{p}^{3}}\]
D. \[s{{p}^{3}}d\]

Answer 1

Hint: As per question we need to find the hybridization of all carbon present compounds formed by the reaction of ethyl chloride with aq. KOH. If carbon bonded with other atoms (four atoms) through single bond then the hybridization is \[s{{p}^{3}}\], if it is bonded through one double bond and two single bond then the hybridization is \[s{{p}^{2}}\] (means one double bond is counted as one single bond). If carbon bonded through one triple bond and one single bond then the hybridization is sp (means triple bond is treated as one single bond).

Complete Step by Step Answer:
The ethyl chloride is defined as the chain of two carbons whose one terminal is bonded with a chloride atom. When the ethyl chloride is made react with aq. KOH then the resultant compound is ethanol (chloride as a good leaving group gets substituted by OH nucleophile). The ethanol again contain two carbon and one terminal is bonded with OH group such as
\[C{{H}_{3}}C{{H}_{2}}OH\].

Now the hybridization of both carbons depends on the bonding of carbon with other atoms. Carbon valence is four (one electron in an orbital, tree unpaired electron in p orbital) so it can form four bonds. In the given compound fist carbon is bonded with three hydrogen and one carbon so we can say it’s valence is fulfilled with four single bond (if there is one double bond then number of bonded atom will decreases to 3 and if one triple bond present so number of bonded atom will be 2). Also the second carbon is forming four bonds, two with hydrogen, one with carbon, and other with oxygen of OH group, so, we can say it is also forming four single bonds (bonded to four atoms directly).
As per hint if atoms form all single bonds as per its valence so it’s valence is \[s{{p}^{3}}\], thus, both carbon in ethanol has \[s{{p}^{3}}\] hybridization.
Hence, the correct option is C.

Additional information: The atomic number of carbon is 6 so two electrons will be filled into an orbital of the first shell such as\[1{{s}^{2}}\]. Now left four electrons filled into s (maximum 2 electron) and p (maximum 6 electron) of the second shell. Now as two electrons will get filled into an orbital of the second shell such as \[2{{s}^{2}}\]so in p orbital only two electrons will get filled into a second shell such as \[2{{p}^{2}}\]. So as per this valence of carbon should be two. But in actual, in excited state one electron from a orbital of second shell jumped to p orbital so that valence of carbon increase to four such as \[2{{s}^{1}},\text{ }2{{p}^{3}}\]. Now as per this in carbon left with one unpaired electron in an orbital and three unpaired electrons in p orbital of the second shell. Now as per question each one electron of hydrogen bonded and one electron of carbon bonded with first carbon of ethanol group, resulting in hybridization of \[s{{p}^{3}}\] because one s orbital filled and three p orbital filled second shell of carbon.

Note: It is important to note that carbon has no d orbital because of the presence of only two shells. One shell has only an orbital and the second shell has a and p orbital only so there is no d orbital so carbon cannot have \[s{{p}^{3}}d\]hybridization. Also as per \[s{{p}^{3}}d\] hybridization number of bond need to be five and this is also not possible in case of carbon so carbon cannot have \[s{{p}^{3}}d\]hybridization because of both reasons.