Number of hybrid orbitals of C atoms in C(CN) | vedclass

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Question

(C) In $C(CN)_4$,the central carbon atom is $sp^3$ hybridized,meaning it has four $sp^3$ hybrid orbitals. Each $sp^3$ orbital has $25\%$ $s$ character

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$0$

Vedclass · Answer

(C) In $C(CN)_4$,the central carbon atom is $sp^3$ hybridized,meaning it has four $sp^3$ hybrid orbitals. Each $sp^3$ orbital has $25\%$ $s$ character and $75\%$ $P$ character.Each of the four cyano group carbon atoms is $sp$ hybridized,meaning each has two $sp$ hybrid orbitals. Each $sp$ orbital has $50\%$ $s$ character and $50\%$ $P$ character.$33\%$ $P$ character corresponds to $sp^2$ hybridization (where $P$ character is $2/3 \approx 66.6\%$ is incorrect,actually $sp^2$ has $66.6\%$ $P$ character,$sp^3$ has $75\%$ $P$ character,and $sp$ has $50\%$ $P$ character).Since there are no $sp^2$ hybridized carbon atoms in $C(CN)_4$,the number of hybrid orbitals with $33\%$ $P$ character is $0$.

Column-$I$	Column-$II$
$A$. Boron in $[B(OH)_4]^-$	$1$. $sp^2$
$B$. Aluminum in $[Al(H_2O)_6]^{3+}$	$2$. $sp^3$
$C$. Boron in $B_2H_6$	$3$. $sp^3d^2$
$D$. Carbon in Buckminsterfullerene
$E$. $Si$ in $SiO_4^{4-}$
$F$. $Ge$ in $[GeCl_6]^{2-}$

Number of hybrid orbitals of $C$ atoms in $C(CN)_4$ which have $33\%$ $P$ character :-

Similar Questions

Which species do not have $sp^3$ hybridization?

Assertion : Among the carbon allotropes,diamond is an insulator,whereas,graphite is a good conductor of electricity.
Reason : Hybridization of carbon in diamond and graphite are $sp^3$ and $sp^2$,respectively.

Choose the correct code of characteristics for the given order of hybrid orbitals of the same atom,$sp < sp^2 < sp^3$:
$(i)$ Electronegativity
$(ii)$ Bond angle between same hybrid orbitals
$(iii)$ Size
$(iv)$ Energy level

$sp^3$ hybridisation is found in

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