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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

$0$

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 $0$.

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

Similar Questions

Which one of the following is a correct set?

$KI + MnO_2 + H_2SO_4 \to$ Vapours $(B) +$ other products
Vapours $(B) + K_2S_2O_8 \to (C) + (D)$
$(D) \xrightarrow{(B)}$ Brown $(E)$
Hybridisation of central atom in $(E)$ is

Non-axial $d-$ orbital is used in hybridisation of

$C-H$ bond length is greatest in

Which of the following is a linear molecule?

Vedclass Test Series

Exam Paper Generator

Online Exam Module