Question

Which of the following is true statement?

• A. All the carbon in $H_3C-C \equiv C-C \equiv C-CH_3$ are in $sp^3$ hybridisation
• B. In $H_3C(CN)_3$ there are six $\sigma$ bonds
• C. In diamond C is in $sp^2$ hybridisation
• D. In $C_3O_2$ all the carbons are in $sp$ hybridisation

Answer 1

Answer: (D)

In $C_3O_2$ all the carbons are in $sp$ hybridisation

Answer 2

• (A) All the carbon in $H_3C-C \equiv C-C \equiv C-CH_3$ are in $sp^3$ hybridisation

  The molecule is $CH_3-C \equiv C-C \equiv C-CH_3$.

  • The two terminal $CH_3$ carbons are $sp^3$ hybridized (bonded to 4 atoms).
  • The four carbons involved in the triple bonds ($C \equiv C$) are $sp$ hybridized (bonded to 2 atoms).

  Therefore, this statement is false.

• (B) In $H_3C(CN)_3$ there are six $\sigma$ bonds

  The molecule is $C(CH_3)(CN)_3$. The central carbon is bonded to one methyl group ($CH_3$) and three cyano groups ($CN$).

  $\sigma$ bonds:

  • 3 C-H $\sigma$ bonds in the $CH_3$ group.
  • 1 C-C $\sigma$ bond between the central carbon and the $CH_3$ carbon.
  • 3 C-C $\sigma$ bonds between the central carbon and the carbons of the three $CN$ groups.
  • 3 C-N $\sigma$ bonds within the three $CN$ groups (each $C \equiv N$ has one $\sigma$ and two $\pi$ bonds).

  Total $\sigma$ bonds = $3 + 1 + 3 + 3 = 10$.

  Therefore, this statement is false.

• (C) In diamond C is in $sp^2$ hybridisation

  In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms. This corresponds to $sp^3$ hybridization.

  Therefore, this statement is false.

• (D) In $C_3O_2$ all the carbons are in $sp$ hybridisation

  The molecule is carbon suboxide, $O=C=C=C=O$.

  Each carbon atom is involved in two double bonds and is bonded to two other atoms.

  The steric number for each carbon is 2 (2 bond pairs, 0 lone pairs), which corresponds to $sp$ hybridization.

  Therefore, this statement is true.