MCQ
The compound having both $sp$ and $s{p^2}$ hybridised carbon atom is
- APropene
- BPropyne
- ✓Propadiene
- DNone of these
✓
Answer
Correct option: C.
Propadiene
c
(c) $\mathop {\mathop {C{H_3}}\limits_{s{p^2}} = \mathop {{C_{}}}\limits_{sp} = \mathop {C{H_2}}\limits_{s{p^2}} }\limits_{{\rm{Propadiene}}} $
(c) $\mathop {\mathop {C{H_3}}\limits_{s{p^2}} = \mathop {{C_{}}}\limits_{sp} = \mathop {C{H_2}}\limits_{s{p^2}} }\limits_{{\rm{Propadiene}}} $
Propadiene has both $sp$ and $s{p^2}$-hybridized carbon atoms.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Ratio of radii of K, L, M shells of hydrogen atom is:
Choose the correct option for the following molecule in view of chemical bonding
How many isomers of ${C_5}{H_{11}}OH$ will be primary alcohols
→Sequence of acidic character is
An element $M$ has an atomic mass $19$ and atomic number $9$. Its ion is represented by
→Which one of the following is present in the chain structure of silicates ?
In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present ?
Which of the following facts given is not correct ?
$(I)$ Bond length order : $H^-_2 = H^+_2 > H_2$
→$(I)$ Bond length order : $H^-_2 = H^+_2 > H_2$
$(II)\, O^+_2 ,NO,N^-_2$ have same bond order of $2 \frac{1}{2}$
$(III)$ Bond order can assume any value including zero upto four
$(IV)\, NO^-_3$ and $BO^-_3$ have same bond order for $X - O$ bond (where $X$ is central atom)
The decreasing order of stability of following anions is
$2CuI\, \to Cu + Cu{I_2}$, the reaction is
→