The ratio of the number of $s p, s p^{2},$ and $s p^{3}$ orbitals in the compound is:
$C H_{3} - C H = C = C H - C \equiv C - C H_{3}$

1 : 1 : 1

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2 : 2 : 1

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3 : 2 : 1

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3 : 3 : 4

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Solution

The correct option is D $3 : 3 : 4$
In the given compound,

There is 2 $s p^{3}$ , 2 $s p^{2}$ and 3 $s p$ hybridized carbon atom.

Hence,number of $s p^{3}$ hybrid orbitals $= 4 + 4$ ,
number of $s p^{2}$ hybrid orbitals $= 3 + 3$ ,
numberof $s p$ hybrid orbitals $= 2 + 2 + 2$ .

Hence ratio of $s p : s p^{2} : s p^{3}$ $= 6 : 6 : 8 = 3 : 3 : 4$

Hence, the option (D) is correct.

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The ratio of the number of sp,sp2, and sp3 orbitals in the compound is:CH3−CH=C=CH−C≡C−CH3

The correct option is D 3:3:4In the given compound,There is 2 sp3 , 2 sp2 and 3 sp hybridized carbon atom. Hence,number of sp3 hybrid orbitals =4+4,number of sp2 hybrid orbitals =3+3,numberof sp hybrid orbitals =2+2+2.Hence ratio of sp:sp2:sp3=6:6:8=3:3:4Hence, the option (D) is correct.

The ratio of the number of $s p, s p^{2},$ and $s p^{3}$ orbitals in the compound is:
$C H_{3} - C H = C = C H - C \equiv C - C H_{3}$