Match List-I with List-II. List-I (Molec | vedclass

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(B) Ethane $(CH_3-CH_3)$ contains a single bond between two carbon atoms which is one $\sigma$-bond. Thus,$A-III$.Ethene $(CH_2=CH_2)$ contains a doub

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$A-III, B-IV, C-II, D-I$

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(B) Ethane $(CH_3-CH_3)$ contains a single bond between two carbon atoms which is one $\sigma$-bond. Thus,$A-III$.Ethene $(CH_2=CH_2)$ contains a double bond between two carbon atoms which consists of one $\sigma$-bond and one $\pi$-bond. Thus,$B-IV$.Carbon molecule $(C_2)$ in its ground state,according to molecular orbital theory,has two $\pi$-bonds and no $\sigma$-bond between the two carbon atoms. Thus,$C-II$.Ethyne $(HC \equiv CH)$ contains a triple bond between two carbon atoms which consists of one $\sigma$-bond and two $\pi$-bonds. Thus,$D-I$.The correct matching is $A-III, B-IV, C-II, D-I$.

List-$I$ (Molecule)	List-$II$ (Number and types of bond/s between two carbon atoms)
$A.$ Ethane	$I.$ One $\sigma$-bond and two $\pi$-bonds
$B.$ Ethene	$II.$ Two $\pi$-bonds
$C.$ Carbon molecule,$C_2$	$III.$ One $\sigma$-bond
$D.$ Ethyne	$IV.$ One $\sigma$-bond and one $\pi$-bond

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