To determine which molecule represents the order of hybridization $sp^2, sp^2, sp, sp$ from left to right, we need to analyze the hybridization of each atom in the given options.$\\$Hybridization is determined by the number of sigma bonds and lone pairs around an atom:$\\$• $sp^3$ hybridization occurs with 4 sigma bonds/lone pairs.$\\$• $sp^2$ hybridization occurs with 3 sigma bonds/lone pairs.$\\$• $sp$ hybridization occurs with 2 sigma bonds/lone pairs.$\\$Let's analyze each option:$\\$Option 1: $$CH$_2=$CH$-$CH$=$CH$_2 \\$• The first and last carbon atoms are $sp^2$ hybridized (each forms 3 sigma bonds: 2 with H and 1 with C).$\\$• The middle two carbon atoms are also $sp^2$ hybridized (each forms 3 sigma bonds: 1 with C and 2 with other C atoms).$\\$This option does not match the required hybridization order.$\\$Option 2: $$CH$_2=$CH$-$C$\equiv$CH$\\$• The first carbon is $sp^2$ hybridized (3 sigma bonds: 2 with H and 1 with C).$\\$• The second carbon is $sp^2$ hybridized (3 sigma bonds: 1 with C and 2 with other C atoms).$\\$• The third carbon is $sp$ hybridized (2 sigma bonds: 1 with C and 1 triple bond with C).$\\$• The fourth carbon is $sp$ hybridized (2 sigma bonds: 1 triple bond with C and 1 with H).$\\$This option matches the required hybridization order: $sp^2, sp^2, sp, sp. \\$Option 3: $$HC$\equiv$C$-$C$\equiv$CH$\\$• All carbon atoms are $sp$ hybridized (each forms 2 sigma bonds: 1 triple bond with C and 1 with H or C).$\\$This option does not match the required hybridization order.$\\$Option 4: $$CH$_3-$CH$=$CH$-$CH$_3 \\$• The first and last carbon atoms are $sp^3$ hybridized (each forms 4 sigma bonds: 3 with H and 1 with C).$\\$• The middle two carbon atoms are $sp^2$ hybridized (each forms 3 sigma bonds: 1 with C and 2 with other C atoms).$\\$This option does not match the required hybridization order.$\\$Therefore, the correct option is Option 2: $$CH$_2=$CH$-$C$\equiv$CH$.$