In this example problem, we look at the hybridization in a larger molecule. Our problem reads, "First, determine the hybridization for the red atoms in arginine. Second, determine the bonding scheme for the bonds in green. Third, determine the total number of sigma and pi bonds." To determine the hybridization of our atoms in red, we count the number of electron groups on the atom in question. For the nitrogen on the left, we have four electron groups, so we need four hybrid orbitals. The hybridization will be sp3. When counting electron groups, we count both bonds and lone pairs. For the nitrogen on the right, we have three electron groups, so we need sp2 hybridization. When counting bonds, a double bond counts as one electron group. For the carbon on the left, we have four electron groups, so we need sp3 hybridization. For the carbon on the right, we have three electron groups, so we need sp2 hybridization. And for the oxygen atom, we have four electron groups, so we need sp3 hybridization. Now we can determine the bonding schemes for the bonds in green. For the single bond on the left, we have a sigma bond created using one sp3 hybridized orbital from the carbon on the left and one sp2 hybridized orbital from the carbon on the right. For the double bond, we have one sigma component created using the sp2 hybridized orbital from carbon and an unhybridized p orbital on oxygen. We also have one pi component created using the unhybridized p orbital on carbon and another unhybridized p orbital on oxygen. For this single bond on the right, we have a sigma bond created using oxygen's sp3 hybridized orbital and the unhybridized 1s orbital of hydrogen. For part c, we can determine the number of sigma bonds, knowing that every bond, whether it's a single, double, or triple bond, has one sigma component. We have a total of 25 bonds, so we have 25 sigma bonds. Each double bond has one sigma component and one pi component, and each triple bond has one sigma component and two pi components. We have a total of two double bonds in the molecule, so we have two pi bonds.