Answer
4A1 + 1A2 + 2B1 + 2B2
Work Step by Step
Step 1: To find the symmetry species of the vibration modes of CH2Cl2 molecule, we exclude the symmetry species of the three rotation modes and the three translation modes for CH2Cl2 molecule from the symmetry species of the 15 modes (3N=15) from the displacement of all 5 atoms (N=5).
Step 2: The problem gives the symmetry species of the 15 modes from the displacement of all 5 atoms: 5A1 + 2A2 + 4B1 + 4B2.
Step 3: To get the symmetry species of the three rotation modes, we can simply look for the symmetry species Rx, Ry and Rz belong to from the right two columns in the character table for C2v point group shown by diagram 1 in the image below. According to the table, the symmetry species of the three rotation modes are 1A2 + 1B1 + 1B2.
Step 4: To get the symmetry species of the three translation modes,we can also simply look for the symmetry species x, y and z belong to from the right two columns in the same character table as shown by diagram 2. According to the table, the symmetry species of the three translation modes are 1A1 + 1B1 + 1B2.
Step 5: Finally, taking away the symmetry species of the three rotation modes we just got (1A2 + 1B1 + 1B2) and those of the three translation modes (1A1 + 1B1 + 1B2) from the 15 modes from the displacement of all 5 atoms (5A1 + 2A2 + 4B1 + 4B2), we would have the symmetry species of the remaining 9 vibration modes: 4A1 + 1A2 + 2B1 + 2B2.
