Let's analyze the bond structure in an ethylene molecule (C₂H₄) to determine the count of $$\sigma$$ and $$\pi$$ bonds.

Ethylene consists of two carbon atoms double-bonded to each other, with each carbon atom also bonded to two hydrogen atoms. Each single bond is a $$\sigma$$ bond. The double bond between the carbon atoms is made up of one $$\sigma$$ bond and one $$\pi$$ bond. Here's how the bonds break down:

1. Each carbon-hydrogen bond is a $$\sigma$$ bond. Since there are four C-H bonds, we have 4 $$\sigma$$ bonds from C-H bonding.
2. The carbon-carbon double bond consists of one $$\sigma$$ bond and one $$\pi$$ bond. Thus, we have 1 additional $$\sigma$$ bond from the C=C double bond.
3. Total $$\sigma$$ bonds = 4 (from C-H) + 1 (from C=C) = 5.
4. There is 1 $$\pi$$ bond present in the carbon-carbon double bond.

Therefore, the correct option is:

Option B: 5 and 1