If we're balancing to whole numbers, we have to make sure the right side oxygen is even so let's start by multiplying that by 2:

___**N**_{2}** + **___**O**_{2}** => **2 **N**_{2}**O**_{5}

From there, we end up with four nitrogen and 10 oxygen left on the right side. We balance that would by dividing both by 2 (since they are both diatomic on the left side) and get the answer below:

2 **N**_{2}** + **5 **O**_{2}** => **2 **N**_{2}**O**_{5}

If fractions are allowed, you can leave the right side along and just solve each element on the left for their respective amount on the right side (2 N to 2 N and 2 O to 5 O):

1 **N**_{2}** + 5/2 O**_{2}** => **1 **N**_{2}**O**_{5}