N has 5 valence electrons, meaning it can triple bond, have one lone pair, and still obey the octet rule.
F can't triple bond - if it does, it would either disobey the octet rule (can't happen, it's in period 2 with no d orbitals) or have a really high positive formal charge, which isn't likely since F is really electronegative and tends to hold on to its electrons.
Pb has 4 valence electrons. If it triple bond, it would leave one floating, unpaired electron. Unlikely.
Finally, S has 6 valence electrons. If it triple bonds, it would have one lone pair, and an extra electron to bond with something else. It's logistically possible, since elements beyond period 2 can go disobey the octet rule, but it's not as likely to occur.
N just fits.
