The Baltimore Ravens have lost their top free agents before free agency has truly begun. On the first day of the legal tampering period, the Ravens saw Matthew Judon agree to terms with the New England Patriots. Now the Oakland Raiders are set to sign Yannick Ngakoue, according to NFL Network’s Ian Rapoport. Per ESPN’s Adam Schefter, it’s a two-year deal worth $26 million.
Ngakoue had a tough 2020 campaign, ultimately getting traded twice. After playing his first four years with the Jacksonville Jaguars, Ngakoue grew upset with his contract status and demanded a trade, eventually getting shipped off to the Minnesota Vikings. The Ravens pulled off another midseason trade, acquiring Ngakoue from the Vikings for a third-round pick and fifth-round pick in the 2021 NFL draft.
Unlike the trade for cornerback Marcus Peters in 2019, Ngakoue didn’t provide nearly the boost Baltimore was hoping for. Over nine games, Ngakoue accounted for just three sacks and three quarterback hits, playing in a very limited role in the Ravens’ defense. While that could be chalked up to a down year for sacks across the entire league and Ngakoue getting traded twice over the course of just a few months, it makes sense that Baltimore wouldn’t be super eager to pay him top dollar in free agency given the limited production.
The Raiders seem to be banking on a return to his previous production. Though he didn’t add much in Baltimore, Ngakoue has put up 45.5 sacks and 96 quarterback hits over his five-year career, including a 12-sack season in 2017. The Ravens, on the other hand, might need to move quickly if they plan on signing the few top free-agent pass rushers remaining.
