Crosby Is AFC Defensive Player of Week
by Tom LaMarre
Rookie defensive end Maxx Crosby of the Oakland Raiders was selected AFC Defensive Player of the Week it was announced on Wednesday, after his dominating performance in a 17-10 victory over the Cincinnati Bengals on Sunday.
The 6-5, 267-pound Crosby, selected in the fourth round (No. 106 overall) out of Eastern Michigan by the Raiders in the 2019 NFL draft, recorded five tackles (four solo), including three for losses, four sacks, five quarterback hits and a forced fumble against the Bengals.
Brian Baldinger, a former NFL lineman who is an analyst for the NFL Channel wrote in a Twitter post: “This kid is something else. #MaddMaxx @CrosbyMaxx is making a name for himself this year all over the football field. Crosby is the definition of a disruptive game-wrecker! #BaldbysBreakdowns #Raiders #JustWinBaby.”
For the season, Crosby has 6.5 sacks, three forced fumbles, three passes defensed and 28 tackles.
The 22-year-old Crosby has one more sack than Khalil Mack, traded by the Raiders to the Chicago Bears last season, while Nick Bosa of the San Francisco 49ers and Josh Allen of the Buffalo Bills are the only rookies in the NFL with more sacks this season.
Crosby’s four sacks in one game, a Raiders record, tied for the second-most by a rookie in NFL history and he became the first rookie to accomplish the feat since Brian Orakpo of the Washington Redskins in 2009.
Crosby became the first player to record three sacks in the fourth quarter since 1991, and the last one pushed the Bengals into a desperate situation deep in their own territory, leading to an interception by fellow rookie Trayvon Mullen that clinched the Raiders’ victory.
