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Report: Rockies, Bryant agree to 7-year, $182M deal

Rob Leiter / Major League Baseball / Getty

The Colorado Rockies and free-agent infielder/outfielder Kris Bryant agreed on a seven-year, $182-million contract, sources told ESPN's Jeff Passan.

The deal includes a full no-trade clause, the New York Post's Joel Sherman reports.

Bryant will make $26 million annually, matching St. Louis Cardinals first baseman Paul Goldschmidt for the 24th-highest in league history, according to The Athletic's Ken Rosenthal.

The $182-million deal for Bryant is the second-richest in Rockies history, according to The Athletic's Nick Groke. The move only trails the eight-year, $260-million extension given to former All-Star third baseman Nolan Arenado in February 2019.

Bryant is no stranger to playing in the NL West, appearing in 51 games with the San Francisco Giants last season after a trade from the Chicago Cubs.

The four-time All-Star authored another strong campaign in 2021, hitting .265/.353/.481 with 25 home runs, 73 RBIs, and 10 stolen bases in 144 games.

The 30-year-old joins Colorado at a period of transition and uncertainty for the franchise. In the past 13 months, the Rockies traded Arenado and $51 million to the Cardinals, and they now appear set to lose stalwart shortstop Trevor Story in free agency.

Bryant is a lifetime .278/.376/.504 hitter with 167 home runs and 487 RBIs in 884 games. He won the NL MVP in 2016, blasting a career-best 39 homers, helping the Cubs to their first World Series title in 108 years.

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