Late last week, the New York Mets reached an agreement with star second baseman and reigning National League batting champion Jeff McNeil on a four-year, $50 million contract extension.
The deal will last him through the 2026 season and includes a club option and $2 million buyout for 2027.
The veteran second baseman hit for a .326 average with nine home runs, 62 RBI, a 5.7 WAR, an on-base percentage of .382, and an OPS+ of 140.
Today, the Mets and McNeil made their new pact official.
On Twitter, MLB insider Jon Heyman revealed the terms of the new contract.
McNeil will earn $6.25 million in 2023, $10.25 million in 2024, and $15.75 million in 2025 and 2026.
Jeff McNeil Mets deal is official. $6.25 2023, $10.25 2024, $15.75 2025 and 2026, $2 mil buyout on $15.75 club option 2027
— Jon Heyman (@JonHeyman) January 31, 2023
The Mets have been busy this winter.
Despite missing out on Carlos Correa, they’ve signed some players to pretty large deals, with McNeil being the latest example.
They inked Justin Verlander, Kodai Senga, and Jose Quintana to bolster the starting rotation and also re-signed outfielder Brandon Nimmo to an eight-year, $162 million deal.
Now, McNeil is locked up for at least the next four seasons, with the potential of the deal even stretching into 2027.
McNeil will receive the $2 million buyout in the event that the Mets decide not to exercise his club option.
The 30-year-old has had a successful career thus far, having made his Major League debut in 2018.
Over the first five seasons of his career, McNeil has an average of .307 and has hit a total of 46 home runs.
He has also driven in 214 runs, posted an on-base percentage of .370, an OPS of .827, and an OPS+ of 128.
He’ll be entering his sixth season as a member of the Mets.
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