EA Sports Released FIFA 2022 Team Of The Year
Thumbnail Picture: Screengrab via EA
EA Sports announced the names of the eleven players that compose the Team Of The Year (TOTY) available on FIFA 2022 Ultimate Team.
The release aims at celebrating the best players of the past year by creating new and upgraded cards for them, which will be wildly contended by FIFA users throughout their journeys into the Ultimate Team universe.
The unveiling of the TOTY was characterized by a wake of praising and criticism as only such a hot and debatable topic could do among soccer fanatics.
This year’s team sees Gianluigi Donnarumma, one of the protagonists of Italy’s triumph at EURO 2020, now defending Paris Saint Germain’s goal, as the goalkeeper.
Hachraf Hakimi (PSG) and Joao Cancelo (Manchester City) are, respectively, right and left-back, while Marquinhos (PSG) and Ruben Dias (Manchester City) form the central defenders couple.
The midfield trio is composed of Jorginho and N’Golo Kante, who both played a crucial role in Chelsea’s Champions League victory, together with Kevin De Bruyne (Manchester City.)
As always, the major hype was reserved for the forward triplet.
Robert Lewandowski, for many, the legit winner of the Balon d’Or, is joined by the actual Balon d’Or winner Leo Messi and French superstar Kylian Mbappe, both PSG’s players.
Credits: Instagram @brfootball
FIFA users selected the team among an initial total of 80 players.
The exclusion that immediately stands out is Cristiano Ronaldo’s one. The Portuguese pays a grey final of season with Juventus and his below expectations come back at Manchester United.
Other eminents exclusions were the ones of Borussia Dortmund striker Erling Haaland, and Italy’s captain Giorgio Chiellini.
Paris Saint Germain is the most represented team (five players). Lewandowski is the only one from German Bundesliga, and the rest of the roster is left to the Premier League.
No Spanish Liga or Italian Serie A representatives made it to the final team.
For Ultimate Team maniacs, the hunt is now on.