French Guiana face action over Malouda

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Florent Malouda won the Premier League, Champions League and three FA Cups during six years at Chelsea

French Guiana are facing disciplinary action after fielding ex-Chelsea and France midfielder Florent Malouda in a Concacaf Gold Cup match in America.

Malouda played in Wednesday’s goalless draw against Honduras despite being ineligible under the Fifa rules in use.

French Guiana-born Malouda, 37, played 80 times for France.

“We are using Fifa rules so a player who’s played in an official match for a different [country] cannot play in the Gold Cup,” said a Concacaf official.

Malouda, who currently plays for Indian Super League team Delhi Dynamos, featured in the 2006 World Cup final for France and, during six years at Chelsea, won three FA Cups, a Premier League and a Champions League.

French Guiana is an overseas department of France.

The national team are not Fifa members and could not, for example, play in the World Cup – but as they are not Fifa members the global governing body’s eligibility rules do not normally apply to them.

As a result Malouda was allowed to play for French Guiana in the 2017 Caribbean Cup, featuring twice as they finished third in Martinique in June.

However, Concacaf’s decision to use Fifa rules means he is ineligible to play in the Gold Cup and Wednesday’s result could be ruled a 3-0 forfeit in favour of Honduras.

“The disciplinary committee will review the case and render its decision in due time,” Concacaf said in a statement.

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