The fate of eight Indian players will be determined in the second-round tiebreaks of the FIDE Chess World Cup 2025, taking place at the Resort Rio Convention Centre in Goa. Of the 17 Indians who entered Round 2, five players- D. Gukesh, Arjun Erigaisi, Pentala Harikrishna, Diptayan Ghosh, and Karthik Venkataraman – have already secured their spots in Round 3, four have been eliminated, and the remaining eight are competing in tiebreaks today.
Eight Indian players who are competing in the second-round tiebreaks include – R. Praggnanandhaa, Vidit Gujrathi, Nihal Sarin, Pranav V, S.L. Narayanan, Karthikeyan Murali, Pranesh M, and Raunak Sadhwani.
After two drawn games in Round 2, Vidit Gujrathi finally defeats Faustino Oro in the tiebreaks to advance to Round 3, eliminating youngest player in the history of this event . Pranav V and Pranesh M have also joined Vidit as qualified Indians for round 3 after winning their set of tiebreaks. Meanwhile, Nihal Sarin and Raunak Sadhwani fall short of the required win and are knocked out of the tournament in the early stages.
In a dramatic turnaround, Praggnanandhaa, last edition’s finalist, is facing tough time against Temur Kuybokarov after 1-1 draw in 10+10 rapid games.
After draw matches in 10+10 rapid games, 3 Indian players – Praggnanandhaa, S.L. Narayanan and Karthikeyan Murali’s fate will be decided in 5+3 blitz tiebreak.
