Iga Swiatek and Elena Rybakina renew their rivalry on Wednesday evening in Rome, facing off for the third time this year. Rybakina has won those two matches fairly easy, not dropping a set in either the Australian Open or Indian Wells. Swiatek will be hoping the transition to clay from hard courts will serve her better, with the conditions definitely more in favour for her than for the Kazakh.
After blasting through her first two opponents, it was a bit more difficult for Swiatek in her fourth round match against Donna Vekic. She joined an elite list of names with the 6-3 6-4 win, becoming the sixth woman to win 14 or more consecutive matches in Rome – a group that includes the likes of Chris Evert and Serena Williams. She was clutch on the big points, taking all three break points and saving six of seven faced on her serve.
Rybakina is yet to drop a set in Rome through three matches, most recently beating Marketa Vondrousova in the Italian capital. Her 6-3 6-3 win was impressive against an opponent who can be very difficult on slow clay, hitting 28 winners in the process. “I’m happy to stay aggressive,” Rybakina said. “There were some moments it was up and down but overall it was a good match, good serving from me. For sure now I need as many matches as I can to be more confident at the French Open.”
Swiatek has to be considered the favourite for this one with the move to her most favoured service and likely Rybakina’s most popular with the slower clay dampening her main strengths. The head-to-head will be a concern and in a night session where matches have been going on late, the Pole may not be at her focused best in the early parts. This may be a contender to go the distance but it should be one where Swiatek does eventually grind out the victory and further establish her spot as the top player on clay.
“I think [clay] changes it a lot,” Rybakina said. “It’s more rallies, it’s more physical, she has more time, I have more time. I think it’s much different than the hard courts for sure.
Prediction – Iga Swiatek in three sets
