Why is this position in the 3.d4 d5 Petroff equal?

Question

I'm trying to learn some theory about the Petroff and was reading about the 3.d4 d5 sideline. Eventually, the position below is reached.

[FEN ""]
1.e4 e5 2.Nf3 Nf6 3.d4 d5 4.exd5 exd4 5.Qxd4 Qxd5 6.Qe3+ Qe6 7.Bb5+ Nc6 8.Bxc6+ bxc6 9.Nc3

What I can't wrap my head around is that Stockfish says that this position is completely equal. Why is that? From my perspective, White looks objectively better here - they are ready to castle and have developed an extra knight, while Black is behind on development and has serious damage to their queenside pawn structure. What's Black's advantage here? What do they have to counter all White's advantages?

Edit: Sorry, I jumped the gun a bit - Stockfish actually evaluates this position as -.07 at depth 22, which feels somehow more baffling to me.

White has developed one more piece, but it's now black's turn. From move parity, that's not too different from the situation after white makes the first move of the game. — SE - stop firing the good guys, Dec 15 '21 at 11:37

score 19 · Accepted Answer · answered Dec 14 '21 at 14:28

19

What I can't wrap my head around is that Stockfish says that this position is completely equal. Why is that?

Black has the bishop pair in a completely open position (no center pawns). This advantage plus the extra open b file, which black is ready to exploit with Rb8, compensates for the doubled isolated pawns.

answered Dec 14 '21 at 14:28

Brian Towers

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Also note that White will have a hard time accessing those weak pawns. For example, the "c" file is close and White no longer has a light-squared bishop to attack c6 – David Dec 14 '21 at 17:45

Why is this position in the 3.d4 d5 Petroff equal?

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