Because he’s usually a step behind White in the opening – Black has to be wary of setting traps.
Krzysztov Pytel White
Oscar Castro Black
Dortmund 1977
Queen’s Gambit Accepted D20
1 d4 d5 2 c4 dxc4 3 e4 e5 4 ♘f3 exd4 5 ♗xc4 ♘c6
White can see a tactical weakness at f7. But 6 ♕b3 can be safely met by 6...♗b4+ (7 ♘bd2 ♕e7).
6 0-0 ♗g4?
This is based on two traps. The first is 7 ♗xf7+? ♔xf7 8 ♘g5+? ♕xg5! (9 ♕b3+ ♗e6) and Black wins.
7 ♕b3! ♕d7
The second trap is 8 ♕xb7? ♖b8. Black would get a powerful attack, 9 ♕a6 ♗xf3 10 gxf3 ♖b6 11 ♕a4 ♗d6.
Question 52: What if 9 ♗xf7+ so that 9...♔xf7 10 ♕xc6 ♔xc6 11 ♘e5+ ?
8 ♗xf7+!
The trapper is trapped. Clearly 8...♔d8 9 ♗xg8 won’t do (9...♗xf3 10 gxf3 ♘a5 11 ♕e6).
8...♕xf7 9 ♕xb7 ♖c8
In a bad position, Black might have gambled on 10...♔d7!?.
The idea is 11 ♕xa8? ♗xf3 12 gxf3 ♗d6.
For example, 13 ♘d2?? ♕f4 and Black wins.
Question 53: What’s wrong with this analysis?
10 ♕xc6+ ♗d7 11 ♕a6 ♘f6 12 e5! ♘e4 13 ♖e1 ♘c5 14 e6!? ♘xe6 15 ♘g5 resigns.
In view of 14...♘xa6 15 exf7+ ♔xf7 16 ♘e5+ ♔e6 17 ♘g6+.
Or 15...♔d8 16 ♗g5+ and mates.
Viktor Korchnoi White
Ilias Nikolopoulos Black
Greece 2002
1 d4 d5 2 c4 dxc4 3 e4 e5 4 ♘f3 exd4 5 ♗xc4 ♘c6 6 0-0 ♗g4? 7 ♕b3! ♘a5? 8 ♗xf7+ ♔e7 9 ♕b5 ♘f6 10 ♕xa5 ♗xf3 11 ♗b3 ♗g4 12 ♕e5+ ♔d7 13 f3! ♗d6 14 ♕b5+ ♔c8 15 fxg4 and White wins.