A) 11 \[\sigma \] bonds and 2 \[\pi \] bonds
B) 13 \[\sigma \] bonds and no \[\pi \] bond
C) 10 \[\sigma \] bonds and 3 \[\pi \] bonds
D) 8 \[\sigma \] bonds and 5\[\pi \] bonds
Correct Answer: C
Solution :
\[HC\equiv CCH=CHC{{H}_{3}}\]\[10\sigma +3\pi -Bonds\] |
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