A) $BeCl_{2},\ sp^{2},$ linear
B) $BeCl_{2},\ sp^{2},\$triangular planar
C) $BCl_{3},\ sp^{2},$ triangular planar
D) $BCl_{3},\ sp^{3},$ tetrahedral
Explanation
The correct answer is Option A: $BeCl_{2},\ sp^{2},$ linear.
A detailed explanation for this question is being prepared.