Question

question 8
1 pts
which atomic orbitals form the $sp^3d$ hybrids
one s, two p, two d
two s, one p, two d
one s, one p, one d
one s, three p, one d

Explanation:

To determine the atomic orbitals forming \( sp^3d \) hybrids, we analyze the notation: the coefficient of \( s \) is 1 (implies 1 \( s \) orbital), the coefficient of \( p \) is 3 (implies 3 \( p \) orbitals), and the coefficient of \( d \) is 1 (implies 1 \( d \) orbital). So we check the options:

• "one s, two p, two d": Incorrect (p count is 2, should be 3).
• "two s, one p, two d": Incorrect (s count is 2, p count is 1).
• "one s, one p, one d": Incorrect (p count is 1, d count is 1 but p should be 3).
• "one s, three p, one d": Matches the counts from \( sp^3d \) (1 s, 3 p, 1 d).

Answer:

D. one s, three p, one d (assuming the last option is D; if options are labeled as per order, the last option with "one s, three p, one d")