Question

question 7
1 pts
the molecular geometry about the carbon atoms in $c_2h_6$ is
trigonal planar
octahedral
linear
tetrahedral
question 8
1 pts
what is the hybridization at the atom indicated by the arrow in the following molecule?
(diethyl ether molecular structure image)
diethyl ether, an anesthetic
$sp^3d^2$
$sp^3$
$sp^3d$
$sp$

Explanation:

Question 7

Step1: Determine the Lewis structure of \( C_2H_6 \) (ethane)

In ethane, each carbon atom is bonded to four other atoms (3 hydrogen atoms and 1 carbon atom). The Lewis structure shows that each carbon has four single bonds.

Step2: Apply VSEPR theory

For a central atom with four bonding pairs (and no lone pairs), the electron - pair geometry and molecular geometry are tetrahedral. The VSEPR formula for each carbon in \( C_2H_6 \) is \( AX_4 \) (where \( A \) is the central atom, \( X \) is the bonded atom), and the molecular geometry for \( AX_4 \) is tetrahedral. Trigonal planar is for \( AX_3 \), octahedral for \( AX_6 \), and linear for \( AX_2 \), so the molecular geometry about the carbon atoms in \( C_2H_6 \) is tetrahedral.

Question 8

Step1: Analyze the oxygen atom (indicated by the arrow)

First, we determine the number of electron groups around the oxygen atom. The oxygen atom has two single bonds (to two carbon atoms) and two lone pairs of electrons. So the total number of electron groups (bonding + lone pairs) is \( 2 + 2=4 \).

Step2: Determine the hybridization based on electron groups

The hybridization of an atom is related to the number of electron groups. For an atom with 4 electron groups, the hybridization is \( sp^3 \). \( sp^3d^2 \) is for 6 electron groups, \( sp^3d \) for 5 electron groups, and \( sp \) for 2 electron groups. So the hybridization of the oxygen atom is \( sp^3 \).

Answer:

D. Tetrahedral