Question

1. carbon dioxide reacts with an aqueous solution of sodium hydroxide to form carbonate ion. what change in the hybridization of carbon occurs in this reaction? a) sp to sp² b) sp² to sp³ c) sp³ to sp³d d) sp³ to sp³d² e) no change 24. one product of the combustion of ethylene, c₂h₄, is carbon dioxide. what change in hybridization of the carbon occurs in this reaction? a) sp³ to sp² b) sp³ to sp c) sp³ to sp³d d) sp² to sp³d² e) sp² to sp 25. sulfur dioxide so₂ reacts with oxygen to form sulfur trioxide. what change in hybridization of the sulfur occurs in this reaction? a) sp² to sp³ b) sp² to sp c) sp³ to sp³d d) sp² to sp³d² e) no change 26. which of the following hybridized atoms is not possible? a) an sp hybridized carbon atom b) an sp² hybridized nitrogen atom c) an sp³ hybridized oxygen atom d) an sp³d hybridized fluorine atom e) an sp³d² hybridized sulfur atom 27. which of the following characteristics apply to pcl₃? 1. nonpolar molecule 2. polar bonds 3. trigonal - pyramidal molecular geometry 4. sp² hybridized a) 1 and 2 b) 2 and 3 c) 3 and 4 d) 1, 2, and 3 e) 1, 2, 3, and 4

Explanation:

Step1: Analyze question 23

Carbon in $CO_2$ is $sp$ hybridized. In the reaction with $NaOH$ to form carbonate ion, the carbon becomes $sp^2$ hybridized. The reaction involves the formation of new bonds and change in electron - pair geometry around carbon.

Step2: Analyze question 24

In the combustion of ethylene ($C_2H_4$), carbon in ethylene is $sp^2$ hybridized. The product carbon dioxide has carbon as $sp$ hybridized. So the change is from $sp^2$ to $sp$.

Step3: Analyze question 25

In $SO_2$, sulfur is $sp^2$ hybridized. When it reacts with oxygen to form $SO_3$, sulfur becomes $sp^2$ hybridized (trigonal - planar geometry around sulfur in $SO_3$ as well), so no change in hybridization.

Step4: Analyze question 26

An $sp$ hybridized carbon atom has two $\sigma$ - bonds and two $\pi$ - bonds. An $sp^2$ hybridized nitrogen atom has three $\sigma$ - bonds and one $\pi$ - bond. An $sp^3$ hybridized oxygen atom has two $\sigma$ - bonds and two lone - pairs. An $sp^3d$ hybridized fluorine atom is not possible as fluorine is a second - period element and does not have $d$ - orbitals available for hybridization. An $sp^3d^2$ hybridized sulfur atom is possible for sulfur in some of its higher - oxidation state compounds.

Step5: Analyze question 27

In $PCl_3$, the phosphorus atom has three bonding pairs and one lone - pair. The molecular geometry is trigonal - pyramidal, the bonds are polar, and the molecule is polar. Phosphorus is $sp^3$ hybridized. So characteristics 1, 2, and 3 apply.

Answer:

1. b) $sp$ to $sp^2$
2. b) $sp^2$ to $sp$
3. e) no change
4. d) an $sp^3d$ hybridized fluorine atom
5. d) 1, 2, and 3