TPR Chem Passage 2 Question

[Perseverance7779311]

Could someone please assist me with the following question in Passage 2 TPR Chem? It says Compounds I and II, two oxides of the same transition metal, are placed separately into two vessels of aqueous media. Compound I produces an acidic solution, and Compound II produces a basic solution. It can be concluded that the metal-oxide bonding in:

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[Teleologist]

Where's the rest of the question?

 

[Lunasly]

I'm also looking for help with this question.

The question asks: Compounds I and II, two oxides of the same transition metal, are placed separately into two vessels of aqueous media. Compound I produces an acidic solution, and compound II produces a basic solution. It can be concluded that the metal oxide bonding in:

Ans: Compound I is more covalent, and the bonding in Compound II is more ionic.

And from TPR back of book: According to the passage, acidic compounds are those in which MO-H bond is the ionic one; basic compounds are those in which the M-OH bond is ionic. Since Compound I is acidic and Compound II is basic, the metal oxide bond in compound I is not ionic (because the O-H bond is), but the metal-oxide bond in compound II must be ionic.

 

[Czarcasm]

I'm also looking for help with this question.

The question asks: Compounds I and II, two oxides of the same transition metal, are placed separately into two vessels of aqueous media. Compound I produces an acidic solution, and compound II produces a basic solution. It can be concluded that the metal oxide bonding in:

Ans: Compound I is more covalent, and the bonding in Compound II is more ionic.

And from TPR back of book: According to the passage, acidic compounds are those in which MO-H bond is the ionic one; basic compounds are those in which the M-OH bond is ionic. Since Compound I is acidic and Compound II is basic, the metal oxide bond in compound I is not ionic (because the O-H bond is), but the metal-oxide bond in compound II must be ionic.

I remember this passage. Basically, the more ionic the bond is, the more likely it is to dissociate at that bond. For acidic oxides, the ionic character is between the oxygen and hydrogen such that the hydrogen will dissociate from the bond. Therefore we expect the M-OH bond to be more covalent and the MO-H bond to be more ionic. For basic oxides it's just the opposite. Basic oxides release -OH into solution. Therefore the M-OH bond is more ionic (causing -OH to dissociate) rather than H+ (the MO-H bond is more covalent). In this way, the -OH ion is strongly attached to each other, but weakly attached to the metal.

It took me a moment to realize MO was a symbol forthe metal and oxygen (not molecular orbital). That threw me off, but once I understood their notation, it made sense. Maybe try re-reading the relevant passage info. It may help you too.

 

[Teleologist]

The more electropositive the metal is, the more basic its oxide. Consider NaOH. Very electropositive metal. Very basic salt. When NaOH is put into water, the hydroxide group leaves, leaving the sodium portion of the salt with a positive charge. This is okay because Na by definition is very electropositive and thus able to stabilize positive charge. If instead you said that a hydrogen proton leaves, then we'd be left with NaO(-) and that's not very "good," stability wise, because the brunt of the charge is borne by the oxygen (it's the more electronegative) with some also being born by the sodium (a no-no). Having an entire hydroxide anion leave is better since then we'd have a solution of Na(+) and HO(-) and that's good; we have an EP element stabilizing a positive charge and an EN complex stabilizing a negative charge.

On the other hand the less electropositive the metal or whatever the counterion is, the less the metal can stabilize a positive charge through the loss of HO- (hydroxide anion). Consider (HO)2SO2. Sulfur is more electronegative than it is electropositive, with an EN of 2.6. Thus, when you put (HO)2SO2 in water, you don't get hydroxide ion. You get hydronium ion. The leaving of the hydrogen proton creates a negative charge on the remaining molecule, and that's okay, since we have an electronegative central element (and other EN elements).

Tl; dr, you don't ever see Na(-) but Na(+) for a reason.

 

[Lunasly]

I remember this passage. Basically, the more ionic the bond is, the more likely it is to dissociate at that bond. For acidic oxides, the ionic character is between the oxygen and hydrogen such that the hydrogen will dissociate from the bond. Therefore we expect the M-OH bond to be more covalent and the MO-H bond to be more ionic. For basic oxides it's just the opposite. Basic oxides release -OH into solution. Therefore the M-OH bond is more ionic (causing -OH to dissociate) rather than H+ (the MO-H bond is more covalent). In this way, the -OH ion is strongly attached to each other, but weakly attached to the metal.

Thanks for the help. So, the M-O bond of an acid metal anhydride is not ionic given that the ionic character is between the O-H bond (the electronegativity difference between the O-H bond is greater than the M-O bond). However, the opposite is true for basic oxides. The ionic character lies within the M-OH bond since the electronegativity difference between M-OH is greater than the O-H bond. Did I say this correctly? Is this information we were supposed to just interpret from the passage or should it be something we conceptually understand. I have not encountered it during my content review yet.

 

[Teleologist]

Is this information we were supposed to just interpret from the passage or should it be something we conceptually understand. I have not encountered it during my content review yet.

It should probably be conceptually understood after taking gen chem.

 

[Czarcasm]

Thanks for the help. So, the M-O bond of an acid metal anhydride is not ionic given that the ionic character is between the O-H bond (the electronegativity difference between the O-H bond is greater than the M-O bond). However, the opposite is true for basic oxides. The ionic character lies within the M-OH bond since the electronegativity difference between M-OH is greater than the O-H bond. Did I say this correctly? Is this information we were supposed to just interpret from the passage or should it be something we conceptually understand. I have not encountered it during my content review yet.

That's exactly right. As far as knowing it, I think conceptually, it's something we probably could reason out. Just the wording of the question was weird. In this instance though, like most passages, the information is right there in the passage. So you have to use a little about what you know, ionic/covalent character, electronegativity, etc. to answer the question based on the info they provided. It's straightforward, the difficult part is understanding what the passage is telling you.

 

[Lunasly]

The more electropositive the metal is, the more basic its oxide. Consider NaOH. Very electropositive metal. Very basic salt. When NaOH is put into water, the hydroxide group leaves, leaving the sodium portion of the salt with a positive charge. This is okay because Na by definition is very electropositive and thus able to stabilize positive charge. If instead you said that a hydrogen proton leaves, then we'd be left with NaO(-) and that's not very "good," stability wise, because the brunt of the charge is borne by the oxygen (it's the more electronegative). Having an entire hydroxide anion leave is better since then we'd have a solution of Na(+) and HO(-) and that's good; we have an EP element stabilizing a positive charge and an EN complex stabilizing a negative charge.

On the other hand the less electropositive the metal or whatever the counterion is, the less the metal can stabilize a positive charge through the loss of HO- (hydroxide anion). Consider (HO)2SO2. Sulfur is more electronegative than it is electropositive, with an EN of 2.6. Thus, when you put (HO)2SO2 in water, you don't get hydroxide ion. You get hydronium ion. The leaving of the hydrogen proton creates a negative charge on the remaining molecule, and that's okay, since we have an electronegative element.

Oh ok. Thanks for the explanation!

So just to get this right, given that Compound I produces an acidic solution (i.e., hydronium ion is produced), the ionic character lies within the bond between the atom and the acidic hydrogen rather than the bond between the metal and the oxide (since the electronegativity difference between the metal and the oxide is lower).

You're explanation makes a lot of sense and this principle is very basic. I am just having a hard time connecting this concept to this particular question.

 

[Teleologist]

Oh ok. Thanks for the explanation!

So just to get this right, given that Compound I produces an acidic solution (i.e., hydronium ion is produced), the ionic character lies within the bond between the atom and the acidic hydrogen rather than the bond between the metal and the oxide (since the electronegativity difference between the metal and the oxide is lower).

You're explanation makes a lot of sense and this principle is very basic. I am just having a hard time connecting this concept to this particular question.

Here's the play-by-play thought process I had:

Compound 1 forms an acidic solution because:

1) Compound one releases hydrogen cations into solution.

2) Charge must be conserved, so in solution, after the release of the hydrogen proton, we have also have a compound of negative charge.

3) Electronegative elements can better stabilize negative charge.

4) Oxygen is electronegative itself.

5) Electronegative element + oxygen = compound with high degree of covalent character.

Do the same for compound 2, which forms a basic solution.

1) Compound 2 releases hydroxide ions into solution.

2) Charge must be conserved; we have also in solution a compound of positive charge.

3) Electropositive elements stabilize positive charge.

4) We're taking about oxides.

5) EP element + oxygen (electronegative) = compound of high ionic character.

 

[Lunasly]

Thanks that makes a lot of sense. I really appreciate the help and the efforts you have put into answering questions in this forum.

 

[Perseverance7779311]

thanks guys for all ur help! i appreciate it!