Chem question modification

Kahynickel SG

Hello! UltimaOnline How are you? I am busy these days finalizing the mock examination paper for A-level Chemistry as CIE exams are approaching near here in Pakistan.

Again i need your guidance for this question.

(b) Oxides of elements can be acidic, basic or amphoteric. Therefore the nature of the oxides depends upon the position of the element in the periodic table.

Following oxides are all white solids. Series of tests were performed on these six oxides to distinguish them from each other.

Al₂O₃, MgO, PbO, PbO₂, SnO₂

The oxides were distributed into two sets.

Read the following description of tests then answer the questions that follow.

Set I: MgO, SnO₂, PbO.

Three samples were tested each containing only one of these oxides.

· Adding H₂SO_4(aq) to the first sample of white powder produces a colorless solution. Another sample of this white powder initially produces a white precipitate in NaOH_(aq) which dissolved when excess NaOH_(aq) was added.

· A second sample of white solid produces a white solid on addition of aqueous H₂SO_4. When dilute sodium hydroxide was added dropwise, a white precipitate was seen which dissolved in excess of the reagent.

· Adding H₂SO_4(aq) to the third sample of white solid produces a colorless solution. Adding NaOH(aq) to the resultant solution gave a white precipitate.

(b)(i) Which two oxides might be present in the first sample of white powder when aqueous H₂SO_4(aq) was added?

………………………………………………………………………………………………………

(ii) Identify the oxide was present in the first sample.

………………………………………………………………………………………………………

Give balanced equations when the oxide reacts with the H₂SO₄ and NaOH_(aq).

……………………………………………………………………………………………………...

…………………………………………………………………………………………………..…..

(iii) Which of these samples first, second or third was MgO? ………………………………………………………………………………………………………

(iv) Suggest the identity of the white precipitate formed in second sample when H₂SO_4(aq) was added and gave an equation for its formation.

………………………………………………………………………………………………………

Set II- contains Al₂O₃, SnO, PbO₂ (All are white solids at r.t.p)

The oxides can be distinguished by the following reaction schemes:

dil H₂SO₄

A Colorless solution

_{Conc: H2SO4}

B White solid + O₂

C

I have put A is SnO. (since it is reducing)

B is PbO2 (since it is oxidizing)

C is Al2O3

Now the thing is that All these oxides are amphoteric I have a choose a reagent or create a situation which claerly distinguishes C from A and B. I was thinking of making an aqueous solution of C since it is an insoluble oxide but that would be very easy.

Secondly I have put PbO2 in separate sets from PbO because PbO2 being a strong oxidizing agent can be identified easily from the rest of Set II, but in combination with PbO it is relatively difficult since both producesa white solid with Con: sulfric acid while oxygen is released in case of PbO2.

Kahynickel SG

This is the correct version of the detail after set II.

A dil H2SO4 Colorless solution

B Conc : H2SO4 White solid + O₂

UltimaOnline SG

Hi Kahynickel, long time no see. I'm confident your students will do well with your good guidance.

I agree with your thoughts that it's better to put lead(II) and lead(IV) oxides in different sets.

Edit :
Ah ok, I see your 2nd post was in response to the formatting problems in your 1st post.

Suggestion :
As you say, of the 3 oxides :
1 is a strong reducing agent, 1 is a strong oxidizing agent, and 1 is neither reducing nor oxidizing. How about using either/both a oxidizing or reducing agent on C, for the student to deduce its identity?

Kahynickel SG

From Sc to Cu the atomic radii remains invariant. This is because the effective nuclear charge increases but the increase in nuclear charge is outweighed by the poor shielding effect of 3d-orbitals.

In view of this expalnation can we summarize the following trend.

Elements effective nuclear charge shielding effect atomic radii

Na to Cl increaeses invariant decreases

Sc to Cu increaeses invariant invariant

F to I decreaeses increases increaeses

Should i state "Invariant" or "deacreses" for Sc to Cu??

UltimaOnline SG

From Sc to Cu the atomic radii remains invariant. This is because the effective nuclear charge increases but the increase in nuclear charge is outweighed by the poor shielding effect of 3d-orbitals.

In view of this expalnation can we summarize the following trend.

Elements effective nuclear charge shielding effect atomic radii

Na to Cl increaeses invariant decreases

Sc to Cu increaeses invariant invariant

F to I decreaeses increases increaeses

Should i state "Invariant" or "deacreses" for Sc to Cu??

Inorganic chemistry, particularly transition metal chemistry, is problematic. Even for 'A' level purposes (which is significantly simpified from University / professional Chemistry levels), transition metal chemistry is still problematic and confusing for students and teachers alike, because we're dealing with forced simplifications of complicated multi-factorial patterns.

(I personally much prefer Organic chemistry over Inorganic chemistry... the love or sexual electrostatic attraction between nucleophiles and electrophiles make for more fascinating observation, exploration and mechanisms.)

In explaining the relatively smaller atomic radii of d block metals compared to the s block metals, we say the 3d orbitals provide relatively poorer shielding, as 3d orbitals are more diffused compared to the less diffused 3s orbitals.

In explaining the relatively invariant atomic radii of d block metals compared to the s block metals, we say the 3d orbitals provide relatively more effective shielding, as each additional electron is added into the penultimate quantum shell's 3d orbitals, which being closer to the nucleus, naturally provides more effective shielding compared to the outer quantum shell's 4s orbital.

Back to your original question; this is rather problematic to set as an MCQ, because there is no one single perfect term to describe the trends of the shielding effect or the atomic radii.

Elements E.N.C shielding effect atomic radii

Sc to Cu increases relatively invariant relatively invariant
(increases slightly) (decreases, then increases)

The initial decrease in atomic radii is caused by the increase in nuclear charge outweighing the (initially gradual) increase in shielding effect from additional inner 3d electrons.

The subsequent increase in atomic radii is caused by the (subsequently steeper) increase in shielding effect from too having many additional inner 3d electrons, repelling the outer 4s electrons.

For 'A' level purposes, it will suffice for the 'A' level student to state "relatively invariant" when it comes to atomic radius of the transition metals, and to provide the simplified explanation as given (several paragprahs) above.

(The atomic radius of Mn is anomalously high, and reasons for this are beyond the scope of the A levels; and A level students are not expected to even know this, let alone explain this).