Which one of the following statements best describes the acidic/basic/amphoteric nature of ZnO and C...

ZnO (Zinc oxide):It can react with both acids and bases.For example:With acid: $ \text{ZnO} + 2HCl \rightarrow ZnCl_2 + H_2O $With base: $ \text{ZnO} + 2NaOH + H_2O \rightarrow Na_2[Zn(OH)_4] $→ Hence,ZnO is amphoteric.CaO (Calcium oxide):It reacts with water to formCa(OH)_2, a strong base, and reacts with acids to form salts, but not with bases.→ Hence,CaO is basic.✅Correct answer: Option AZnO is amphoteric, while CaO is basic.

$$ \text { What are the correct orders of stability for the following compounds? } $$...

1️⃣ Comparison of$ \mathrm{VF_5}$and$ \mathrm{VCl_5}$Vanadium pentahalidesinvolve the+5 oxidation stateof vanadium ($V^{5+}$).As we move fromFtoI, the+5 oxidation statebecomesless stablebecause:Larger halides ($Cl^-$, $Br^-$, $I^-$) are less electronegative and less able to stabilize a high positive oxidation state on the metal.Fluorineis the most electronegative and forms stronger, more covalent, and more stable bonds with highly oxidized metals.✅ Thus,$ \mathrm{VF_5 > VCl_5} \text{ (in stability)} $2️⃣ Comparison of$ \mathrm{CuCl_2}$and$ \mathrm{CuI_2}$InCu(+2)compounds, we must consider redox tendencies:$ \mathrm{Cu^{2+} + I^- → Cu^+ + \frac{1}{2} I_2} $The above reaction indicates that $ \mathrm{CuI_2} $ isunstable; it disproportionates to $ \mathrm{CuI} $ (Cu⁺) and $ \mathrm{I_2} $.$ \mathrm{CuCl_2} $ is stable because $ \mathrm{Cl^-} $ is not easily oxidized, hence no such disproportionation.✅ So,$ \mathrm{CuCl_2 > CuI_2} \text{ (in stability)} $✅ Final conclusion$ \boxed{\mathrm{VF_5 > VCl_5; \, CuCl_2 > CuI_2}} $Correct option: A

D And F Block Elements - IAT-IISER Chemistry PYQs

1️⃣ Comparison of

%5Cmathrm%7BVF_5%7D

and

%5Cmathrm%7BVCl_5%7D

Vanadium pentahalidesinvolve the+5 oxidation stateof vanadium (

V%5E%7B5%2B%7D

).As we move fromFtoI, the+5 oxidation statebecomesless stablebecause:Larger halides (

Cl%5E-

Br%5E-

I%5E-

) are less electronegative and less able to stabilize a high positive oxidation state on the metal.Fluorineis the most electronegative and forms stronger, more covalent, and more stable bonds with highly oxidized metals.✅ Thus,

%5Cmathrm%7BVF_5%20%3E%20VCl_5%7D%20%5Ctext%7B%20(in%20stability)%7D

2️⃣ Comparison of

%5Cmathrm%7BCuCl_2%7D

and

%5Cmathrm%7BCuI_2%7D

InCu(+2)compounds, we must consider redox tendencies:

%5Cmathrm%7BCu%5E%7B2%2B%7D%20%2B%20I%5E-%20%E2%86%92%20Cu%5E%2B%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20I_2%7D

The above reaction indicates that

%5Cmathrm%7BCuI_2%7D

isunstable; it disproportionates to

%5Cmathrm%7BCuI%7D

(Cu⁺) and

%5Cmathrm%7BI_2%7D

%5Cmathrm%7BCuCl_2%7D

is stable because

%5Cmathrm%7BCl%5E-%7D

is not easily oxidized, hence no such disproportionation.✅ So,

%5Cmathrm%7BCuCl_2%20%3E%20CuI_2%7D%20%5Ctext%7B%20(in%20stability)%7D

✅ Final conclusion

%5Cboxed%7B%5Cmathrm%7BVF_5%20%3E%20VCl_5%3B%20%5C%2C%20CuCl_2%20%3E%20CuI_2%7D%7D

Correct option: A

D And F Block Elements

High-Yield Trend

Questions
2 MCQs

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Official Solution

D And F Block Elements

High-Yield Trend

Questions 2 MCQs

Official Solution

Official Solution

Questions
2 MCQs