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Chemistry: Post your doubts here!

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But then would HCN give a different product?
it will just form HBr instead of KBr right?
And suppose we said that happened then what would the HCl do? It will also form an acidic hydrolysis - COOH??
No ... i think HCN is too weak to even be used for this reaction ... i'm not sure if that is the reason but option C is perfectly correct, so why choose A instead?
 
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Help please!!!!

Use of the Data Booklet is relevant to this question. Which ions contain one or more unpaired electrons?
1 Cu2+
2 Mn3+
3 V3+

The Ans is A

V3+ has unpaired electrons??? :O
 
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upload_2015-6-8_20-32-28-png.54838

Why is 1 wrong?
Will the pressure be equal?
It will be equal. It can't be greater or it will keep expanding until equal

vDDjUTu.png

All will make aldehyde or carboxylic acid but which one is the one with low BP?
D? Besides, none of them are going to form aldehydes, they will form ketones. Carboxylic acids have higher boiling points bwecause of H-bonds. A and C form carboxylic acids, B forms nothing, and D forms ketone.
 
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Help please!!!!

Use of the Data Booklet is relevant to this question. Which ions contain one or more unpaired electrons?
1 Cu2+
2 Mn3+
3 V3+

The Ans is A

V3+ has unpaired electrons??? :O
The electronic configuration of Vanadium is [Ar] 4s2 3d3 ...
In general, electrons are removed from the valence-shell s orbitals before they are removed from valence d orbitals when transition metals are ionized.
So, you have the following electronic configuration for V3+ : [Ar] 4s0 3d2
therefore, u got 2 unpaired electrons ...
 
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2 moles of metallic salt react with 1 mole of sodium sulfite, oxidising sulfite needs 2 electrons. So, the original oxidation number of metal is +3, and there are two moles, so the new oxidation will be +2.

RTP apparently


Magnesium is second in it's group, so it would have the second smallest radius (because of more electron shells) and it has the third lowest ionisation after sodium and aluminium.. IE increases across the period, but Aluminium has an outer p subshell so it's IE is lower ^^
Sorry but i didnt get the Ionisation part :/
 
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The electronic configuration of Vanadium is [Ar] 4s2 3d3 ...
In general, electrons are removed from the valence-shell s orbitals before they are removed from valence d orbitals when transition metals are ionized.
So, you have the following electronic configuration for V3+ : [Ar] 4s0 3d2
therefore, u got 2 unpaired electrons ...
Omg i seriously forgot this sis :p A2 topic :p
Thank you sooo much :D JazakAllah khairan :)
 
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Sorry but i didnt get the Ionisation part :/
IE naturally increases across the period because the nuclear charge is higher, so Mg wil lhavve more IE than Na. It will also have more IE than Al, because Al has an electron in the outer p shell which has more energy and is also farther than the nucleus. (Mg outer shell => 3s2, Al outer shell => 3s2 3p1)
 
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upload_2015-6-8_20-33-7-png.54840

Help! it's 1 and 2 only
How would each react and what will it give in each case? And what would 3 give since it's wrong??
 
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