Question 6 [Inorganic chemistry> group7, group 2]
Two elements, V and W, are in adjacent groups
in the Periodic Table.
Reference: Past Exam Paper – 9701 March 2016 Paper 42 Q6
Solution:
(a)
Explanations:
-
V’s oxide
and W’s oxide is an acidic gas => from right hand side of the periodic
table.
-
V forms an
anion in the form of VOm-
single negative charge on anion is only observed in Group15 and
Group 17. Hence Nitrogen and Chlorine are acceptable answers.
-
X are
common oxides that you need to pen down from memory. Nitrogen is undoubtedly
the easier option.
-
For m take
common oxidation states to calculate. For example, N is commonly +5, doing the
math to get overall charge as -1 gives m to be 3 [ 1(+5)+m(-2)= -1]
-
W forms
anion with net charge +2 => that element iss from Group 14 or Group 16.
-
Anion
forms precipitate with Ba2+, from memory we know that Barium
Sulphate is insoluble/ ppt. Therefore W must be Sulphur.
-
Y can be
either of the very common sulphur di and trioxide gas.
-
Anion can
be SO42- or SO32- thus, n can be 4
or 3
(b) (white
precipitate is BaSO4) descending the group ∆Hsol becomes more endothermic /
positive;
any two from:
∆Hlatt
decreases /becomes more endothermic / becomes less exothermic
∆Hhyd decreases / becomes more endothermic /
becomes less exothermic
∆Hhyd decreases
more than ∆Hlatt
Solutions provided by Kashish Varshney, India
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