Q.5 Which one of fluoride of
Vanadium (Oxidation State) exist
in stable form *
V(III)
V(IV)
V(V)
V(VI)
Answers
Explanation:
CHEMISTRY
(A) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows oxidation state of +4.
(ii) Cr
2+
is a strong reducing agent.
(iii) Cu
2+
salts are coloured, while Zn
2+
salts are white.
(B) Complete the following equations:
(i) 2MnO
2
+4KOH+O
2
→
Δ
(ii) Cr
2
O
7
2−
+14H
+
+6e
−
→
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VIDEO EXPLANATION
ANSWER
(A)-(i)
Element Name and Symbol Atomic Number Common Oxidation States
Scandium (se) 21 +3
Titanium (Ti) 22 +4
Vanadium (V) 23 +2, +3, +4, +5
Chromium (Cr) 24 +2, +3, +6
Manganese (Mn) 25 +2, +3, +4, +6, +7
Iron (Fe) 26 +2, +3
Cobalt (Co) 27 +2, +3
So Manganese (Mn) shows highest number of oxidation state.
Element Sc Ti V Cr Mn Fe Co
M.P. (
o
C) 1540 1668 1890 1875 1245 1537 1495
(ii) Scandium shows only +3 oxidation state.
Zn has electric cofiguration =3d
10
4s
2
Zn
2+
=3d
10
Cu=3d
5
4s
1
Cu
2+
=3d
6
In case of Zn fully filled d orbital is present therefore no d−d transition can be possible in this case it is colorless.
In case of copper because of d−d transition electrons emits light in the visible range and hence they are colored compounds.
(iii) Cr
2+
gets converted to Cr
3+
as the +3 oxidation state has half filled t
2g
orbitals - thus it is a good reducing agent.
(B)
(i) 2MnO
2
+4KOH+O
2
→2K
2
MnO
4
+2H
2
O
(ii) Cr
2
O7
2−
+14H
+
+6e
−
→2Cr
3+
+7H
2
O