Give reasons for the following: (a) Transition metals show variable oxidation states. (b) {\E}^{\circ} value for ( {\Zn}^{2+} \/ {\Zn}) is negative while that of ( {\Cu}^{2+} \/ {\Cu}) is positive. (c) Higher oxidation state of {\Mn} with fluorine is +4 whereas with oxygen is +7 .

CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

Question : 16 of 27

Marks: +1, -0

Give reasons for the following:
(a) Transition metals show variable oxidation states.
(b)

{\E}^{∘}

value for

( {\Zn}^{2+} \/ {\Zn})

is negative while that of

( {\Cu}^{2+} \/ {\Cu})

is positive.
(c) Higher oxidation state of

{\Mn}

with fluorine is +4 whereas with oxygen is +7 .

Solution:

(a) Because of availability of partially filled orbitals and comparable energies of

{\ns}

and

( {\n}-1)

d-orbitals.
(b)

{\E}^{∘}

value for

( {\Zn}^{2+} \/ {\Zn})

is negative due to stable completely filled

d^{10}

configuration in

{\Zn}^{2+}

. The positive value of

( {\Cu}^{2+} \/ {\Cu})

accounts for its ability to liberate

{\H}_2

from acids due to its high enthalpy of atomization and low hydration energy.
(c) Mn can form multiple bonds with oxygen by using

2 p

-orbital of oxygen and

3 d

-orbital of

{\Mn}

because of which it shows highest oxidation state of +7 with fluorine, Mn cannot form multiple bonds thus shows an oxidation state of +4 .

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