Q. 25

(a) Account for t

Answer :

(a)

(i) Mn has +7 oxidation state in Mn₂O₇ and +2 in MnO. Transition metals are basic in lowest oxide and acts as acidic in highest oxide. Thus, MnO is basic whereas Mn2O7 is acidic in nature.

(ii) Cu has electronic configuration 3d104s1. On losing two electrons it forms Cu2+ and its electronic configuration becomes 3d9 which is partially filled. It means Cu has partially filled d-orbitals in its stable oxidation state and thus, Cu is a transition element.

(iii) Actinoids show wide range of oxidation states due to comparable energies of 5f, 6d, 7s orbitals. Moreover, 5f orbitals are not as buried as 4f and hence, 5f electrons can participate in bonding.


(b)


OR


(a)


(i) Cr has highest melting point because it has half-filled orbitals (3d54s1). More the number of unpaired electrons, stronger is the metallic bonding because the strength of metallic bonding depends on the no. of unpaired d-electrons. Stronger the metallic bonding, higher will be the boiling point.


(ii) Manganese is a strong oxidizing agent in +3 oxidation state as Mn3+ has 3d4 configuration, it will oxidize others and itself will get reduced to Mn2+ which has stable 3d5 configuration because it is exactly half-filled.


(iii) Zn is a soft element because it has d10 configuration that’s why it does not form metallic bonds and is soft.


(b) The equations involved in the preparation of Potassium dichromate from Sodium chromate (Na2CrO4) is


2 Na2CrO4 + 2H+ Na2Cr2O7 + H2O + 2Na+


Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl

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