The elements of 3d transition series are gives as:

Sc Ti V Cr Mn Fe Co Ni Cu Zn

Answer the following:

A. Write the element which shows maximum number of oxidation states. Give reason.

B. Which element has the highest m.p?

C. Which element shows only +3 oxidation state?

D. Which element is a strong oxidizing agent in +3 oxidation state and why?

a) Manganese shows maximum number of oxidation states.

Theatomic number of manganese (Mn) is Z = 25.

The electronic configuration of ₂₅Mn= [Ar] 4s² 3d⁵

As all the 7 electrons of 3d and 4s can participate in bonding, hence Mn maximum number of variable oxidation states.

b) Cr has highest melting point due to the following reasons:

⇒ The strength of metallic bond depends upon the number of unpaired d-electrons (half-filled d-orbitals).

⇒ Greater is the number of unpaired electrons (half-filled orbitals), stronger is the metallic bonding.

⇒ Because of the stronger metallic bonding, the element has high melting point.

For chromium

Theatomic number of chromium (Cr) is Z = 24.

The electronic configuration of ₂₄Cr= [Ar] 3d⁵ 4s¹

As there are maximum number of unpaired electrons, hence Cr has highest melting point.

c) Scandium shows only +3 oxidation state.

Theatomic number of Scandium (Sc) is Z = 21.

The electronic configuration of ₂₁Sc= [Ar] 3d¹ 4s²

d) Manganese is a strong oxidizing agent in +3 oxidation state due to the following reasons:

⇒ Mn³⁺ has 3d⁴ configuration.

⇒ It can gain electron to form Mn²⁺ which stable 3d⁵ configuration (as it is exactly half-filled)

⇒ Hence, Mn²⁺ is a stronger oxidizing agent.

Note: Oxidizing agent is an element which gains(accepts) electrons.