Since,1st transition series belongs to 3d orbital and 2nd transition series belongs to 4d-orbital. According to the given option, atomic number 39 to 48 (5s²4d^1-10) belongs to second transition series.

Mn²⁺/ Mn have - 1.18SRPvalue.

Ford-block elements, group number is equal to the number of electrons in (n - 1)d subshell + number of electrons in valence shell (nth shell).

[Kr]4d¹⁰5s⁰

Sc (3d¹ 4s²)has 631 kJ mol^-1 and Zn (3d¹⁰ 4s²) has 906 kJ mol^-1 

The element lying above Z = 42 has Z value 24, i.e. chromium. Hence, Cr configuration is [Ar]3d⁵4s¹

The correct answer is option C

Ce exist in the oxidation state other than +3 .

(a) B - Boron has +1 and +3 oxidation state.

(b) Al - Aluminium has +1,+2 and +3 oxidation state.

(c) Ce - Cerium has +3 and +4 oxidation state.

(d) Ga - Gallium has +1 and +3 oxidation state.

Cerium (Ce) have [Xe] 4f²6s² electronic configuration.

Fe (Z = 26) in + 3 oxidation state show Fe³⁺ ([Ar] 3d⁵) configuration. Ni(28) in +2 oxidation state show Ni²⁺([Ar]3d⁸) configuration.

Due to lanthanide contraction, Ag and Au have nearly same size.

(i) → (s), (ii) → (r), (iii) → (q), (iv) → (p)

The correct answer is option C
In Electrolytic refining, the impure metal is made to act as an anode. A strip of the same metal in pure form is used as a cathode. 
They are put in a suitable electrolytic bath containing a soluble salt of the same metal. 
The more basic metal remains in the solution and the less basic ones go to the anode mud.
Copper is refined using an electrolytic method.
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold, and platinum; recovery of these elements may meet the cost of refining.

Osmium has the electronic configuration 5de⁶s². As 5d and 6s are close in energy, all the 8 electron can participate in bonding. highest oxidation state of osmium is +8 due to its ability to expand their octet by using its all 8 electrons (2 from 6s and 6 from 5d).