(i) Answer the following questions
(a) Which element of the first transition series has highest second ionisation enthalpy?
(b) Which element of the first transition series has highest third ionisation enthalpy?
(c) Which element of the first transition series has lowest enthalpy of atomisation?
(ii) Identify the metal and justify your answer. Carbonyl M(CO)₅
(iii) Although F is more electronegative than O, the highest fluoride of Mn is MnF₄, whereas the highest oxide is Mn₂O₇. Give reason.

(i) (a) Cu.
The electronic configuration of Cu is 3d¹⁰ 4s¹.
Second electron is removed from completely filled d-orbital which is very difficult.
(b) Zinc.
Electronic configuration of Zn is 3d¹⁰ 4s²
3^rd electron is removed from fully filled orbital which is very stable.
(c) Zinc.
It has completely filled 3d subshell and no unpaired electron is available for metallic bonding.
(ii) Fe(CO)₅ .
According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic number of nearest inert gas
EAN is calculated as;
EAN = number of electrons in metal + 2× (CO)
= atomic number of nearest inert gas
In M(CO)₅ =x+2× (5 )=36 (Kr is the nearest inert gas)
x = 26 (atomic number of metal)
So, the metal is Fe(iron).
(iii) Oxygen can form multiple bonds while fluorine can only form single bonds with metals.