Answer the following questions
(i) Which element of the first transition series has highest second ionisation enthalpy?
(ii) Which element of the first transition series has highest third ionisation enthalpy?
(iii) Which element of the first transition series has lowest enthalpy of atomisation?
(b) Identify the metal and justify your answer.
(i) Carbonyl `M(CO)_(5)`
(ii) `MO_(3)F`

Explanation

(i) Cu, because the electronic configuration of Cu is `3d^(10)4s^(1)`. So, second electron needs to be removed from completely filled d-orbital which is very difficult.
(ii) Zinc, because of electronic configuration of Zn = `3d^(10)4s^(2)` and `Zn^(2+) = 3d^(10)` which is fully filled and hence is very stable. Removal of third electron requires very high energy.
(iii) Zinc, because of it has completely filled 3d subshell and no unpaired electron is available for metallic bonding.
(b) (i) Carbonyl `M(CO)_(5)` is `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 of metal +2 `xx` (CO)
= atomic number of nearest inert gas
In `M(CO)_(5)` = x + `2 xx (5)` = 36 (Kr is the nearest inert gas)
x = 26 (atomic number of metal)
So, the metal is Fe(iron).
(ii) `MO_(3)F` is `MnO_(3)F`.
In `MO_(3)F`
Let us assume that oxidation state of M is x
`x + 3 xx (-2) + (-1) = 0`
or, x = + 7 i.e., M is in +7 oxidation state of +7. Hence, the given compound is `MnO_(3)F`.