Match the properties given in Column $I$ with the metals given in Column $II$.

	Column I (Property)		Column II (Metal)
(A)	Element with highest second ionisation enthalpy	(i)	$Co$
(B)	Element with highest third ionisation enthalpy	(ii)	$Cr$
(C)	$M$ in $M(CO{)}_6$ is	(iii)	$Cu$
(D)	Element with highest heat of atomisation	(iv)	$Zn$
		(v)	$Ni$

Part (A)

$C{u}^{+}$ ion has a $3d^{10}$ electronic configuration. It has a fully filled stable $d^{10}$ configuration, thus it requires high energy to remove an electron from the fully filled $d$ orbital. Therefore, $Cu$ has the highest second ionisation enthalpy.

$\left(A\right)\to \left(iii\right)$

Part (B)

$Z{n}^{2+}$ has a $3d^{10}$ electronic configuration. $Z{n}^{2+}$ ion is highly stable due to the fully filled $d^{10}$ configuration and thus it requires high energy to remove an electron from the fully filled d orbital. Therefore, $Zn$ has the highest third ionization energy.

$\left(B\right)\to \left(iv\right)$

Part (C)

The $M$ in $M(CO{)}_6$is Chromium. A stable complex is formed when its $EAN$ (Effective atomic number) is equal to the atomic number of the nearest noble gas. The nearest noble gas is $Kr$ having atomic number $36$.

$EAN=$ Total electrons of metal + number of electrons given by ligand.

Each $CO$ ligand contributes two electrons.

$36=x+2\left(6\right)$

$\left(x\right)=36-12=24$

$24$ is the atomic number of chromium.

$\left(C\right)\to \left(ii\right)$

Part (D)

Heat of atomisation is defined as the energy required to break metal lattice into isolated gaseous atoms.

As the number of unpaired electrons increases, the metallic bond strength increases and thus atoms come closer and are held more tightly.

Thus, a metal with more number of unpaired electrons and higher effective nuclear charge will have higher enthalpy of atomisation.

Hence, among the given options $Ni$ will have highest enthalpy of atomisation.

Electronic configuration of $Ni:\left[Ar\right]3d^{8}4s^2$ It has $2$ unpaired electrons.

$\left(D\right)\to \left(v\right)$

$\left(A\right)\to \left(iii\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(ii\right),$
$\left(D\right)\to \left(v\right)$.