Write a note on halide compounds of transition elements.

Oxidation Number	$\mathrm{Ti}$	$\mathrm{v}$	$\mathrm{cr}$	$\mathrm{Mn}$	$\mathrm{fe}$	$\mathrm{co}$	$\mathrm{ni}$	$\mathrm{cu}$	$\mathrm{zn}$
$+6$			$\mathrm{CrF}_{6}$
$+5$		$\mathrm{VF}_{5}$	$\mathrm{CrF}_{5}$
$+4$	$\mathrm{TiX}_{4}$	$\mathrm{VX}_{4}^{\mathrm{I}}$	$\mathrm{CrX}_{4}$	$\mathrm{MnF} 4$
$+3$	$\mathrm{TiX}_{3}$	$\mathrm{VX}_{3}$	$\mathrm{CrX}_{3}$	$\mathrm{MnF}_{3}$	$\mathrm{FeX}_{3}^{\mathrm{I}}$	$\mathrm{CoF}_{3}$
$+2$	$\mathrm{TiX}_{2}$	$\mathrm{VX}_{2}$	$\mathrm{CrX}_{2}$	$\mathrm{MnX}_{2}$	$\mathrm{FeX}_{2}$	$\mathrm{CoX}_{2}$	$\mathrm{NiX}_{2}$	$\mathrm{CuX}_{2}^{\mathrm{II}}$	$\mathrm{ZnX}_{2}$
$+1$								$\mathrm{CuX}^{\mathrm{III}}$

Key : $\mathrm{X}=\mathrm{F} \rightarrow \mathrm{I} ; \mathrm{X}^{\mathrm{I}}=\mathrm{F} \rightarrow \mathrm{Br} ; \mathrm{X}^{\mathrm{II}}=\mathrm{F}, \mathrm{CI} ; \mathrm{X}^{\mathrm{III}}=\mathrm{CI} \rightarrow \mathrm{I}$

The transition elements form ionic halides with fluorine and covalent halides with chlorine, bromine and iodine. As fluorine is most electro-negative, the elements exist in maximum oxidation state with fluorine.Ex. : $\mathrm{Tix}_{4},(\mathrm{X} \rightarrow \mathrm{I}), \mathrm{VF}_{5}, \mathrm{CrF}_{6}$ etc.

The fluorine stabilizes the high oxidation state because of either high lattice enthalpy as in case of $\mathrm{CoF}_{3}$ or high bond enthalpy of covalent bonds as in case of $\mathrm{VF}_{6}, \mathrm{VF}_{5}$ etc.

The Manganese is not known in (+7) oxidation state with fluorine i.e., $\mathrm{MnF}_{7}$ is not known. However, $\mathrm{MnO}_{3} \mathrm{~F}$ is known because oxygen stabilizes the system more effectively than fluorine due to its tendency to form a multiple bonds. Hence highest oxidation state of Mn is $(+4)$ with fluorine.

After manganese only trihalides of $\mathrm{Fe}$ and $\mathrm{Co}$ are known. Ex. : $\mathrm{FeCl}_{3}, \mathrm{FeBr}_{3}, \mathrm{CoF}_{3}, \mathrm{CoCl}_{3}$ etc.

The metal halides with a metal having high oxidation states easily hydrolyse to produce oxohalides. Ex. : VF $_{5}$ when hydrolysed produces VOF $_{3}$ and $\mathrm{HF} . \mathrm{VF}_{5}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{VOF}_{3}+2 \mathrm{HF}$

As a result, the solution of metal halides are acidic as they produce acid on hydrolysis.

All halides of copper, such as $\mathrm{CuF}_{2}, \mathrm{CuCl}_{2}$ and $\mathrm{CuBr}_{2}$ are known. However, CuI $_{2}$ is not known because $\mathrm{Cu}^{2+}$ is good oxidizing agent while $\mathrm{I}^{-}$is good reducing agent. So, they form $\mathrm{Cu}_{2} \mathrm{I}_{2}$ on reaction with each other.

$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2(\mathrm{~s})}+\mathrm{I}_{2}$

In aqueous solution, $\mathrm{Cu}(\mathrm{I})$ compounds are not known and undergoes disproportionation reaction to form $\mathrm{Cu}^{2+}$.

$2 \mathrm{Cu}_{\text {(aq) }}^{+} \rightarrow \mathrm{Cu}+\mathrm{Cu}_{\text {(aq) }}^{2+}$

$\mathrm{Cu}^{2+}$ is stable in aqueous medium than $\mathrm{Cu}^{+}$because of its high hydration enthalpy which is more than compensates for the second ionization enthalpy of copper. Amongst the halides of copper only $\mathrm{CuF}_{2}$ is ionic.

The covalent nature of metal halides increases with the increase in size of halogen atom (greater polarization) or with increases in oxidation state of metal atom (greater polarizing power).