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).
Nickel steel contain $\%$ of $ Ni$
Metal used for making joints in jewellery is
The spin-only magnetic moment value of the compound with strongest oxidizing ability among $MnF _{4}, MnF _{3}$ and $MnF _{2}$ is $.............$ $B.M.$ [nearest integer]
Which of the following statement(s) is(are) correct with reference to the ferrous and ferric ions
An oxide of copper which is red in colour has the formula