Question
Write a note on halide compounds of Transition elements.
✓
Answer
$\rightarrow$ The Transition elements form Ionic halides with fluorine and co$-$valent halides with chlorine, bromine and iodine.
$\rightarrow$ The highest oxidation numbers are achieved in $TiX _{4} ($tetrahalides$), VF_5$ and $CrF_{6.}$
$\rightarrow$ Mn is not known in $+7$ oxidation state with fluorine i.e. $MnF,$ is not known. But $MnO_3F$ is known because oxygen stabilizes the compound due to its tendency to form a multiple bonds.
$\rightarrow$ The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of $CoF_3,$ or higher bond enthalpy terms for the higher covalent compounds, e.g., $VF_5$ and $CrF_6$
$\rightarrow$ Although $V ^{ V }$ is represented only by $VF _5$, the other halides, however, undergo hydrolysis to give oxohalides, $VOX _3$. Another feature of fluorides
$VF _5+ H _2 O $\rightarrow$ VOF _3+2 HF$
$\rightarrow$ The solution of metal halides are acidic as they produce acid on hydrolysis.
$\rightarrow$ All halides of copper such as $CuF _2, CuCl _2$ and $CuBr _2$ are known. However $CuI _2$ is not know because $Cu ^{2+}$ is good oxidising agent while $1$ is good reducing agent. So they form $Cu _2 I _2$ on reaction with each other.
$2 Cu ^{2+}+4 I ^{-} \rightarrow Cu _2 I _{2(s)}+ I _2$
$\rightarrow$ However, many copper $(I)$ compounds are unstable in aqueous solution and undergo disproportionation.
$2 Cu ^{+} \rightarrow Cu ^{2+}+ Cu$
$\rightarrow$ The stability of $Cu ^{2+} (aq)$ rather than $Cu ^{+}(aq)$ is due to the much more negative $\Delta_{\text {hyd }} H ^{\ominus}$ of $Cu ^{2+} (aq)$ than $Cu ^{+}$, which more than compensates for the second ionization enthalpy of $Cu .$
Formulas of Halides of $3d$ Metals
$K e y: X=F \rightarrow I ; X^1=F \rightarrow B r ; X^{11}=F, C l ; X^{111}=C l \rightarrow I $
$\rightarrow$ The highest oxidation numbers are achieved in $TiX _{4} ($tetrahalides$), VF_5$ and $CrF_{6.}$
$\rightarrow$ Mn is not known in $+7$ oxidation state with fluorine i.e. $MnF,$ is not known. But $MnO_3F$ is known because oxygen stabilizes the compound due to its tendency to form a multiple bonds.
$\rightarrow$ The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of $CoF_3,$ or higher bond enthalpy terms for the higher covalent compounds, e.g., $VF_5$ and $CrF_6$
$\rightarrow$ Although $V ^{ V }$ is represented only by $VF _5$, the other halides, however, undergo hydrolysis to give oxohalides, $VOX _3$. Another feature of fluorides
$VF _5+ H _2 O $\rightarrow$ VOF _3+2 HF$
$\rightarrow$ The solution of metal halides are acidic as they produce acid on hydrolysis.
$\rightarrow$ All halides of copper such as $CuF _2, CuCl _2$ and $CuBr _2$ are known. However $CuI _2$ is not know because $Cu ^{2+}$ is good oxidising agent while $1$ is good reducing agent. So they form $Cu _2 I _2$ on reaction with each other.
$2 Cu ^{2+}+4 I ^{-} \rightarrow Cu _2 I _{2(s)}+ I _2$
$\rightarrow$ However, many copper $(I)$ compounds are unstable in aqueous solution and undergo disproportionation.
$2 Cu ^{+} \rightarrow Cu ^{2+}+ Cu$
$\rightarrow$ The stability of $Cu ^{2+} (aq)$ rather than $Cu ^{+}(aq)$ is due to the much more negative $\Delta_{\text {hyd }} H ^{\ominus}$ of $Cu ^{2+} (aq)$ than $Cu ^{+}$, which more than compensates for the second ionization enthalpy of $Cu .$
Formulas of Halides of $3d$ Metals
| Oxidation Number | |||||||||
| $+6$ | $CrF_6$ | ||||||||
| $+5$ | $VF_5$ | $CrF_5$ | |||||||
| $+4$ | $TiX_4$ | $VX_4^1$ | $CrX_4$ | $MnF_4$ | |||||
| $+3$ | $TiX_3$ | $VX _3$ | $CrX_3$ | $MnF_3$ | $FeX_3^1$ | $CoF_3$ | $CuX _2{ }^{11}$ | $ZnX _2$ | |
| $+2$ | $TiX _2{ }^{ III }$ | $VX _2$ | $CrX_2$ | $MnF_2$ | $FeX _2$ | $CoX _2$ | $NiX _2$ | $CuX ^{111}$ | |
| $+1$ |
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