Question
Write a note on halide compounds of Transition elements.
✓
Answer
→ The Transition elements form Ionic halides with fluorine and co-valent halides with chlorine, bromine and iodine.
→ The highest oxidation numbers are achieved in TiX 4 (tetrahalides), VF5 and CrF6.
→ Mn is not known in +7 oxidation state with fluorine i.e. MnF, is not known. But MnO3F is known because oxygen stabilizes the compound due to its tendency to form a multiple bonds.
→ The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6
→ 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
$
→ The solution of metal halides are acidic as they produce acid on hydrolysis.
→ 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
$
→ However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.
$
2 Cu ^{+} \rightarrow Cu ^{2+}+ Cu
$
→ 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
$K e y: X=F \rightarrow I ; X^1=F \rightarrow B r ; X^{11}=F, C l ; X^{111}=C l \rightarrow I$
→ The highest oxidation numbers are achieved in TiX 4 (tetrahalides), VF5 and CrF6.
→ Mn is not known in +7 oxidation state with fluorine i.e. MnF, is not known. But MnO3F is known because oxygen stabilizes the compound due to its tendency to form a multiple bonds.
→ The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6
→ 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
$
→ The solution of metal halides are acidic as they produce acid on hydrolysis.
→ 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
$
→ However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.
$
2 Cu ^{+} \rightarrow Cu ^{2+}+ Cu
$
→ 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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