Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
(N/A) Both oxide $(O^{2-})$ and fluoride $(F^-)$ ions are highly electronegative and have a very small size. Due to these properties,they are able to oxidize the metal to its highest oxidation state.
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Although $Zr$ belongs to the $4d$ and $Hf$ belongs to the $5d$ transition series, it is quite difficult to separate them. Why?
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View SolutionMatch List-$I$ with List-$II$:
List-$I$ ($\text{Transition metal ion}$) List-$II$ ($\text{Spin only magnetic moment (B.M.)}$) $A$. $Ti^{3+}$ $I$. $3.87$ $B$. $V^{2+}$ $II$. $0.00$ $C$. $Ni^{2+}$ $III$. $1.73$ $D$. $Sc^{3+}$ $IV$. $2.84$
Choose the correct answer from the options given below:
| List-$I$ ($\text{Transition metal ion}$) | List-$II$ ($\text{Spin only magnetic moment (B.M.)}$) |
|---|---|
| $A$. $Ti^{3+}$ | $I$. $3.87$ |
| $B$. $V^{2+}$ | $II$. $0.00$ |
| $C$. $Ni^{2+}$ | $III$. $1.73$ |
| $D$. $Sc^{3+}$ | $IV$. $2.84$ |
Choose the correct answer from the options given below:
Which of the following transition elements have nearly the same atomic radii?
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View SolutionIf $A$ is $[Rn]\,6d^2\,7s^2$,then identify the block,period,and group.
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