Due to covalent bonding, the transitional metals are
Due to covalent bonding, the transitional metals are
- A
Lustrous
- B
Conductor
- C
Hard and brittle
- D
Ductile
Similar Questions
In the following members of transition elements, the lowest ionization energy is of
In the following members of transition elements, the lowest ionization energy is of
Which occludes hydrogen
Which occludes hydrogen
Consider the following reaction
$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O} .$
Product ' $A$ ' in neutral or acidic medium disproportionate to give products ' $\mathrm{B}$ ' and ' $\mathrm{C}$ ' along with water. The sum of spin-only magnetic moment values of $B$ and $C$ is . . . . . .$BM$. (nearest integer)
(Given atomic number of $\mathrm{Mn}$ is $25$ )
Consider the following reaction
$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O} .$
Product ' $A$ ' in neutral or acidic medium disproportionate to give products ' $\mathrm{B}$ ' and ' $\mathrm{C}$ ' along with water. The sum of spin-only magnetic moment values of $B$ and $C$ is . . . . . .$BM$. (nearest integer)
(Given atomic number of $\mathrm{Mn}$ is $25$ )
- [JEE MAIN 2024]
Although $Zr$ belongs to $4d$ and $Hf$ belongs to $5d$ transition series but it is quite difficult to separate them. Why ?
Although $Zr$ belongs to $4d$ and $Hf$ belongs to $5d$ transition series but it is quite difficult to separate them. Why ?
In the first transition series, the highest $B.P. $ and $M.P.$ is of
In the first transition series, the highest $B.P. $ and $M.P.$ is of