The set having ions which are coloured and paramagnetic both is :
- A$Sc^{3+}, V^{5+}, Ti^{4+}$
- B$Cu^{2+}, Cr^{3+}, Sc^{+}$
- C$Ni^{2+}, Mn^{7+}, Hg^{2+}$
- D$Cu^{+}, Zn^{2+}, Mn^{4+}$
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Similar Questions
Statement-$I$: Mercury is liquid at room temperature.
Statement-$II$: In mercury,there is no unpaired $d$-electron and thus,metallic bonding is weakest.
Statement-$II$: In mercury,there is no unpaired $d$-electron and thus,metallic bonding is weakest.
Medium
View SolutionMatch the following:
List-$I$ (Transition metal,$M$) List-$II$ ($E_{M^{2+}/ M}^{\ominus}$ in $V$) $A$. $Ni$ $I$. $-1.18$ $B$. $Mn$ $II$. $-0.91$ $C$. $Fe$ $III$. $-0.25$ $D$. $Cr$ $IV$. $-0.44$
The correct answer is
| List-$I$ (Transition metal,$M$) | List-$II$ ($E_{M^{2+}/ M}^{\ominus}$ in $V$) |
| $A$. $Ni$ | $I$. $-1.18$ |
| $B$. $Mn$ | $II$. $-0.91$ |
| $C$. $Fe$ | $III$. $-0.25$ |
| $D$. $Cr$ | $IV$. $-0.44$ |
The correct answer is
Match the species in List-$I$ with their correct electronic configuration in List-$II$.
List-$I$ List-$II$ $(1) \ \text{Cr}$ $(A) \ [\text{Ar}] \ 3d^8 4s^0$ $(2) \ \text{Fe}^{2+}$ $(B) \ [\text{Ar}] \ 3d^{10} 4s^1$ $(3) \ \text{Ni}^{2+}$ $(C) \ [\text{Ar}] \ 3d^6 4s^0$ $(4) \ \text{Cu}$ $(D) \ [\text{Ar}] \ 3d^5 4s^1$ $(E) \ [\text{Ar}] \ 3d^6 4s^2$
| List-$I$ | List-$II$ |
| $(1) \ \text{Cr}$ | $(A) \ [\text{Ar}] \ 3d^8 4s^0$ |
| $(2) \ \text{Fe}^{2+}$ | $(B) \ [\text{Ar}] \ 3d^{10} 4s^1$ |
| $(3) \ \text{Ni}^{2+}$ | $(C) \ [\text{Ar}] \ 3d^6 4s^0$ |
| $(4) \ \text{Cu}$ | $(D) \ [\text{Ar}] \ 3d^5 4s^1$ |
| $(E) \ [\text{Ar}] \ 3d^6 4s^2$ |
Medium
View SolutionThe atomic number of $Pd$ is $46$. The atomic numbers of the elements lying immediately above and below it in the same group are,respectively,.....................
Medium
View SolutionWhich of the following represents the decreasing order of the second ionization enthalpy $(I.P_2)$?
Easy
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