JEE MAIN - Chemistry (2025 - 24th January Evening Shift - No. 2)
Match List - I with List - II.
| List - I (Transition metal ion) |
List - II (Spin only magnetic moment (B.M.)) |
||
|---|---|---|---|
| (A) | $\mathrm{Ti}^{3+}$ | (I) | 3.87 |
| (B) | $\mathrm{V}^{2+}$ | (II) | 0.00 |
| (C) | $\mathrm{Ni}^{2+}$ | (III) | 1.73 |
| (D) | $\mathrm{Sc}^{3+}$ | (IV) | 2.84 |
Choose the correct answer from the options given below :
$\mathrm{(A) -(III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})$
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{III})$
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{IV})$
$(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{I})$
Explanation
$$\begin{array}{ll} \mathrm{Sc}^{+3}=3 \mathrm{~d}^{\circ} & \therefore \mu_{\text {spin }}=0 \\ \mathrm{~V}^{+2}=3 \mathrm{~d}^3 & \therefore \mu_{\text {spin }}=3.87 \text { B.M. } \\ \mathrm{Ni}^{+2}=3 \mathrm{~d}^8 & \therefore \mu_{\text {spin }}=2.84 \text { B.M. } \end{array}$$
$\mathrm{Ti}^{+3}=3 \mathrm{~d}^1 \quad \therefore \mu_{\mathrm{spin}}=1.73 \mathrm{~B} . \mathrm{M}$
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