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The number of paramagnetic ions among Ti...

The number of paramagnetic ions among `Ti^(3+), Cr^(3+), TI^(4+), Fe^(2+), Co^(2+), Zn^(2+), Cu^+, V^(3+), Mn^(2+)` is

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To determine the number of paramagnetic ions among the given ions \( Ti^{3+}, Cr^{3+}, Ti^{4+}, Fe^{2+}, Co^{2+}, Zn^{2+}, Cu^{+}, V^{3+}, Mn^{2+} \), we need to analyze the electronic configurations of each ion and check for unpaired electrons. Paramagnetic substances have unpaired electrons, while diamagnetic substances have all paired electrons. ### Step-by-step Solution: 1. **Determine the electronic configurations of the neutral atoms:** - \( Ti: [Ar] 3d^2 4s^2 \) - \( Cr: [Ar] 3d^5 4s^1 \) - \( Fe: [Ar] 3d^6 4s^2 \) - \( Co: [Ar] 3d^7 4s^2 \) - \( Zn: [Ar] 3d^{10} 4s^2 \) - \( Cu: [Ar] 3d^{10} 4s^1 \) - \( V: [Ar] 3d^3 4s^2 \) - \( Mn: [Ar] 3d^5 4s^2 \) 2. **Calculate the electronic configurations for each ion:** - **\( Ti^{3+} \)**: Remove 3 electrons (2 from 4s and 1 from 3d) - Configuration: \( [Ar] 3d^1 \) (1 unpaired electron) - **\( Cr^{3+} \)**: Remove 3 electrons (1 from 4s and 2 from 3d) - Configuration: \( [Ar] 3d^3 \) (3 unpaired electrons) - **\( Ti^{4+} \)**: Remove 4 electrons (2 from 4s and 2 from 3d) - Configuration: \( [Ar] \) (no unpaired electrons) - **\( Fe^{2+} \)**: Remove 2 electrons (2 from 4s) - Configuration: \( [Ar] 3d^6 \) (4 unpaired electrons) - **\( Co^{2+} \)**: Remove 2 electrons (2 from 4s) - Configuration: \( [Ar] 3d^7 \) (3 unpaired electrons) - **\( Zn^{2+} \)**: Remove 2 electrons (2 from 4s) - Configuration: \( [Ar] 3d^{10} \) (no unpaired electrons) - **\( Cu^{+} \)**: Remove 1 electron (1 from 4s) - Configuration: \( [Ar] 3d^{10} \) (no unpaired electrons) - **\( V^{3+} \)**: Remove 3 electrons (2 from 4s and 1 from 3d) - Configuration: \( [Ar] 3d^2 \) (2 unpaired electrons) - **\( Mn^{2+} \)**: Remove 2 electrons (2 from 4s) - Configuration: \( [Ar] 3d^5 \) (5 unpaired electrons) 3. **Count the number of paramagnetic ions:** - \( Ti^{3+} \): 1 unpaired electron (paramagnetic) - \( Cr^{3+} \): 3 unpaired electrons (paramagnetic) - \( Ti^{4+} \): 0 unpaired electrons (diamagnetic) - \( Fe^{2+} \): 4 unpaired electrons (paramagnetic) - \( Co^{2+} \): 3 unpaired electrons (paramagnetic) - \( Zn^{2+} \): 0 unpaired electrons (diamagnetic) - \( Cu^{+} \): 0 unpaired electrons (diamagnetic) - \( V^{3+} \): 2 unpaired electrons (paramagnetic) - \( Mn^{2+} \): 5 unpaired electrons (paramagnetic) ### Conclusion: The total number of paramagnetic ions is **6**: \( Ti^{3+}, Cr^{3+}, Fe^{2+}, Co^{2+}, V^{3+}, Mn^{2+} \).
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