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Which of the followig ions will show hig...

Which of the followig ions will show highest magnetic moment `(Z` values for nautral atoms are as follows: `N = 7, Cr = 24, Fe = 26 & Co = 27)`

A

`Fe^(3+)`

B

`Cr^(3+)`

C

`N^(3+)`

D

`Co^(3+)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given ions (N, Cr, Fe, Co) will show the highest magnetic moment, we need to calculate the number of unpaired electrons for each ion. The magnetic moment can be calculated using the formula: \[ \text{Magnetic Moment} = \sqrt{n(n + 2)} \] where \( n \) is the number of unpaired electrons. The greater the number of unpaired electrons, the higher the magnetic moment. ### Step-by-Step Solution: 1. **Determine the electronic configuration of each ion:** - **Fe³⁺ (Iron):** - Atomic number of Fe = 26 - Electronic configuration of neutral Fe = [Ar] 3d⁶ 4s² - For Fe³⁺, we remove 3 electrons (2 from 4s and 1 from 3d): - Configuration of Fe³⁺ = [Ar] 3d⁵ - Unpaired electrons = 5 - **Cr³⁺ (Chromium):** - Atomic number of Cr = 24 - Electronic configuration of neutral Cr = [Ar] 3d⁵ 4s¹ - For Cr³⁺, we remove 3 electrons (1 from 4s and 2 from 3d): - Configuration of Cr³⁺ = [Ar] 3d³ - Unpaired electrons = 3 - **N³⁻ (Nitrogen):** - Atomic number of N = 7 - Electronic configuration of neutral N = 1s² 2s² 2p³ - For N³⁻, we add 3 electrons (to fill the 2p orbital): - Configuration of N³⁻ = 1s² 2s² 2p⁶ - Unpaired electrons = 0 - **Co³⁺ (Cobalt):** - Atomic number of Co = 27 - Electronic configuration of neutral Co = [Ar] 3d⁷ 4s² - For Co³⁺, we remove 3 electrons (2 from 4s and 1 from 3d): - Configuration of Co³⁺ = [Ar] 3d⁶ - Unpaired electrons = 4 2. **Count the unpaired electrons:** - Fe³⁺: 5 unpaired electrons - Cr³⁺: 3 unpaired electrons - N³⁻: 0 unpaired electrons - Co³⁺: 4 unpaired electrons 3. **Identify the ion with the highest number of unpaired electrons:** - Fe³⁺ has the highest number of unpaired electrons (5). 4. **Calculate the magnetic moment for Fe³⁺:** \[ \text{Magnetic Moment} = \sqrt{n(n + 2)} = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \] ### Conclusion: The ion that shows the highest magnetic moment is **Fe³⁺** with 5 unpaired electrons.
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