Question

Which one of following has least magnetic moment

• A. $Cu^{2+}$
• B. $Ni^{2+}$
• C. $Co^{2-}$
• D. $Fe^{2}$

Answer 1

Answer: (A)

A

Answer 2

To determine which ion has the least magnetic moment, we need to find the number of unpaired electrons (n) for each ion. The magnetic moment ($\mu$) is given by the spin-only formula:

$\mu = \sqrt{n(n+2)}$ BM (Bohr Magnetons)

A smaller number of unpaired electrons (n) results in a smaller magnetic moment.

Let's analyze each option, assuming the common stable oxidation states for transition metals and interpreting the given options as such (e.g., $Co^{2-}$ as $Co^{2+}$ and $Fe^{2}$ as $Fe^{2+}$).

1. $Cu^{2+}$

   • Atomic number of Cu = 29
   • Electronic configuration of Cu: $[Ar] 3d^{10} 4s^1$
   • Electronic configuration of $Cu^{2+}$: $[Ar] 3d^9$ (one electron removed from 4s, one from 3d)
   • For $3d^9$, there is 1 unpaired electron. d-orbitals: $\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow$
   • Number of unpaired electrons (n) = 1
   • Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM

2. $Ni^{2+}$

   • Atomic number of Ni = 28
   • Electronic configuration of Ni: $[Ar] 3d^8 4s^2$
   • Electronic configuration of $Ni^{2+}$: $[Ar] 3d^8$ (two electrons removed from 4s)
   • For $3d^8$, there are 2 unpaired electrons. d-orbitals: $\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow$
   • Number of unpaired electrons (n) = 2
   • Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ BM

3. $Co^{2+}$ (assuming $Co^{2-}$ is a typo for $Co^{2+}$)

   • Atomic number of Co = 27
   • Electronic configuration of Co: $[Ar] 3d^7 4s^2$
   • Electronic configuration of $Co^{2+}$: $[Ar] 3d^7$ (two electrons removed from 4s)
   • For $3d^7$, there are 3 unpaired electrons (in high spin configuration, which is typical for free ions). d-orbitals: $\uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \uparrow$
   • Number of unpaired electrons (n) = 3
   • Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM

4. $Fe^{2+}$ (assuming $Fe^{2}$ is a typo for $Fe^{2+}$)

   • Atomic number of Fe = 26
   • Electronic configuration of Fe: $[Ar] 3d^6 4s^2$
   • Electronic configuration of $Fe^{2+}$: $[Ar] 3d^6$ (two electrons removed from 4s)
   • For $3d^6$, there are 4 unpaired electrons (in high spin configuration). d-orbitals: $\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow$
   • Number of unpaired electrons (n) = 4
   • Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM

Comparing the number of unpaired electrons:

• $Cu^{2+}$: n = 1
• $Ni^{2+}$: n = 2
• $Co^{2+}$: n = 3
• $Fe^{2+}$: n = 4

The ion with the least number of unpaired electrons is $Cu^{2+}$ (n=1). Therefore, $Cu^{2+}$ has the least magnetic moment.