To determine which pairs of ions have the same spin-only magnetic moment values, let's calculate the spin-only magnetic moment for each ion using the formula:
\(\mu = \sqrt{n(n+2)}\)
where \(n\) is the number of unpaired electrons.
- For Zn2+:
- Electronic configuration of Zn: [Ar] 3d10 4s2
- Zn2+: [Ar] 3d10
- Unpaired electrons: 0
- Spin-only magnetic moment, \(\mu = 0\)
- For Ti2+:
- Electronic configuration of Ti: [Ar] 3d2 4s2
- Ti2+: [Ar] 3d2
- Unpaired electrons: 2
- Spin-only magnetic moment, \(\mu = \sqrt{2(2+2)} = 2.83 \, \text{BM}\)
- For Cr2+:
- Electronic configuration of Cr: [Ar] 3d5 4s1
- Cr2+: [Ar] 3d4
- Unpaired electrons: 4
- Spin-only magnetic moment, \(\mu = \sqrt{4(4+2)} = 4.90 \, \text{BM}\)
- For Fe2+:
- Electronic configuration of Fe: [Ar] 3d6 4s2
- Fe2+: [Ar] 3d6
- Unpaired electrons: 4
- Spin-only magnetic moment, \(\mu = \sqrt{4(4+2)} = 4.90 \, \text{BM}\)
- For Ti3+:
- Electronic configuration of Ti: [Ar] 3d2 4s2
- Ti3+: [Ar] 3d1
- Unpaired electrons: 1
- Spin-only magnetic moment, \(\mu = \sqrt{1(1+2)} = 1.73 \, \text{BM}\)
- For Cu2+:
- Electronic configuration of Cu: [Ar] 3d10 4s1
- Cu2+: [Ar] 3d9
- Unpaired electrons: 1
- Spin-only magnetic moment, \(\mu = \sqrt{1(1+2)} = 1.73 \, \text{BM}\)
- For V2+:
- Electronic configuration of V: [Ar] 3d3 4s2
- V2+: [Ar] 3d3
- Unpaired electrons: 3
- Spin-only magnetic moment, \(\mu = \sqrt{3(3+2)} = 3.87 \, \text{BM}\)
- For Cu+:
- Electronic configuration of Cu: [Ar] 3d10 4s1
- Cu+: [Ar] 3d10
- Unpaired electrons: 0
- Spin-only magnetic moment, \(\mu = 0\)
Now, comparing the calculated values to find ions with the same spin-only magnetic moment:
- Cr2+ and Fe2+ both have \(\mu = 4.90 \, \text{BM}\).
- Ti3+ and Cu2+ both have \(\mu = 1.73 \, \text{BM}\).
Thus, the correct answer is B and C only.