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\newline ‘Spin only’ magnetic moment is same for which of the following ions? \newline \newline (A) Ti³⁺ \newline (B) Cr²⁺ \newline (C) Mn²⁺ \newline (D) Fe²⁺ \newline (E) Sc³⁺ \newline Choose the most appropriate answer from the options given below:
hard
Coordination Compounds
2024
chemistry
Explanation To solve this problem, we need to calculate the 'spin only' magnetic moment for each of the given ions.
\newline The 'spin only' magnetic moment ( μ ) (\mu) ( μ ) is given by the formula: μ = n ( n + 2 ) μ B
\newline \mu = \sqrt{n(n+2)} \, \mu_B
\newline μ = n ( n + 2 ) μ B
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where
is the number of unpaired electrons and
is the Bohr magneton.
Let's determine the number of unpaired electrons for each ion:
• Ti
Titanium has an atomic number of 22. The electronic configuration is
Ar
] 3 d 1 . ] \, 3d^1.
\newline ] 3 d 1 . Therefore, Ti
has 1 unpaired electron.
• Cr
Chromium has an atomic number of 24. The electronic configuration is
Ar
] 3 d 4 . ] \, 3d^4.
\newline ] 3 d 4 . Therefore, Cr
has 4 unpaired electrons.
• Mn
Manganese has an atomic number of 25. The electronic configuration is
Ar
] 3 d 5 . ] \, 3d^5.
\newline ] 3 d 5 . Therefore, Mn
has 5 unpaired electrons.
• Fe
Iron has an atomic number of 26. The electronic configuration is
Ar
] 3 d 6 . ] \, 3d^6.
\newline ] 3 d 6 . Therefore, Fe
has 4 unpaired electrons.
• Sc
Scandium has an atomic number of 21. The electronic configuration is
Ar
] 3 d 0 . ] \, 3d^0.
\newline ] 3 d 0 . Therefore, Sc
has 0 unpaired electrons.
Now, calculate the magnetic moment for ions with unpaired electrons:
• For Cr
and Fe
(both have 4 unpaired electrons):
μ = 4 ( 4 + 2 ) = 24 ≈ 4.90 μ B
\newline \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \mu_B
\newline μ = 4 ( 4 + 2 ) = 24 ≈ 4.90 μ B • For Ti
3 + : μ = 1 ( 1 + 2 ) = 3 ≈ 1.73 μ B ^{3+}: \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \mu_B
\newline 3 + : μ = 1 ( 1 + 2 ) = 3 ≈ 1.73 μ B • For Mn
2 + : μ = 5 ( 5 + 2 ) = 35 ≈ 5.92 μ B ^{2+}: \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \mu_B
\newline 2 + : μ = 5 ( 5 + 2 ) = 35 ≈ 5.92 μ B • For Sc
3 + : μ = 0 ( 0 + 2 ) = 0 μ B ^{3+}: \mu = \sqrt{0(0+2)} = 0 \, \mu_B
\newline 3 + : μ = 0 ( 0 + 2 ) = 0 μ B The ions with the same 'spin only' magnetic moment are Cr
and Fe
Therefore, the correct option is Option 3: B and D only.