Q.Calculate the magnetic moment of the following ions: a) Ti^2+ (Z = 22) b) Zn^2+ (Z = 30) c) Cr^2+ (Z = 24)

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Write the formula for calculating 'spin only' magnetic moment.

   ›Reveal solutionSolution

   The spin-only formula estimates a transition metal ion's magnetic moment purely from its number of unpaired electrons, ignoring orbital contribution.

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 mark
   Q.Give one example of a complex having tetrahedral geometry and paramagnetic in nature.

   ›Reveal solutionSolution

   $[NiCl_4]^{2-}$ is the standard example of a tetrahedral, paramagnetic complex, arising from $sp^3$ hybridisation of $Ni^{2+}$ with the weak-field $Cl^-$ ligand.

   Why $[NiCl_4]^{2-}$ fits

   Ni has configuration $[Ar]3d^84s^2$; in $Ni^{2+}$, this becomes $3d^8$. $Cl^-$ is a weak-field ligand (low in the spectrochemical series), so it does not force pairing of the $3d$ electrons. With four ligands and no $d$-orbital freed by pairing, nickel uses one $4s$ and three $4p$ orbitals — $sp^3$ hybridisation — giving a **tetrahedr …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 mark
   Q.According to VBT, which one has the highest paramagnetic character? $[Cr(H_2O)_6]^{3+}$ or $[Fe(H_2O)_6]^{2+}$

   ›Reveal solutionSolution

   Counting unpaired $d$-electrons for each ion under VBT shows $Fe^{2+}$ ($d^6$, high-spin, 4 unpaired) is more paramagnetic than $Cr^{3+}$ ($d^3$, always 3 unpaired).

   $[Cr(H_2O)_6]^{3+}$

   $Cr$ ($Z=24$) is $[Ar]3d^54s^1$; $Cr^{3+}$ removes 3 electrons to give $3d^3$. With only 3 electrons for the three $t_{2g}$ orbitals, Hund's rule places one electron in each — $t_{2g}^3$ — giving 3 unpaired electrons, regardless of whether the ligand is weak- or strong-field (there's no way to pair up 3 electrons across 3 orbitals to reduce this further).

   $[Fe(H_2O)_6]^{2+}$

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 mark
   Q.Write True or False: Value of magnetic moment of a divalent ion in aqueous solution having atomic number 25, will be 5.92 B.M.

   ›Reveal solutionSolution

   Mn2+ has 5 unpaired electrons, giving a spin-only moment of 5.92 B.M., so the statement is true.

   Atomic number 25 = manganese, [Ar] 3d5 4s2. The divalent ion Mn2+ = [Ar] 3d5, which has 5 unpaired electrons.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which one of the following metal ions is likely to have a magnetic moment of 1.73 BM?

   (a) Fe²⁺

   (b) Mn²⁺

   (c) Cr²⁺

   (d) Cu²⁺

   ›Reveal solutionSolution

   Using μ = √(n(n+2)) BM, 1.73 BM means n = 1 unpaired electron; Cu²⁺ (d⁹) is the only ion with one unpaired electron — option (D).

   The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. A value of $1.73$ BM gives $\sqrt{1(1+2)} = \sqrt{3} = 1.73$, so $n = 1$ unpaired electron.

   Now count unpaired electrons for each ion:

   • $Fe^{2+}$: $3d^6$ → 4 unpaired ($\mu \approx 4.9$ BM). …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set JZ1 markMCQ
   Q.Magnetic moment of a bivalent ion in aqueous solution will be, if its atomic number is 25

   (a) 1.73 BM

   (b) 2.83 BM

   (c) 4.96 BM

   (d) 5.92 BM

   ›Reveal solutionSolution

   Mn$^{2+}$ ($3d^5$) has 5 unpaired electrons, so $\mu=\sqrt{5(5+2)}=5.92$ BM — option (d).

   Concept. The magnetic moment of a transition-metal ion depends only on the number of unpaired d-electrons ($n$), through the spin-only formula $\mu=\sqrt{n(n+2)}$ BM.

   Step 1 — identify the ion. Atomic number 25 → manganese (Mn), configuration $[Ar]3d^5 4s^2$. A bivalent ion Mn$^{2+}$ loses the two $4s$ electrons: Mn$^{2+}=[Ar]3d^5$.

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The spin magnetic moment of Co3+ ion is:

   (a) sqrt(3) BM

   (b) sqrt(8) BM

   (c) sqrt(15) BM

   (d) sqrt(24) BM

   ›Reveal solutionSolution

   Co3+ has the configuration [Ar]3d6; in the high-spin (free-ion) state this places 4 electrons unpaired, giving a spin-only magnetic moment of √(n(n+2)) = √24 BM.

   Cobalt (Z = 27) has ground state configuration [Ar]3d7 4s2. Removing 3 electrons to form Co3+ removes the two 4s electrons first and then one 3d electron, giving Co3+: [Ar]3d6.

   Filling the five d orbitals with 6 electrons by Hund's rule (maximum multiplicity, i.e., high-spin, as would apply to the free gaseous ion or in a weak field):

   ↑↓ ↑ ↑ ↑ ↑ → one orbital doubly occupied, four orbitals singly occupied → 4 unpaired electrons (n = 4).

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following is a paramagnetic complex?

   (a) [Ni(H2O)6]2+

   (b) [Ni(CO)4]

   (c) [Zn(NH3)4]2+

   (d) [Co(NH3)6]

   ›Reveal solutionSolution

   Ni2+ (d8) with the weak-field ligand H2O keeps 2 electrons unpaired; the other three complexes all have a d10 or strong-field-paired d-count and are diamagnetic.

   [Ni(H₂O)₆]²⁺: Ni²⁺ is d⁸; H₂O is a weak-field ligand and cannot force pairing, so 2 electrons remain unpaired — paramagnetic (octahedral, sp³d² outer-orbital complex).

   [Ni(CO)₄]: here nickel is in the zero oxidation state, Ni(0), configuration 3d¹⁰4s⁰ — a completely filled d-subshell regardless of ligand field, so it is diamagnetic (sp³, tetrahedral).

   [Zn(NH₃)₄]²⁺: Zn²⁺ is always 3d¹⁰ (fully filled) in its only common oxidation state, so it is diamagnetic (sp³, tetrahedral) irrespective of the ligand.

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The magnetic moment of $Mn^{+2}$ in aqueous solution is –

   (a) 2.84 B.M

   (b) 3.87 B.M

   (c) 4.90 B.M

   (d) 5.92 B.M

   ›Reveal solutionSolution

   Mn²⁺ has a half-filled d⁵ configuration with 5 unpaired electrons, and the spin-only formula gives a magnetic moment of 5.92 B.M.

   $Mn^{2+}$ has the configuration $[Ar]3d^5$ — a half-filled d-subshell, with all 5 electrons unpaired (by Hund's rule, each of the 5 d-orbitals holds one electron).

   Using the spin-only formula:

   $\mu = \sqrt{n(n+2)}\ B.M., \quad n = 5$

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2023Set ANNUAL1 mark
    Q.Calculate the spin only magnetic moment of $M^{2+}(aq)$ ion ($Z = 27$).

    ›Reveal solutionSolution

    $Z=27$ corresponds to cobalt; $Co^{2+}(aq)$ has the configuration $3d^7$ with 3 unpaired electrons, giving a spin-only magnetic moment of $\sqrt{15} \approx 3.87\ BM$.

    Identify the ion: $Z = 27$ is cobalt (Co), with ground-state configuration $[Ar]\,3d^7 4s^2$. Removing 2 electrons (always from $4s$ first) to form $Co^{2+}$ gives:

    $Co^{2+}: [Ar]\,3d^7$

    Count unpaired electrons: Distributing 7 electrons among the five $3d$ orbitals following Hund's rule (each orbital singly filled first, before pairing) for the aqua ion (a weak-field, high-spin case):

    $\underbrace{\uparrow\downarrow}_{}\ \underbrace{\uparrow\downarrow}_{}\ \underbrace{\uparrow}_{}\ \underbrace{\uparrow}_{}\ \underbrace{\uparrow}_{}$

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2020Set 56/2/11 markMCQ
    Q.Total number of unpaired electrons present in $Co^{3+}$ (Atomic number = 27) is (A) 2 (B) 7 (C) 3 (D) 5

    ›Reveal solutionSolution

    Cobalt loses three electrons to form $\text{Co}^{3+}$, leaving an electronic configuration of $[\text{Ar}] \, 3d^6$. In the $d^6$ configuration, pairing depends on ligand field strength, but the question asks for the ground-state free ion, which follows Hund's rule and has 4 unpaired electrons.

    The number of unpaired electrons in a transition metal ion determines its magnetic properties. To find this, we need the electronic configuration of the ion and then apply Hund's rule of maximum multiplicity.

    Understanding the Configuration

    Cobalt has atomic number 27. The neutral atom's electronic configuration is:

    $[\text{Ar}] \, 3d^7 \, 4s^2$

    When cobalt forms $\text{Co}^{3+}$, it loses three electrons. Electrons are always removed from the outermost shell first—both $4s$ electrons go first, then one $3d$ electron:

    $\text{Co}^{3+}: [\text{Ar}] \, 3d^6$

    Applying Hund's Rule

    The five $3d$ orbitals can hold up to 10 electrons. With 6 electrons to place, Hund's rule tells us to:

    1. Maximize unpaired electrons first by placing one electron in each orbital with parallel spin.
    2. Then pair up any remaining electrons.

    Let me show the filling pattern for $3d^6$:

    $d_{xy}$	$d_{yz}$	$d_{zx}$	$d_{x^2-y^2}$	$d_{z^2}$
    ↑↓	↑	↑	↑	↑

    The first five electrons occupy all five orbitals singly (all spin-up). The sixth electron must pair with one of them.

    Result: 4 unpaired electrons and 1 paired set. …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2019Set ANNUAL1 mark
    Q.Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

    ›Reveal solutionSolution

    The divalent ion of element 25 is $Mn^{2+}$, a $3d^5$ ion with all five d-orbitals singly occupied; the spin-only formula then gives $\mu \approx 5.92\ BM$.

    Element with atomic number 25 is manganese (Mn): $[Ar]3d^5 4s^2$.

    Forming the divalent ion $Mn^{2+}$ removes the two $4s$ electrons first:

    $Mn^{2+}:\ [Ar]3d^5$

    By Hund's rule, all five 3d electrons occupy the five d-orbitals singly (maximum multiplicity), giving $n = 5$ unpaired electrons.

    …

    🔒 Start your 14-day free trial to see the full solution →