Although Cr³⁺ and Co²⁺ ions have same number of unpaired electrons but the magnetic moment of Cr³⁺ is 3.87 B.M

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Cr^{3+}

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds: …

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The magnetic moment depends on both the number of unpaired electrons and the orbital contribution. $Cr^{3+}$ has no orbital contribution (quenched by crystal field), while $Co^{2+}$ retains a significant orbital contribution, giving a higher magnetic moment despite the same number of unpaired electrons.

The key here is that magnetic moment is not simply a function of the number of unpaired electrons — it also depends on whether the orbital angular momentum contributes to the total moment. For transition metal ions in complexes, the crystal field can "quench" the orbital contribution, but not always.

Let’s break this down.

The spin-only formula For an ion with $n$ unpaired electrons, the spin-only magnetic moment is:

\mu_{s.o.} = \sqrt{n(n+2)} \, \text{B.M.}

For $n = 3$ , this gives $\sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ B.M.

So $Cr^{3+}$ matches the spin-only value exactly. That tells us immediately that its orbital angular momentum is completely quenched.

Why is $Cr^{3+}$ spin-only?

$Cr^{3+}$ has the electronic configuration $[Ar]\,3d^3$ . In an octahedral crystal field, the three electrons occupy the $t_{2g}$ orbitals ( $d_{xy}, d_{yz}, d_{zx}$ ) with parallel spins (Hund’s rule).

The $t_{2g}$ set is triply degenerate, but in a perfect octahedral field, the orbital angular momentum is quenched because the orbitals are real (not complex) and the ground term is $^4A_{2g}$ — an orbital singlet. No orbital degeneracy means no orbital contribution.
Now $Co^{2+}$ : same $n$ , different story

$Co^{2+}$ has the configuration $[Ar]\,3d^7$ . In an octahedral field, this is $t_{2g}^5 e_g^2$ . The ground term is $^4T_{1g}(F)$ — an orbital triplet.

The $T_{1g}$ term has orbital angular momentum that is not fully quenched. The magnetic moment is therefore higher than the spin-only value. The experimental value of 4.87 B.M. is close to what is expected when spin-orbit coupling mixes in some orbital contribution.

Tip

A quick way to check: if the ground term symbol has $A$ (as in $^4A_{2g}$ ), orbital contribution is zero. If it has $T$ (as in $^4T_{1g}$ ), orbital contribution is present and the magnetic moment will exceed the spin-only value.

Quantifying the difference For $Co^{2+}$ in an octahedral field, the effective magnetic moment is given by:

\mu_{\text{eff}} = \mu_{s.o.} \left(1 - \frac{\alpha \lambda}{\Delta} \right)

where $\lambda$ is the spin-orbit coupling constant (negative for $d^7$ ), $\Delta$ is the crystal field splitting, and $\alpha$ is a factor depending on the term. The negative $\lambda$ and the mixing cause $\mu_{\text{eff}}$ to be larger than $\mu_{s.o.}$ . …

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✗ Mistake 1: Assuming same number of unpaired electrons means same magnetic moment

Why it's wrong:

Magnetic moment ( $\mu$ ) depends not just on the number of unpaired electrons ( $n$ ), but also on the orbital contribution (via the spin–orbit coupling). The formula $\mu = \sqrt{n(n+2)}$ assumes only spin contribution — this works for most first-row transition metals, but not always.

How to avoid:

Always check if the ion is in a weak field or strong field environment, and whether orbital angular momentum is quenched or not. For $Co^{2+}$ in an octahedral field, there is an orbital contribution that increases $\mu$ beyond the spin-only value.

✗ Mistake 2: Forgetting the electronic configuration details

Why it's wrong:

Students often write $Cr^{3+}$ as $[Ar] 3d^3$ and $Co^{2+}$ as $[Ar] 3d^7$ — correct so far. But then they assume both have 3 unpaired electrons and stop there. They miss that the arrangement of electrons in $t_{2g}$ and $e_g$ orbitals affects orbital angular momentum.

How to avoid:

Draw the d-orbital splitting diagram for each ion in an octahedral field:

$Cr^{3+}$ ( $d^3$ ): All three electrons go into $t_{2g}$ (one each) — all unpaired, but no orbital degeneracy → no orbital contribution.
$Co^{2+}$ ( $d^7$ ): High-spin configuration: $t_{2g}^5 e_g^2$ — this has orbital degeneracy in the $t_{2g}$ set → orbital angular momentum is not fully quenched.

✗ Mistake 3: Using the spin-only formula blindly

Why it's wrong:

For $Cr^{3+}$ : $n = 3$ , so $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M. — matches.

For $Co^{2+}$ : $n = 3$ , same formula gives $3.87$ B.M., but the actual value is 4.87 B.M. — the formula fails here.

How to avoid:

Remember: The spin-only formula works well when orbital angular momentum is quenched (common for $d^3$ , $d^5$ , $d^8$ high-spin, $d^6$ low-spin). For $d^7$ high-spin (like $Co^{2+}$ ), orbital contribution is significant — use the formula $\mu = \sqrt{4S(S+1) + L(L+1)}$ or simply memorise the experimental trend.

✗ Mistake 4: Confusing high-spin vs low-spin

Why it's wrong: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$Ce^{4+}, Yb^{2+}$
›Reveal solutionSolution
This tests which unusual lanthanide oxidation states behave as reducing vs oxidising agents; $Eu^{2+}$ and $Yb^{2+}$ are the textbook pair of reducing agents, so the answer is (D).

Concept and Intuition

Lanthanides normally exist as $Ln^{3+}$ . A few elements also show +2 or +4 states when that unusual state happens to correspond to an especially stable $f^0$ , $f^7$ (half-filled) or $f^{14}$ (fully-filled) configuration. But "isolable" is not the same as "thermodynamically preferred" — the normal +3 state is still the most stable overall for the element, so:

An unusual +4 ion (higher than normal +3) tends to gain an electron and fall back to +3 — it therefore acts as an oxidising agent (itself gets reduced). Examples: $Ce^{4+}$ ( $4f^0\to 4f^1$ ), $Tb^{4+}$ ( $4f^7\to 4f^8$ ).

An unusual +2 ion (lower than normal +3) tends to lose an electron and rise back to +3 — it therefore acts as a reducing agent (itself gets oxidised). Examples: $Eu^{2+}$ ( $4f^7\to 4f^6$ ), $Yb^{2+}$ ( $4f^{14}\to 4f^{13}$ ), and (more weakly) $Sm^{2+}$ .

Step-by-Step Solution

Identify each ion's usual/unusual character: $Ce^{4+}$ (unusual +4, oxidiser), $Tb^{4+}$ (unusual +4, oxidiser), $Eu^{2+}$ (unusual +2, reducer), $Yb^{2+}$ (unusual +2, reducer).

Option (A) $Ce^{4+}, Yb^{2+}$ — mixes an oxidiser with a reducer, not a matching pair.

Option (B) $Ce^{4+}, Tb^{4+}$ — both are oxidising agents, not reducing. …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Ce^{4+}, Tb^{4+}$
›Reveal solutionSolution
$Eu^{2+}$ and $Yb^{2+}$ are strong reducing agents because oxidising to the common $+3$ lanthanide state gives them extra electronic stability.

Concept and Intuition

Lanthanides are overwhelmingly found in the $+3$ oxidation state. Ions that deviate from $+3$ tend to revert to it if doing so gives a specially stable electron configuration (empty, half-filled, or fully-filled f subshell). $Eu^{2+}$ ( $4f^7$ , half-filled — already quite stable) still readily loses an electron to $Eu^{3+}$ ( $4f^6$ ) — actually, more precisely, the drive is that $+2$ ions with configurations one electron short of a stable count are pushed to lose an electron toward +3, or +4 ions are pulled to gain one toward +3. The exam-relevant memorised fact: $Eu^{2+}$ and $Sm^{2+}$ (and here, $Yb^{2+}$ , which is $4f^{14}$ , fully filled) act as reducing agents by oxidising to $+3$ , while $Ce^{4+}$ and $Tb^{4+}$ act as oxidising agents by being reduced to the stable $+3$ state ( $Ce^{3+}$ is $4f^0$ ; $Tb^{3+}$ is $4f^8$ /near half-filled +1 pattern that's more stable than $Tb^{4+}$ ).

Step-by-Step Solution

Recall the standard exceptions to $+3$ among lanthanides: $Ce^{4+}, Pr^{4+}, Tb^{4+}$ (oxidising agents, reduce to $+3$ ) and $Eu^{2+}, Sm^{2+}, Yb^{2+}$ (reducing agents, oxidise to $+3$ ).

Option (A) $Ce^{4+}, Tb^{4+}$ — both oxidising agents, not reducing. …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\mathrm{Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy}$
›Reveal solutionSolution
Tests which lanthanides depart from the dominant +3 state to also show +4, based on electronic-configuration stability.

Concept and Intuition

Lanthanides are overwhelmingly +3 because that oxidation state matches a filled 6s/5d loss while leaving a reasonably stable 4f configuration. A few members show +4 (or +2) only when losing (or gaining) one more electron gets them to, or close to, an empty ( $f^0$ ), half-filled ( $f^7$ ), or fully-filled ( $f^{14}$ ) 4f sub-shell — extra stability that offsets the higher ionisation energy.

Step-by-Step Solution

Ce (4f¹5d¹6s²): losing one more electron beyond +3 gives Ce⁴⁺ with $f^0$ — very stable, +4 is in fact its best-known state (e.g., $\mathrm{CeO_2}$ ).

Pr and Nd: Pr⁴⁺ ( $f^1$ ) and Nd⁴⁺ ( $f^2$ ) are known, though less stable than Ce⁴⁺ (e.g., $\mathrm{PrO_2}$ , and Nd⁴⁺ in a few solid oxides).

Pm: radioactive, poorly characterised; +4 not established.

Sm, Eu: these instead favour +2 ( $f^6$ close to half-filled for Sm²⁺, and $f^7$ half-filled exactly for Eu²⁺) — the opposite direction, not +4. …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\mathrm{Yb^{2+}}$
›Reveal solutionSolution
Of the four statements, only "CrO is basic" is correct — the others misstate Yb $^{2+}$ 's redox role, Lu $^{3+}$ 's magnetism, and brass's composition.

Concept and Intuition

This question tests several standard d- and f-block facts together: (i) lanthanide ions in unusual oxidation states tend to revert to the characteristic, most stable +3 state, so $\mathrm{Eu^{2+}}$ / $\mathrm{Yb^{2+}}$ act as reducing agents (they get oxidized to +3) while $\mathrm{Ce^{4+}}$ / $\mathrm{Tb^{4+}}$ act as oxidizing agents (they get reduced to +3);

(ii) magnetism of lanthanide ions follows from unpaired 4f electrons — a fully-filled or fully-empty 4f subshell is diamagnetic;

(iii) the acid-base character of transition-metal oxides follows their oxidation state — low oxidation states give basic oxides, high oxidation states give acidic oxides, with intermediate ones amphoteric;

(iv) common alloy compositions (brass vs bronze) are a factual recall point.

Step-by-Step Solution

(A) $\mathrm{Yb^{2+}}$ : has configuration $4f^{14}$ (fully filled, stable), so it readily loses an electron to attain the general lanthanide-favoured $\mathrm{Yb^{3+}}$ state — meaning $\mathrm{Yb^{2+}}$ itself gets oxidized, i.e. it is a reducing agent, not an oxidant. Statement false.

(B) $\mathrm{Lu^{3+}}$ : Lu (Z=71) is $[\mathrm{Xe}]4f^{14}5d^16s^2$ ; removing 3 electrons ( $5d^1,6s^2$ ) gives $\mathrm{Lu^{3+}} = [\mathrm{Xe}]4f^{14}$ — completely filled 4f, hence diamagnetic, not paramagnetic. Statement false. …
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AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$Q. For which of the following +3 oxidation state is highly oxidizing in character? (A) Al (B) Ga (C) In (D) Tl$
›Reveal solutionSolution
The inert pair effect is strongest for the heaviest Group 13 element, thallium, making its +3 oxidation state a strong oxidizer that is readily reduced to the more stable +1 state.

Concept and Intuition

Going down Group 13 (B, Al, Ga, In, Tl), the ns² electron pair becomes increasingly reluctant to participate in bonding due to poor shielding by intervening d/f electrons and relativistic contraction of the ns orbital — this is the "inert pair effect". Consequently, the +1 oxidation state (where the ns² pair stays un-ionized) becomes progressively more stable relative to +3 as we move down the group. For Al and Ga, the +3 state is the dominant, stable state. For Tl, the inert pair effect is so pronounced that +1 is actually the more stable oxidation state, which means Tl(III) compounds are strong oxidizing agents — they are readily reduced to Tl(I), releasing energy in the process.

Step-by-Step Solution

Rank the inert pair effect across Al, Ga, In, Tl — it strengthens down the group, being negligible for Al and dominant for Tl.

For Al, +3 is essentially the only common, stable oxidation state (Al³⁺ is not oxidizing).

For Ga and In, +3 is still the more stable state, with +1 being a minor, less common state. …
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AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$SnO_2$
›Reveal solutionSolution
The inert-pair effect destabilises Pb(IV) relative to Pb(II), making $PbO_2$ a strong oxidising agent — the strongest among $SnO_2$ , $SiO_2$ , $GeO_2$ , $PbO_2$ .

Concept and Intuition

Down group 14, the ns² electron pair becomes increasingly reluctant to participate in bonding (the inert-pair effect), so the lower oxidation state ( $+2$ ) becomes progressively more stable relative to the group oxidation state ( $+4$ ) as you go from C to Pb. This means $Pb^{4+}$ compounds readily oxidise other species while being reduced to the more stable $Pb^{2+}$ .

Step-by-Step Solution

$SiO_2$ , $GeO_2$ : silicon and germanium show little inert-pair effect; $+4$ is their stable, common state, so these oxides are not strong oxidisers.

$SnO_2$ : tin shows a mild inert-pair effect; $Sn^{4+}$ is reasonably stable, only a weak oxidiser. …
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AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
$NCl_5$
›Reveal solutionSolution
The inert-pair effect makes lead prefer the $+2$ (divalent) oxidation state, not $+4$ (tetravalent) — so the claim that "Pb prefers to form tetravalent compounds" is the incorrect statement.

Concept and Intuition

Going down Group 14, the heavier elements' $ns^2$ electron pair becomes increasingly reluctant to participate in bonding (the inert-pair effect, due to poor shielding by intervening $d$ / $f$ electrons and relativistic contraction of the $ns$ orbital). This makes the lower oxidation state progressively more stable for heavier members: $Sn$ shows both $+2$ and $+4$ fairly readily, but $Pb$ strongly favours $+2$ ( $PbO$ , $PbCl_2$ , $Pb(NO_3)_2$ are common; $Pb^{4+}$ compounds like $PbO_2$ are comparatively strong oxidizers/less stable).

Step-by-Step Solution

(A) $NCl_5$ does not exist (N has no accessible $d$ orbitals to expand its octet beyond 4 bonds) while $PCl_5$ does (P can use $3d$ orbitals) — this is a correct/true statement.

(B) " $Pb$ prefers to form tetravalent compounds" — false; due to the inert-pair effect $Pb$ actually prefers divalent compounds. This is the incorrect statement. …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$Ga < In < Al < Tl$
›Reveal solutionSolution
The inert-pair effect grows down group 13, making the +1 oxidation state progressively more stable: least for Al, most for Tl.

Concept and Intuition

The inert-pair effect describes the reluctance of the outermost $ns^2$ electron pair to participate in bonding as atomic number increases down a group, due to poor shielding by intervening $d$ / $f$ electrons and relativistic effects for heavier elements. In group 13, this makes the $+1$ oxidation state (retaining the $ns^2$ pair, only losing the single $p$ electron) increasingly favoured relative to $+3$ as you go down the group.

Step-by-Step Solution

Al: +3 is overwhelmingly the stable/common oxidation state; +1 compounds are rare and unstable.

Ga: +1 exists but is less stable than +3; +3 still dominant.

In: +1 and +3 are both reasonably common, with +1 gaining stability. …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$Fe$
›Reveal solutionSolution
Manganese exhibits the broadest range of oxidation states among first-row transition metals. Answer: Mn.

Concept and Intuition

The number of oxidation states a transition metal can adopt tends to be maximised near the middle of the 3d series, where there are enough d-electrons to support both low and (through loss of many/all valence electrons) very high oxidation states, while still having partially-filled d-orbitals available for a range of intermediate states. Manganese, with configuration $3d^54s^2$ , is the textbook example, spanning $+2$ (as $Mn^{2+}$ ) all the way to $+7$ (as $MnO_4^-$ ).

Step-by-Step Solution

$V$ (Z=23): common oxidation states $+2, +3, +4, +5$ — 4 common states.

$Cr$ (Z=24): common oxidation states $+2, +3, +6$ (and $+4, +5$ less commonly) — fewer well-established states than Mn. …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$Q. The most common oxidation state among lanthanoids is ______ (A) +4 (B) +3 (C) +2 (D) +1$
›Reveal solutionSolution
Almost all lanthanoids show a dominant $+3$ oxidation state because losing the $6s^2$ (and $5d^1$ where present) electrons is comparatively easy, while removing a 4th electron from the deeply shielded $4f$ subshell requires a very high ionisation energy.

Concept and Intuition

The chemistry of lanthanoids is governed by the filling of the inner $4f$ subshell, which is well shielded from the surrounding chemical environment by the outer $5s^2 5p^6$ shells. Because the $4f$ electrons are so shielded, they don't participate readily in bonding, and it is always the outermost $6s^2$ electrons (plus the occasional $5d^1$ ) that ionise first.

Step-by-Step Solution

General configuration: $[Xe]4f^{1-14}5d^{0,1}6s^2$ .

First two ionisations remove $6s^2$ readily; where a $5d^1$ electron is present it is the third to go — giving the $Ln^{3+}$ ion with configuration $[Xe]4f^n$ .

The 4th ionisation energy (removing an electron from $4f^n$ ) is very large, because the $4f$ subshell is already at a comparatively stable, contracted low-energy configuration once the outer electrons are gone. …
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Q.Although Cr3+Cr^{3+}Cr3+ and Co2+Co^{2+}Co2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+Cr^{3+}Cr3+ is 3.87 B.M. and that of Co2+Co^{2+}Co2+ is 4.87 B.M. Why?

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Q.Although $Cr^{3+}$ and $Co^{2+}$ ions have same number of unpaired electrons but the magnetic moment of $Cr^{3+}$ is 3.87 B.M. and that of $Co^{2+}$ is 4.87 B.M. Why?

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