Q.The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of $Cr^{3+}$ ion is ___________.

Classify each species by its unpaired electrons:

• $\text{O}_2$ — molecular-orbital theory places 2 electrons singly in the degenerate $\pi^*$ orbitals $\Rightarrow$ paramagnetic.
• $\text{C}_6\text{H}_6$ (benzene) — all bonding and $\pi$ electrons paired $\Rightarrow$ diamagnetic.
• $\text{Cu}^{2+}$ — $[\text{Ar}]3d^9$, one unpaired electron $\Rightarrow$ paramagnetic.
• $\text{Cu}^{+}$ — $[\text{Ar}]3d^{10}$, fully filled $\Rightarrow$ diamagnetic.
• $\text{NaCl}$ — $\text{Na}^+$ ($[\text{Ne}]$) and $\text{Cl}^-$ ($[\text{Ar}]$), both closed shells $\Rightarrow$ diamagnetic.
• $\text{O}_2^+$ — removing one $\pi^*$ electron leaves one unpaired electron (bond order 2.5) $\Rightarrow$ paramagnetic. …

Concept and intuition

Group‑14 elements (C, Si, Ge, Sn, Pb) can expand their coordination number beyond four by using empty d‑orbitals. However, forming a stable octahedral $\mathrm{[MX_6]^{2-}}$ complex requires the central atom to be large enough to sterically accommodate six bulky ligands. Fluorine is very small, so even silicon (atomic radius ~111 pm) can host six fluorides. Chlorine is much larger (ionic radius ~181 pm), so only larger central atoms like Ge or Sn can fit six chlorides around them. Silicon is too small for six chlorides — the ligand‑ligand repulsion destabilises the complex.

Step‑by‑step reasoning

1. Identify the central atoms and their sizes

   The central atoms are Si, Ge, and Sn. Atomic radii increase down the group:

   Si ≈ 111 pm, Ge ≈ 125 pm, Sn ≈ 145 pm. Larger central atoms can accommodate more or larger ligands.

2. Consider the ligand sizes

   Fluoride ion (F⁻) is very small (ionic radius ~133 pm). Chloride ion (Cl⁻) is much larger (~181 pm). Hydroxide (OH⁻) is intermediate but flexible.

3. Check each complex

   • (B) $\left[\mathrm{SiF}_6\right]^{2-}$: Si is small, but F⁻ is tiny. Six fluorides fit comfortably — this complex is well‑known (e.g., hexafluorosilicate).
   • (A) $\left[\mathrm{GeCl}_6\right]^{2-}$: Ge is larger than Si, so it can accommodate six chlorides. This complex exists (e.g., as the salt $\mathrm{(NH_4)_2[GeCl_6]}$).
   • (D) $\left[\mathrm{Sn(OH)}_6\right]^{2-}$: Sn is even larger, and OH⁻ is not too bulky. This complex exists (e.g., in stannate salts). …

Step 1 — How many unpaired electrons give $2.86$ BM?

So the target species must have $n=2$.

Step 2 — Test each complex

• (A) $[\text{CoF}_6]^{3-}$: $\text{Co}^{3+}$ is $d^6$; $\text{F}^-$ is a weak-field ligand → high-spin octahedral $t_{2g}^{4}e_g^{2}$ → 4 unpaired ($\mu\approx4.9$ BM).
• (B) $[\text{Co(NH}_3)_6]^{3+}$: $\text{Co}^{3+}$ ($d^6$); $\text{NH}_3$ is strong-field → low-spin $t_{2g}^{6}$ → 0 unpaired ($\mu=0$). …

Concept and intuition:

Unpaired electrons in a complex depend on the metal ion's d-electron configuration and the ligand field strength. Weak-field ligands (like $\mathrm{F}^-$ and $\mathrm{Cl}^-$) cause a small splitting, so electrons fill all five d-orbitals singly before pairing (high-spin, Hund's rule maximized across all five orbitals). Strong-field ligands cause a large splitting, forcing pairing in the lower-energy $t_{2g}$ orbitals first (low-spin). We find the oxidation state of the metal, then its d-electron count, then apply the appropriate spin state.

1. $[\mathrm{FeF}_6]^{3-}$

   • Iron is +3 (since $x + 6(-1) = -3 \Rightarrow x = +3$).
   • $\mathrm{Fe}^{3+}$: $[\mathrm{Ar}]\,3d^5$.
   • $\mathrm{F}^-$ is weak-field → high-spin: all five d-electrons occupy the five orbitals singly.
   • Unpaired electrons = 5.

2. $[\mathrm{MnCl}_6]^{3-}$

   • Manganese is +3 ($x + 6(-1) = -3 \Rightarrow x = +3$).
   • $\mathrm{Mn}^{3+}$: $[\mathrm{Ar}]\,3d^4$.
   • $\mathrm{Cl}^-$ is weak-field → high-spin: the four electrons occupy four different orbitals ($t_{2g}^3 e_g^1$), all unpaired.
   • Unpaired electrons = 4.

3. $[\mathrm{CoF}_6]^{3-}$

   • Cobalt is +3 ($x + 6(-1) = -3 \Rightarrow x = +3$).
   • $\mathrm{Co}^{3+}$: $[\mathrm{Ar}]\,3d^6$.
   • $\mathrm{F}^-$ is weak-field → high-spin: $t_{2g}^4 e_g^2$ — $t_{2g}$ has one paired orbital and two singly occupied (2 unpaired), $e_g$ has two singly occupied orbitals (2 unpaired).
   • Unpaired electrons = 4.

4. $[\mathrm{Co(C_2O_4)_3}]^{3-}$

   • Same $\mathrm{Co}^{3+}$, $3d^6$, but oxalate is a stronger-field ligand that favours pairing for $\mathrm{Co}^{3+}$ → low-spin: $t_{2g}^6 e_g^0$.
   • Unpaired electrons = 0.

5. Compare the pairs

   • (A) $[\mathrm{FeF}_6]^{3-}$ (5) and $[\mathrm{MnCl}_6]^{3-}$ (4) — not equal. …

The concept here is inner orbital vs. outer orbital complexes, a classification based on whether the metal ion uses its (n−1)d orbitals (inner) or nd orbitals (outer) for hybridisation with ligands. Inner orbital complexes are typically low-spin (strong-field ligands) and can be paramagnetic if unpaired electrons remain. Outer orbital complexes are high-spin (weak-field ligands). We must check each complex’s electronic configuration, ligand field strength, and resulting hybridisation.

1. Determine the oxidation state and d-electron count for each metal.

   • I: $[\mathrm{Mn(CN)_6}]^{3-}$ — Mn is +3 (since CN⁻ is −1, total charge −3: Mn + 6(−1) = −3 → Mn = +3). Mn³⁺ has $d^4$ configuration.
   • II: $[\mathrm{FeF_6}]^{3-}$ — Fe is +3 (F⁻ is −1, total −3: Fe + 6(−1) = −3 → Fe = +3). Fe³⁺ has $d^5$.
   • III: $[\mathrm{MnCl_6}]^{3-}$ — Mn is +3 again, $d^4$.
   • IV: $[\mathrm{CoF_6}]^{3-}$ — Co is +3 (F⁻ is −1, total −3: Co + 6(−1) = −3 → Co = +3). Co³⁺ has $d^6$.
   • V: $[\mathrm{Fe(CN)_6}]^{3-}$ — Fe is +3, $d^5$.
   • VI: $[\mathrm{Co(C_2O_4)_3}]^{3-}$ — Oxalate (C₂O₄²⁻) is −2 each, total −6; Co must be +3 to give overall −3. Co³⁺, $d^6$.
   • VII: $[\mathrm{Fe(CN)_6}]^{4-}$ — Fe is +2 (CN⁻ six gives −6, total −4 → Fe = +2). Fe²⁺ has $d^6$.

2. Classify ligands as strong-field or weak-field.

   • CN⁻ is a strong-field ligand (causes large splitting, favours low-spin).
   • F⁻ and Cl⁻ are weak-field ligands (small splitting, favour high-spin).
   • C₂O₄²⁻ (oxalate) is intermediate but generally considered moderate to strong; for Co³⁺ it is strong enough to cause low-spin.

3. Determine spin state and hybridisation for each.

   • I: $d^4$, strong-field CN⁻ → low-spin. Configuration: $t_{2g}^4 e_g^0$. All four electrons paired in $t_{2g}$? No — $t_{2g}$ holds 6 electrons max, so with 4 electrons, two are paired and two unpaired. Actually, $t_{2g}^4$ means one orbital doubly occupied, two singly occupied → 2 unpaired electrons. Hybridisation: inner $d^2sp^3$ (uses two 3d orbitals). Inner orbital, paramagnetic.
   • II: $d^5$, weak-field F⁻ → high-spin. Configuration: $t_{2g}^3 e_g^2$ → 5 unpaired electrons. Hybridisation: $sp^3d^2$ (uses outer 4d orbitals). Outer orbital, paramagnetic.
   • III: $d^4$, weak-field Cl⁻ → high-spin. Configuration: $t_{2g}^3 e_g^1$ → 4 unpaired electrons. Hybridisation: $sp^3d^2$. Outer orbital, paramagnetic.
   • IV: $d^6$, weak-field F⁻ → high-spin. Configuration: $t_{2g}^4 e_g^2$ → 4 unpaired electrons. Hybridisation: $sp^3d^2$. Outer orbital, paramagnetic. …

The question tests your understanding of how ligand field strength influences geometry and magnetic properties in nickel and cobalt complexes. The key is to determine the oxidation state and d-electron configuration of the central metal ion, then see how the ligand (strong or weak field) affects electron pairing and geometry.

Let's examine each statement one by one.

1. Statement A: Geometry of $[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$

   Nickel in both complexes is in the +2 oxidation state. Ni has atomic number 28, so Ni$^{2+}$ has the electron configuration $[Ar]\,3d^8$.

   For $[Ni(CN)_4]^{2-}$, CN$^-$ is a strong field ligand. With 8 d-electrons, a strong field causes pairing in the lower energy orbitals, leaving one d-orbital empty. This favours square planar geometry (dsp$^2$ hybridisation), not tetrahedral.

   For $[NiCl_4]^{2-}$, Cl$^-$ is a weak field ligand. The 8 d-electrons remain unpaired as much as possible, leading to tetrahedral geometry (sp$^3$ hybridisation).

   Since one is square planar and the other tetrahedral, statement A is false.

2. Statement B: Magnetic behaviour of $[Ni(CO)_4]$

   CO is a strong field ligand. Nickel in $[Ni(CO)_4]$ is in the 0 oxidation state. Ni(0) has configuration $[Ar]\,3d^8 4s^2$. In the presence of strong field CO, the 4s electrons pair with the 3d electrons, resulting in complete pairing of all 10 d-electrons (d$^{10}$ configuration). The complex is diamagnetic (no unpaired electrons), not paramagnetic.

   So statement B is false.

3. Statement C: Magnetic behaviour of $[Co(NH_3)_6]^{3+}$ …

Concept & Intuition

Bond order tells us the net number of chemical bonds between two atoms. For molecular oxygen species, we use the molecular orbital (MO) diagram for homonuclear diatomic molecules of the second period (with $2s$ and $2p$ orbitals). The key ordering for $O_2$ and its ions is:

For oxygen (Z=8), each atom contributes 8 electrons, so $O_2$ has 16 valence electrons. Ions add or remove electrons from the highest occupied molecular orbitals (HOMOs), which are the $\pi^*$ orbitals. The bond order formula is:

We are told that $O_2^+$ has bond order $x$. We compute the bond orders of $O_2^-$ and $O_2^{2+}$ numerically, then express them as multiples of $x$.

Step-by-step solution

1. Find $x$ (bond order of $O_2^+$)

   • $O_2^+$ has 15 valence electrons (lost one from neutral $O_2$).
   • Electron configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^1$
   • Bonding electrons: $2+2+4 = 8$ (the $\sigma_{2s}$ and $\sigma_{2s}^*$ cancel each other, but we count all bonding and antibonding explicitly). Actually, careful: Bonding: $\sigma_{2s}$ (2), $\sigma_{2p_z}$ (2), $\pi_{2p}$ (4) → total 8 bonding. Antibonding: $\sigma_{2s}^*$ (2), $\pi_{2p}^*$ (1) → total 3 antibonding.
   • Bond order $= \frac{8 - 3}{2} = \frac{5}{2} = 2.5$.
   • So $x = 2.5$.

2. Bond order of $O_2^-$

   • $O_2^-$ has 17 valence electrons (one extra).
   • Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^3$
   • Bonding electrons: still 8.
   • Antibonding electrons: $\sigma_{2s}^*$ (2) + $\pi_{2p}^*$ (3) = 5.
   • Bond order $= \frac{8 - 5}{2} = \frac{3}{2} = 1.5$.

3. Bond order of $O_2^{2+}$

   • $O_2^{2+}$ has 14 valence electrons (lost two).
   • Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^0$ …

Concept & Intuition

Magnetic properties (paramagnetic vs. diamagnetic) arise from unpaired electrons. Paramagnetic compounds have at least one unpaired electron; diamagnetic compounds have all electrons paired.

To decide, we need:

• The oxidation state of the metal.
• The number of d‑electrons.
• Whether the ligands are strong‑field (causing low‑spin, more pairing) or weak‑field (causing high‑spin, fewer pairings). Carbonyl ($\mathrm{CO}$) is a very strong‑field ligand; cyanide ($\mathrm{CN}^-$) is also strong‑field; water ($\mathrm{H}_2\mathrm{O}$) is intermediate; chloride ($\mathrm{Cl}^-$) is weak‑field. We’ll check each complex step by step.

1. Option (A): $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Br}_2$

   • The complex cation is $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ (since two $\mathrm{Br}^-$ counterions).
   • Chromium in +2 oxidation state: atomic number 24, $\mathrm{Cr}^{2+}$ has $3d^4$ configuration.
   • $\mathrm{H}_2\mathrm{O}$ is a weak‑field ligand (spectrochemical series: $\mathrm{H}_2\mathrm{O}$ < $\mathrm{NH}_3$ < $\mathrm{CN}^-$).
   • For $d^4$ in an octahedral weak field: high‑spin arrangement, $t_{2g}^3 e_g^1$ → 4 unpaired electrons.
   • Paramagnetic — matches the statement. ✓

2. Option (B): $\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_6]$

   • The complex anion is $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ (four $\mathrm{Na}^+$ counterions).
   • Iron in +2 oxidation state: $\mathrm{Fe}^{2+}$ has $3d^6$ configuration.
   • $\mathrm{CN}^-$ is a very strong‑field ligand → low‑spin octahedral.
   • For $d^6$ low‑spin: $t_{2g}^6$ → all electrons paired (0 unpaired).
   • Diamagnetic — matches the statement. ✓

3. Option (C): $[\mathrm{Ni}(\mathrm{CO})_4]$

   • Nickel in zero oxidation state: $\mathrm{Ni}^0$ has $3d^8 4s^2$? Actually, neutral Ni is $[\mathrm{Ar}]\,3d^8 4s^2$, but in complexes the 4s electrons are lost first; for $\mathrm{Ni}(0)$, the effective configuration is $3d^{10}$ after reorganisation due to strong‑field ligands.
   • $\mathrm{CO}$ is an extremely strong‑field ligand, and the complex is tetrahedral (not octahedral).
   • In tetrahedral geometry, the splitting is smaller and inverted, but for $\mathrm{Ni}(\mathrm{CO})_4$, the strong field causes complete pairing: all 10 d‑electrons are paired → $d^{10}$ configuration. …

The key here is to recognize that $\mathrm{SO}_2$ is a well-known reducing agent, and $\mathrm{KMnO}_4$ is a powerful oxidizing agent whose reduction product depends heavily on the pH of the medium. In aqueous solution, $\mathrm{SO}_2$ dissolves to form sulfurous acid ($\mathrm{H}_2\mathrm{SO}_3$), which creates an acidic environment. This is the crucial clue — the medium is acidic, not neutral or alkaline.

In acidic medium, permanganate ion ($\mathrm{MnO}_4^-$) always undergoes a 5-electron reduction to give $\mathrm{Mn}^{2+}$ (colourless or pale pink). The half-reaction is:

Meanwhile, $\mathrm{SO}_2$ (or $\mathrm{H}_2\mathrm{SO}_3$) gets oxidized to sulfate ion ($\mathrm{SO}_4^{2-}$):

Let’s walk through the reasoning step by step.

1. Identify the nature of the medium. When $\mathrm{SO}_2$ gas is passed into an aqueous solution of $\mathrm{KMnO}_4$, it dissolves and reacts with water to form $\mathrm{H}_2\mathrm{SO}_3$. This acidifies the solution. So the reaction medium is distinctly acidic.

2. Recall the pH-dependent behaviour of $\mathrm{KMnO}_4$. This is a classic point in transition metal chemistry. Permanganate shows different reduction products depending on the medium:

   • In acidic medium: $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}$ (pink/violet to colourless)
   • In neutral or faintly alkaline medium: $\mathrm{MnO}_4^- \rightarrow \mathrm{MnO}_2$ (brown precipitate, $\mathrm{Mn}^{4+}$)
   • In strongly alkaline medium: $\mathrm{MnO}_4^- \rightarrow \mathrm{MnO}_4^{2-}$ (green, $\mathrm{Mn}^{6+}$)

   Since $\mathrm{SO}_2$ creates an acidic solution, the reduction must go all the way to $\mathrm{Mn}^{2+}$.

3. Balance the redox reaction to confirm. Combining the two half-reactions:

   • Multiply the permanganate half-reaction by 2 and the $\mathrm{SO}_2$ half-reaction by 5 to balance electrons: $2\mathrm{MnO}_4^- + 16\mathrm{H}^+ + 10e^- \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$ …

The question is about the thermal decomposition of potassium permanganate. When you heat $\text{KMnO}_4$ strongly — around 513 K — it doesn't just melt or vaporise; it undergoes a redox reaction where the manganese in the +7 oxidation state gets both reduced and oxidised (disproportionation is not quite the right word here, because it's a thermal decomposition that yields two different manganese products). The reaction is:

Let’s check the oxidation states. In $\text{KMnO}_4$, Mn is +7. In $\text{K}_2\text{MnO}_4$, Mn is +6. In $\text{MnO}_2$, Mn is +4. So the product with the higher oxidation state is potassium manganate, $\text{K}_2\text{MnO}_4$, where Mn is +6.

Now the question asks about the physical properties of that product — specifically its magnetic behaviour and colour. To answer, you need to think about the electronic configuration of Mn in the +6 state.

1. Electronic configuration of Mn in +6 oxidation state

   Mn (atomic number 25) has the ground-state configuration $[\text{Ar}]\,3d^5 4s^2$. When it loses six electrons (to become Mn$^{6+}$), the configuration becomes $[\text{Ar}]\,3d^1$. That’s a single unpaired electron in the 3d subshell.

2. Magnetic property

   Any species with one or more unpaired electrons is paramagnetic. Since Mn$^{6+}$ has one unpaired electron, $\text{K}_2\text{MnO}_4$ is paramagnetic. (Diamagnetic would require all electrons paired, which is not the case here.)

3. Colour …

The question asks which of these hexaaqua complexes does not exist. All four ions belong to Group 13 (boron family) and carry a +3 charge. The difference lies in their ionic sizes, which dramatically affect their chemistry.

1. The concept: polarising power and hydration.

   A small, highly charged cation strongly attracts the lone pairs on water molecules. This is called high polarising power (charge-to-size ratio). For $\mathrm{B}^{3+}$, the ionic radius is tiny (about 27 pm). The intense positive field pulls electron density so strongly from the coordinated water molecules that the O–H bonds weaken, and a proton can be lost — hydrolysis occurs. Instead of a stable $[\mathrm{B}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ ion, boron in water forms boric acid, $\mathrm{B}(\mathrm{OH})_3$, or related species.

2. Compare the ions.

   As we go down Group 13, ionic size increases:

   $\mathrm{B}^{3+} < \mathrm{Al}^{3+} < \mathrm{Ga}^{3+} < \mathrm{In}^{3+}$.

   The charge remains +3, so polarising power decreases down the group.

   • $\mathrm{Al}^{3+}$ is large enough to form a stable hexaaqua complex; $[\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ is well-known in acidic solutions.
   • $\mathrm{Ga}^{3+}$ and $\mathrm{In}^{3+}$ are even larger, so their hexaaqua complexes also exist.

3. The specific case of boron. …