Q.Explain on the basis of valence bond theory that $[Ni(CN)_4]^{2-}$ ion with square planar structure is diamagnetic and the $[NiCl_4]^{2-}$ ion with tetrahedral geometry is paramagnetic.

Concept & Intuition

In coordination chemistry, primary valency refers to the oxidation state of the central metal ion — it’s the charge the metal would have if all ligands were removed as neutral molecules or ions. Secondary valency is the coordination number — the total number of ligand donor atoms directly bonded to the metal. The trick is that some ligands (like oxalate, $\mathrm{C}_2\mathrm{O}_4^{2-}$) are bidentate, so they count as two donor atoms even though they are one ligand molecule.

1. Determine the oxidation state of chromium (primary valency $x$).

   The complex ion is $[\mathrm{CrCl}_2(\mathrm{C}_2\mathrm{O}_4)_2]^{3-}$.

   • Each chloride ligand ($\mathrm{Cl}^-$) carries a charge of $-1$.
   • Each oxalate ligand ($\mathrm{C}_2\mathrm{O}_4^{2-}$) carries a charge of $-2$.
   • The overall charge on the complex is $-3$.

   Let the oxidation state of Cr be $n$. Then:

So primary valency $x = 3$.

2. Determine the coordination number (secondary valency $y$).

   • Each $\mathrm{Cl}^-$ is monodentate (donates one pair of electrons), so two chlorides contribute $2$ donor atoms.
   • Each $\mathrm{C}_2\mathrm{O}_4^{2-}$ is bidentate (donates two pairs of electrons via two oxygen atoms), so two oxalates contribute $2 \times 2 = 4$ donor atoms.
   • Total donor atoms = $2 + 4 = 6$.

   Hence secondary valency $y = 6$.

3. Sum $x$ and $y$.

The question asks which artificial sweetener contains a chlorine atom ($-\text{Cl}$) in its structure and remains stable even at cooking temperatures. Let’s break down the reasoning.

Concept and intuition:

Artificial sweeteners are often modified sugars or synthetic compounds. Stability at cooking temperature means the molecule does not break down when heated (e.g., during baking). Chlorine substitution can enhance thermal stability because the C–Cl bond is strong and resists hydrolysis or thermal degradation. Among the options, only one is known to be a chlorinated sugar derivative.

Step-by-step reasoning:

1. Aspartame is a dipeptide (aspartic acid + phenylalanine methyl ester). It contains no chlorine atoms and decomposes at high temperatures, so it is unsuitable for cooking.

   → Eliminate (A).

2. Alitame is also a dipeptide (aspartic acid + alanine amide), but it does not contain chlorine. It is more stable than aspartame but still not ideal for prolonged high heat.

   → Eliminate (B).

3. Saccharin is a sulfonimide (benzosulfimide) with no chlorine in its structure. It is heat-stable, but the question specifically requires a chlorine atom.

   → Eliminate (C).

4. Sucralose is made by chlorinating sucrose: three hydroxyl groups ($-\text{OH}$) are replaced by chlorine atoms ($-\text{Cl}$). This gives it the formula $\text{C}_{12}\text{H}_{19}\text{Cl}_3\text{O}_8$. The chlorine atoms make it exceptionally stable at high temperatures, so it can be used in baking and cooking.

   → This matches both conditions.

Concept & Intuition

Ambidentate ligands are like molecular “two‑handed” tools — they have two different atoms that can donate a lone pair to a metal. For example, the thiocyanate ion ($\ce{SCN^-}$) can bind through sulfur or through nitrogen. The key is that the two possible donor atoms must be different elements (or at least chemically distinct). If a ligand has only one type of donor atom (even if it has many of them), it is not ambidentate — it’s just polydentate. Here, we check each ion for two chemically distinct binding sites.

Step‑by‑step reasoning

1. $\ce{CN^-}$ (cyanide)

   • The carbon and nitrogen atoms both have lone pairs.
   • Cyanide usually binds through carbon (giving a cyano complex), but it can also bind through nitrogen (giving an isocyano complex).
   • Because it can coordinate via two different atoms (C or N), it is ambidentate.

2. $\ce{SCN^-}$ (thiocyanate)

   • Sulfur and nitrogen each have lone pairs available.
   • It can bind through sulfur (thiocyanato) or through nitrogen (isothiocyanato).
   • Hence it is ambidentate.

3. $\ce{SO4^{2-}}$ (sulfate)

   • All four oxygen atoms are equivalent in terms of donor ability; the sulfur has no lone pair to donate.
   • Sulfate can bind as a monodentate ligand (through one oxygen) or as a bidentate ligand (through two oxygens), but every binding site is an oxygen atom.
   • Since there is only one type of donor atom, it is not ambidentate — it is simply polydentate (or monodentate). This is the classic pitfall: students confuse “multiple binding sites” with “different binding atoms.”

4. $\ce{NO_2^-}$ (nitrite)

   • It can bind through the nitrogen atom (nitro complex) or through one of the oxygen atoms (nitrito complex).
   • Two different donor atoms (N and O) → is ambidentate.

Concept & Intuition

In a cubic close-packed (ccp) arrangement of oxygen atoms, each unit cell contains 4 O atoms. The key void relationships are:

• Number of tetrahedral voids = 2 × (number of atoms) = 8 per unit cell.
• Number of octahedral voids = 1 × (number of atoms) = 4 per unit cell.

The molecular formula AB₂O₄ tells us the ratio of A : B : O atoms in the crystal. Since O atoms form the lattice, we can determine how many A and B atoms are present per unit cell, then see how many voids they occupy. The question asks for the fraction of octahedral voids that remain empty.

Step-by-step reasoning

1. Determine the number of O atoms per unit cell.

   In a ccp lattice, there are 4 atoms per unit cell. So O atoms per unit cell = 4.

2. Relate the formula to the unit cell contents.

   The formula AB₂O₄ means for every 4 O atoms, there is 1 A atom and 2 B atoms. Since we have exactly 4 O atoms per unit cell, the unit cell contains:

   • A atoms = 1
   • B atoms = 2

3. Find how many tetrahedral voids are occupied.

   Total tetrahedral voids in the unit cell = 2 × 4 = 8.

   A occupies $\frac{1}{8}$ of these:

This matches the 1 A atom we need — so all A atoms are in tetrahedral voids.

4. Find how many octahedral voids are occupied by B.

   Total octahedral voids = 4.

   B atoms needed = 2.

   So the fraction of octahedral voids occupied = $\frac{2}{4} = \frac{1}{2}$.

5. Calculate the fraction of vacant octahedral voids.

   Vacant fraction = $1 - \frac{1}{2} = \frac{1}{2}$.

Concept & Intuition

Oxidation state is the formal charge on the central atom after assigning all bonding electrons to the more electronegative atom. Covalency is simply the number of covalent bonds the central atom forms with its ligands. In coordination complexes, each ligand donates a pair of electrons to the metal, so the number of ligands (times their denticity) gives the covalency. Here, water is a neutral ligand, and chloride is a monovalent anionic ligand. The overall charge of the complex tells us the oxidation state of the metal.

Step-by-step reasoning

1. Identify the ligands and their charges

   • Water ($\text{H}_2\text{O}$) is a neutral ligand: charge = 0.
   • Chloride ($\text{Cl}^-$) carries a charge of $-1$.

2. Set up the charge balance equation

   Let the oxidation state of Al be $x$. The complex ion has an overall charge of $+2$.

   The sum of charges from the metal and all ligands equals the complex charge:

This simplifies to:

So the oxidation state of Al is $+3$.

3. Determine the covalency Covalency is the number of bonds formed by the central atom.
   • Each $\text{H}_2\text{O}$ donates one lone pair to Al, forming one coordinate bond. Five water molecules → 5 bonds.
   • The $\text{Cl}^-$ also donates a lone pair, forming one bond. Total bonds = $5 + 1 = 6$. Hence, the covalency of Al in this complex is 6.

4. Match with the options Oxidation state $+3$, covalency $6$ corresponds to option (B).

Concept and Intuition

A chloro-bridged structure occurs when a chlorine atom is shared between two metal centers, forming a bridge (e.g., $\mathrm{M-Cl-M}$). This typically happens when the metal is electron-deficient (small, high charge density) and can achieve a higher coordination number by sharing halide ligands. In contrast, metals that are too large or too electropositive tend to form ionic lattices with discrete $\mathrm{Cl^-}$ ions, not bridges. At room temperature, the solid-state structure is key: we look for a polymeric network rather than isolated molecules or simple ionic packing.

Step-by-Step Reasoning

1. Identify the nature of each compound

   • $\mathrm{BeCl_2}$: Beryllium is small (ionic radius ~31 pm) and has high charge density. It is electron-deficient and tends to form covalent polymers.
   • $\mathrm{MgCl_2}$: Magnesium is larger (ionic radius ~72 pm) and more electropositive; it forms an ionic lattice with $\mathrm{Mg^{2+}}$ and $\mathrm{Cl^-}$ ions.
   • $\mathrm{PbCl_2}$: Lead is large and polarizable; it forms an ionic structure (though with some covalent character) but not chloro-bridged chains.
   • $\mathrm{ZnCl_2}$: Zinc is intermediate; in the solid state, it can form a layered structure, but at room temperature it is typically ionic (or molecular in the gas phase), not chloro-bridged.

2. Recall known solid-state structures

   • $\mathrm{BeCl_2}$: In the solid state, it forms infinite chains where each beryllium is tetrahedrally coordinated by four chlorine atoms, and each chlorine bridges two beryllium atoms. This is a classic example of a chloro-bridged polymer.
   • $\mathrm{MgCl_2}$: Adopts the cadmium chloride ($\mathrm{CdCl_2}$) layer structure, which is ionic with octahedral $\mathrm{Mg^{2+}}$ ions, but no bridging chlorines between separate $\mathrm{Mg}$ centers in the sense of a discrete bridge — it’s a close-packed ionic lattice.
   • $\mathrm{PbCl_2}$: Has a complex ionic structure (cotunnite type) with $\mathrm{Pb^{2+}}$ in a nine-coordinate environment, but again not a simple chloro-bridged chain.
   • $\mathrm{ZnCl_2}$: At room temperature, it exists in several polymorphs; the common α-form has a tetrahedral network but with $\mathrm{ZnCl_4}$ units sharing corners, not edges (i.e., not chloro-bridged in the same sense as $\mathrm{BeCl_2}$). In fact, $\mathrm{ZnCl_2}$ is often described as having a layered structure with some bridging, but it is not the classic linear chloro-bridged polymer.

3. Focus on the defining feature

   A “chloro bridged structure” in this context typically refers to a chain or ring where each chlorine is bonded to two metal atoms, as seen in $\mathrm{BeCl_2}$ (and also in $\mathrm{AlCl_3}$ dimer in gas phase, but here we consider solid). $\mathrm{BeCl_2}$ is the only one among the options that is well-known for this polymeric chain at room temperature.

4. Eliminate the others

   • $\mathrm{MgCl_2}$: Ionic, no bridging.
   • $\mathrm{PbCl_2}$: Ionic, no bridging.
   • $\mathrm{ZnCl_2}$: While it can form dimers in the gas phase, the solid at room temperature is not a simple chloro-bridged chain; it’s a network solid with tetrahedral coordination but not the same edge-sharing chain as $\mathrm{BeCl_2}$.

The question asks which of the listed metal carbonyl complexes contain bridging CO ligands — that is, carbonyl groups that are bonded to more than one metal atom, not just one. This is a structural feature that arises from the need to satisfy the 18-electron rule and the coordination preferences of the metal atoms.

Bridging CO ligands are common in polynuclear carbonyls where metal–metal bonds also exist. The CO group can act as a two-electron donor to two metals simultaneously (μ₂-bridge) or sometimes to three metals (μ₃-bridge). The key is to know the known structures of these classic carbonyls.

Let’s examine each complex one by one.

1. \left[\mathrm{Co}_2(\mathrm{CO})_8 — Dicobalt octacarbonyl. This molecule has a Co–Co bond. In the solid state, it has two bridging CO groups and six terminal CO groups (three on each Co). In solution, it can equilibrate with a non-bridged isomer, but the bridged form is well-established. So I contains CO bridges.

2. \left[\mathrm{Fe}_3(\mathrm{CO})_{12} — Triiron dodecacarbonyl. Its structure is a triangle of iron atoms with an Fe–Fe bond along each edge. One CO group bridges each edge of the triangle (three μ₂-CO bridges), and the remaining nine CO groups are terminal (three per Fe). So II contains CO bridges.

3. \left[\mathrm{Mn}_2(\mathrm{CO})_{10} — Dimanganese decacarbonyl. This molecule has a direct Mn–Mn bond, but all ten CO groups are terminal — five on each Mn, arranged in an octahedral-like geometry around each metal. There are no bridging CO ligands. So III does NOT contain CO bridges.

4. \left[\mathrm{Fe}_2(\mathrm{CO})_9 — Diiron nonacarbonyl. This is a classic example: it has three bridging CO groups between the two iron atoms, and six terminal CO groups (three on each Fe). The Fe–Fe bond is also present. So IV contains CO bridges.

Thus, the complexes with CO bridges are I, II, and IV.

The electronic configuration of an atom describes how electrons are distributed among its atomic orbitals. This distribution generally follows the Aufbau principle, Hund's rule, and the Pauli exclusion principle. However, certain elements, particularly transition metals like Chromium (Cr) and Copper (Cu), exhibit exceptions to these rules due to the enhanced stability associated with half-filled or fully-filled subshells.

The key concept here is that orbitals that are exactly half-filled (e.g., $d^5$) or completely filled (e.g., $d^{10}$) have extra stability. This stability arises from two main factors:

1. Symmetry: A half-filled or fully-filled subshell has a more symmetrical distribution of electrons, which leads to a lower energy state.
2. Exchange Energy: Electrons with the same spin in degenerate orbitals can exchange their positions. The more exchange possibilities, the greater the stabilization energy (exchange energy). Half-filled and fully-filled subshells maximize these exchange possibilities.

Let's determine the valency shell electronic configurations for Cr and Cu.

1. Chromium (Cr, Atomic Number $Z=24$)

   • The general electronic configuration for an atom with $Z=24$ would typically follow the Aufbau principle, filling orbitals in increasing order of energy. The noble gas core for Cr is Argon ($[Ar]$), which has 18 electrons.
   • After $[Ar]$, the next orbitals to fill are $3d$ and $4s$. According to the Aufbau principle, $4s$ fills before $3d$.
   • The expected configuration would be $[Ar] 3d^4 4s^2$.
   • However, a $3d^5$ configuration (half-filled $d$ subshell) is significantly more stable than $3d^4$. To achieve this, one electron from the $4s$ orbital shifts to the $3d$ orbital.
   • Therefore, the actual valency shell electronic configuration for Cr is $[Ar] \mathbf{3d^5 4s^1}$.
   Watch out

   Always remember that $4s$ and $3d$ orbitals are very close in energy. This proximity allows for the electron shift to achieve the more stable half-filled or fully-filled $d$ subshell.

2. Copper (Cu, Atomic Number $Z=29$)

   • Similar to Chromium, we start with the Argon core ($[Ar]$) for Copper, which has 18 electrons.
   • The expected configuration, following the Aufbau principle, would be $[Ar] 3d^9 4s^2$.
   • However, a $3d^{10}$ configuration (fully-filled $d$ subshell) is significantly more stable than $3d^9$. To achieve this, one electron from the $4s$ orbital shifts to the $3d$ orbital.
   • Therefore, the actual valency shell electronic configuration for Cu is $[Ar] \mathbf{3d^{10} 4s^1}$.

   The general rule for these exceptions is:

   $d^4 s^2 \rightarrow d^5 s^1$ (for half-filled stability)

   $d^9 s^2 \rightarrow d^{10} s^1$ (for fully-filled stability)

3. Comparing with the options:

   • We found the configuration for Cr to be $3d^5 4s^1$.
   • We found the configuration for Cu to be $3d^{10} 4s^1$.
   • Let's check the given options: (A) $3d^4 4s^2 ; 3d^{10} 4s^1$ (Cr is incorrect) (B) $3d^5 4s^1 ; 3d^{10} 4s^1$ (Both Cr and Cu are correct) (C) $3d^4 4s^1 ; 3d^9 4s^2$ (Both Cr and Cu are incorrect) (D) $3d^4 4s^2 ; 3d^9 4s^2$ (Both Cr and Cu are incorrect, these are the expected configurations without considering stability)

The option that correctly lists the valency shell electronic configurations for Cr and Cu is (B).

The key idea is the barycentre (centre of gravity) rule: in an octahedral field, the total energy of all five d‑orbitals must remain the same as in the free ion. The splitting energy $\Delta_o$ is the difference between the $e_g$ and $t_{2g}$ levels. If we set the barycentre at zero, the three $t_{2g}$ orbitals are lowered by some amount and the two $e_g$ orbitals are raised by some amount, such that the weighted sum is zero.

1. Set up the energy balance Let the energy of each $t_{2g}$ orbital be $-x$ (lowered) and each $e_g$ orbital be $+y$ (raised). There are three $t_{2g}$ and two $e_g$ orbitals. The barycentre condition is:

2. Use the definition of $\Delta_o$ The splitting energy $\Delta_o$ is the energy difference between an $e_g$ orbital and a $t_{2g}$ orbital:

3. Solve the system From step 1: $y = \frac{3}{2}x$. Substitute into step 2:

Then

4. Interpret the result Each $e_g$ orbital is raised by $\frac{3}{5}\Delta_o$ above the barycentre. The question asks: “the energy of two $e_g$ orbitals will be increased by” — this phrasing means the increase per orbital (since the options are fractions of $\Delta_o$). So the answer is $\frac{3}{5}\Delta_o$.

Concept & Intuition

Crystal field theory treats the ligands as point negative charges that repel the d‑electrons. In an octahedral field, the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals (the $e_g$ set) point directly at the ligands, so they feel stronger repulsion and rise in energy. The other three orbitals ($d_{xy}, d_{xz}, d_{yz}$ — the $t_{2g}$ set) point between the ligands, so they are less repelled and drop in energy.

The key constraint is energy conservation: the total energy of all five d‑orbitals must remain the same as in the free ion (where they are degenerate). So the total destabilization of the $e_g$ orbitals must exactly balance the total stabilization of the $t_{2g}$ orbitals.

Let the energy splitting between the two sets be $\Delta_o$ (often called $10Dq$). If we set the average energy of the five orbitals to zero, then the $t_{2g}$ orbitals are each lowered by some amount $x$, and the $e_g$ orbitals are each raised by some amount $y$, with $y - x = \Delta_o$.

Step‑by‑Step Reasoning

1. Set up the energy conservation equation. There are three $t_{2g}$ orbitals and two $e_g$ orbitals. The total energy change relative to the average must be zero:

2. Relate $x$ and $y$ to the splitting $\Delta_o$. By definition, the gap between the $e_g$ and $t_{2g}$ levels is $\Delta_o$:

3. Solve the system. From $2y = 3x$ we get $x = \frac{2}{3}y$. Substitute into $y + x = \Delta_o$:

Then $x = \frac{2}{3} \cdot \frac{3}{5}\Delta_o = \frac{2}{5}\Delta_o$.

4. Interpret the result. Each $e_g$ orbital is raised by $\frac{3}{5}\Delta_o$ above the average (barycenter), and each $t_{2g}$ orbital is lowered by $\frac{2}{5}\Delta_o$ below it. The question asks for the increase in energy of the two $e_g$ orbitals — that is $\frac{3}{5}\Delta_o$ per orbital.