Q34

1 markSection D

(c) $[NiCl_4]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral. Why? [Atomic No.: $Ni = 28$ ]

OR

(c) Write hybridization and magnetic behaviour of the complex $[Fe(CN)_6]^{3-}$ . [Atomic No.: $Fe = 26$ ]

Q: (c) $[NiCl_4]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral. Why? [Atomic No.: $Ni = 28$] OR (c) Write hybridization and magnetic behaviour of the complex $[Fe(CN)_6]^{3-}$. [Atomic No.: $Fe = 26$]

**Primary answer — $[NiCl_4]^{2-}$ vs $[Ni(CO)_4]$** The difference is the **oxidation state of Ni and the field strength of the ligand**. - **$[NiCl_4]^{2-}$:** Ni is +2 → **$3d^8$**. $Cl^-$ is a **weak-field** ligand and does not pair the d-electrons. $Ni^{2+}$ uses **$sp^3$** hybridisation, leaving **2 unpaired electrons** → **paramagnetic** (tetrahedral). - **$[Ni(CO)_4]$:** Ni is in the **0** oxidation state → $3d^8 4s^2$. CO is a **strong-field** ligand; it forces the 4s electrons into the 3d set giving **$3d^{10}$** (all paired). Ni(0) then uses **$sp^3$** hybridisation with **0 unpaired electrons** → **diamagnetic** (tetrahedral). So both are tetrahedral, but weak-field $Cl^-$ leaves unpaired electrons while strong-field CO removes them. **OR answer — $[Fe(CN)_6]^{3-}$** - Fe is +3 → **$3d^5$**. $CN^-$ is a **strong-field** ligand → pairs the electrons ($t_{2g}^5 e_g^0$). - Hybridisation: **$d^2sp^3$** (inner-orbital, octahedral). - **1 unpaired electron** → **paramagnetic** (low-spin).

Q: Why can two tetrahedral complexes differ in magnetic behaviour?

Geometry alone does not fix magnetism. $[NiCl_4]^{2-}$ ($Ni^{2+}$, weak-field $Cl^-$) keeps 2 unpaired electrons and is paramagnetic, whereas $[Ni(CO)_4]$ ($Ni^0$, strong-field CO) pairs all electrons to $d^{10}$ and is diamagnetic — the ligand field strength and oxidation state decide.

Q: What is the hybridisation of $[Fe(CN)_6]^{3-}$?

$Fe^{3+}$ is $d^5$; strong-field $CN^-$ pairs the electrons to $t_{2g}^5$, giving inner-orbital $d^2sp^3$ hybridisation, octahedral geometry, and 1 unpaired electron (paramagnetic).

Coordination Compounds

VBT: magnetic behaviour and hybridisation of tetrahedral/octahedral complexes

Official Answer

Primary answer — $[NiCl_4]^{2-}$ vs $[Ni(CO)_4]$

The difference is the oxidation state of Ni and the field strength of the ligand.

$[NiCl_4]^{2-}$ : Ni is +2 → $3d^8$ . $Cl^-$ is a weak-field ligand and does not pair the d-electrons. $Ni^{2+}$ uses $sp^3$ hybridisation, leaving 2 unpaired electrons → paramagnetic (tetrahedral).
$[Ni(CO)_4]$ : Ni is in the 0 oxidation state → $3d^8 4s^2$ . CO is a strong-field ligand; it forces the 4s electrons into the 3d set giving $3d^{10}$ (all paired). Ni(0) then uses $sp^3$ hybridisation with 0 unpaired electrons → diamagnetic (tetrahedral).

So both are tetrahedral, but weak-field $Cl^-$ leaves unpaired electrons while strong-field CO removes them.

OR answer — $[Fe(CN)_6]^{3-}$

Fe is +3 → $3d^5$ . $CN^-$ is a strong-field ligand → pairs the electrons ( $t_{2g}^5 e_g^0$ ).
Hybridisation: $d^2sp^3$ (inner-orbital, octahedral).
1 unpaired electron → paramagnetic (low-spin).

NiCl4 2- paramagneticNi(CO)4 diamagneticsp3 hybridisationweak field CO strong fieldoxidation state Nid²sp³ inner orbitalFe(CN)6 3-magnetic behaviour

Marking Scheme

1Primary: ½ mark — $[NiCl_4]^{2-}$ is $Ni^{2+}$ ( $d^8$ ) with weak-field $Cl^-$ , $sp^3$ , 2 unpaired → paramagnetic; ½ mark — $[Ni(CO)_4]$ is Ni(0) with strong-field CO giving $d^{10}$ (all paired), $sp^3$ → diamagnetic.
2OR: ½ mark — $[Fe(CN)_6]^{3-}$ has $d^2sp^3$ hybridisation (strong-field $CN^-$ , inner orbital); ½ mark — paramagnetic with 1 unpaired electron.

Hint

Compare the oxidation state of Ni and the field strength of the ligand — $Cl^-$ is weak, CO is strong.

Quick Oral Answer

In $[NiCl_4]^{2-}$ nickel is +2 ( $d^8$ ) and weak-field $Cl^-$ leaves two unpaired electrons, so it is $sp^3$ and paramagnetic; in $[Ni(CO)_4]$ nickel is zero-valent and strong-field CO pairs everything to $d^{10}$ , so it is $sp^3$ and diamagnetic — both tetrahedral.

Analysis & Explanation

Concept

Valence Bond Theory explains geometry and magnetism through hybridisation and electron pairing driven by the ligand.

In $[NiCl_4]^{2-}$ , $Ni^{2+}$ ( $d^8$ ) with the weak-field $Cl^-$ keeps two unpaired electrons; $sp^3$ hybridisation gives a tetrahedral, paramagnetic complex.
In $[Ni(CO)_4]$ , zero-valent Ni has a $3d^8 4s^2$ arrangement; the strong π-acceptor CO promotes pairing so the 4s electrons collapse into $3d^{10}$ , leaving no unpaired electrons — $sp^3$ , tetrahedral, diamagnetic.

Exam trap

The examiner is testing whether you notice the different oxidation states ( $Ni^{2+}$ vs $Ni^0$ ) AND the ligand strengths ( $Cl^-$ weak, CO strong). Quoting only one factor loses the mark. For the OR part, remember $CN^-$ is strong-field, so $[Fe(CN)_6]^{3-}$ is inner-orbital $d^2sp^3$ , not $sp^3d^2$ .

Real-world

$[Ni(CO)_4]$ is central to the Mond process for purifying nickel; its volatility and easy thermal decomposition let nickel be separated from impurities.

Common Mistakes

1Assuming Ni has the same oxidation state in both complexes — it is +2 in $[NiCl_4]^{2-}$ but 0 in $[Ni(CO)_4]$ .
2Writing $sp^3d^2$ for $[Fe(CN)_6]^{3-}$ ; the strong-field $CN^-$ gives inner-orbital $d^2sp^3$ hybridisation.
3Concluding that geometry decides magnetism — both Ni complexes are tetrahedral yet differ in magnetic behaviour because of ligand field strength.

Interesting Facts

$[Ni(CO)_4]$ is a colourless, highly toxic, volatile liquid (b.p. ~43 °C) used industrially in the Mond process to refine nickel.

The strong π-back-bonding from filled metal d-orbitals into CO's empty $\pi^*$ orbitals is what makes CO such a strong-field ligand.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why can two tetrahedral complexes differ in magnetic behaviour?

Geometry alone does not fix magnetism. $[NiCl_4]^{2-}$ ( $Ni^{2+}$ , weak-field $Cl^-$ ) keeps 2 unpaired electrons and is paramagnetic, whereas $[Ni(CO)_4]$ ( $Ni^0$ , strong-field CO) pairs all electrons to $d^{10}$ and is diamagnetic — the ligand field strength and oxidation state decide.

What is the hybridisation of $[Fe(CN)_6]^{3-}$ ?

$Fe^{3+}$ is $d^5$ ; strong-field $CN^-$ pairs the electrons to $t_{2g}^5$ , giving inner-orbital $d^2sp^3$ hybridisation, octahedral geometry, and 1 unpaired electron (paramagnetic).