Coordination Compounds

hybridization

hybridization

hybridization

hybridization

diamagnetic,

hybridisation, number of unpaired electrons

paramagnetic,

hybridisation, number of unpaired electrons

(A) Strength of anionic ligands cannot be explained by CFT instead LFT i.e., ligand field theory explains the strength of ligands.

(B) VBT does not give a quantitative interpretation of kinetic stability of coordination compounds.

(C) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \rightarrow \mathrm{Ni}^{2+}$ in presence of $\mathrm{CN}^{-}$ligand.

Square planar geometry since $\rightarrow 4, d s p^2$ hybridised orbitals created.

(D) cis- $\left[\mathrm{PtCl}_2(\mathrm{en})_2\right] \Rightarrow$ It has two possible isomers

Rosy red/Bright Red precipitate

Let's examine each statement for correctness: (A) Ethane-1, 2-diamine is a chelating ligand.

Ethane-1,2-diamine, also known as ethylenediamine (en), has two nitrogen atoms that can coordinate to a metal ion, forming a ring structure in the process.

Because it can form these two bonds, it can "chelate" a metal ion, thus it is correctly identified as a chelating ligand.

(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.

Aluminium is indeed produced industrially by the Hall-Héroult process, which involves the electrolysis of aluminium oxide (

) dissolved in molten cryolite (

).

Cryolite acts as a solvent for the aluminium oxide and reduces the melting point of the mixture, thus decreasing energy consumption during electrolysis.

This statement is correct.

(C) Cyanide ion is used as a ligand for leaching of silver.

Gold and silver are often extracted from their ores via a leaching process using a cyanide solution.

The cyanide ion (

) complexes with the metal ions to form soluble complexes like [Ag(CN)$_2$]$^-$, enabling the separation of silver from the ore.

So, this statement is correct as well.

(D) Phosphine act as a ligand in Wilkinson's catalyst.

Wilkinson's catalyst is

, where PPh$_3$ stands for triphenylphosphine, a type of phosphine ligand.

Phosphines are indeed ligands in Wilkinson's catalyst, and they play an important role in its catalytic activity, particularly in hydrogenation reactions.

Therefore, this statement is correct.

(E) The stability constants of

and

are similar with EDTA complexes.

EDTA (ethylenediaminetetraacetic acid) forms strong complexes with many metal ions including

and

.

However, the stability constants of their complexes with EDTA are not similar; the stability constant for the calcium complex is notably higher than that for the magnesium complex.

Thus, this statement is incorrect.

With all the information above, we can conclude: (A) is correct, (B) is correct, (C) is correct, (D) is correct, and (E) is incorrect.

Therefore, the correct statements are (A), (B), (C), and (D), making Option D—(A), (B), (C) only—the correct choice.

The coordination environment of the

ion when it forms a complex with

(ethylenediaminetetraacetic acid) is octahedral.

Ethylenediaminetetraacetic acid (EDTA) is a hexadentate ligand, which means it has six donor atoms that can bind to a central metal ion.

In the case of EDTA, these donor atoms are four oxygen atoms from its four carboxyl groups and two nitrogen atoms from its two amine groups.

When

forms a complex with

, all six donor atoms from EDTA coordinate with the calcium ion.

As a result, the coordination number of the calcium ion is 6, leading to an octahedral geometry.

This is because the octahedral geometry is the most common and energetically favorable arrangement for a coordination number of 6.

In such a geometry, the six ligands are placed at equal distances from the central ion and at 90° angles relative to adjacent ligands, maximizing the distance between all ligands to minimize repulsion.

The correct answer is Option B, octahedral.

To evaluate the assertion and the reason, let's first analyze the given complex ion

. 1. Assertion (A): The total number of geometrical isomers shown by

complex ion is three. 2. Reason (R):

complex ion has an octahedral geometry. First, we know that

is a coordination complex with

(Cobalt) in the center coordinated to two ethylenediamine (en) ligands and two chlorides (Cl).

Ethylenediamine is a bidentate ligand, meaning it binds through two donor atoms, giving the overall complex an octahedral geometry around the central cobalt ion.

To verify the assertion about the geometrical isomers: In an octahedral complex, the two chlorides and the two

ligands can have different spatial arrangements relative to each other. Specifically, for

, the possible geometrical isomers are: cis-isomer: where the two chlorides are adjacent to each other. trans-isomer: where the two chlorides are opposite each other.

These two configurations are the typical geometrical isomers for this type of complex.

Thus, the assertion that there are three geometrical isomers appears incorrect, as the common understanding is that there are only two geometrical isomers for this type of octahedral complex.

Now, let’s consider the reason: The reason states that

has an octahedral geometry.

This statement is indeed correct because the coordination number of 6 (from the four donor atoms of the two

ligands and the two chloride ions) leads to an octahedral geometry.

Therefore, the most appropriate answer is: Option B: (A) is not correct but (R) is correct

unpaired electrons

unpaired electrons Above 2 complex have even number of unpaired electrons.

First, let us determine the $d$-electron count of the cobalt center in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$.

1.

Oxidation state and $d$-electron count Neutral cobalt (Co) has an atomic number of 27 and an electronic configuration: $\mathrm{Co} \; \bigl[\mathrm{Ar}\bigr]\, 3d^7\, 4s^2.$ In the +3 oxidation state, cobalt has lost a total of 3 electrons: First two electrons are removed from the $4s$ orbital.

The third electron is removed from the $3d$ orbitals.

Thus, $\mathrm{Co}^{3+}$ has $3d^{7-1} = 3d^6.$ So in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$, the cobalt center is $d^6$.

2.

High-spin vs Low-spin for $\mathrm{Co}^{3+}$ Even though $\mathrm{H}_2\mathrm{O}$ is generally considered a weak field ligand, the $\mathrm{Co}^{3+}$ ion has a relatively high charge (+3).

A higher charge on the metal center usually increases the ligand-field splitting ($\Delta_\mathrm{o}$) significantly compared to lower oxidation states of the same metal.

In most typical octahedral complexes of $\mathrm{Co}^{3+}$, the splitting $\Delta_\mathrm{o}$ is large enough that the complex ends up being low spin.

Electronic configuration in an octahedral field For a $d^6$ octahedral low-spin complex, all six electrons pair up in the lower-energy $\mathrm{t_{2g}}$ orbitals: $\mathrm{t_{2g}^6}\, \mathrm{e_g^0}$ That leaves 0 unpaired electrons.

3.

Final answer Therefore, the number of unpaired $d$-electrons in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$ is $\boxed{0}.$ Correct Option: A (0)

Hybridisation of

in

is

but the hybridisation of

in

is

. So, statement-I is false. Hybridisation of

in

is

but the hybridisation of

in

is

as

is a strong field ligand forcing the unpaired electrons to pair up. So, statement-II is false.