Coordination Compounds

The IUPAC name for [Ni(NH$_3$)$_4$] [NiCl$_4$] is : Tetraamminenickel (II) - tetrachloronickelate (II)

The name of the complex cation is written first followed by the name of the complex anion. Complex cation contains four ammine ligands and complex anion contains four chloride ligands.

The oxidation state of Fe in the complex, [Fe(H$_2$O)$_5$NO]SO$_4$ is +1 [NO has +1 charge]

Fe$^+$: [Ar] 3d$^+$4s$^1$

When the weak ligand H$_2$O and strong ligand NO attacks, the configuration changes.

NO$^+$ causes pairing of 4s electrons inside. Thus the configuration is 3d$^7$ and number of unpaired electrons = 3.

Fe$^+$ : [Ar] 3d$^7$4s$^0$

$\therefore$ Fe$^+$ has 3 unpaired electrons.

For a complex to be optically active, it should not possess either centre of symmetry or axis of symmetry or plane of symmetry. Here, the given complex is an octahedral complex that possesses plane of symmetry as well as axis of symmetry and thus is optically inactive.

The $cis$ and $trans$ both form of complex [M(NH$_3$)$_4$Cl$_2$] are optically inactive.

(A) $\mathrm{Mn^+=3}d^5 4s^1$; in presence of CO, effective configuration = $3d^64s^0$

Three lone pair of electrons in metal 3d orbitals for back bonding with vacant ($\pi$) pi* orbital of C in CO.

(B) $\mathrm{Fe^0}=3d^6 4s^2$; in presence of CO, effective configuration = $3d^8$

Four lone pair for back bonding with CO.

(C) $\mathrm{Cr^0}=3d^5 4s^1$; effective configuration = $3d^6$

Three lone pair for back bonding with CO.

(D) $\mathrm{V^-}=3d^4 4s^2$; effective configuration = $3d^6$

Three lone pair for back bonding with CO.

Since, Vanadium has a negative charge, it is present in lower oxidation state and can easily back donate the electrons from metal $d$ orbitals to ($\pi$) pi* orbital of carbonyl ligand. Hence, it forms a much stronger bond with carbon or carbonyl thereby weakening the carbon oxygen bond.

(A) The complex [Co(NH$_3$)$_4$(H$_2$O)$_2$]Cl$_2$ shows geometrical isomers (cis-trans isomers). It is paramagnetic due to presence of unpaired electrons. The central metal (Co) ion has +2 oxidation state.

(B) The complex [Pt(NH$_3$)$_2$Cl$_2$] shows geometrical isomers (cis-trans isomers). The complex is diamagnetic as all the electrons are paired. The Pt metal ion has +2 oxidation state.

(C) The complex [Co(H$_2$O)$_5$Cl]Cl cannot show geometrical isomerism. It is paramagnetic due to presence of unpaired electrons. The cobalt ion has +2 oxidation state.

(D) The complex [Ni(H$_2$O)$_6$]Cl$_2$ is paramagnetic in nature due to presence of unpaired electrons. In [Ni(H$_2$O)$_6$]Cl$_2$, oxidation state of nickel is +2

Ni$^{2+}$ : [Ar] 3d$^8$ 4s$^2$;

2 unpaired electrons.

The IUPAC name for the complex $[\text{Co(NO}_2)(\text{NH}_3)_5]\text{Cl}_2$ is: Pentaammine(nitrito-$\text{N,O}$)cobalt(III) chloride

Here's how to arrive at the name:

The complex is a cobalt complex, and cobalt has a +3 charge in this compound (since it has one negatively charged NO$_2$ ligand and one negatively charged Cl ion, for a total of -2, to balance the +3 charge on cobalt).

The ligands are NH$_3$ (ammonia) and NO$_2$ (nitrito).

Ammonia is a neutral ligand, so its name is just "ammine" and it is designated by the prefix "penta" because there are five of them.

Nitrito is a negatively charged ligand, so its name is "nitrito-$\text{N,O}$" (the "-$\text{N,O}$" indicates that the NO$_2$ ligand is bound through both the nitrogen and oxygen atoms).

Finally, the compound has a chloride ion, so its name is "chloride" and it is designated by the prefix "di" because there are two of them.

Putting it all together, we get the name "pentaammine(nitrito-$\text{N,O}$)cobalt(III) chloride".

The first step is to find out the chemical formula for complex A and B.

The complete reaction for the formation of A and B are given below:

$\mathrm{NiCl}_{2}+\mathrm{KCN}\left(\right.$ excess) $\rightarrow \mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]+4 \mathrm{KCl}$

The complex A is Tetra cyano nickelate (II) $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$.

$\mathrm{NiCl}_{2}+$ Conc. $\mathrm{HCl}$ (excess) $\rightarrow$ $\mathop {{{[\mathrm{NiC{l_4}}]}^{2 - }}}\limits_{\mathrm{Chloro\,complex}}$

$\mathrm{K_2[Ni(CN)_4]+4~Conc.~HCl(excess)\to K_2[Ni(Cl)_4]^{2-}+4HCN}$

The complex B is $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$.

The name of the complex A is Potassium tetracyanonickelate (II).

The name of the complex B is Potassium tetrachloronickelate (II).

The complex A is $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$.

The electronic configuration of $\mathrm{Ni}^{2+}$ ion is given below.

Electronic configuration of $\mathrm{Ni}^{2+}$ :

$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 $

The $3 d$ orbital of $\mathrm{Ni}^{2+}$ can be shown as:

Since, cyanide is a strong ligand, electrons in $3 d$ orbitals are paired up. One $3 d$, one $4 s$ and two $4 p$ orbitals undergo hybridization to form four $d s p^2$ hybrid orbitals. Hence, it is diamagnetic in nature. The $d s p^2$ hybridisation occur. The four cyanide ions $\left(\mathrm{CN}^{-}\right)$ions donate electrons to vacant $d s p^2$ hybrid orbitals.

In $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$, potassium is a counter ion and complex ion is $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$. In $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}, \mathrm{Ni}$ exists as $\mathrm{Ni}^{2+}$ ion.

The electronic configuration of $\mathrm{Ni}^{2+}$ ion is given below.

Electronic configuration of $\mathrm{Ni}^{2+}$ :

$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 $

Since, chloride is a weak ligand, electrons in $3 d$ orbitals remains unpaired. One $4 s$ and three $4 p$ orbitals under hybridisation to form four $s p^3$ hybridised orbitals. The $4 \mathrm{Cl}^{-}$ligands will accommodate electrons in empty $s p^3$ hybrid orbitals.

As shown above, there are 2 unpaired electrons present in $\mathrm{Ni}^{2+}$. Hence, it is paramagnetic in nature.

Therefore, A is diamagnetic and B is paramagnetic with two unpaired electrons.

The complex A has $d s p^2$ hybridisation and complex B has $s p^3$ hybridisation as discussed above.

The reaction involving neutron capture by $\mathrm{N}-14$ to form $\mathrm{C}-14$ of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ is given as. The proportion of living matter is $1: 10^{12}$.

The structure of metal carbonyl complex, $\mathrm{Fe}(\mathrm{CO})_5$ is shown below:

In $\mathrm{Fe}(\mathrm{CO})_5$ complex, CO groups are ligands attached to central metal ion, Fe. This happens by donation of lone pair of electrons on carbonyl carbon to vacant d orbitals of metal forming a sigma bond and back donation of electron by metal d orbitals to pi ( $\pi$ ) molecular orbitals of carbonyl group forming pi ( $\pi$ ) bond. The back bonding between Fe and CO groups generates a synergic effect which leads to the strengthening of bond between metal ion and ligands and weakening the bond between carbon and oxygen.

This also results in increase in bond length of $\mathrm{C} \equiv \mathrm{O}$ slightly.

On comparing the bond length given in options A to D, option A shows bond length as $1.15 \mathop {\rm{A}}\limits^{\rm{o}}$. This is slightly higher than that of previous bond length that is $1.128 \mathop {\rm{A}}\limits^{\rm{o}}$. The option C shows bond length as $1.72 \mathop {\rm{A}}\limits^{\rm{o}}$ which is much higher than $1.128\mathop {\rm{A}}\limits^{\rm{o}}$.

Hence, Option C is incorrect.

To determine the correct option, we should evaluate the oxidation states of the metals in the given compounds to see in which cases the metals are in their highest oxidation states.

Option A: MnO₂, FeCl₃

For MnO₂, manganese (Mn) has an oxidation state of +4. While this is a high oxidation state for Mn, it is not its highest, as it can go up to +7 (as in KMnO₄).

For FeCl₃, iron (Fe) has an oxidation state of +3, which is its highest stable oxidation state.

So, this option is not entirely correct.

Option B: [MnO₄]^-, CrO₂Cl₂

For [MnO₄]^-, Mn is in the +7 oxidation state, which is indeed its highest oxidation state.

In CrO₂Cl₂, chromium (Cr) is in the +6 oxidation state, which is its highest oxidation state.

Both Mn and Cr are in their highest oxidation states, making this option correct.

Option C: [Fe(CN)₆]^3-, [Co(CN)₆]^3-

For [Fe(CN)₆]^3-, Fe is in the +3 oxidation state, its highest stable oxidation state.

However, [Co(CN)₆]^3- seems to be incorrectly provided as it suggests an oxidation number that does not fit standard cobalt complexes or could be a typographical error because typically, cobalt in a cyanide complex like [Co(CN)₆]^3- would suggest an oxidation state of +3 for cobalt. Cobalt can exist in +2 and +3 oxidation states commonly, with +3 being very stable in this coordination compound. But it isn't necessarily the highest oxidation state Co can achieve (it could theoretically go up to +4, but such compounds are very rare and not stable).

Therefore, there seems to be a misunderstanding in the elements present or typo in this option for Co. Considering the usual cobalt complex and common oxidation states, though Co is in a high oxidation state (+3), the context seems mixed up, and noting the rarity and instability of Co(IV) compounds, the focus should rather be on the correctness of available options against the question's criteria.

Option D: [NiCl₄]^2-, [CoCl₄]^-

For [NiCl₄]^2-, Ni is in the +2 oxidation state, which is not its highest (it can go up to +3).

The [CoCl₄]^- mistakenly listed suggests cobalt in an oxidation state lower than its maximum, specifically +2, considering the anion's charge and typical coordination chemistry. Like Ni, Co in +2 is common but not its highest.

This option does not have metals in their highest oxidation states either.

Conclusion

Option B, with [MnO₄]^- and CrO₂Cl₂, correctly presents the metals (Mn and Cr) in their highest oxidation states, making it the correct choice.

The complex compound Co(NH₃)₅Cl₃ dissociates in water to form three moles of ions. This dissociation can be represented as:

$ \text{Co(NH}_3\text{)}_5\text{Cl}_3\,\,\leftrightharpoons\,\,\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]^{+2} + 2\text{Cl}^- $

Therefore, the structure of the complex is:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 $

When this complex reacts with two moles of AgNO₃ solution, it produces:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 + 2\text{AgNO}_3 \to \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right](\text{NO}_3)_2 + 2\text{AgCl} $

This reaction yields two moles of AgCl, confirming the complex structure as:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 $

The brown ring complex compound is given as [Fe(H₂O)₅(NO)⁺]SO₄. To determine the oxidation state of iron (Fe) in this complex, we need to analyze the components and their charges:

1. Let the oxidation state of Fe be $x$.

2. The water molecules (H₂O) are neutral ligands; hence, they do not contribute to the overall charge.

3. The nitrosyl ligand (NO) in this complex is positively charged, $NO^+$, contributing a +1 charge.

4. The sulfate ion (SO₄) is doubly charged, with a -2 charge.

The overall charge of the complex compound, [Fe(H₂O)₅(NO)⁺], must balance the charge of the sulfate ion (SO₄):

Since the sulfate ion $SO_4^{2-}$ has a charge of -2, the complex ion must have a charge of +2 to balance it out:

Therefore:

$ x + 5(0) + 1 = +2 $

$ x + 1 = +2 $

$ x = +1 $

Thus, the oxidation state of iron (Fe) in the brown ring complex compound is +1.

The correct answer is:

Option A

1