Coordination Compounds
Match the complexes in Column I with their properties listed in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | $\mathrm{[Co(NH_3)_4(H_2O)_2]Cl_2}$ | (P) | geometrical isomers |
| (B) | $\mathrm{[Pt(NH_3)_2Cl_2]}$ | (Q) | paramagnetic |
| (C) | $\mathrm{[Co(H_2O)_5Cl]Cl}$ | (R) | diamagnetic |
| (D) | $\mathrm{[Ni(H_2O)_6]Cl_2}$ | (S) | metal ion with +2 oxidation state |
The IUPAC names of $\mathbf{A}$ and $\mathbf{B}$ are
Predict the magnetic nature of $\mathbf{A}$ and $\mathbf{B}$.
Both are diamagnetic.
A is diamagnetic and B is paramagnetic with one unpaired electron.
A is diamagnetic and B is paramagnetic with two unpaired electrons.
Both are paramagnetic.
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.
The hybridisation of $A$ and $B$ are :
$d s p^2, s p^3$
$s p a, s p^3$
$d s p^2, d s p$
$s p^3 d^2, d^2 s p^3$
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 $1: 10^{12}$.If the bond length of CO bond in carbon monoxide is $1.128 \mathop {\rm{A}}\limits^{\rm{o}}$, then what is the value of CO bond length in $\mathrm{Fe}(\mathrm{CO})_5$?
$1.15 \mathop {\rm{A}}\limits^{\rm{o}}$
$1.128\mathop {\rm{A}}\limits^{\rm{o}}$
$1.72 \mathop {\rm{A}}\limits^{\rm{o}}$
$1.118\mathop {\rm{A}}\limits^{\rm{o}}$
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.$F{e^{3 + }}\buildrel {SN{C^ - }(excess)} \over \longrightarrow $ blood red $F{e^{3 + }}\buildrel {SN{C^ - }(excess)} \over \longrightarrow $ colourless (B):
(A) Identify compound A and B and write their IUPAC names.
(B) Determine the spin only magnetic moment of B.
Explanation:
(A) When Fe$^{3+}$ reacts with excess of SCN$^-$ ions, a blood red coloured complex is formed, which is compound A.
$\mathrm{F{e^{3 + }} + \mathop {SC{N^ - }}\limits_{excess} \to \mathop {{{[Fe(SCN){{({H_2}O)}_5}]}^{2 + }}}\limits_{A\,(blood\,red)}}$
The IUPAC name of compound A is Pentaaquathiocyanatoferrate (III) ion.
When compound A reacts with excess of fluoride ions, all water and SCN$^-$ ligands are replaced by fluoride ions producing compound B.
$\mathrm{\mathop {{{[Fe(SCN){{({H_2}O)}_5}]}^{2 + }}}\limits_A + 6{F^ - } \to \mathop {{{[Fe{F_6}]}^{3 - }}}\limits_{B\,(complex)} + SC{N^ - } + 5{H_2}O}$
The IUPAC name of compound B is Hexafluoroferrate (III) ion.
(B) The coordination number of Fe in hexafluoroferrate (III) is 6. The Fe exists in +3 oxidation state in compound B.
The magnetic moment ($\mu$) for compound B can be calculated using following formula,
$\mu = \sqrt {n(n + 2)} $ BM ...... (i)
Here, n is the number of unpaired electrons.
The number of unpaired electrons in Fe$^{3+}$ can be determined as shown.
${}_{26}Fe = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^2}$
$F{e^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5}$
Therefore, $n=5$ for Fe$^{3+}$
Substitute the value of n in equation (i),
$\mu = \sqrt {5(5 + 2)} $ BM
$\mu = \sqrt {5(7)} $ BM $ = \sqrt {35} $ BM
$\therefore$ $\mu = 5.92$ BM
Final Answer :
(A) Compound A: [Fe(SCN)(H$_2$O)$_5$]$^{2+}$ IUPAC name: Pentaaquathiocyanatoferrate (III) ion Compound B: [FeF$_6$]$^{3-}$ IUPAC name: Hexafluoroferrate (III) ion.
(B) $\mu=5.92$ BM
Hints :
Determine the oxidation state of Fe in compound B and determine the electronic configuration of Fe$^{3+}$ From that determine the number of unpaired electrons in it. The magnetic moment of compound B can be determined using following formula.
$\mu=\sqrt{n(n+2)}$
The n represents the number of unpaired electrons.
In the following reaction sequence, M is a transition metal.
Identify the metal M and MCl$_4$. Explain the difference in colours of MCl$_4$ and A.
Explanation:
As given, M is a transition metal. The MCl$_4$ is a colourless compound which on reaction with Zn gives a purple-coloured compound A. Also, MCl$_4$ on reaction with moist air gives compound B which has white fumes and pungent smell.
The metal M must be titanium, (Ti) and MCl$_4$ is TiCl$_4$.
In TiCl$_4$, Ti exists in +4 oxidation state which does not contain electrons in its d orbital.
Ti$^{4+}$ : [Ar]
When TiCl$_4$ reacts with Zn, it produces [Ti(H$_2$O)$_6$]$^{3+}$ which is purple in colour.
Therefore, compound A is [Ti(H$_2$O)$_6$]$^{3+}$.
In [Ti(H$_2$O)$_6$]$^{3+}$, Ti exists in +3 oxidation state and it has only one electron in its d orbital.
Ti$^{4+}$ : [Ar]3d$^1$ 4s$^0$
The presence of unpaired electrons in d orbital in [Ti(H$_2$O)$_6$]$^{3+}$ and the absence of d electrons in TiCl$_4$ causes the colour difference in these two compounds.
When TiCl$_4$ reacts with moist air, white fumes with pungent smell of TiO$_2$ is produced. Hence, the compound B is TiO$_2$.
Final Answer
$\mathrm{M:Ti}$
$\mathrm{MCl_4:TiCl_4}$
$\mathrm{A:[Ti(H_2O)_6]^{3+}}$
$\mathrm{B:TiO_2}$
Hints :
The chloride of titanium, TiCl$_4$ is a colourless compound. Check the oxidation state of metal M in both MCl$_4$ and compound A. Check the number of unpaired electrons in d orbital in each case.