Coordination Compounds

Chemistry

NCERT

1   Write the formulas for the following coordination compounds:$\\$ (i) Tetraamminediaquacobalt (III) chloride$\\$ (ii) Potassium tetracyanonickelate(II)$\\$ (iii) Tris(ethane-1,2-diamine) chromium(III) chloride$\\$ (iv) Amminebromidochloridonitrito-N-platinate(II)$\\$ (v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate$\\$ (vi) Iron(III) hexacyanoferrate(II)

Solution :

(i)$[CO(H_2O)(NH_3)_4]Cl_3$$\\$ (ii)$K_2[Ni(CN)_4]$$\\$ (iii)$[Cr(en)_3]Cl_3$$\\$ (iv)$[Pt(NH)_3 BrCl(NO_2)]$$\\$ (v)$[PtCl_2(en)_2(NO_3)_2]$$\\$ (vi)$Fe_4[Fe(CN)_6]_3$

2   Write the $IUPAC$ names of the following coordination compounds:$\\$ (i) $[CO ( NH_3 )_8 ] Cl_3$$\\$ (ii) $[CO ( NH_3 )_6 Cl ] Cl_3$$\\$ (iii) $K_3 [ Fe ( CN )_8 ]$$\\$ (iv) $K_3 [ Fe ( C_2 N_4 )_3 $$\\$ (v) $K_2 [ PdCl_4 ]$$\\$ (vi) $[Pt ( NH_3 )_2 Cl ( NH_2 CH_3 ) ] Cl$

Solution :

(i) Hexaamminecobalt(III) chloride$\\$ (ii) Pentaamminechloridocobalt (III) chloride$\\$ (iii) Potassium hexacyanoferrate(III)$\\$ (iv) Potassium trioxalatoferrate(III)$\\$ (v) Potassium tetrachloridopalladate(II)$\\$ (vi) Diamminechlorido(methylamine)platinum(II) chloride

3   Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:$\\$ (i)$K[Cr(H_2O)_2(C_2O_4)_2]$$\\$ (ii)$[CO(en)_2]Cl_2$$\\$ (iii)$[Co(NH_3)_6(NO_2)](NO_3)_2$$\\$ (iv)$[Pt(NH_2)(H_2O)Cl_2]$

Solution :

(i) Both geometrical (cis-, trans-) isomers for $K [Cr ( H_2 O )_2 ( C_2 O_4 )_2 ]$ can exist. Also, optical isomers for cis-isomer exist.$\\$ Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active.$\\$ (ii) Two optical isomers for $[ CO ( en )_2 ] Cl_2$ exist.$\\$ Two optical isomers are possible for this structure.$\\$ (iii)$[CO(NH_3)(NO_2)](NO_3)_2$$\\$ A pair of optical isomers:$\\$ It can also show linkage isomerism.$9741621450$$\\$ $[CO ( NH_3 )_5 ( NO_2 ) ] ( NO_3 )_2$ and $[ CO ( NH_3 )_5 ( ONO ) ] ( NO _3 )_ 2$$\\$ It can also show ionization isomerism.$\\$ $[CO ( NH_3 ) _5 ( NO _2 ) ] ( NO _3 ) _2$ and $ [CO ( NH_3 )_ 5 ( NO_3 ) ] ( NO _3 )( NO _2 )$$\\$ (iv) Geometrical (cis-, trans-) isomers of $[ Pt ( NH _3 )( H _2 O ) Cl _2 ]$ can exist.

4   Give evidence that $[Co(NH_3 )_5 Cl]SO _4$ and $[Co(NH_3 )_5 SO_4 ]Cl$ are ionization isomers.

Solution :

When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products.

$[CO(NH_3)_5Cl]SO_4+Ba^{2+}\longrightarrow \underset{\text{White precipitate}}{BaSO_4 \downarrow}$ $\\$ $[CO(NH_3)_5Cl]SO_4+Ag^+ \longrightarrow $ No Reaction $\\$ $[CO(NH_3)_5SO_4]Cl+Ba^{2+}\longrightarrow $ No Reaction $\\$ $[CO(NH_3)_5SO_4]Cl+Ag^+ \longrightarrow \underset{\text{White precipitate}}{AgCl \downarrow}$ $\\$

5   Explain on the basis of valence bond theory that $Ni ( CN )_4 $ ion with square planer structure is diamagnetic and the $Ni ( Cl )_4$ ion with tetrahedral geometry is paramagnetic.

Solution :

Ni is in the $+2$ oxidation state i.e., in $d^ 8$ configuration.$\\$ There are $4 CN ^-$ ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since $CN ^- $ ion is a strong field ligand, it causes the pairing of unpaired $3d$ electrons. $\\$ It now undergoes $\text{dsp }^2$ hybridization. Since all electrons are paired, it is diamagnetic. In case of $[NiCl_4] ^{2-}, Cl ^-$ ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired $3d$ electrons.

6   $[NiCl_ 4 ]^{ 2-}$ is paramagnetic while $[Ni(CO) _4 ]$ is diamagnetic though both are tetrahedral. Why?

Solution :

Though both $[NiCl_ 4 ]^{ 2-}$ and $[Ni(CO) _4 ]$ are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. $Cl ^-$ is a weak field ligand and it does not cause the pairing of unpaired $3$d electrons. Hence,$ [NiCl_ 4 ]^{ 2-}$ is paramagnetic.$\\$

In $ [Ni(CO) _4 ], Ni $ is in the zero oxidation state i.e., it has a configuration of $3$d $^8 4s^ 2 $.$\\$

But $CO$ is a strong field ligand. Therefore, it causes the pairing of unpaired $3$d electrons. Also, it causes the $4s$ electrons to shift to the $3d$ orbital, thereby giving rise to sp $3$ hybridization. Since no unpaired electrons are present in this case, $[Ni(CO)_ 4 ]$ is diamagnetic.

7   $[Fe(H_ 2 O)_ 6 ]^{ 3+ } $ is strongly paramagnetic whereas $[Fe(CN)_ 6 ]^{ 3+}$ is weakly paramagnetic. Explain.

Solution :

In both $[Fe(H _2 O) _6 ]^{ 3+}$ and $[Fe(CN)_ 6 ]^{ 3+}$ , Fe exists in the $+3$ oxidation state i.e., in d $ 5$ configuration

Since $CN ^-$ is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.$\\$

Therefore magnetic moment is given by,$\\$ $\mu = n ( n + 2)$ $\\$ $= 1(1 + 2)$ $\\$ $= 3 $ $\\$ $= 1.732$ BM $\\$ On the other hand,$H _2 O$ is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is $5.$ Therefore, magnetic moment is given by,$\\$ $\mu = n + ( n + 2)$ $\\$ $= 5(5 + 2)$ $\\$ $= 35 $ $\\$ $= 5.91$ BM $\\$ Thus, it is evident that $ [Fe(H_ 2 O) _6 ]^{ 3+}$ is strongly paramagnetic, while $[Fe(CN)_ 6 ] ^{3+ } $is weakly paramagnetic

8   Explain $[Co ( NH _3 )_ 6 ]$ is an inner orbital complex whereas $[Ni ( NH_ 3 )_ 6 ]$ is an outer orbital complex.

Solution :

$\color{CadetBlue}{[Co(NH _3 ) _6 ]^{ 3+}}$$\\$ Oxidation state of cobalt =$ +3$ $\\$ Electronic configuration of cobalt = $d^ 6$ $\\$

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$NH _3$ being a strong field ligand causes the pairing. Therefore, Cobalt can undergo $d^ 2 sp^ 3$ hybridization.

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Hence, it is an inner orbital complex.

$\color{CadetBlue }{[Ni(NH _3 ) _6 ] ^{2+}}$ $\\$ Oxidation state of $Ni = +2$$\\$ Electronic configuration of nickel =$ d^ 8$

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If $NH_ 3$ causes the pairing, then only one $3d$ orbital is empty. Thus, it cannot undergo $d^ 2 sp ^3$ hybridization. So in this complex$ NH _3$ acts as weak field ligand. Therefore, it undergoes $sp^3d ^2$ hybridization.

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Hence, it forms an outer orbital complex.

9   Predict the number of unpaired electrons in the square planar $[Pt(CN _4 )]^{ 2-}$ ion.

Solution :

$[Pt(CN_ 4 )]^{ 2-}$ $\\$ In this complex, Pt is in the $+2$ state. It forms a square planar structure. This means that it undergoes dsp 2 hybridization. Now, the electronic configuration$ Pd (+2)$ of is $5d^ 8$ .

$CN ^-$ being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in $[Pt(CN _4 )] ^{2-}$ .

10   The hexaquomanganese (II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.

Solution :

$\color{CadetBlue}{[ Mn ( H _2 O )_ 6 ]^{2+}}$ $\\$ Mn is in the $+2$ oxidation state.$\\$ The electronic configuration is $d^ 5$ $\\$. The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in $[ Mn ( H _2 O )_ 6 ]^{2+}$is $t _2 g ^3 eg^ 2 $.$\\$ $\color{CadetBlue}{[ Mn ( CN )_ 6 ]^{4+}}$ $\\$ Mn is in the $+2$ oxidation state$\\$ The electronic configuration is $d^ 5$ .$\\$ The crystal field is octahedral. Cyanide is a strong field ligand and hence force pairing occurs. Therefore, the arrangement of the electrons in $[ Mn ( CN )_6 ]^{4-}$ is $t^ 2 g ^5 eg ^0 .$ $\\$ Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron.