Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry

$$SC{N^ - }$$  is a monodentate ligand hence cannot show chelation.

$$\mathop {\text{N}}\limits^{.\,.} {{\text{H}}_3} + {H^ + }\left( {{\text{acidic}}\,{\text{medium}}} \right) \rightleftharpoons \mathop N\limits^ + {H_4}$$

Key concept Wavelength $$\left( \lambda \right)$$  of absorption is inversely proportional to $$CFSE$$  $$\left( {{\vartriangle _o}\,{\text{value}}} \right)$$   of ligands attached with the central metal ion
$${\text{i}}{\text{.e}}{\text{.}}\,\,\lambda \propto \frac{1}{{{\Delta _O}}}$$
According to spectrochemical series.
$${I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - }$$     $$ < C{l^ - } < {F^ - } < O{H^ - } < {C_2}O_4^{2 - }$$     $$ < {O^{2 - }} < {H_2}O < NS{S^ - } < N{H_3}$$       $$ < en < NO_2^ - < C{N^ - }$$

The $$CFSE$$   of ligands attached with $$C{O^{3 + }}$$  $$ion$$  is in the order $$en > N{H_3} > {H_2}O$$     ( From spectrochemical series )
$$\because $$  Wavelength of absorbed light $$\left( \lambda \right) \propto \frac{1}{{{\vartriangle _o}}}$$
$$\therefore $$  For ligand the order of wavelength of absorption in the visible region will be : $$en < N{H_3} < {H_2}O$$
$${\text{or,}}{\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }} < {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ < {\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$

Magnetic moment, $$\sqrt {n\left( {N + 2} \right)} = 5.92\,B.M \Rightarrow n$$
( unpaired electrons ) $$= 5$$

$$s{p^3}$$  hybridization (tetrahedral)

$${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }}$$   is $${d^2}s{p^3}$$ - hybridised and octahedral complex. In $${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }},$$   $$Mn$$  is in +3 oxidation state $$M{n^{3 + }} = 3{d^4}4{s^0}$$
Otbitals of $$M{n^{3 + }}$$  $$ion$$  =  
$${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }}$$   =  

The number of geometrical isomers
$$\eqalign{ & {\left[ {Co{{\left( {N{H_3}} \right)}_2}{{\left( {Cl} \right)}_4}} \right]^ - } - 2 \cr & AuC{l_2}B{r_2} - 2 \cr} $$
$${\left[ {Co\left( {N{O_2}} \right){{\left( {N{H_3}} \right)}_5}} \right]^{2 + }} - $$     No isomerism

Complex is paramagnetic due to presence of unpaired electron at $$O_2^ - $$  i.e., superoxide acting as ligand.

In both states (paramagnetic and diamagnetic) of the given complex, $$Ni$$  exists as $$N{i^{2 + }}$$  whose electronic configuration is $$\left[ {Ar} \right]3{d^8}4{s^0}.$$

In the above paramagnetic state the geometry of the complex is $$s{p^3}$$ giving tetrahedral geometry. The diamagnetic state is achieved by pairing of electrons in $$3d$$ orbital.

Thus the geometry of the complex will be $$ds{p^2}$$ giving square planar geometry.

In $$Fe{\left( {CO} \right)_5},$$   $$Fe$$ - atom is in $$ds{p^3}$$  hybridised state, therefore the shape of molecule is trigonal bipyramidal. The hybridisation is as follows :
$$_{26}Fe = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^2}4{p^0}$$
In presence of strong field ligand $$(CO),$$  the electrons of $$4s$$  are pushed in $$3d$$  orbital and get paired up.
In $$Fe{\left( {CO} \right)_5},$$   the $$Fe$$ - atom is

$$C{N^ - }$$  is a strong field ligand and form low spin complexes thus $${\Delta _ \circ } > P.$$