Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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21.
Which of the following does not depict the correct name of the compound?
A
$${K_2}\left[ {Zn{{\left( {OH} \right)}_4}} \right]:$$ Potassiumtetrahydroxidozincate(II)
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}C{O_3}} \right]Cl:$$ Pentaammine carbonatochloridocobaltate(III)
C
$$N{a_3}\left[ {Co{{\left( {N{O_2}} \right)}_6}} \right]:$$ Sodium hexanitrocobaltate(III)
D
$${K_3}\left[ {Cr{{\left( {CN} \right)}_6}} \right]:$$ Potassiumhexacyanidochromate(III)
22.
The hybridisation involved in complex $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is ( Atomic number of $$Ni=28$$ )
A
$$ds{p^2}$$
B
$$s{p^3}$$
C
$${d^2}s{p^2}$$
D
$${d^2}s{p^3}$$
Answer :
$$ds{p^2}$$
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
Let oxidation state of $$Ni$$ in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is $$x.$$
$$\eqalign{
& \therefore \,\,x - 4 = - 2 \cr
& {\text{or}}\,\,\,\,x = 2 \cr} $$
Now, $$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$
$$\because \,\,C{N^ - }$$ is a strong field ligand. Hence, all unpaired electrons are paired up.
∴ Hybridisation of $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is $$ds{p^2}$$
23.
Which of the following paramagnetic ions would exhibit a magnetic moment ( spin only ) of the order of $$5\,BM?$$
$${\text{(}}\,{\text{At}}{\text{. Nos}}{\text{. }}Mn = 25,Cr = 24,V = 23,{\text{ }}Ti = 22\,)$$
A
$$M{n^{2 + }}$$
B
$$T{i^{2 + }}$$
C
$${V^{2 + }}$$
D
$$C{r^{2 + }}$$
Answer :
$$C{r^{2 + }}$$
Hence $$C{r^{2 + }}$$ has magnetic moment of the order of $$5\,B.M.$$
24.
Which of the following isomers will give white precipitate with $$BaC{l_2}$$ solution?
A
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$
C
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {S{O_4}} \right)}_2}} \right]Br$$
D
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}Br\left( {S{O_4}} \right)} \right]$$
26.
Hexacyano complexes of metals in their +2 oxidation state are usually yellow while the corresponding hexaaqua compounds are often blue or green. This
is so because
A
hexacyano complexes absorb orange or red light thus appear yellow while hexaaqua complexes absorb indigo thus appear yellow
B
hexacyano complexes absorb indigo thus appearing yellow while hexaaqua complexes absorb orange or red light thus appear blue or green
C
hexacyano complexes absorb yellow light while hexaaqua complexes absorb blue light
D
$$C{N^ - }$$ ions are yellow in colour while aqua ions are blue or green in colour
Answer :
hexacyano complexes absorb indigo thus appearing yellow while hexaaqua complexes absorb orange or red light thus appear blue or green
No explanation is given for this question. Let's discuss the answer together.
27.
Native silver metal forms a water soluble complex with a dilute aqueous solution of $$NaCN$$ in the presence of
A
nitrogen
B
oxygen
C
carbon dioxide
D
argon
Answer :
oxygen
In the presence of oxygen, $$Ag$$ metal forms a water soluble complex $$Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$ with dilute solution of $$NaCN$$
$$4Ag + 8NaCN + 2{H_2}O + {O_2} \to \mathop {4Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]}\limits_{\left( {{\text{Soluble}}} \right)} + 4NaOH$$
28.
Consider the following isomers.
$$\eqalign{
& \left( {\text{i}} \right)\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]B{r_2} \cr
& \left( {{\text{ii}}} \right)\left[ {Pt{{\left( {N{H_3}} \right)}_4}B{r_2}} \right]C{l_2} \cr
& \left( {{\text{iii}}} \right)\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]N{O_2} \cr} $$
Which of the following observations is correct?
A
(i) will give a pale yellow and (ii) will give a white precipitate with $$AgN{O_3}$$ solution.
B
(iii) will give a white precipitate with $$AgN{O_3}$$ solution.
C
(i), (ii) and (iii) will give white precipitate with $$AgN{O_3}$$ solution.
D
None of the above isomers will give white precipitate with $$AgN{O_3}$$ solution.
Answer :
(i) will give a pale yellow and (ii) will give a white precipitate with $$AgN{O_3}$$ solution.
29.
The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?
A
$$\left[ {Fe{{\left( {CO} \right)}_5}} \right]$$
B
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
C
$${\left[ {Fe{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$$
D
$${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
$${C_2}O_4^{2 - }$$ (oxalate) is didentate ligand and forms chelate
complexes which are more stable than similar complexes containing unidentate ligands.
30.
The octahedral complex of a metal ion $${M^{3 + }}$$ with four monodentate ligands $${L_1},{L_2},{L_3}$$ and $${L_4}$$ absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is:
A
$${L_4} < {L_3} < {L_2} < {L_1}$$
B
$${L_1} < {L_3} < {L_2} < {L_4}$$
C
$${L_3} < {L_2} < {L_4} < {L_1}$$
D
$${L_1} < {L_2} < {L_4} < {L_3}$$
Answer :
$${L_1} < {L_3} < {L_2} < {L_4}$$
For a given metal ion, weak field ligands create a complex with smaller $$\vartriangle ,$$ which will absorbs light of longer $$\lambda $$ and thus lower frequency. Conservely, stronger field ligands create a larger $$\vartriangle ,$$ absorb light of shorter $$\lambda $$ and thus higher vi.e. higher energy.
$$\mathop {{\text{Red}}}\limits_{\lambda = 650nm} < \mathop {{\text{Yellow}}}\limits_{570\,nm} < \mathop {{\text{Green}}}\limits_{490\,nm} < \mathop {{\text{Blue}}}\limits_{450\,nm} $$
So order of ligand strength is
$${L_1} < {L_3} < {L_2} < {L_4}$$