Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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331.
Few isomers are given below. Mark the correct statement regarding them.
$$\eqalign{
& \left( {\text{i}} \right)\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3} \cr
& \left( {{\text{ii}}} \right)\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2} \cdot {H_2}O \cr
& \left( {{\text{iii}}} \right)\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl \cdot 2{H_2}O \cr} $$
A
(i), (ii) and (iii) are hydrate isomers.
B
(i), (ii) and (iii) are coordination isomers.
C
(i), (ii) and (iii) are ionisation isomers.
D
(i) and (ii) are stereoisomers.
Answer :
(i), (ii) and (iii) are hydrate isomers.
No explanation is given for this question. Let's discuss the answer together.
332.
Among the following, which are ambidentate ligands.
$$\eqalign{
& \left( {\text{i}} \right)SC{N^ - } \cr
& \left( {{\text{ii}}} \right)NO_3^ - \cr
& \left( {{\text{iii}}} \right)NO_2^ - \cr
& \left( {{\text{iv}}} \right){C_2}O_4^{2 - } \cr} $$
A
(i) and (iii)
B
(i) and (iv)
C
(ii) and (iii)
D
(ii) and (iv)
Answer :
(i) and (iii)
No explanation is given for this question. Let's discuss the answer together.
333.
Incorrect match is :
A
$$\left[ {Rh{{\left( {PP{h_3}} \right)}_3}Cl} \right]:$$ Wilkinson's catalyst
B
$${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }:$$ Bond order of $$Co—CO$$ bond is greater than one
$${\left[ {Cr{{\left( {CN} \right)}_6}} \right]^{3 - }}:$$ Hybridization : $${d^2}s{p^3}$$ High or low spin is not defined because $$C{r^{3 + }}$$ is an example of $${d^3}$$ system of configuration.
334.
Which one of the following complexes will consume more equivalents of aqueous solution of $$AgN{O_3}?$$
A
$$N{a_2}\left[ {CrC{l_5}\left( {{H_2}O} \right)} \right]$$
B
$$N{a_3}\left[ {CrC{l_6}} \right]$$
C
$$\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}$$
D
$$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$
More equivalents of $$AgN{O_3}$$ aqueous solution will be consumed if complex will furnish more $$C{l^ - }\,ions$$ in solution. Hence complex $$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$ will consume more equivalents of aqueous solution of $$AgN{O_3}.$$
335.
In the coordination compound, $${K_4}\left[ {Ni{{\left( {CN} \right)}_4}} \right],$$ the oxidation state of nickel is
336.
Mark the correct statements regarding the geometry of complex ions.
(i) The geometry of the complex ion depends upon the coordination number.
(ii) If coordination number is 6, the complex is octahedral.
(iii) If coordination number is 4, the geometry of the complex may be tetrahedral or square planar.
A
(i), (ii) and (iii)
B
(i) and (ii) only
C
(i) and (iii) only
D
(ii) and (iii) only
Answer :
(i), (ii) and (iii)
No explanation is given for this question. Let's discuss the answer together.
337.
Which one of the following is expected to exhibit optical isomerism ? ( $$en =$$ ethylenediamine )
A
$$Cis - \left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]$$
B
$$Trans - {\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$$
C
$$Trans - \left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]$$
D
$$Cis - {\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$$
$$Cis - {\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$$ is able to show the
phenomenon of optical isomerism because it can form a superimposable mirror image.
It gives super imposable structure. but $$trans$$ - form is optically inactive.
338.
Which of the following pairs represent linkage isomers?
The $$SC{N^ - }$$ ion can coordinate through $$S$$ or $$N$$ atom giving rise to linkage isomerism
$$M \leftarrow SCN$$ thiocyanato
$$M \leftarrow NCS$$ isothiocyanato.
339.
Which of the following type of isomerism is shown by given complex compound ?
A
Facial
B
Meridional
C
Cis
D
Both (B) and (C)
Answer :
Meridional
Given compound shows meridional isomerism.
340.
$${\left[ {Co{F_6}} \right]^{3 - }}$$ is
A
paramagnetic and undergoes $$s{p^3}{d^2}$$ hybridisation
B
diamagnetic and undergoes $${d^2}s{p^3}$$ hybridisation
C
paramagnetic and undergoes $$s{p^3}d$$ hybridisation
D
diamagnetic and undergoes $$s{p^3}$$ hybridisation
Answer :
paramagnetic and undergoes $$s{p^3}{d^2}$$ hybridisation