Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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201.
Consider the following reaction and statements:
$${\left[ {Co{{\left( {N{H_3}} \right)}_4}B{r_2}} \right]^ + } + B{r^ - } \to \left[ {Co{{\left( {N{H_3}} \right)}_3}B{r_3}} \right] + N{H_3}$$
(I) Two isomers are produced if the reactant complex ion is a $$cis - $$ isomer.
(II) Two isomers are produced ifthe reactant complex ion is a$$trans - $$ isomer
(III) Only one isomer is produced if the reactant complex ion is a $$trans - $$ isomer
(IV) Only one isomer is produced if the reactant complex ion is a $$cis - $$ isomer.
The correct statements are:
A
(I) and (II)
B
(I) and (III)
C
(III) and (IV)
D
(II) and (IV)
Answer :
(I) and (III)
202.
The spin magnetic moment of cobalt in the compound $$Hg\left[ {Co{{\left( {SCN} \right)}_4}} \right]$$ is
A
$$\sqrt 3 $$
B
$$\sqrt 8 $$
C
$$\sqrt {15} $$
D
$$\sqrt {24} $$
Answer :
$$\sqrt {15} $$
TIPS/FFORMULAE:
$$\mu = \sqrt {n\left( {n + 2} \right)} {\text{BM}}\,\,\,\left( {\mu = {\text{spin}}\,{\text{magnetic}}\,{\text{moment}}} \right)$$
Here $$Co$$ is present as $$C{o^{2 + }}$$ ion which has $$3$$ unpaired electrons. So the spin magnetic moment will be
$$\sqrt {3\left( {3 + 2} \right)} ,\,{\text{i}}{\text{.e}}\,\,\sqrt {15} \,{\text{BM}}.$$
203.
Which of the following has an optical isomer
A
$${\left[ {Co\left( {en} \right){{\left( {N{H_3}} \right)}_2}} \right]^{2 + }}$$
B
$${\left[ {Co{{\left( {{H_2}O} \right)}_4}\left( {en} \right)} \right]^{3 + }}$$
C
$${\left[ {Co{{\left( {en} \right)}_2}{{\left( {N{H_3}} \right)}_2}} \right]^{3 + }}$$
D
$${\left[ {Co{{\left( {N{H_3}} \right)}_3}Cl} \right]^ + }$$
204.
The coordination number ofa central metal atom in a complex is determined by
A
the number of ligands around a metal ion bonded by sigma and $$pi - {\text{bonds}}$$ both
B
the number of ligands around a metal ion bonded by $$pi - {\text{bonds}}$$
C
the number of ligands around a metal ion bonded by sigma bonds
D
the number of only anionic ligands bonded to the metal ion.
Answer :
the number of ligands around a metal ion bonded by sigma bonds
The coordination number of central metal atom in a complex is equal to number of monovalent ligands, twice the number of bidentate ligands and so on, around the metal ion bonded by coordinate bonds. Hence coordination number = no. of $$\sigma $$ bonds formed by metals with ligands
205.
Which one is the most likely structure of $$CrC{l_3}.6{H_2}O$$ if $$\frac{1}{3}$$ of total chlorine of the compound is precipitated by adding $$AgN{O_3}$$
A
$$CrC{l_3}.6{H_2}O$$
B
$$\left[ {Cr{{\left( {{H_2}O} \right)}_3}C{l_3}} \right].{\left( {{H_2}O} \right)_3}$$
C
$$\left[ {CrC{l_2}{{\left( {{H_2}O} \right)}_4}} \right]Cl.2{H_2}O$$
D
$$\left[ {CrCl{{\left( {{H_2}O} \right)}_5}} \right]C{l_2}.{H_2}O$$
The ions present in the ionisation sphere are precipited Hence $$\left[ {CrC{l_2}{{\left( {{H_2}O} \right)}_4}} \right]Cl.2{H_2}O$$ contains $$\frac{1}{3}Cl$$ in ionisation sphere to be precipited by $$AgN{O_3}$$ as $$AgCl$$
206.
For complex ion/compound formation reactions
$$\left( {\text{i}} \right)C{o^{3 + }}\left( {aq} \right) + EDT{A^{4 - }} \to P$$
$$\left( {{\text{ii}}} \right)N{i^{2 + }}\left( {aq} \right) + dmg{\text{(excess)}}$$ \[\xrightarrow{N{{H}_{4}}OH}Q\]
$$\left( {{\text{iii}}} \right)Z{n^{2 + }}\left( {aq} \right) + gly{\text{(excess)}} \to R$$
$$\left( {{\text{iv}}} \right)P{t^{4 + }}\left( {aq} \right) + en{\text{(excess)}} \to S$$
which of the following complex ion/compound does not exhibit optical activity ?
A
$$P$$
B
$$Q$$
C
$$R$$
D
$$S$$
Answer :
$$Q$$
$$N{i^{2 + }}\left( {aq} \right)$$ with $$DMG$$ (dimethylglyoxime) form
square planar complex which is obtically inactive. While $${\left[ {Co\left( {EDTA} \right)} \right]^ - },{\left[ {Zn{{\left( {gly} \right)}_2}} \right]^{2 + }}$$ and $${\left[ {Pt{{\left( {en} \right)}_3}} \right]^{4 + }}$$ are optically active as they
do not have $$P.O.S.$$
207.
Which one of the following complexes shows optical isomerism?
$$\left( {en = {\text{ethylenediamine}}} \right)$$
A
$$trans\,\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl$$
C
$$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$$
D
$$cis\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$
Optical isomerism occurs when a molecule is non-superimposable with its mirror image hence the complex $$cis\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$ is optically active.
208.
$$C{H_3} - Mg - Br$$ is an organo metallic compound due to
A
$$Mg - Br\,{\text{bond}}$$
B
$$C - Mg\,{\text{bond}}$$
C
$$C - Br\,{\text{bond}}$$
D
$$C - H{\text{ bond}}$$
Answer :
$$C - Mg\,{\text{bond}}$$
Compounds that contain at least one carbon metal bond are known as organometallic compounds. In $$C{H_3} - Mg - Br$$ (Grignard's reagent) a bond is present between carbon and $$Mg$$ (Metal) hence it is an organometallic compound.
209.
Match the column I with column II and mark the appropriate choice.
No explanation is given for this question. Let's discuss the answer together.
210.
Using valence bond theory, the complex $${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$ can be described as
A
$$s{p^3}{d^2},$$ outer orbital complex, paramagnetic
B
$$ds{p^2},$$ inner orbital complex, diamagnetic
C
$${d^2}s{p^3},$$ inner orbital complex, paramagnetic
D
$${d^2}s{p^3},$$ outer orbital complex, diamagnetic
Answer :
$${d^2}s{p^3},$$ inner orbital complex, paramagnetic
In $${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},$$ oxidation state of $$Cr = + 3,C{r^{3 + }} = 3{d^3}$$
It is an inner orbital complex. Due to presence of three unpaired electrons, it is paramagnetic.