Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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351.
Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type $$\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]$$ and $$\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]$$ are
A
linkage isomers
B
coordination isomers
C
ionisation isomers
D
geometrical isomers
Answer :
linkage isomers
No explanation is given for this question. Let's discuss the answer together.
352.
Which of the following complexes will show maximum paramagnetism?
A
$$3{d^4}$$
B
$$3{d^5}$$
C
$$3{d^6}$$
D
$$3{d^7}$$
Answer :
$$3{d^5}$$
$$3{d^5}$$ has maximum number of unpaired electrons.
353.
According to IUPAC nomenclature sodium nitroprusside is named as
A
sodium pentacyanonitrosyl ferrate (II)
B
sodium pentacyanonitrosyl ferrate (III)
C
sodium nitroferricyanide
D
sodium nitroferrocyanide
Answer :
sodium pentacyanonitrosyl ferrate (III)
IUPAC name of sodium nitroprusside $$N{a_2}\left[ {Fe{{\left( {CN} \right)}_5}NO} \right]$$ is sodium pentacyanoni trosyl ferrate (III) because in it $$NO$$ is neutral ligand and the oxidation number of $$Fe$$ is + 3. Which is calculated as
$$\eqalign{
& N{a_2}\left[ {Fe{{\left( {CN} \right)}_5}NO} \right] \cr
& 2 \times \left( { + 1} \right) + x + 5 \times \left( { - 1} \right) + 1 \times 0 = 0 \cr
& \,\,\,\,\,\,{\text{[where }}x{\text{ }} = {\text{ oxidation state of }}Fe{\text{]}} \cr
& 2 + x - 5 = 0 \cr
& \,\,\,\,\,\,\,\,\,x - 3 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 3 \cr} $$