12th Grade > Chemistry
COORDINATION COMPOUNDS MCQs
Total Questions : 29
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Answer: Option A. -> yellow
:
A
The colour of the complex is complimentary to that which it absorbs.
:
A
The colour of the complex is complimentary to that which it absorbs.
Answer: Option D. -> All K+,Al3+ and SO2−4
:
D
Potash alum is K2SO4.Al2(SO4)3.24H2O
:
D
Potash alum is K2SO4.Al2(SO4)3.24H2O
Answer: Option D. -> t52g,e2g
:
D
High spin complex means ligand is weak, hence, Δ0 < P (pairing energy)
:
D
High spin complex means ligand is weak, hence, Δ0 < P (pairing energy)
Answer: Option C. -> Cis−[Co(NH3)2(en)2]3+
:
C
Only Cis-octahedral complexes show optical activity.
:
C
Only Cis-octahedral complexes show optical activity.
Answer: Option B. -> Both σ and π character
:
B
There is dπ−ρπ metal to ligand back bonding in Fe - C bond of Fe(CO)5. So, it possesses both σ and π character.
:
B
There is dπ−ρπ metal to ligand back bonding in Fe - C bond of Fe(CO)5. So, it possesses both σ and π character.
Answer: Option C. -> [Co(en)3]3+
:
C
[Co(en)3]3+ does not have plane of symmetry.
:
C
[Co(en)3]3+ does not have plane of symmetry.
Answer: Option C. -> Platinum
:
C
Zeise’s salt is K[PtCl3(C2H4)]
:
C
Zeise’s salt is K[PtCl3(C2H4)]
Answer: Option A. -> [Fe(CN)6]4−
:
A
The solid obtained is
FeSO4+KCN→K4[Fe(CN)6]⇒4K++[Fe(CN)6]4−
:
A
The solid obtained is
FeSO4+KCN→K4[Fe(CN)6]⇒4K++[Fe(CN)6]4−
Answer: Option B. -> 1
:
B
EDTA is a hexadentate ligand and therefore only one EDTA molecule is required to form octahedral complex.
:
B
EDTA is a hexadentate ligand and therefore only one EDTA molecule is required to form octahedral complex.
Answer: Option A. -> [Co(NH3)3Cl3]
:
A
Complexes of the type MA3B3do not show optical isomerism but they show mer-fac isomerism.
:
A
Complexes of the type MA3B3do not show optical isomerism but they show mer-fac isomerism.