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NEET • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

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211 Medium
211 of 320
Remembering +4 | -1
Complexes with bidentate ligands are called:
View answer and explanation
Correct option: B
Chelates
Option C. Both NH3  and  Cl−\mathrm{N}{\mathrm{H}_3}\mathrm{\;and\;C}{\mathrm{l}^ - } is correct because both ammonia and chloride ions are directly bonded to the central chromium ion as ligands.
212 Medium
212 of 320
Remembering +4 | -1
The colour of CoCl3⋅5NH3⋅H2O\mathrm{CoC}{\mathrm{l}_3} \cdot 5\mathrm{N}{\mathrm{H}_3} \cdot {\mathrm{H}_2}\mathrm{O} is:
View answer and explanation
Correct option: D
Pink
The correct answer is D (Pink) because CoCl3⋅5NH3⋅H2O\mathrm{CoCl}_3 \cdot 5\mathrm{NH}_3 \cdot {\mathrm{H}_2}\mathrm{O} exhibits a pink color due to its specific coordination environment.
213 Hard
213 of 320
Understanding +4 | -1
In the complex ion [Fe(EDTA)]−[\mathrm{Fe}(\mathrm{EDTA})]^{-}, the coordination number and oxidation state of central metal ion is
View answer and explanation
Correct option: A
CN=6,ON=+3\mathrm{CN}=6, \mathrm{ON}=+3
Option A (CN=6,ON=+3\mathrm{CN}=6, \mathrm{ON}=+3) is correct because EDTA is a hexadentate ligand, leading to a coordination number of 6, and the overall charge of the complex allows for the calculation of the iron's oxidation state as +3.
214 Medium
214 of 320
Application +4 | -1
Which ion is paramagnetic?
View answer and explanation
Correct option: A
[Ni(H2O)6]2+{\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}
Option A ([Ni(H2O)6]2+) is correct because Ni2+ has a d⁸ configuration, and H2O is a weak field ligand, resulting in 2 unpaired electrons, making it paramagnetic.
215 Medium
215 of 320
Remembering +4 | -1
Choose the option which show correct preferential order of groups among the following
View answer and explanation
Correct option: A
−COOH,  −CHO,  −OH,  −NH2- \mathrm{COOH},\mathrm{\;} - \mathrm{CHO},\mathrm{\;} - \mathrm{OH},\mathrm{\;} - \mathrm{N}{\mathrm{H}_2}
Option C (Coordination number) is correct because the number of unidentate ligands directly bonded to the central metal ion defines the coordination number.
216 Medium
216 of 320
Analyzing +4 | -1
Assertion: C−C{C}-{C} bond length in zeise's salt is same as ethylen.

Reason: Double bond is shorter as compaire to single bond.
View answer and explanation
Correct option: D
Assertion is false, Reason is true
Option D (Assertion is false, Reason is true) is correct because the C-C bond length in Zeise's salt is longer than that in ethylene due to back bonding, and the reason correctly states that double bonds are shorter than single bonds.
217 Easy
217 of 320
Application +4 | -1
The effective atomic number of Cr(At.No.=24)in[Cr(NH3)6]Cl3  Cr\mathrm{ }\left( {\mathrm{At}\mathrm{. No}\mathrm{.} = 24} \right)\mathrm{in }\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}\mathrm{\;} is
View answer and explanation
Correct option: C
33
The correct answer is 33 because the EAN of Cr in [Cr(NH3)6]Cl3\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3} is calculated as 24 (atomic number) - 3 (oxidation state) + 2(6) (electrons from ligands) = 33.
218 Medium
218 of 320
Remembering +4 | -1
The donar sites of [EDTA]−4\mathrm{[EDTA}]^{-4} are:
View answer and explanation
Correct option: C
Two N\mathrm{N} atoms and four O\mathrm{O} atoms
Option A (Inert gas configuration) is correct because the Effective Atomic Number (EAN) rule states that the central metal atom tends to achieve the electronic configuration of the nearest noble gas.
219 Easy
219 of 320
Analyzing +4 | -1
How many unpaired electrons are present in [Fe(H2O)5NO]SO4\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4 complex?
View answer and explanation
Correct option: A
3
The correct answer is 3 because in [Fe(H2O)5NO]SO4\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4, NO is present as NO+NO^+ which is a strong field ligand, and Fe is in +1 oxidation state, resulting in 3 unpaired electrons.
220 Hard
220 of 320
Analyzing +4 | -1
Which of the following statements are correct?

(i) Square planar complexes of MABXLMABXL type show three isomers-two cis and one trans.

(ii) Complexes of Ma3B3{Ma}_3 {B}_3 type show three isomers-two cis and one trans.

(iii) Optical isomerism is common in octahedral complexes involving bidentate ligands.

(iv) [Co(NH3)4Cl(NO2)]Cl\left[{Co}\left({NH}_3\right)_4 {Cl}\left({NO}_2\right)\right] {Cl} show linkage isomerism.

(v) Hydrate isomerism is another name of solvate isomerism.
View answer and explanation
Correct option: B
(i), (iii) and (iv)
Option B ((i), (iii) and (iv)) is correct because these statements accurately describe properties of coordination complexes, including isomerism and linkage.
221 Easy
221 of 320
Remembering +4 | -1
[Co(NH3)6]Cl3\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3} is called:
View answer and explanation
Correct option: A
Hexaammine cobalt (III) chloride
Option A (Hexaammine cobalt (III) chloride) is correct because it follows the IUPAC nomenclature rules for coordination compounds.
222 Medium
222 of 320
Analyzing +4 | -1
Which of the following complex species is not expected to exhibit optical isomerism?
View answer and explanation
Correct option: C
[Co(NH)3Cl3][Co{(NH)_3}C{l_3}]
Option C ([Co(NH)3Cl3][Co{(NH)_3}C{l_3}]) is correct because it can exist as fac and mer isomers, but neither are chiral and thus do not show optical isomerism.
223 Hard
223 of 320
Analyzing +4 | -1
In which of the following compound, C – O bond length is maximum
View answer and explanation
Correct option: A
[Fe(CO)5]−{\left[ {Fe{{\left( {CO} \right)}_5}} \right]^ - }
Option A ([Fe(CO)5]-) is correct because the negative charge on the complex increases back-bonding from the metal to the CO ligand, weakening the C-O bond and thus increasing the C-O bond length.
224 Hard
224 of 320
Understanding +4 | -1
For kinetic study of the reaction of iodide ion with H2O2\mathrm{H}_2 \mathrm{O}_2 at room temperature :

(A) Always use freshly prepared starch solution.

(B) Always keep the concentration of sodium thiosulphate solution less than that of KI\mathrm{KI} solution.

(C) Record the time immediately after the appearance of blue colour.

(D) Record the time immediately before the appearance of blue colour.

(E) Always keep the concentration of sodium thiosulphate solution more than that of KI\mathrm{KI} solution.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(A), (B), (C) only
Option A ((A), (B), (C) only) is correct because it lists the correct procedures for the kinetic study of the reaction of iodide ion with H2O2\mathrm{H}_2 \mathrm{O}_2.
225 Easy
225 of 320
Remembering +4 | -1
Which order is correct in spectrochemical series of ligands?
View answer and explanation
Correct option: A
Cl−<F−<[C2O4]2−<NO2−<CN−\mathrm{C}{\mathrm{l}^ - } < {\mathrm{F}^ - } < {\left[ {{\mathrm{C}_2}{\mathrm{O}_4}} \right]^{2 - }} < \mathrm{NO}_2^ - < C{\mathrm{N}^ - }
Option A is correct because it represents the ligands arranged in increasing order of their ability to cause d-orbital splitting in a complex.
226 Medium
226 of 320
Understanding +4 | -1
The compound which does not show paramagnetism is
View answer and explanation
Correct option: C
[Ag(NH3)2]Cl\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}
The correct answer is C ([Ag(NH3)2]Cl\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}) because Ag+ has a d10 configuration, resulting in no unpaired electrons and thus diamagnetism.
227 Medium
227 of 320
Remembering +4 | -1
The shape of Wilkinson’s catalyst is
View answer and explanation
Correct option: B
Square planar
Option B is correct because Wilkinson's catalyst, [RhCl(PPh3)3], has a square planar geometry with the rhodium center coordinated to one chloride and three triphenylphosphine ligands.
228 Medium
228 of 320
Understanding +4 | -1
The hybridization of central metal ion and shape of Wilkinson’s catalyst is
View answer and explanation
Correct option: C
dsp2,ds{p^2}, squre planar
Option C (dsp2,ds{p^2}, squre planar) is correct because Wilkinson's catalyst, [RhCl(PPh3)3][RhCl(PPh_3)_3], has Rh in a +1 oxidation state, leading to a d8d^8 configuration and dsp2ds{p^2} hybridization, resulting in a square planar geometry.
229 Easy
229 of 320
Analyzing +4 | -1
Which of the following ions forms most stable complex compound?
View answer and explanation
Correct option: B
Mn2+\mathrm{M}{\mathrm{n}^{2 + }}
Option D (Cu2+\mathrm{C}{\mathrm{u}^{2 + }}) is correct because Cu2+Cu^{2+} has a high charge density and a suitable ionic radius, leading to a higher polarizing power and thus forming more stable complexes.
230 Medium
230 of 320
Remembering +4 | -1
The ligand called π\pi acid is :
View answer and explanation
Correct option: A
CO{CO}
Option A (CO{CO}) is correct because CO is a well-known π-acid ligand, capable of accepting electron density from the metal through back-bonding into its π* antibonding orbitals.
231 Easy
231 of 320
Remembering +4 | -1
CO ligand is
View answer and explanation
Correct option: A
Carbonyl
Option A (Carbonyl) is correct because CO is the ligand name for Carbonyl.
232 Medium
232 of 320
Application +4 | -1
Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper(II) complexes will show the maximum structural distortion?

(en - ethylenediamine; H2 N−CH2−CH2−NH2\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2 )
View answer and explanation
Correct option: D
trans-[Cu(en )2Cl2]\left.)_2 \mathrm{Cl}_2\right]
Option D (trans-[Cu(en)2Cl2]) is correct because the trans isomer exhibits the greatest distortion due to the arrangement of ligands around the central Cu(II) ion.
233 Easy
233 of 320
Understanding +4 | -1
K3[Al(C2O4)3]{\mathrm{K}_3}\left[ {\mathrm{Al}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right] is called:
View answer and explanation
Correct option: D
Potassium trioxalatoaluminate(III)
Option D (Potassium trioxalatoaluminate(III)) is correct because it correctly names the complex, indicating the potassium cation, the trioxalato ligands, the aluminate anion, and the oxidation state of aluminum.
234 Easy
234 of 320
Remembering +4 | -1
Which of the following is potassiumferricyanide?
View answer and explanation
Correct option: B
K3[Fe(CN)6]{K}_3\left[{Fe}({CN})_6\right]
Option B (K3[Fe(CN)6]{K}_3\left[{Fe}({CN})_6\right]) is correct because it represents potassium ferricyanide, where iron is in the +3 oxidation state and is complexed with six cyanide ligands, balanced by three potassium cations.
235 Medium
235 of 320
Analyzing +4 | -1
Which have octahedral shape (d2sp3)({d^2}s{p^3}) hybridization of central atom?
View answer and explanation
Correct option: D
All are correct
Option D (All are correct) is correct because all the given complexes, [Cr(NH3)6]2+{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}, [Fe(CN)6]3−{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}, and [Cu(NH3)6]+{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^ + }, possess octahedral shape with d2sp3d^2sp^3 hybridization.
236 Easy
236 of 320
Analyzing +4 | -1
Which is not true of the coordination compound [Co(en)2Cl2]Cl\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}?
View answer and explanation
Correct option: C
Exhibits ionisation isomerism
Option C (Exhibits ionisation isomerism) is correct because the counter ion (Cl-) is already outside the coordination sphere, so exchanging it with a ligand inside the sphere does not create a different compound.
237 Easy
237 of 320
Understanding +4 | -1
[Co(NH3)5Br]SO4  \left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Br}} \right]\mathrm{S}{\mathrm{O}_4}\mathrm{\;} and [Co(NH3)5SO4]\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]Br\mathrm{Br} are examples of which type of isomerism?
View answer and explanation
Correct option: D
Ionisation
The correct answer is D (Ionisation) because the complexes [Co(NH3)5Br]SO4  {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Br}} \right]\mathrm{S}{\mathrm{O}_4}\mathrm{\;}} and [Co(NH3)5SO4]Br{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]\mathrm{Br}} exchange ions (Br- and SO42-) in the coordination sphere and outside it.
238 Medium
238 of 320
Understanding +4 | -1
The correct increasing order of trans-effect of the following species is
View answer and explanation
Correct option: C
CN−>  C6H5−>Br−> NH3C{N^ - } > \,\,{C_6}{H_5}^ - > B{r^ - } > \,N{H_3}
Option C (CN−> C6H5−>Br−> NH3C{N^ - } > \,{C_6}{H_5}^ - > B{r^ - } > \,N{H_3}) is correct because it represents the correct increasing order of trans-effect based on the ligand's ability to weaken the bond trans to itself.
239 Easy
239 of 320
Analyzing +4 | -1
Which of the following will give maximum number of isomers?
View answer and explanation
Correct option: D
[Cr(SCN)2(NH3)4]+{\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }
Option D ([Cr(SCN)2(NH3)4]+{\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }) is correct because SCN- is an ambidentate ligand, leading to linkage isomers in addition to geometric isomers.
240 Medium
240 of 320
Analyzing +4 | -1
Which of the following does not have optical isomer?
View answer and explanation
Correct option: C
[Co(NH3)3Cl]\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]
Option C ([Co(NH3)3Cl]{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]}) is correct because it is not a complex; it is missing the counter ion to balance the charge, and it is not chiral.
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