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NEET • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

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301 Medium
301 of 320
Analyzing +4 | -1
Metal carbon bond is strongest in
View answer and explanation
Correct option: D
[Co(CO)4]−{\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }
Option D ([Co(CO)4]−{\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }) is correct because the negative charge on the complex increases the electron density on the metal, enhancing back-bonding and strengthening the metal-carbon bond.
302 Easy
302 of 320
Remembering +4 | -1
The hardness of water is estimated by:
View answer and explanation
Correct option: B
EDTA method
Option B (EDTA method) is correct because the EDTA method is a complexometric titration used to determine the total hardness of water by measuring the concentration of calcium and magnesium ions.
303 Hard
303 of 320
Analyzing +4 | -1
Complex XX of composition Cr(H2O)6ClnCr{\left( {{H_2}O} \right)_6}C{l_n} has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3AgN{O_3} and shows geometrical isomerism. The IUPAC nomenclature of XX is
View answer and explanation
Correct option: D
tetraaquadichlorido - chromium (III) chloride dihydrate
Option D (tetraaquadichlorido - chromium (III) chloride dihydrate) is correct because it matches the given information about magnetic moment, reaction with AgNO3AgNO_3, and geometrical isomerism.
304 Hard
304 of 320
Analyzing +4 | -1
For the following molecules; if xx is the total number of pp-orbitals used in hybridisation of each central atom, then what is the value of ∑x6\frac{\sum {x}}{6} ?

XeO2 F2, B2H6,[NiCl4]−2,[PtCl4]−2, N(SiH3)3,H3BO3{XeO}_2 {~F}_2, {~B}_2 {H}_6,\left[{NiCl}_4\right]^{-2},\left[{PtCl}_4\right]^{-2}, {~N}\left({SiH}_3\right)_3, {H}_3 {BO}_3
View answer and explanation
Correct option: A
3
Option A (3) is correct because the sum of p-orbitals used in hybridization for each central atom is 18, and 18/6 = 3.
305 Medium
305 of 320
Application +4 | -1
The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2\left[ {\mathrm{Co}\left( {\mathrm{N}{\mathrm{O}_2}} \right){{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}} \right]\mathrm{C}{\mathrm{l}_2} is
View answer and explanation
Correct option: D
Pentaammine nitrito-N- cobalt (III) chloride
The correct answer is D (Pentaammine nitrito-N- cobalt (III) chloride) because it follows the IUPAC naming conventions for coordination compounds, correctly identifying the nitro ligand as nitrito-N.
306 Easy
306 of 320
Application +4 | -1
The compound having the lowest oxidation state of iron is
View answer and explanation
Correct option: D
Fe(CO)5\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}
Option D (Fe(CO)5) is correct because the oxidation state of Fe in this complex is 0, which is the lowest among the given options.
307 Hard
307 of 320
Analyzing +4 | -1
Among the following complexes (A−FA-F),
K3[Fe(CN)6](A),[Co(NH3)6]Cl3(B)Na3[Co( oxalate )3](C),[Ni2(H2O)6]Cl2(D), K2[Pt(CN)4](E) and [Zn(H2O)6](NO3)2( F)\begin{aligned} & {K}_3\left[{Fe}({CN})_6\right] ({A}),\left[{Co}\left({NH}_3\right)_6\right] {Cl}_3 (B)\\ & {Na}_3\left[{Co}(\text { oxalate })_3\right]({C}),\left[{Ni}_2\left({H}_2 {O}\right)_6\right] {Cl}_2 (D),\\ & {~K}_2\left[{Pt}({CN})_4\right]({E}) \text { and }\left[{Zn}\left({H}_2 {O}\right)_6\right]\left({NO}_3\right)_2({~F}) \end{aligned}

The diamagnetic complexes are:
View answer and explanation
Correct option: C
B,C,E,FB, C, E, F
Option C is correct because B, C, E, and F are all diamagnetic complexes, meaning they have no unpaired electrons.
308 Hard
308 of 320
Remembering +4 | -1
[EDTA]4−[{EDTA}]^{4-} is a
View answer and explanation
Correct option: D
Hexadentate ligand
Option D (Hexadentate ligand) is correct because EDTA can bind to a central metal ion through six donor atoms.
309 Medium
309 of 320
Understanding +4 | -1
The IUPAC name of Na3[Co(ONO)6]\mathrm{N}{\mathrm{a}_3}\left[ {\mathrm{Co}{{\left( {\mathrm{ONO}} \right)}_6}} \right] is:
View answer and explanation
Correct option: B
Sodium hexanitritocobaltate(III)
Option B (Sodium hexanitritocobaltate(III)) is correct because it correctly names the compound, indicating the sodium cation, the hexanitrito ligands, the cobaltate anion, and the oxidation state of cobalt.
310 Easy
310 of 320
Remembering +4 | -1
Among the ligands NH3,en,CN−and CO,N{H_3},en,C{N^ - }and\,CO, the correct order of their increasing field strength, is
View answer and explanation
Correct option: B
NH3<en<CN−<CON{H_3} < en < C{N^ - } < CO
Option B (NH3<en<CN−<CON{H_3} < en < C{N^ - } < CO) is correct because it accurately reflects the increasing order of field strength for the given ligands based on the spectrochemical series.
311 Medium
311 of 320
Understanding +4 | -1
Which one is organometalllic compound?
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Correct option: D
Methyl lithium
The correct answer is D (Methyl lithium). An organometallic compound contains a direct metal-carbon bond. Methyl lithium (CH3Li) has a direct Li-C bond.
312 Medium
312 of 320
Understanding +4 | -1
Which of the following shell, form an outer octahedral complex?
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Correct option: B
d8{d^8}
Option B (d8{d^8}) is correct because d8{d^8} configuration favors outer orbital octahedral complexes due to electronic repulsions and the need to accommodate all electrons.
313 Easy
313 of 320
Analyzing +4 | -1
Which among the following statements are true for the complex [Co(NH3)6][Cr(CN)6]\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]?
(a) It is anon electrolyte
(b) The magnitude of the charge on each complex ion is 3
(c) The complex will not conduct current.
(d) The complex will exhibit coordination isomerism
(e) The magnitude of the charge on each complex ion is 1
View answer and explanation
Correct option: C
b and d
Option C (b and d) is correct because the magnitude of the charge on each complex ion is 3, and the complex can exhibit coordination isomerism.
314 Medium
314 of 320
Application +4 | -1
The total number of possible isomers of the complex compound [CuII(NH3)4][PtIICl4]\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_4\right]\left[{Pt}^{\mathrm{II}} {Cl}_4\right] is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the complex can exhibit linkage and coordination isomerism, leading to a total of 4 isomers.
315 Hard
315 of 320
Understanding +4 | -1
For a d4{d^4} metal ion in an octahedral field, the correct electronic configuration is
View answer and explanation
Correct option: A
t2g3eg1   when  Δ0<Pt_{2g}^3e_g^1\,\,\,\mathrm{when}\,\,\mathrm{ }{\Delta _0} < P
Option A is correct because when Δ₀ < P, pairing energy is higher, leading to high spin configuration with electrons occupying both t2g and eg orbitals.
316 Medium
316 of 320
Analyzing +4 | -1
The shortest C−O\mathrm{C} - \mathrm{O} bond order exists in:
View answer and explanation
Correct option: D
[V(CO)6]−{\left[ {\mathrm{V}{{\left( {\mathrm{CO}} \right)}_6}} \right]^ - }
Option D ([V(CO)6]−{\left[ {\mathrm{V}{{\left( {\mathrm{CO}} \right)}_6}} \right]^ - }) is correct because the negative charge on the complex increases back-bonding, which weakens the C-O bond and thus shortens the C-O bond length.
317 Hard
317 of 320
Understanding +4 | -1
K4[Fe(CN)6]{K}_4\left[{Fe}({CN})_6\right] reacts with FeCl3{FeCl}_3 to form:
View answer and explanation
Correct option: B
KFe[Fe(CN)6]{KFe}\left[{Fe}({CN})_6\right]
Option B (KFe[Fe(CN)6]{KFe}\left[{Fe}({CN})_6\right]) is correct because K4[Fe(CN)6]{K}_4\left[{Fe}({CN})_6\right] reacts with FeCl3{FeCl}_3 to form Prussian blue, which has the formula KFe[Fe(CN)6]{KFe}\left[{Fe}({CN})_6\right].
318 Easy
318 of 320
Application +4 | -1
The IUPAC name of K2[Cr(CN)2O2  (O)2(NH3)]  {\mathrm{K}_2}\left[ {\mathrm{Cr}{{\left( {\mathrm{CN}} \right)}_2}{\mathrm{O}_2}\mathrm{\;}{{\left( \mathrm{O} \right)}_2}\left( {\mathrm{N}{\mathrm{H}_3}} \right)} \right]\mathrm{\;} is
View answer and explanation
Correct option: A
Potassium ammine dicyano dioxoperoxochromate (VI)
Option A (Potassium ammine dicyano dioxoperoxochromate (VI)) is correct because it follows IUPAC naming conventions for coordination compounds.
319 Hard
319 of 320
Analyzing +4 | -1
Assertion : In a mixture of Cd(II){\text{Cd(II)}} and Cu(II){\text{Cu(II)}}, (Cd+2){\text{(C}}{{\text{d}}^{ + 2}}{\text{)}} gets precipitated in presence of KCN\mathrm{KCN} by H2S.

Reason : The stability constant of [(Cu(CN)4]−3{\left[ {{\text{(Cu(CN}}{{\text{)}}_{\text{4}}}} \right]^{ - 3}} is greater than [(Cu(CN)4]−2{\left[ {{\text{(Cu(CN}}{{\text{)}}_{\text{4}}}} \right]^{ - 2}}
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
Option A is correct because in the presence of KCN, Cu(II) forms a stable complex, preventing its precipitation as CuS, while Cd(II) remains free to precipitate as CdS.
320 Easy
320 of 320
Remembering +4 | -1
Lithium tetrahydridoaluminate is correctly represented as:
View answer and explanation
Correct option: C
Li[AlH4]\mathrm{Li}\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]
Option C (Li[AlH4]\mathrm{Li}\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]) is correct because it accurately represents the structure of lithium tetrahydridoaluminate, showing lithium as the cation and the tetrahydroaluminate as the anion.
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