NCERT Solution Class 12th Chemistry Chapter – 9 Coordination Compounds
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Chemistry |
| Chapter | Chapter – 9 |
| Chapter Name | Coordination Compounds |
| Category | Class 12th Chemistry Question & Answer |
| Medium | English |
| Source | last doubt |
NCERT Solution Class 12th Chemistry Chapter – 9 Coordination Compounds
?Chapter – 9?
✍Coordination Compounds✍
?Question & Answer?
Q1. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer: ( a ) A metal shows two kinds of valencies viz primary valency and secondary valency. Negative ions satisfy primary valencies and secondary valencies are filled by both neutral ions and negative ions.
( b ) A metal ion has a fixed amount of secondary valencies about the central atom. These valencies also orient themselves in a particular direction in the space provided to the definite geometry of the coordination compound.
( c ) Secondary valencies cannot be ionized, while primary valencies can usually be ionized.
Q2. FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why.
Answer: FeSO4 solution when mixed with (NH4)2SO4in 1: 1 molar ratio produces a double salt FeSO4(NH4)2SO4-6H2O. This salt is responsible for giving the Fe2+.CuSO4 mixed with aqueous ammonia in the ratio of 1:4 gives a complex salt. The complex salt does not ionize to give Cu2+, hence failing the test.
Q3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer: ( a ) Ligands – they are neutral molecules or negative ions bound to a metal atom in the coordination entity. Example- Cl–, –OH
( b ) Coordination entity – they are electrically charged radicals or species. They constitute a central ion or atom surrounded by neutral molecules or ions. Example – [ Ni(CO)4] , [COCL3(NH3)3]
( c ) Coordination number– it is the number of bonds formed between ligands and central atom/ion.
Example : ( i ) In K2[PtCl6], 6 chloride ions are attached to Pt in the coordinate sphere. Thus, 6 is the coordination number of Pt.
( ii ) In [Ni(NH3)4]Cl2, the coordination number of the central metal ion (Ni) is 4.
( d ) Coordination polyhedron – it is the spatial positioning of ligands that are directly connected to the central atom in the coordination sphere. Example –
(v) Homoleptic complexes: These are those complexes in which the metal ion is bound to only one kind of a donor group. For eg: [Co(NH3)6]3+ , [PtCl4]2- etc.
(vi) Heteroleptic complexes: Heteroleptic complexes are those complexes where the central metal ion is bound to more than one type of a donor group. For e.g.: [Co(NH3)4 Cl2]+ , [Co(NH3)5 Cl]2+
Q4. What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.
Answer: ( i ) Unidentate ligands: these are ligands with one donor site. Example Cl–, NH3
( ii ) Ambidentate ligands: these are ligands that fasten themselves to the central metal ion/ atom via two different atoms.
Example NO–2or ONO–, CN– or NC–
( iii ) Bidentate – these are ligands with two donor sites.
Example – Ethane-1,2-diamine , Oxalate ion ( C2O42- )
Q5. Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+
(ii) [CoBr2(en)2]+
(iii) [PtCl4]2–
(iv) K3[Fe(CN)6]
(v) [Cr(NH3)3Cl3]
Answer:
( i ) [Co (H2O) (CN) (en)2 ] 2+
=> x + 0 + ( -1) + 2 (0) = +2
x -1 = +2
x = +3
( ii ) [ Co Br2 (en)2 ] 1+
=> x + 2 ( -1 ) + 2 ( 0 ) = +1
x – 2 = +1
x = +3
( iii) [ PtCl4 ] 2-
=> x + 4 ( -1 ) = -2
x = +2
( iv ) K3 [Fe (CN)6 ]
=> [ Fe ( CN )6]3-
=> x + 6 ( -1 ) = -3
x = +3
( v ) [ Cr(NH3)3Cl3 ]
=> x + 3( 0 ) + 3 ( -1 ) = 0
x – 3 = 0
x = 3
Q6. Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxidozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanidonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt(III)
Answer:
( i ) [Zn(OH)4]2 –
( ii ) K2[ Pd Cl4]
( iii ) [ Pt ( NH3)2Cl2]
( iv ) K2[ Ni(CN )4]
( v ) [ Co (NO2) ( NH3)5] 2+
( vi) [ Co( NH3)6]2 (SO4)3
( vii ) K3 [ Cr ( C2O4)3]
( viii ) [ Pt (NH3)6] 4+
( ix ) [ Cu (Br)4] 2−
( x ) [Co ( ONO )( NH3)5] 2+
Q7. Using IUPAC norms write the systematic names of the following:
(i) [Co(NH3)6]Cl3
(ii) [Pt(NH3)2Cl(NH2CH3)]Cl
(iii) [Ti(H2O)6]3+
(iv) [Co(NH3)4Cl(NO2)]Cl
(v) [Mn(H2O)6]2+
(vi) [NiCl4]2–
(vii) [Ni(NH3)6]Cl2
(viii) [Co(en)3]3+
(ix) [Ni(CO)4]
Answer:
( i ) Hexaamminecobalt(III) chloride
( ii ) Diamminechlorido(methylamine) platinum(II) chloride
( iii ) Hexaquatitanium(III) ion
( iv ) Tetraamminichloridonitrito-N-Cobalt(III) chloride
( v ) Hexaquamanganese(II) ion
( vi ) Tetrachloridonickelate(II) ion
( vii ) Hexaamminenickel(II) chloride
( viii ) Tris(ethane-1, 2-diammine) cobalt(III) ion
( ix ) Tetracarbonylnickel(0)
Q8. List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:
(a) Geometric isomerism:
This type of isomerism is common in heteroleptic complexes. It arises due to the different possible geometric arrangements of the ligands. For example:
(b) Optical isomerism:
This type of isomerism arises in chiral molecules. Isomers are mirror images of each other and are non-superimposable.
(c) Linkage isomerism:
This type of isomerism is found in complexes that contain ambidentate ligands. For example:
[Co(NH3)5 (NO2)]Cl2 and [Co(NH3)5 (ONO)Cl2
Yellow form Red form
(d) Coordination isomerism:
This type of isomerism arises when the ligands are interchanged between cationic and anionic entities of differnet metal ions present in the complex.
[Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]
(e) Ionization isomerism:
This type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. Thus, complexes that have the same composition, but furnish different ions when dissolved in water are called ionization isomers. For e.g., Co(NH3)5SO4)Br and Co(NH3)5Br]SO4.
(f) Solvate isomerism: Solvate isomers differ by whether or not the solvent molecule is directly bonded to the metal ion or merely present as a free solvent molecule in the crystal lattice.
[Cr[H2O)6]Cl3 [Cr(H2O)5Cl]Cl2 . H2O [Cr(H2O)5Cl2]Cl . 2H2O
Violet Blue-green Dark green
Q9. How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3–
(ii) [Co(NH3)3Cl3]
Answer:
( i ) In [ Cr(C2O4)3] 3− no geometric isomers are present because it is a bidentate ligand.
( ii ) In [ Co( NH3)3 Cl3 ]two isomers are possible.
Q10. Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3–
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Answer: ( i ) [ Cr( C2O4 )3 ] 3−
Q11. Draw all the isomers (geometrical and optical) of:
(i) [CoCl2(en)2]+
(ii) [Co(NH3)Cl(en)2]2+
(iii) [Co(NH3)2Cl2(en)]+
Answer:
( i ) [ CoCl2 (en)2 ] +
( ii ) [ Co(NH3) Cl( en)2 ] 2+
Q12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
[Pt(NH3)(Br)(Cl)(py)
From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerization. They do so only in the presence of unsymmetrical chelating agents.
Q13. Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium fluoride and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results
Answer: The blue colour of aqueous CuSO4 solution is because of the presence of [ Cu( H2O)4 ] 2+ ions.
( i ) So when KF is added, H2O ligands are replaced by F– ligands which yield green coloured [ CuF4 ] 2+ ions.
[Cu(H_{2}O)_{4}]^{2+} + 4F^{-}\rightarrow [CuF_{4}]^{2-} + 4H_{2}O
( ii ) So when KCL is added, H2O ligands are replaced by Cl– ligands which yield bright green coloured [ CuCl4 ] 2+ ions.
[Cu(H_{2}O)_{4}]^{2+} + 4Cl^{-}\rightarrow [CuCl_{4}]^{2-} + 4H_{2}O
Q14. What is the coordination entity formed when an excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?
Answer: CuSO4(aq) + 4KCN(aq) → K2[Cu(CN)4](aq) + K2SO4(aq)
i.e., [Cu(H2O)4]2+ + 4CN– → [Cu(CN)4]2- + 4H2O
Thus, the coordination entity formed in the process is K2[Cu(CN)4]. K2[Cu(CN)4 is a very stable complex, which does not ionize to give Cu2+ ions when added to water. Hence, Cu2+ ions are not precipitated when H2S(g) is passed through the solution.
Q15. Discuss the nature of bonding in the following coordination entities on the
basis of valence bond theory:
(i) [Fe(CN)6]4–
(ii) [FeF6]3–
(iii) [Co(C2O4)3]3–
(iv) [CoF6]3-
Answer:
( i ) [ Fe(CN)6] 4−
In this coordination complex, the oxidation state of Fe is +3.
Fe 2+ : Electronic configuration is 3d6
Orbitals of Fe2+ ion :
Since CN− is a strong field ligand, it causes the unpaired 3d electrons to pair up:
As there are six ligands around the central metal ion, the most practical hybridization is d2sp3, d2sp3 hybridized orbitals of Fe2+ are:
6 electron pairs from CN − ions take the place of the six hybrid d2sp3 orbitals.
Then,
Thus, the geometry of the complex is octahedral and it is a diamagnetic complex (since all the electrons are paired).
( ii ) [FeF6] 3−
In this coordinate entity, the oxidation state of iron is +3.
Orbitals of Fe +3 ion:
There are 6 F− ions. Hence, it will go through d2sp3 or sp3d2 hybridization.
Since F− is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Thus, the most practical hybridization is sp3d2. sp3d 2 hybridized orbitals of Fe are :
Thus, the geometry of this coordinate entity is octahedral.
( iii ) [ Co( C2O4)3] 3−
In this complex, the oxidation state of cobalt is +3.
Orbitals of Co 3+ ion :
Oxalate is a weak field ligand. Thus, it will not cause the pairing of the 3d orbital electrons.
As there are 6 ligands, hybridization has to be either sp3d 2 or d 2sp3 hybridization.
sp3d2 hybridization of Co 3+ :
The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d 2 orbitals
Thus, the complex shows octahedral geometry.
( iv )) [CoF6] 3−
In this complex, Cobalt has an oxidation state of +3.
Orbitals of Co3+ ion:
As fluoride ion is a weak field ligand it will not cause the 3d electrons to pair. Hence, the Co3+ ion will go through sp3d2 hybridization.
sp3d2 hybridized orbitals of Co3+ ion are :
Thus, the complex has a geometric configuration of an octahedral and it is paramagnetic.
Q16. Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Answer:
The splitting of the d orbitals in an octahedral field takes palce in such a way that dx2y2 , dz2 experience a rise in energy and form the eg level, while dxy, dyzand dzx experience a fall in energy and form the t2g level.
Q17. What is the spectrochemical series? Explain the difference between a weak
field ligand and a strong field ligand.
Answer: A series of common ligands in ascending order of their crystal-field splitting energy (CFSE) is termed as the Spectrochemical series.
Strong field ligands have larger values of CFSE. Whereas, weak field ligands have smaller values of CFSE.
Q18. What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity?
Answer: Crystal-field splitting energy is the difference in the energy between the two levels ( i.e., t2g and eg ) that have split from a degenerated d orbital because of the presence of a ligand. It is symbolized as ∆o.
Once the orbitals split up, electrons start filling the vacant spaces. An electron each goes into the three t2g orbitals, the fourth electron, however, can enter either of the two orbitals:
( 1 ) It can go to the eg orbital ( producing t2g3eg 1 like electronic configuration), or
( 2 ) it can go to the t2g orbitals ( producing t2g4eg 0 like electronic configuration).
This filling of the fourth electron is based on the energy level of ∆o. If a ligand has an ∆o value smaller than the pairing energy, then the fourth electron enters the eg orbital. However, if the value of ∆o is greater than the value of pairing energy, the electron enters t2gorbital.
Q19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why.
Answer:
In [ Ni ( CN)4 ] 2−, Ni has an oxidation state of +2. Thus, it has d 8configuration.
Ni 2+ :
CN − being a strong field ligand causes the electrons in 3d orbitals to pair. This causes, Ni 2+ to undergo dsp2 hybridization.
Since all the electrons are paired, it is diamagnetic in nature.
Cr has an oxidation state of +3. Thus, it has a d 3 configuration. As NH3 is not a strong field ligand it does not cause the electrons in the 3d orbital to pair.
Cr3+ :
It undergoes d2sp3 hybridization and the 3d orbital electrons remain unpaired. Thus, [ Ni ( CN)4 ] 2− is paramagnetic in nature.
Q20. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain
Answer: In [Ni(H2O)6]2+, H2Ö is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d – d transition is present. Hence, [Ni(H2O)6]2+ is coloured.
In [Ni(CN)4]2 – , the electrons are all paired as CN– is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)4]2 – . Hence, it is colourless.
Q21. [Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Answer: The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]4- and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now, CN– is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also differs.
Q22. Discuss the nature of bonding in metal carbonyls.
Answer: The metal-carbon bonds in metal carbonyls have both σ and π characters. A σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal. A π bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti-bonding π orbital (also known as back bonding of the carbonyl group). The σ bond strengthens the π bond and vice-versa. Thus, a synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and the metal.
Q23. Give the oxidation state, d orbital occupation and coordination number of
the central metal ion in the following complexes:
(i) K3[Co(C2O4)3] (iii) (NH4)2[CoF4]
(ii) cis-[CrCl2(en)2]Cl (iv) [Mn(H2O)6]SO4
Answer: (i) K3[Co(C2O4)3]
The central metal ion is Co.
Its coordination number is 6.
The oxidation state can be given as:
x – 6 = -3
x = + 3
The d orbital occupation for Co3+ is t2g6eg0.
(ii) cis-[Cr(en)2Cl2]Cl
The central metal ion is Cr.
The coordination number is 6.
The oxidation state can be given as:
x + 2(0) + 2(-1) = +1
x – 2 = +1
x = +3
The d orbital occupation for Cr3+ is t2g3.
(iii) (NH4)2[CoF4]
The central metal ion is Co.
The coordination number is 4.
The oxidation state can be given as:
x – 4 = -2
x = + 2
The d orbital occupation for Co2+ is eg4 t2g3.
(iv) [Mn(H2O)6]SO4
The central metal ion is Mn.
The coordination number is 6.
The oxidation state can be given as:
x + 0 = +2
x = +2
The d orbital occupation for Mn is t2g3 eg2.
Q24. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also, give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H2O)2(C2O4)2].3H2O
(ii) [Co(NH3)5Cl]Cl2
(iii) [CrCl3(py)3]
(iv) Cs[FeCl4]
(v) K4[Mn(CN)6]
Answer: (i) Potassium diaquadioxalatochromate (III) trihydrate.
Oxidation state of chromium = 3
Electronic configuration: 3d3 : t2g3
Coordination number = 6
Shape: octahedral
Stereochemistry:
(ii) [Co(NH3)5Cl]Cl2
IUPAC name: Pentaamminechloridocobalt(III) chloride
Oxidation state of Co = +3
Coordination number = 6
Shape: octahedral.
Electronic configuration: d6: t2g6.
Stereochemistry:
Magnetic Moment = 0
(iii) CrCl3(py)3
IUPAC name: Trichloridotripyridinechromium (III)
Oxidation state of chromium = +3
Electronic configuration for d3 = t2g3
Coordination number = 6
Shape: octahedral.
Stereochemistry:
Both isomers are optically active. Therefore, a total of 4 isomers exist.
(iv) Cs[FeCl4]
IUPAC name: Caesium tetrachloroferrate (III)
Oxidation state of Fe = +3
Electronic configuration of d6 = eg2t2g3
Coordination number = 4
Shape: tetrahedral
Stereochemistry: optically inactive
Magnetic moment:
(v) K4[Mn(CN)6]
Potassium hexacyanomanganate(II)
Oxidation state of manganese = +2
Electronic configuration: d5+: t2g5
Coordination number = 6
Shape: octahedral.
Streochemistry: optically inactive
Q25. Explain the violet colour of the complex [Ti(H2O)6]3+ on the basis of crystal field theory
Answer: The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.
M + 3L ↔ ML3
Stability Constant, β = [ML3] / [M][L3]
For this reaction, the greater the value of the stability constant, the greater is the proportion of ML3 in the solution.
Stability can be of two types:
(a) Thermodynamic stability:
The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.
(b) Kinetic stability:
This helps in determining the speed with which the transformation will occur to attain the state of equilibrium.
Factors that affect the stability of a complex are:
(a) Charge on the central metal ion: Thegreater the charge on the central metal ion, the greater is the stability of the complex.
(b) Basic nature of the ligand: A more basic ligand will form a more stable complex.
(c) Presence of chelate rings: Chelation increases the stability of complexes.
Q26. What is meant by the chelate effect? Give an example.
Answer: When a polydentate or a bidentate ligand fastens itself to a metal ion in such a way that it assumes the shape of a ring, the metal-ligand bond becomes more stable. These rings are called chelate rings.
From here we can infer that complexes with chelate rings are more stable than complexes without the rings. This phenomenon is termed the chelate effect.
Ni2+ (aq) + 6NH3 ↔ [ Ni( NH3)6 ]2+ (aq)
logβ = 7.99
Ni2+ (aq) + 3en (aq) ↔ [ Ni( en)3 ]2+ (aq)
logβ = 18.1 ( more stable )
Q27. Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems (iii) analytical chemistry
(ii) medicinal chemistry and (iv) extraction/metallurgy of metals.
Answer: (i) Role of coordination compounds in biological systems:
We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen-carrier of blood, i.e., haemoglobin, is a coordination compound of iron.
(ii) Role of coordination compounds in medicinal chemistry:
Certain coordination compounds of platinum (for example, cis- platin) are used for inhibiting the growth of tumours.
(iii) Role of coordination compounds in analytical chemistry:
During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.
(iv) Role of coordination compounds in extraction or metallurgy of metals:
The process of extraction of some of the metals from their ores involves the formation of complexes. For example, in aqueous solution, gold combines with cyanide ions to form [Au(CN)2]. From this solution, gold is later extracted by the addition of zinc metal.
Q28. How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6 (ii) 4 (iii) 3 (iv) 2
Answer: (iii) The given complex can be written as Co(NH3)6Cl2.
Thus, [Co(NH3)6]+ along with two Cl– ions are produced.
Q29. Amongst the following ions which one has the highest magnetic moment value?
(i) [Cr(H2O)6]3+
(ii) [Fe(H2O)6]2+
(iii) [Zn(H2O)6]2+
Answer: (i) No. of unpaired electrons in [Cr(H2O)6]3+ = 3
(ii) No. of unpaired electrons in[Fe(H2O)6]2+ = 4
Then, μ
(iii) No. of unpaired electrons in [Zn(H2O)6]2+ = 0
Hence, [Fe(H2O)6]2+ has the highest magnetic moment value.
Q30. What is the oxidation number of cobalt in K[Co(CO)4]?
Answer: K[Co( CO )4] = K+[Co( CO )4]–
We know,
=> x + 0 = – 1 [ Where x is the oxidation number.]
x = -1
Q31. Amongst the following, the most stable complex is
(i) [Fe(H2O)6]3+
(ii) [Fe(NH3)6]3+
(iii) [Fe(C2O4)3]3–
(iv) [FeCl6]3–
Answer: We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C2O4)3]3-.
Q32. What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO2)6]4–, [Ni(NH3)6]2+, [Ni(H2O)6]2+?
Answer: The central metal ion in all the three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series is as follows:
H2O < NH3 < NO2 –
Thus, the amount of crystal-field splitting observed will be in the following order:
Hence, the wavelengths of absorption in the visible region will be in the order:
[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4-