What is the oxidation number of Nickel in K4[Ni(CN)4]?

What is the oxidation number of Nickel in K4[Ni(CN)4]?
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Zero
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The oxidation number of nickel in K4[Ni(CN)4] is - (1) Zero (2) +4 (3) -4 (4) +8

Description : The oxidation number of nickel in K4[Ni(CN)4] is - (1) Zero (2) +4 (3) -4 (4) +8

Last Answer : (1) Zero Explanation: The oxidation number of nickel in K4[Ni(CN)4] is zero.

The oxidation number of nickel in K4 [Ni(CN)4 ] is (1) Zero (2) +4 (3) –4 (4) +8

Description : The oxidation number of nickel in K4 [Ni(CN)4 ] is (1) Zero (2) +4 (3) –4 (4) +8

Last Answer : Zero

A complex compound in which the oxidation number of a metal is zero is - (1) K4 [Fe (CN)6] (2) K3 [Fe (CN)6] (3) [Ni (CO)4] (4) [Pl (NH3)4]Cl2

Description : A complex compound in which the oxidation number of a metal is zero is - (1) K4 [Fe (CN)6] (2) K3 [Fe (CN)6] (3) [Ni (CO)4] (4) [Pl (NH3)4]Cl2

Last Answer : (3) [Ni (CO)4]

The oxidation state of nickel in `K_(4)Ni(CN)_(4)` is:

Description : The oxidation state of nickel in `K_(4)Ni(CN)_(4)` is:

Last Answer : The oxidation state of nickel in `K_(4)Ni(CN)_(4)` is: A. `-1` B. `0` C. `+1` D. `+2`

Both `[Ni(CO)_(4)] and [Ni(CN)_(4)]^(2-)` are diamagnetic. The hybridisation of nickel in these complexes, respectively are :

Description : Both `[Ni(CO)_(4)] and [Ni(CN)_(4)]^(2-)` are diamagnetic. The hybridisation of nickel in these complexes, respectively are :

Last Answer : Both `[Ni(CO)_(4)] and [Ni(CN)_(4)]^(2-)` are diamagnetic. The hybridisation of nickel in these complexes, ... (2),sp^(3)` D. `dsp^(2),sp^(2)`

Out of NaCl, MgSO4 , Al2 (SO4 ) 3 , K4 [Fe(CN)6 ], which one will bring about the coagulation of a gold sol quickest and in the least of concentration ?

Description : Out of NaCl, MgSO4 , Al2 (SO4 ) 3 , K4 [Fe(CN)6 ], which one will bring about the coagulation of a gold sol quickest and in the least of concentration ?

Last Answer : Ans. Al2 (SO4 ) 3 .

What is expected value of ‘i’ for K4[Fe(CN)6] in dilute solution?

Description : What is expected value of ‘i’ for K4[Fe(CN)6] in dilute solution?

Last Answer : Ans:- Expected value of n in K4(Fe(CN)6) is 5. 

Assertion :- Oxidation number of `Ni` in `[Ni(CO)_(4)]` is zero. ltbr. Reason :- Nickel is bonded to neutral ligand, carbonyl.

Description : Assertion :- Oxidation number of `Ni` in `[Ni(CO)_(4)]` is zero. ltbr. Reason :- Nickel is bonded to neutral ligand, carbonyl.

Last Answer : Assertion :- Oxidation number of `Ni` in `[Ni(CO)_(4)]` is zero. ltbr. Reason :- Nickel is ... is False. D. If both Assertion & Reason are False.

Compare the following complexes’ with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units : (i) [Ni(CN)4]2- (ii) [NiCl4]2- (iii) [CoF6]3- [At. Nos. : Ni = 28; Co = 27] -Chemistry

Description : Compare the following complexes’ with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units : (i) [Ni(CN)4]2- (ii) [NiCl4]2- (iii) [CoF6]3- [At. Nos. : Ni = 28; Co = 27] -Chemistry

Last Answer : (i) [Ni(CN)4]2- Shape : Octahedral outer orbital complex Hybridisation : sp3d2 Magnetic behaviour : Paramagnetic (4 unpaired electrons)

Describe the shape and magnetic behaviour of following complexes : (i) [CO(NH3)6]3+ (ii) [Ni(CN)4]2- (At. No. Co = 27, Ni = 28) -Chemistry

Description : Describe the shape and magnetic behaviour of following complexes : (i) [CO(NH3)6]3+ (ii) [Ni(CN)4]2- (At. No. Co = 27, Ni = 28) -Chemistry

Last Answer : (i) [CO(NH3)6]3+ : Orbitals of CO3+ ion : Hybridization : d2sp3 Shape : Octahedral Magnetic behaviour : Diamagnetic (absence of unpaired electrons) (ii) [Ni(CN)4]2-

Among the following complexes : `K_(3)[Fe(CN)_(6)],[Co(NH_(3))_(6)]Cl_(3)` , `Na_(3)[Co ( o x )_(3)],[Ni(H_(2)O)_(6)]Cl_(2)`, `K_(2)[Pt(CN)_(4)]` and

Description : Among the following complexes : `K_(3)[Fe(CN)_(6)],[Co(NH_(3))_(6)]Cl_(3)` , `Na_(3)[Co ( o x )_(3)],[Ni(H_(2)O)_(6)]Cl_(2)`, `K_(2)[Pt(CN)_(4)]` and

Last Answer : Among the following complexes : `K_(3)[Fe(CN)_(6)],[Co(NH_(3))_(6)]Cl_(3)` , `Na_(3)[Co ( o x )_(3)],[Ni(H_( ... ,M,N B. K,N,O,P C. L,M,O,P D. L,M,N,O

Geometrical shapes of the complex formed by the reaction of `Ni^(2+)` with `Cl^(Θ),CN^(Θ)` and `H_(2)O` are :

Description : Geometrical shapes of the complex formed by the reaction of `Ni^(2+)` with `Cl^(Θ),CN^(Θ)` and `H_(2)O` are :

Last Answer : Geometrical shapes of the complex formed by the reaction of `Ni^(2+)` with `Cl^(Θ) ... and octahedral D. octahedral, square planar and octahedral

Amongst `Ni(CO)_(4),[Ni(CN)_(4)]^(2-) and NiCl_(4)^(2-)`:

Description : Amongst `Ni(CO)_(4),[Ni(CN)_(4)]^(2-) and NiCl_(4)^(2-)`:

Last Answer : Amongst `Ni(CO)_(4),[Ni(CN)_(4)]^(2-) and NiCl_(4)^(2-)`: A. `Ni(CO)_(4) and NiCl_(4)^(2-)` ... NiCl_(4)^(2-) and [Ni(CN)_(4)]^(2-)` are paramagnetic.

What will be the oxidation number of Fe in `K_(3)[Fe(CN)_(6)]` :-

Description : What will be the oxidation number of Fe in `K_(3)[Fe(CN)_(6)]` :-

Last Answer : What will be the oxidation number of Fe in `K_(3)[Fe(CN)_(6)]` :- A. `+4` B. `+3` C. `+2` D. Zero

The oxidation number of iron in potassium ferricyanide `[K_(3)Fe(CN)_(6)]` is :-

Description : The oxidation number of iron in potassium ferricyanide `[K_(3)Fe(CN)_(6)]` is :-

Last Answer : The oxidation number of iron in potassium ferricyanide `[K_(3)Fe(CN)_(6)]` is :- A. Two B. Six C. Three D. Four

A blue colour complex is obtained in the analysis of `Fe^(+3)` having formula `Fe_(4)[Fe(CN)_(6)]_(3)` Let a= oxidation number of Iron in the corrdina

Description : A blue colour complex is obtained in the analysis of `Fe^(+3)` having formula `Fe_(4)[Fe(CN)_(6)]_(3)` Let a= oxidation number of Iron in the corrdina

Last Answer : A blue colour complex is obtained in the analysis of `Fe^(+3)` having formula `Fe_(4)[Fe(CN)_ ... coordination sphere. Then find the value of (c+a-2b)

The oxidation state of Cr in `[Cr(H_(2)O)_(6)]Cl_(3), [Cr(C_(6)H_(6))_(2)], and K_(2)[Cr(CN)_(2)(O)_(2)(O_(2))(NH_(3))]` respectively are :

Description : The oxidation state of Cr in `[Cr(H_(2)O)_(6)]Cl_(3), [Cr(C_(6)H_(6))_(2)], and K_(2)[Cr(CN)_(2)(O)_(2)(O_(2))(NH_(3))]` respectively are :

Last Answer : The oxidation state of Cr in `[Cr(H_(2)O)_(6)]Cl_(3), [Cr(C_(6)H_(6))_(2)], and K_(2)[Cr(CN)_(2)(O)_(2 ... and +4` C. `+3,+4 and +6` D. `+3,+2 and +4`

Sulphide ion reacts with `[Fe(CN)_(5)NO]` to form a purple coloured compound (X). In this reaction oxidation state of iron .

Description : Sulphide ion reacts with `[Fe(CN)_(5)NO]` to form a purple coloured compound (X). In this reaction oxidation state of iron .

Last Answer : Sulphide ion reacts with `[Fe(CN)_(5)NO]` to form a purple coloured compound (X). In this reaction ... . changes from +2 to +4 D. does not change.

Methemoglobin is formed as a result of the oxidation of haemoglobin by oxidation agent: (A) Oxygen of Air (B) H2O2 (C) K4Fe(CN)6 (D) KMnO4

Description : Methemoglobin is formed as a result of the oxidation of haemoglobin by oxidation agent: (A) Oxygen of Air (B) H2O2 (C) K4Fe(CN)6 (D) KMnO4

Last Answer : Answer : C

(xiv) Following equation represents a method of purification of nickel by `underset ("Impure")(Ni) + 4CO overset (320 K)rarr Ni(CO)_(4) overset(420 K)

Description : (xiv) Following equation represents a method of purification of nickel by `underset ("Impure")(Ni) + 4CO overset (320 K)rarr Ni(CO)_(4) overset(420 K)

Last Answer : (xiv) Following equation represents a method of purification of nickel by `underset ("Impure")(Ni) + ... Arkel B. Zone refining C. Mond D. Cupellation

Assertion : Nickel can be purified by Mond process. Reason : `Ni(CO)_(4)` is a volatile compound which deomposed at `460K` to give pure `Ni`.

Description : Assertion : Nickel can be purified by Mond process. Reason : `Ni(CO)_(4)` is a volatile compound which deomposed at `460K` to give pure `Ni`.

Last Answer : Assertion : Nickel can be purified by Mond process. Reason : `Ni(CO)_(4)` is a volatile compound which deomposed at `460K` to give pure `Ni`.

The number of different spanning trees in complete graph, K4 and bipartite graph K2,2 have .......... and .....…. respectively. (A) 14, 14 (B) 16, 14 (C) 16, 4 (D) 14, 4

Description : The number of different spanning trees in complete graph, K4 and bipartite graph K2,2 have .......... and .....…. respectively. (A) 14, 14 (B) 16, 14 (C) 16, 4 (D) 14, 4

Last Answer : (C) 16, 4 

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The thermal radiative flux from a surface of emissivity = 0.4 is 22.68 kW/m2 . The approximate surface temperature (K) is (Stefan-Boltzmann constant = 5.67 × 10 -8 W/m2 .K4 )(A) 1000(B) 727(C) 800(D) 1200

Description : The thermal radiative flux from a surface of emissivity = 0.4 is 22.68 kW/m2 . The approximate surface temperature (K) is (Stefan-Boltzmann constant = 5.67 × 10 -8 W/m2 .K4 ) (A) 1000 (B) 727 (C) 800 (D) 1200

Last Answer : (A) 1000

Determine the equivalent spring constant of the system shown in FigK1=10 N/m, K2=20 N/m. K3=30 N/m, K4=34 N/m, K5=40 N/m a) 10N/m b)20N/m c)30N/m d) None

Description : Determine the equivalent spring constant of the system shown in FigK1=10 N/m, K2=20 N/m. K3=30 N/m, K4=34 N/m, K5=40 N/m a) 10N/m b)20N/m c)30N/m d) None

Last Answer : b)20N/m

In Ni(CO)4 the oxidation state of Ni is?

Description : In Ni(CO)4 the oxidation state of Ni is?

Last Answer : Need answer

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Last Answer : (A) V2O5

The compound in which nickel has the lower oxidation states is :

Description : The compound in which nickel has the lower oxidation states is :

Last Answer : The compound in which nickel has the lower oxidation states is : A. `Ni(CO)_(4)` B. `(CH_(3)COO)_(2)Ni` C. `NiO` D. `NiCl_(2)(PPh_(3))_(2)`

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Last Answer : (C) Ferromagnetic at room temperature

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Last Answer : (A) Alumina

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Last Answer : (A) Cause cholesterol build up and blood clotting

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Last Answer : (D) Platinum

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Description : What is the atomic number of Cn ?

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`Na_(2)S + Na_(2)[Fe(CN)_(5)NO]to " X"` (Violet colour) The total number of possible isomers for complex " X" is , provided the ambident behaviour of

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BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

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Last Answer : We know that area of a triangle = 1/2 × base × altitude ∴ ar(△ABD) = 1/2 × BD × AM and ar(△BCD) = 1/2 BD × CN Now, ar(quad. ABCD) = ar(△ABD) + ar(△BCD) = 1/2 × BD × AM + 1/2 × BD × CN = 1/2 × BD × (AM + CN)

BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Description : BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Last Answer : We know that area of a triangle = 1/2 × base × altitude ∴ ar(△ABD) = 1/2 × BD × AM and ar(△BCD) = 1/2 BD × CN Now, ar(quad. ABCD) = ar(△ABD) + ar(△BCD) = 1/2 × BD × AM + 1/2 × BD × CN = 1/2 × BD × (AM + CN)

In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Description : In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Last Answer : Solution :-

In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain. (At. no. Fe = 26) -Chemistry

Description : [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain. (At. no. Fe = 26) -Chemistry

Last Answer : In both the cases, Fe is in oxidation state +3. Outer electronic configuration of Fe+3 is : In the presence of CN-, the 3d electrons pair up leaving only one unpaired electron. The ... sp3d2 forming an outer orbital complex. As it contains five unpaired electrons so it is strongly paramagnetic. .

I hv 300 oman baisa where cn I change it plzz?

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Last Answer : Diamond is a very large molecule made up of carbon atoms. Numerous carbon atoms are attached to each other in a single bond. So it is an object with a three-dimensional network. In its ... is also very high (3960). Since the molecule has no circulating electrons, it is electrically conductive.

If `aN = {ax|x in N}` and `bN nn cN = dN`, where `b,c epsilon N`, then

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What would be the equivalent weight of the reductant in the reaction : `[Fe(CN)_(6)]^(-3)+H_(2)O_(2)+2OH^(-)rarr2[Fe(CN)_(6)]^(4-)+2H_(2)O+O_(2)` [Giv

Description : What would be the equivalent weight of the reductant in the reaction : `[Fe(CN)_(6)]^(-3)+H_(2)O_(2)+2OH^(-)rarr2[Fe(CN)_(6)]^(4-)+2H_(2)O+O_(2)` [Giv

Last Answer : What would be the equivalent weight of the reductant in the reaction : `[Fe(CN)_(6)]^(-3)+H_(2)O_(2)+2OH^(-) ... O=16,H=1]` A. 17 B. 212 C. 34 D. 32

Which of the following expression is not applicable on the hydrolysis equilibrium : `CN^(-)H_(2)OrArrHCN+OH^(-)`

Description : Which of the following expression is not applicable on the hydrolysis equilibrium : `CN^(-)H_(2)OrArrHCN+OH^(-)`

Last Answer : Which of the following expression is not applicable on the hydrolysis equilibrium : `CN^(-)H_(2)OrArrHCN+OH^(-)` A. ... H+] =sqrt((K_(w).K_(a))/(C))`

The major product H in the given reaction sequence is `CH_(3)-CH_(2)-CO-CH_(3) overset(._(Theta)CN)toG overset(95% H_(2)SO_(4)) underset("Heat")toH`

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Last Answer : The major product H in the given reaction sequence is `CH_(3)-CH_(2)-CO-CH_(3) overset(._(Theta)CN) ... CH=underset(CH_(3))underset(|)(C)-CO-NH_(2)`

Hydrogen peroxide oxidises `[Fe(CN)_(6)]^(4-)` to `[Fe(CN)_(6)]^(3-)` in acidic medium but reduces `[Fe(CN)_(6)]^(3-)` to `[Fe(CN)_(6)]^(4-)` in alkal

Description : Hydrogen peroxide oxidises `[Fe(CN)_(6)]^(4-)` to `[Fe(CN)_(6)]^(3-)` in acidic medium but reduces `[Fe(CN)_(6)]^(3-)` to `[Fe(CN)_(6)]^(4-)` in alkal

Last Answer : Hydrogen peroxide oxidises `[Fe(CN)_(6)]^(4-)` to `[Fe(CN)_(6)]^(3-)` in acidic medium but reduces `[Fe(CN)_( ... 2) O + O_(2)) and (H_(2)O + OH^(-))`