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

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

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

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

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`

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.

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)`

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 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

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 

Which of the following metals is least subject to oxidation? w) silver x) gold y) magnesium z) nickel

Description : Which of the following metals is least subject to oxidation? w) silver x) gold y) magnesium z) nickel

Last Answer : ANSWER: X -- GOLD

How to on or off network speed indicator in Lenovo K4 note?

Description : How to on or off network speed indicator in Lenovo K4 note?

Last Answer : Follow these steps to toggle on off network speed on your Lenovo Note 1) Go to settings tab on your phone 2) Tap on Notification Centre 3) Go through Status Bar icons tab 4) Find network speed ... tap on the toggle button there. Then the network speed should show on the status bar of your mobile.

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

Catalyst used in the oxidation of benzene to produce Maleic anhydride is (A) V2O5 (B) Pt (C) Ni (D) Cr

Description : Catalyst used in the oxidation of benzene to produce Maleic anhydride is (A) V2O5 (B) Pt (C) Ni (D) Cr

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)`

Pure nickel is (A) Ferromagnetic above its curie point (i.e., 415°C)(B) Having h.c.p. crystal lattice(C) Ferromagnetic at room temperature(D) Not resistant to oxidation at high temperature

Description : Pure nickel is (A) Ferromagnetic above its curie point (i.e., 415°C) (B) Having h.c.p. crystal lattice (C) Ferromagnetic at room temperature (D) Not resistant to oxidation at high temperature

Last Answer : (C) Ferromagnetic at room temperature

The catalyst used in the production of elemental sulphur from H2S (by oxidation-reduction) is (A) Alumina (B) Silica gel(C) Platinum(D) Nickel

Description : The catalyst used in the production of elemental sulphur from H2S (by oxidation-reduction) is (A) Alumina (B) Silica gel (C) Platinum (D) Nickel

Last Answer : (A) Alumina

Oils are partially hydrogenated (not fully) to manufacture Vanaspati, because fully saturated solidified oils (A) Cause cholesterol build up and blood clotting (B) Are prone to rancid oxidation (C) Always contain some amount of nickel (as their complete removal is very difficult) (D) Have affinity to retain harmful sulphur compounds

Description : Oils are partially hydrogenated (not fully) to manufacture Vanaspati, because fully saturated solidified oils (A) Cause cholesterol build up and blood clotting (B) Are prone to rancid oxidation (C ... (as their complete removal is very difficult) (D) Have affinity to retain harmful sulphur compounds

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

Oxidation of ortho-xylene in presence of __________ catalyst is done to produce phthalic anhydride on commercial scale. (A) Nickel (B) Vanadium (C) Alumina (D) Iron

Description : Oxidation of ortho-xylene in presence of __________ catalyst is done to produce phthalic anhydride on commercial scale. (A) Nickel (B) Vanadium (C) Alumina (D) Iron

Last Answer : (B) Vanadium

. Catalyst used in the catalytic converter employed in automobiles to convert CO into CO2 and for complete oxidation of un-burnt hydrocarbons is (A) Nickel (B) Cobalt (C) Vanadium (D) Platinum

Description : . Catalyst used in the catalytic converter employed in automobiles to convert CO into CO2 and for complete oxidation of un-burnt hydrocarbons is (A) Nickel (B) Cobalt (C) Vanadium (D) Platinum

Last Answer : (D) Platinum

The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex i

Description : The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex i

Last Answer : The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2 ... . D. It produces three-fold freezing point depression.

The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex i

Description : The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex i

Last Answer : The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and ... (H_(2)O)_(2)](NO_(3))_(2)`, octahedral

The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex i

Description : The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex i

Last Answer : The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero ... two B. zero, zero C. two, zero D. two ,two

45. When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from (1) Zero to + 1 and zero to –3 (2) Zero to +1 and zero to –5 (3) Zero to –1 and zero to + 5 (4) Zero –1 and zero to + 3

Description : 45. When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from (1) Zero to + 1 and zero to –3 (2) Zero to +1 and zero to –5 (3) Zero to –1 and zero to + 5 (4) Zero –1 and zero to + 3

Last Answer : Zero to –1 and zero to + 5

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. .

Why is Zinc and not Copper used for the recovery of Silver from the complex [Ag (CN)2 ] ?

Description : Why is Zinc and not Copper used for the recovery of Silver from the complex [Ag (CN)2 ] ? 

Last Answer : Ans. Zinc is stronger reducing agent and more electropositive than Copper. (E° = + 0.34V) 

Which of the following is a nucleophile? (a) AlCl3 (b) H3O+ (c) BF3 (d) CN–

Description : Which of the following is a nucleophile? (a) AlCl3 (b) H3O+ (c) BF3 (d) CN–

Last Answer : CN–

A steel alloy containing 36% nickel is called __________, which has a zero co-efficient of expansion. (A) Austenitic stainless steel (B) Heat resisting steel (C) Invar (D) High speed steel

Description : A steel alloy containing 36% nickel is called __________, which has a zero co-efficient of expansion. (A) Austenitic stainless steel (B) Heat resisting steel (C) Invar (D) High speed steel

Last Answer : (C) Invar

Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved : (i) [CoF4]2- (ii) [Cr(H2O)2(C2O4)2]– (iii) [Ni(CO)4] (Atomic number : Co = 27, Cr = 24, Ni = 28) -Chemistry

Description : Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved : (i) [CoF4]2- (ii) [Cr(H2O)2(C2O4)2]– (iii) [Ni(CO)4] (Atomic number : Co = 27, Cr = 24, Ni = 28) -Chemistry

Last Answer : (i) [COF4]2_ : Tetrafluorido cobalt (III) ion Coordination number = 4 Shape = Tetrahedral Hybridisation = sp3 Important Questions for Class 12 Chemistry Chapter 9 Coordination Compounds Class 12 Important Questions 7

What is the atomic number of Cn ?

Description : What is the atomic number of Cn ?

Last Answer : The atomic number of Cn is 112

`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

Description : `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

Last Answer : `Na_(2)S + Na_(2)[Fe(CN)_(5)NO]to " X"` (Violet colour) The total number of possible ... the ambident behaviour of `CN^(-)` is not considered.

Assertion :- Oxidation number of Cr in `Cr(CO)_(6)` is zero. Reason :- Cr is a metal.

Description : Assertion :- Oxidation number of Cr in `Cr(CO)_(6)` is zero. Reason :- Cr is a metal.

Last Answer : Assertion :- Oxidation number of Cr in `Cr(CO)_(6)` is zero. Reason :- Cr is a metal. A. ... Reason is False. D. If both Assertion & Reason are False.

Assertion :- Oxidation number of carbon in `CH_(2)O` is zero. Reason :- `CH_(2)O` (formaldehyde) is a covalent compound.

Description : Assertion :- Oxidation number of carbon in `CH_(2)O` is zero. Reason :- `CH_(2)O` (formaldehyde) is a covalent compound.

Last Answer : Assertion :- Oxidation number of carbon in `CH_(2)O` is zero. Reason :- `CH_(2)O` (formaldehyde ... is False. D. If both Assertion & Reason are False.

Which has zero oxidation number :-

Description : Which has zero oxidation number :-

Last Answer : Which has zero oxidation number :- A. Cd B. `Cl_(2)` C. (1) & (2) D. None