Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Solubility of earbon dioxide gas in cold water can be increased by

Answer»

increase in pressure
decrease in pressure
increase in volume
none of these

Solution :It is because due to increase in INTRA molecular FORCE of ALTRACTION, Solubility of carbon dioxide GAS in cold water is INCREASED.
Discussion
2.

Solubility of carbon dioxide gas in cold water can be increased by.............

Answer»

increase in pressure
decrease in pressure
increase in volume
none of these

Solution :`H_2O(L)+CO_2(G)HARR underset(1 mol e)(H_(2)CO_(3)(l)`
Discussion
3.

Solubility of carbon dioxide gas in cold water can be increased by ……………… .

Answer»

increase in pressure
decrease in pressure
increase in volume
NONE of these

Solution :`UNDERSET(" 2 moles")(H_2O (l) + CO_2(g)) hArr underset("1 mole")(H_2CO_3(l))`
increase in pressure, favours the forward REACTION.
Discussion
4.

Solubility of an ionic compound in water is mainly dependent on A) Lattice enthalpy "" B) Hydration enathalpy Both these factors oppose each otherand the resultant of these determines thhe solubility of an ionic compound in water. If lattice enthalpy has greater value, the compound has greater value, the compound is less soluble. In case hydration enthaply has greater value, the compound is highly soluble is water BeF_(2) is soluble in water while fluorides of other alkalie earth metals are isoluble because of

Answer»

Covalent NATURE of `BeF_(2)`
Ionic nature of `BeF_(2)`
Greater hydration enthalpy of `Be^(2+)` ion
Greater lattice enthalpy of `Be^(2+)` ion

Solution :`BeF_(2)` is SOLUBLE while other fluorides of IIA GROUP elements are insoluble because `Be^(+2)` has more hydration energy DUE to it.s small size
Discussion
5.

Solubility of an ionic compound in water is mainly dependent on A) Lattice enthalpy "" B) Hydration enathalpy Both these factors oppose each otherand the resultant of these determines thhe solubility of an ionic compound in water. If lattice enthalpy has greater value, the compound has greater value, the compound is less soluble. In case hydration enthaply has greater value, the compound is highly soluble is water Compounds of alkalie earth metals are less soluble than alkali metals, due to

Answer»

Theirhigh hydration enthalpy
Their high lattice enthalpy
Their increased covalent character
Their high ionisation enthalpy

Solution :DUE to high lattice enthalpy II A GROUP compounds are less soluble than IA group elements
Discussion
6.

Solubility of an ionic compound in water is mainly dependent on A) Lattice enthalpy "" B) Hydration enathalpy Both these factors oppose each otherand the resultant of these determines thhe solubility of an ionic compound in water. If lattice enthalpy has greater value, the compound has greater value, the compound is less soluble. In case hydration enthaply has greater value, the compound is highly soluble is water Which of the following is more soluble in water

Answer»

`MgSO_(4)`
`CaSO_(4)`
`SrSO_(4)`
`BaSO_(4)`

Solution :The solubility of SULPHATES. Carbonates decreases down the group so `MgSO_(4)` is more soluble than other COMPOUNDS
Discussion
7.

Solubility of AgCl in 0.2 M NaCl is x and that in 0.1 M AgNO_3 is y then which of the following is correct ?

Answer»

` x= y`
` x gt y`
` x LT y`
we cannot predict

Solution :`[AG^(+) ] [CL^(-) ] =Ksp,`
` (i)x (0.2 ) = Ksp (II)(0.1 )y= Ksp `
Discussion
8.

Solubility of Ag CN is maximum in:

Answer»

ACIDIC buffer solution
basic buffer solution
in pure water
EQUAL in all solution

Solution :In acidic buffer , EQUILIBRIUM shifts TOWARDS RIGHT. So, solubilityincreases.
Discussion
9.

Solubilityof solid solutein a liquidsolventincreases withincreasein ______.

Answer»

Solution :When the temperature is increased, the AVERAGE kinetic energy of the MOLECULES of the solute and the solvent increases. The INCREASE in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that KEEP the solute molecules together and hence the solubility increases.
Discussion
10.

Solubility depends on the lattice energy of the solid and the hydration energy of the ions. Some lattice values for Group - II compounds are much lighter than the values for Group - I compounds, because of the effect of the increased charge on the ions in the Boron - Lande equation. Taking any one particular negative ion the lattice energy decreases as the size of the metal increases. The hydration energy also decreases as the metal ions become larger. For a substance to dissolve, the hydration energy must exceed thelattice energy The correct order of hydration energies is

Answer»

`Be^(2+)ltCa^(2+)ltCr^(2+)ltBa^(2+)`
`Be^(2+)ltSr^(2+)ltCa^(2+)ltBa^(2+)`
`Be^(2+)gtSr^(2+)gtCa^(2+)GTBA^(2+)`
`Be^(2+)gtCa^(2+)gtSr^(2+)gtBa^(2+)`

Solution :The order of hydration ENERGIES `Be^(+2) gt Ca^(+2)gt SR^(+2) gt Ba^(+2)`
The hydration energy depends on size of the ion, lesser the size higher will be the hydration energy
Discussion
11.

Solubility depends on the lattice energy of the solid and the hydration energy of the ions. Some lattice values for Group - II compounds are much lighter than the values for Group - I compounds, because of the effect of the increased charge on the ions in the Boron - Lande equation. Taking any one particular negative ion the lattice energy decreases as the size of the metal increases. The hydration energy also decreases as the metal ions become larger. For a substance to dissolve, the hydration energy must exceed thelattice energy The correct order of solubility is

Answer»

`BeF_(2)ltCaF_(2)gtSrF_(2)gtBaF_(2)`
`BeF_(2)gtCaF_(2)ltSrF_(2)ltBaF_(2)`
`CaF_(2)ltSrF_(2)ltBaF_(2)ltBeF_(2)`
`CaF_(2)gtSrF_(2)gtBaF_(2)gtBeF_(2)``

SOLUTION :The correct order of solublity is `CaF_(2) lt SrF_(2) lt BaF_(2) lt BeF_(2)`
The high VALUE of solubility of `BeF_(2)` is due to high hydration energy of `Be^(+2)` ION
Discussion
12.

Solubilirty product constants (K_(sp)) of salts of types MX, MX_(2), and M_(3)X at temperature T are 4.0xx10^(-8), 3.2xx10^(-14), and salts of temperature T is in the order

Answer»

`MX_(2)gtM_(3)XgtMX`
`M_(3)XgtMX_(2)gtMX`
`MXgtM_(3)XgtMX_(2)`
`MXgtMX_(2)gtM_(3)X`

Solution :Let US assume that S is the SOLUBILITY.
`MX(s)hArr M^(+)underset(S)((aq.))+X^(-)underset(S)((aq.))`
`K_(sp)=C_(M^(+))C_(X^(-))=(S)(S)`
or `S=sqrt(K_(sp))=sqrt(4.0xx10^(-8))`
`=2xx10^(-4)M`
`MX_(2)(s)hArrM^(2+)underset(S)((aq.))+2X^(-)underset(2S)((aq.))`
`K_(sp)=C_(M^(2+))C_(X^(-))^(2)=(S)(2S)^(2)=4S^(3)`
or `S=((K_(sp))/(4))^(1//3)= ((3.2xx10^(-4))/(4))^(1//3)`
`=2xx10^(-5)M`
`M_(3)X(s)hArr underset(3S)(3M^(+)(aq.))+X^(3-)underset(S)((aq.))`
`K_(sp)=C_(M^(+))^(3)C_(X^(3-))`
`=(3S)^(3)(S)=27S^(4)`
or`S=((K_(sp))/(27))^(1//4) = ((2.7xx10^(-15))/(27))^(1//4)`
`=10^(-4)M`
Thus, solubility order is
`MXgt M_(3)XgtMX_(2)`
Discussion
13.

Solids possess a ........... range order while liquids possess a ........... range order.

Answer»

SOLUTION :LONG, SHORT
Discussion
14.

Solid NH_(4)I on rapid heating in a closed vessel at 357^(@)C dvelops a constant pressure of 275 mm Hg owing to partial decomposition of NH_(4)I into NH_(3) and HI but the pressure gradually increases further (when the excess solid residue remains in the vessel) owing to the dissociation of HI. Calculate the final pressure developed at equilibrium. K_(p) for HI dissociation is 0.015 at 357^(@)C.

Answer»

SOLUTION :`307.46 MM`,
Discussion
15.

Solid NaOH dissolve in water and if solution is stirred than temperature of solution increases. Now the heating of this solution is suitable ?

Answer»

SOLUTION :NaOH when dissolve in water the process is exothermic. `NaOH_((s)) + (AQ) toNaOH_((aq))`+ heatso HEATING is not SUITABLE.
Discussion
16.

Solid H_2O_2has non planar and non linear structure based on

Answer»

Dipolemoment
X-ray STUDY
Both 1 & 2
CHEMICAL method

Answer :C
Discussion
17.

Solid component of the carth consisting of soil, rocks and mountains is called

Answer»

Hydrosphere
Lithosphere
ATMOSPHERE
BIOSPHERE

ANSWER :B
Discussion
18.

Solid CO_(2) is used as

Answer»

Poison
Anaesthesia
Refrigerant
Artificial respirant 

ANSWER :C
Discussion
19.

Solid carbon dioxide is also known as_____________

Answer»


ANSWER :DRY ICE.
Discussion
20.

Solid barium oxide has ions, but it does not conduct electricity. Why ?

Answer»

Solution :Electrical conductivity isdue to the movement of free electrons or freeions. Barium OXIDE is an ionic SOLID. It has no free electrons. Although solid barium oxide has ions, `Ba^(2+) and O^(2-`), the ions are not free to move in solid STATE, due to a definite crystal lattice.
Discussion
21.

Solid ammonium carbamate it taken in an empty closed container and allowed to attain equilibrium as, NH_2COONH_4(s)hArr2NH_3(g)+CO_2(g),K_(P)=500 choose the incorrect statements :

Answer»

EQUILIBRIUM total PRESSURE is 15 atm
On addition of more solid at equilibrium total pressure DECREASES at NEW equilibrium.
On increasing the volume of container, total pressure decreases at new equilibrium.
On increasing temperature,total pressure increases at new eqilibrium.

Answer :B::C
Discussion
22.

Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows: NH_(3)COONH_(4)(s)hArr2NH_(2)(g)+CO_(2)(g) At equilibrium, ammonia is added such that partial pressures of NH_(3) at new equilibrium equals the original total pressure (at previous equilibrium). If the ratio of the total pressures now to the original total pressure is a//b then report a+b.

Answer»


SOLUTION :`NH_(2)COONH_(4)(s)hArr2NH_(2)(g)+CO_(2)(g)`
`2P_(@) P_(@) K_(P)=(2P_(@))^(2)xxP_(@)=4P_(@)^(2)`
`3P_(@) P' K_(P)=(3P_(@))^(2)xxP=9P_(@)^(2).P`
`P=(4)/(9)P_(@)`
`(P^(t))/(P^(t))=(3P_(@)+(4)/(9)P_(@))/(3P_(@))=(31)/(27)`
Discussion
23.

Solid ammonium cabamate dissociae to give ammonia and carbon dioxide as folows NH_(2)COOMH_(4)(s)hArr2NH_(2)(g)+CO_(2)(g) which of the following graph correctly represents the equilibrium.

Answer»




SOLUTION :N//A
Discussion
24.

Solid AgNO_(3) is added to a solution which 0.1 M in Cl^(-) and 0.1 M in CrO_(4)^(2-). K_(sp) values for AgCl and Ag_(2)CrO_(4) are 1.7 xx 10^(-10) and 1.9 xx 10^(-12) respectively. The concentration of Cl^(-) when Ag_(2)CrO_(4) starts precitating will be

Answer»

`3.9 xx 10^(-5) M`
`2.9 xx 10^(-5) M`
`3.9 xx 10^(-3) M`
`1.9 xx 10^(-2) M`

SOLUTION :`:' K_(sp) (AgCl) = [AG^(+)][Cl^(-)]`
To precipitate `AgCl`,
`:. [Ag^(+)] = (1.7 xx 10^(-10))/(0.1) = 1.7 xx 10^(-9)`
`:' K_(sp) (Ag_(2)CrO_(4)) = [Ag^(+)]^(2)[CrO_(4)^(2-)]`
To precipitate `Ag_(2)CrO_(4)`
`:. [Ag^(+)] = SQRT((1.9 xx 10^(-12))/(0.1)) = 4.34 xx 10^(-6) M`
Since `[Ag^(+)]` required to precinpitate `AgCl` is less than that required to precipitate `Ag_(2)CrO_(4), AgCl` will precipitate first. Now since
`[Ag^(+)] = 4.34 xx 10^(-6)` when `Ag_(2)CrO_(4)` starts can be CALCULATED as
`K_(sp)[AgCl] = [Ag^(+)] [Cl^(-)]`
or `1.7 xx 10^(-10) = (4.34 xx 10^(-6))[Cl^(-)]`
or `[Cl^(-)] = 3.9 xx 10^(-5) M`.
Discussion
25.

Solid A overset("on")underset("heating")rarr anhydrous A A on strong heating produces compound(s) has/have:

Answer»

chain structure
tetrahedral structure
both (a) and (B)
none of the above

Solution :`Na_(2)S_(2)O_(3)` is the only thiosulphate which is hydrated.
`{:(Na_(2)S_(2)O_(3).5H_(2)O OVERSET(Delta)rarr,Na_(2)S_(2)O_(3)+5H_(2)O),(,""darrDelta),(,Na_(2)SO_(4)+Na_(2)S_(5)):} `
`Na_(2)SO_(4): Na^(+) .^(-)O-overset(O)overset(||)underset(O)underset(||)(S)-O^(-)Na^(+)`,

`S_(2)O_(3)^(2-) + CN^(-) rarr SCN^(-) +SO_(3)^(2-)`
`Fe^(3+) +3SCN^(-) rarr Fe(SCN)_(3)`
Discussion
26.

Solid A overset("on")underset("heating")rarr anhydrous A A is:

Answer»

`K_(2)S_(2)O_(3)`
`CaS_(2)O_(3)`
`PbS_(2)O_(3)`
`Na_(2)S_(2)O_(3)`

Solution :`Na_(2)S_(2)O_(3)` is the only THIOSULPHATE which is hydrated.
`{:(Na_(2)S_(2)O_(3).5H_(2)O overset(DELTA)rarr,Na_(2)S_(2)O_(3)+5H_(2)O),(,""darrDelta),(,Na_(2)SO_(4)+Na_(2)S_(5)):} `
`Na_(2)SO_(4): NA^(+) .^(-)O-overset(O)overset(||)underset(O)underset(||)(S)-O^(-)Na^(+)`,

`S_(2)O_(3)^(2-) + CN^(-) rarr SCN^(-) +SO_(3)^(2-)`
`Fe^(3+) +3SCN^(-) rarr Fe(SCN)_(3)`
Discussion
27.

Solid A is very hard, electrical insulator in solid as well as molten state and melts at extremely high temperature. What type of solid is it?

Answer»

Solution :Covalent Network SOLID like `SiO_2` (QUARTZ) or SIC or C (diamond)
Discussion
28.

Solid A is very hard, electrical insulator in solid as well as molten state and melts at extremely high temperature. What type of solid is it ?

Answer»

SOLUTION :COVALENT Network solid like ` SiO_(2)` (quartz) or SIC or C(diamond).
Discussion
29.

Solid A "and" B are taken in a closed container at a certain temperature. These two solids decompose are following equilibria are established simultaneously A(s)hArrX(g)+Y(g) K_(P_(1)=250atm^(2) B(s)hArrY(g)+Z(g) K_(P_(2)=? If the total pressure developed over the solid mixture is 50 atm. Then the value of K_(P) for the 2^(nd) reaction

Answer»

`375`
`625`
`225`
`250`

Solution :`A(s)X+Y`
`alpha alpha+beta`
`B(s)hArrY+Z`
`beta+alpha beta`
`rArrK_(P_(1)=alpha(alpha+beta)`
`K_(P_2)=beta(alpha+beta)`
`P_(total)=(alpha+beta)+alpha+beta=2(alpha+beta)`
`RARR 2(alpha+beta)=50rArr alpha+beta=25`
`rArr 250=25alpha rArr alpha=10,beta=15`
`rArr K_(P_(2)=beta(alpha+beta)=15xx25=375`
Discussion
30.

Soil salinity can be measured by

Answer»

Calorimeter
Potometer
Porometer
CONDUCTIVITY meter

SOLUTION :Soil SALINITY can be MEASURED by conductivity meter
Discussion
31.

Soft water gives lather with Soap.

Answer»


ANSWER :1
Discussion
32.

Soduim acetate can be converted to ethane by

Answer»

Heating with `LiAlH_(4)`
Electrolysis its AQUEOUS solution
Heating with sodalime
Heating with calcium acetate

Solution :`UNDERSET("Sodium acetate")(2CH_(3)COONa) +2H_(2)O OVERSET("Electrolysis")rarr CH_(3)-CH_(3) + 2CO_(2) + 2NaOH + H_(2)`
Discussion
33.

In the reaction, I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(-)+S_(4)O_(6)^(2-).

Answer»


Solution :`2S_(2)O_(3)^(2-) to S_(4)O_(6)^(2-) + 2E^(-)` J
Discussion
34.

Sodium thiosulphate reacts with iodine to give iodide and tetrathionate. In this reaction S_(2)O_(3)^(2-) undergoes

Answer»

Oxidation
Reduction
Disproportionation
Comproportionation

Answer :A
Discussion
35.

Sodium sulphate is soluble in water, whereas barium sulphate is sparingly soluble because

Answer»

The HYDRATION energy of sodium sulphate is more than its lattice energy.
The lattice energy of BARIUM sulphate is more than its hydration energy.
The lattice energy has no role to play in solubility.
The hydration energy of sodium sulphate is LESS than its lattice energy.

Solution :When the hydration energy is more than the lattice energy, the salt is SOLUBLE in water and VICE versa.
Discussion
36.

Sodium sulphate is soluble in water but barium sulphate is sparingly soluble because

Answer»

The hydration energy of `Na_(2)SO_(4)` is more than its LATTICE energy
The lattice energy of `BaSO_(4)` is LESS than its hydration energy
The lattice energy has no ROLE to play in solubility
The lattice energy of `Na_(2)SO_(4)` is more than its hydration energy 

Answer :A
Discussion
37.

Sodium sulphate in soluble in water but barium sulphate is sparingly soluble because

Answer»

The hydration energy of `Na_(2)SO_(4)` is more than its LATTICE energy while the lattice energy of `BaSO_(4)` is more than its hydration energy
The lattice energy has no role to play in solubility
the lattice energy of `Na_(2)SO_(4)` is more tha its hydration energy
None of these.

Solution :`Na_(2)SO_(4)` is soluble in water because its hydration energy is MUCH more than its lattice energy. `BaSO_(4)` is INSOLUBLE in water because its lattice energy is much more than its hydration energy.
Discussion
38.

Sodium street lamp gives off a characteristic yellow light of wavelength 588 nm. Calculate the energy mole (in kJ/mol) of these photons.

Answer»


Solution :`E = N_(0) hv = N_(0) h (c)/(lamda) = ((6.02 XX 10^(23) mol^(-1)) xx (6.63 xx 10^(-34) Js) xx (3 xx 10^(8) ms^(-1)))/(588 xx 10^(-9) m)`
`= 2.036 xx 10^(5) J mol^(-1)`
Discussion
39.

Sodium salt of which acid will be needed for preparation of propane ? Write chemical equation for the reaction.

Answer»

SOLUTION :Since one carbon ATOM is LOST as `CO_2` during decarboxylation, therefore , the ACID needed must CONTAIN one carbon atom more than propane , i.e., butanoic acid or 2-methylpropanoic acid
Discussion
40.

Sodium salt of which acid will be needed for the preparation of propane ? Write chemical equation for the reaction ?

Answer»

SOLUTION : `{:(""CH(CH_(3))_(2)),("|"),(OVERSET(1)CH_(3)-overset(2)CH_(2)-overset(3)CH_(2)-overset(4)CH-overset(5)CH-overset(6)CH_(2)-overset(7)CH_(2)-overset(8)CH_(2)-overset(9)CH_(2)-overset(10)CH_(3)),("|"),(""H_(3)C-CH-CH_(2)-CH_(3)):}`
Name : 5 -SECONDARY -butyl-4-isopropyldecane
(Where -`CH(CH_(3))_(2)` is isopropyl GROUP and `CH_(3)-underset(|)(CH)-CH_(2)-CH_(3)` means secondary butyl group)
Discussion
41.

Sodium salt of benzoic acid when heated with which of the following gives benzene

Answer»

Sodamide 
Sodalime 
SODIUM CHLORIDE 
Soda ash

Answer :B
Discussion
42.

Sodium reacts with water more vigorously than lithium because

Answer»

It has high atomic mass
It is more electronegative
It is more ELECTROPOSITIVE
It is a METAL

Answer :C
Discussion
43.

Sodium readily formsNa^(+)ion but never forms Na^(2+) ion. Explain.

Answer»

Solution :Due to STABLE closed SHELL electronic CONFIGURATION of `Na^(+)` ion.
Discussion
44.

Sodium reacts more vigorously than lithium because, it.....

Answer»

has HIGHER ATOMIC weight.
is a metal.
is more electropositive.
more electronegative.

Answer :C
Discussion
45.

Sodium pyrophosphate is represented by which of the following formula

Answer»

`Na_(2)P_(2)O_(4)`
`Na_(4)P_(2)O_(5)`
`Na_(4)P_(2)O_(7)`
`Na_(2)P_(2)O_(5)`

SOLUTION :Sodium pyrophosphate is represented by `Na_(4)P_(2)O_(7)`.It is sodium salt of pyrophosphoric acid `(H_(4)P_(2)O_(7))`,which may be considered to be made up by two molecules of ORTHO phosphoric acid `(H_(3)PO_(4))`eliminating one molecule of `H_(2)O`.
`2H_(3)PO_(4) underset("Pyrophosphoric acid")(overset(-H_(2)O)toH_(4)P_(2)O_(7)"")`
`underset("Sodium pyrophosphate")(O larr underset(NA)underset(|)underset(O)underset(|)overset(Na)overset(|)overset(O)overset(|)P-O-underset(Na)underset(|)underset(O)underset(|)overset(Na)overset(|)overset(O)overset(|)P RARRO)`
Discussion
46.

Sodium phenoxide reacts with CO_(2) at 400 K and 4.7 atm pressure to give

Answer»

SODIUM salicylate
Salicyl aldehyde
Catechol
Banzoic acid

Solution :The REACTION is CALLED KOLBE's SCHMIDT reaction.
Discussion
47.

Sodium peroxide which is yellow solid , when exposed to air becomes white due to the formation of

Answer»

`H_(2)O_(2)`
`Na_(2)O`
`Na_(2)O and O_(3)`
`NAOH and Na_(2)CO_(3)`

Solution :In presence of MOISTURE and `CO_(2)` present in AIR , `Na_(2)O_(2)` is CONVERTED into NaOH and `Na_(2)CO_(3)` respectively both of which are white
`Na_(2)O_(2) + H_(2) O to 2 NaOH + H_(2)O_(2)`
`2 Na_(2) O_(2) + 2 CO_(2) to 2 Na_(2) CO_(3) + O_(2)`
or `Na_(2)O_(2) + H_(2)O +CO_(2) to Na_(2)CO_(3) + H_(2)O_(2)`
Discussion
48.

Sodium peroxide which is yellow solid, when exposed to air becomes white due to the formation of

Answer»

`H_(2)O_(2)`
`Na_(2)O`
`Na_(2)O +O_(3)`
NaOH and `Na_(2)CO_(3)`

Solution :`CaCO_(3)OVERSET(DELTA)rarrunderset(("basic"))(CAO)+underset(("acidic"))(CO_(2))`
Discussion
49.

Sodium peroxide which is a yellow solid, when exposed to air becomes white due to the formation of

Answer»

`Na_(2)O` and `O_(3)`
NaO
NAOH and `Na_(2)CO_(3)`
NaOH and `H_(2)O_(2)`

SOLUTION :`Na_(2)O_(2)+2H_(2)O rarr H_(2)O_(2)+2NaOHoverset(CO_(2))rarrNa_(2)CO_(3)+H_(2)O`
Discussion
50.

Sodium nitropruside reacts with sulphide ion to give a purple colour due to the formation of ...

Answer»

A)`[FE(CN)_(5)NO]^(3-)`
B)`[Fe(NO)_(5)CN]^(+)`
C)`[Fe(CN)_(5)NOS]^(4-)`
D)`[Fe(CN)_(5)NOS]^(3-)`

Answer :C
Discussion