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.

Consider the following alkyl halides ( 1) (CH_3)_3CBr , (2)CH_3Br , (3)C_2H_5Br , (4)CH_3CHBr CH_3 Arrange these alkyl halides in decreasing order of reactivity in Williamson reaction.

Answer»

1 gt 4 gt 3 gt 2
1 gt 2 gt 3 gt 4
4 gt 3 gt 2 gt1
2 gt 3 gt 4 gt 1

Solution :The Williamson REACTION involves NUCLEOPHILIC SUBSTITUTION of ALKOXIDE ion. It is strictly analogous to the preparation of alcohols by treatment of alkyl halide with aqueous NaOH. Alkyl halides undergoes substitution reaction most readily , t-butyl bromide are inert to nucleophilic substitution because they show maximum steric hindrance to the approaching nucleophile. Thus, reactivity of alkyl halides TOWARDS Williamson reaction is , `1^@ RX gt 2^@ RX gt 3^@ RX `
Discussion
2.

Consider the following abbreviations for hydrated alkali ions X=[Li(H_(2)O)_(n)]^(+), Y=[K(H_(2)O)_(n)]^(+), Z=[Cs(H_(2)O)_(n)]^(+) Which is the correct order of size of these hydrated alkali ions

Answer»

XgtYgtZ
ZgtYgtX
X=Y=Z
ZgtXgtY

Solution :The smaller the CATION, the GREATER is the degree of HYDRATION, HENCE the degree of hydration DECREASES from `Li^(+)` to `Cs^(+)`
Discussion
3.

Consider the followign statements 1. Isotonic solutions have the same molar concentration at a given temperature 2. The molal elevation constant K_b is a characteristic of a solvent, and is independent of the solute added 3. The freezing point of a 0.1 M aqueous AlCl_3 solution. Which of these statements is correct

Answer»

1 and 2
2 and 3
1 and 3
1,2 and 3

Solution :All the STATEMENTS are CORRECT
Discussion
4.

Consider the floowing reaction: Which of the following is correct combination of major products, correspondingly ?

Answer»

`I - a, II - p`
`I - a, II - q`
`I - b, II - q`
`I - b, II - p`

SOLUTION :REACTION `(I)` UNDERGOES `NGP, S_(N)2` and `S_(N)1`, while reaction (II) follows mainly `S_(N)1` PATH.
Discussion
5.

Consider the follow redox reaction and select the correct option (s).2SO_2 O_3 ^(2-)+I_2toS_4O_(6) ^(2-) + 2I^(-)

Answer»

`S_2 O_(3)^(2-)`gets reducedto`S_4O_(6)^(2-)`
`S_2O_(3)^(2-)` getsoxidisedto `S_4O_(6)^(2-)`
`I_2` getsreducedto `I^(-)`
`I_2 ` getsoxidisedto `I^(-)`

Solution :Here `S_2 O_(3)^(2-)`getsoxidisedto `S_4 O_(6)^(2-)` while`I_2`is reducedto`I^(-)`ions
Discussion
6.

Consider the first ionization energies of the elements whose electronic configurations correspond to the following. I. [He]2s^(2) 2p^3 II. [He]2s^(2) 2p^4 III. [Ne]3s^(2)3p^(1)[Ar]3d^(10)4p^1 Which of the above have almost the same first ionization energies?

Answer»

I and II
I and III
III and IV
II and IV

Solution :AL & Ga have approx. same first I.E. due to similar size (due to poor SHIELDING of d orbitals in Ga).
Discussion
7.

Consider the figure given alongside and answer the questions (i) to (vi) given below (i) Give the direction of electron flow. (ii) Is silver plate the anode or cathode ? (iii) What will happen if salt bridge is removed ? (iv) When will the cell stop functioning ? Zinc plate ,, (v) How will concentration of Zn^(2+) ions and Ag^(+) ions be affected when the cell functions ? (vi) How will the concentration of Zn^(2+) ions and Ag^(+) ions be affected after the cell becomes 'dead' ?

Answer»

Solution : (i) Electrons move from Zn to Ag.
(ii) Ag is the cathode.
(iii) CELL will stop functioning.
(iv) When `E_("cell")=0`
(v) Concentration of `Zn^(2+)` ions will increase and concentration of `Ag^(+)` ions will DECREASE.
(vi) When `E_("cell")=0` equilibrium is reached and concentration of `Zn^(2+)` ions and `Ag^(+)` ions will not change.
Discussion
8.

Consider the figure below and answer the following questions : (i) Cell 'A' has E_("cell") = 2V and Cell 'B' has E_("cell") = 1.1 V which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reaction will occur in this cell? (ii) If cell 'A' has E_("cell") = 0.5 V and cell 'B' has E_("cell") =1.1 V then what will be the reaction at anode and cathode?

Answer»

Solution :(i) Cell .B. will act as electrolytic cell as it has lower emf.
The electrode reactions will be :
`ZN^(2+) + 2e^(-) to Zn` at cathode
`Cu to Cu^(2+) + 2e^(-)` at ANODE.
(ii) Now cell .B. acts as GALVANIC cell as it has higher emf and will push electrons into cell .A.. The electrode reaction will be :
`{:("At anode", Zn to Zn^(2+) + 2e^(-)),("At cathode", Cu^(2+) + 2e^(-) to Cu):}`
Discussion
9.

Consider the figure below and answer the following questions : (i) Cell 'A' has E_(cell)=2" V " and Cell B has E_(cell) =1.1" V " which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell ? (ii) If cell 'A' has E_(cell)=0.5" V " and cell 'B' has E_(cell) =1.1" V " then what will be the reactions at anode and cathode ?

Answer»

Solution :(i) SINCE the cell (B) has low e.m.f. , it will act as electrolytic cell. In the cell, the ELECTRODE reaction that takes PLACE is :
`Zn^(2+)(AQ)+2e^(-) to Zn(s)""("At cathode")`
`Cuoverset((aq)) rarr Cu^(2+)(aq)+2e^(-)`
(ii) The data shows that the cell 'B' will act as electrochemical cell since its `E_(cell)`. (1.1V) is mre than that of cell 'A' (0.5V).The electrode reactions which TAKE place are :
`Zn(s) to Zn^(2+)(aq)+2e^(-)"" ("At anode")`
`Cu^(2+)(aq)+2e^(-) to Cu(s)"" ("At cathode")`
Discussion
10.

Consider the figure and mark the correct option.

Answer»

Water will move from SIDE (A) to side (B) id a PRESSURE lower osmitic pressure is applied on piston(B)
Water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on postion (B)
Water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on pistion (B)
Water will more from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A)

Solution :We know that, if a pressure higher than the osmotic pressure is applied on the solution the SOLVENT will flow from the solution into the PURE solvent through the semi-permeable membrane. This process is called REVERSE osmosis.
Thus in this case, water will move form side (B) to side (A) it a pressure greater than osmotic pressure is applied on piston (B).
Discussion
11.

Consider the Figure and mark the correct option.

Answer»

WATER will move from side (A) to side (B) if a presure LOWER than OSMOTIC pressure is APPLIED on piston (B).
Water will move from side (B) to side (A) if a presure greater than osmotic presure is applied on piston (B).
Water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B).
Water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A).

Solution :Explanation : Water will move from side (B) to side (A) if pressure greater than osmotic presure is applied on piston (B) due to reverse osmosis.
Discussion
12.

Consider the figure and answer the questions (i) to (vi) gives below. (i) Redraw the diagram to show the direction of electron flow. (ii) Is silver plate the anode or cathode ? (iii) What will happen if salt bridge is removed ? (iv) When will the cell stop functioning ? (v) How will concentration of Zn^(2+) ions and Ag^(+) ion be affected when the cell functions ? (vi) How will the concentration of Zn^(2+) ions and Ag^(+) ions be affected after the cell becomes 'dead' ?

Answer»

Solution :(i) Electrons move from ZN to Ag.

(II) It acts as cathode.
(ii) Cell will stop functioning.
(iv) When equilibrium is attained i.E., `E_(cell)=0`.
(v) CONCENTRATION of `Zn^(2+)` will increase and `Ag^(+)` ions will decrease.
(vi) When `E_(cell)=0` (equilibrium is REACHED), concentration of `Zn^(2+)` ions and `Ag^(+)` ions will not change.
Discussion
13.

Considerthe figure and answer the questions (i) to (vi) given below : (i) Redraw the diagram to show the direction of electron flow. (ii) Is silver plate the anode or cathode ? (iii) What will happen if salt bridge is removed ? (iv) When will the cell stop functioning ? (v) How will concentration of Zn^(2+) ions and Ag^(+) ions be affected when the cell functions ? How will the concentration of Zn^(2+) ions and Ag^(+) ions be affected after the cell becomes 'dead' ?

Answer»

Solution :(i) The flow of eectrons takes place from zinc plate towards COPPERPLATE in the outer circuit.
(ii) Silver plate acts as cathode.
(iii) Cell will stop working because the inner circuit does not exist any more.
(iv) When `E_(cell)`=0
(v) Concentration of `Zn^(2+)`(AQ) ions will gradually decrease while that of `Ag^(+)`(`mu`aq) ions will increase.
(vi) When the cell becomes dead (`E_(cell)`=0), the concentration of `Zn^(2+)` (aq) and `Ag^(+)` (aq) ions will not change any more.
Discussion
14.

Considerthefigureandmark the correctoption .

Answer»

water will move from SIDE (A) to side (B) if a PRESSURE lower than OSMOTIC pressure is applied on piston (B)

water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B)
water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B)
water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A)

Solution :(b) represents REVERSE OSMOSIS.
Discussion
15.

Consider the figure and answer the following questions. (i) Cell 'A' has E_(cell)2V and cell 'B' has E_(cell)=1.1V which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell ? (ii) If cell 'A' has E_(cell)=0.5Vand cell 'B' has E_(cell)=1.1V then what will be the reaction at anode and cathode ?

Answer»

Solution :(i) Cell .B. will act as electrolytic cell as it has lower EMF
`therefore` The electrode reactions will be:
Reduction at cathode : `Zn^(2+)+2E^(-) to Zn`
Oxidation at anode : `CU to Cu^(2+) +2e^(-)`
(ii) Now cell .B. ACTS as galvanic cell as it has higher emf and will push ELECTRONS into cell .A..
The electrode reaction will be :
At anode : `Zn to Zn^(2+) + 2e^(-)`
At cahotde : `CutoCu^(2+)+2e^(-)`
Discussion
16.

Consider the figure and answer the following question. (i) Cell 'A' h as E_("cell")=2V and Cell 'B' has E_("cell")=1.1V which of the two cell 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell? (ii) If cell 'A' has E_("cell")= 0.5V and cell 'B' has E_("cell")=1.1V then what will be the reactions at anode and cathode?

Answer»

SOLUTION : (i) Cell 'B' act as electrolytic cell due to its lesser value of EMF.
Theelectrode REACTIONS will be
Atcathode `Zn^(2+)+2e^(-)rarrZn`
At anode `CurarrCu^(2+)+2e^(-)`
(ii) If cell 'B' are higher emf, it acts as galvanic cell. Now it will push electrons into cell 'A'
In this CASE, the reactions will be
`ZnrarrZn^(2+)+2e^(-)` (At anode)
`Cu^(2+)+2e^(-)rarrCu`(At cathode)
Discussion
17.

Consider the expression Delta G= - RT " In " K_(p)+RT " In " Q_(p) and select the correct statement at equilibrium where Q_(p) and K_(p) term refer to reaction quotient and eq uilibrium constant at constant pressure respectively.

Answer»

`Delta G= 0, Q_(p) GT K_(p)`the EQUILIBRIUM reaction will SHIFT from left to right
`Delta G=0, Q_(p)=K_(p)` the equilibrium reaction will shift from left to right
`Delta G= oo, Q_(p) lt K_(p)` the equilibrium reaction will shift from right to left
`DeltaG lt 0, Q_(p) gt K_(p)` the equilirbium reaction will shift from right to left

Solution :Van't Hoff reaction isothenn is `Delta G= Delta G^(@)+RT` In Qp When the reaction is in a state of equilibrium `Delta G=0` M
Then `Delta G^(@)= - RT " In " Q_(p)= - RT " In" K_(p)`
Discussion
18.

Consider the Fig. and mark the correct option.

Answer»

WATER will move from SIDE (A) to side (B) if a pressure lower than OSMOTIC pressure is applied on PISTON (B).
water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B).
water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B).
water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A).

Answer :B
Discussion
19.

Consider the figure and mark the correct option.

Answer»

Water will move form side (A) to side (B) if a PRESSURE lower than OSMOTIC pressure is APPLIED on poston (B).
water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piton (B).
water will move from side (B) to side (A) if a pressure EQUAL too osmotic pressure is applied on piston (B).
water will move from side (A) to dide (B) if pressure equal to osmotic pressure is applied on piston (A).

Solution :is the correct option. This is because of revarseosmosis which is taking PLACE.
Discussion
20.

Consider the equilibrium SO_(3) (g) hArr SO_(2) (g) + (1)/(2) O_(2) (g) ""K_(C ) = 1 What should be the initial concentration so that at equilibrium [SO_(3)] =[O_(2)]

Answer»

`(4)/(3) M`
`(1)/(3)M`
`(1)/(4)M`
`(3)/(4)M`

Solution :`{:(,SO_(3) (G),hArr,SO_(2) (g),+,(1)/(2) O_(2) (g),K_(C) = 1,),(t = 0,C_(0),,0,,0,,),("conc.",,,,,,,),(t = t_("eqm"),C_(0) - x,,x,,(x)/(2),,),("conc.",,,,,,,):}`
GIVEN `[SO_(3)] = (O_(2)]`
`C_(0) - x = (x)/(2)`
`(3x)/(2) = C_(0) RARR x = (2C_(0))/(3)`
at equilibrium `[O_(2)] = (x)/(2) = (C_(0))/(3), [SO_(2)] = x = (2C_(0))/(3), [SO_(3)] = C_(0) - x`
`[SO_(3)] = C_(0) - (2C_(0))/(3)`
`[SO_(3)] = (C_(0))/(3)`
`K_(C) = 1 = ([SO_(2)] [O_(2)]^(1//2))/([SO_(3)]) = ((2C_(0))/(3) xx ((C_(0))/(3))^(1//2))/((C_(0))/(3))`
`1 = 4 (C_(0))/(3)`
`C_(0) = (3)/(4) M`
Discussion
21.

Consider the equilibrium HgO(s)+4I^(-)(aq)+H_(2)O(l)hArrHg I_(4)^(2-)(aq)+2OH^(-) (aq), which changes will decrease the equilibrium concentration of HgI_(4)^(2-)

Answer»

Addition of 0.1 M HI (aq)
Addition of HgO(s)
Addition of `H_(2)O` (L)
Addition of KOH (aq)

Answer :C::D
Discussion
22.

Consider the electronic configuration: 1s^(2)2s^(2)2p_(x)^(2) It violates

Answer»

Aufbalu principle
Hund's rule
Pauli's EXCLUSION principle
It is correct.

Answer :B
Discussion
23.

Consider the diamagnetic complex [M(en)_(3)](ClO_(4))_(30, Where M is a first row transition metal except Sc. Find the value of x+y of M in above complex. If x=oxidation state y= number of d electrons present

Answer»


SOLUTION :x=+3,y=6
Discussion
24.

Consider the DeltaG_(f)^(@) (kJ/mol) for the following oxides. Which oxide can be most easily decomposed to form the metal and oxygen gas?

Answer»

`ZNO (DELTAG^(@)=-318.4 , DELTAH^(@)=-348.3)`
`Cu_(2)O (DeltaG^(@)=-146.0 , DeltaH^(@)=-168.8)`
`HgO (DeltaG^(@)=-58.5 , DeltaH^(@)=-90.8)`
`PbO(DeltaG^(@)=-187.9 , DeltaH^(@)=-217.3)`

Solution :N//A
Discussion
25.

Consider the decomposition reaction : 2H_2O_2 overset(OH^(-)//I^-)to 2H_2O + O_2 This reaction takes place in two steps as given below : Step 1. H_2O_2 + I^(-) to H_2O + IO^(-) (slow) Step 2. H_2O_2 +IO^(-) to H_2O + I^(-) + O_2 (fast) Determine rate law expression.

Answer»

SOLUTION :Rate = `K[H_2O_2][I^(-)]` because SECOND step is rate determining step.
Discussion
26.

Consider the decomposition reaction : 2H_2O_2 overset(OH^(-)//I^-)to 2H_2O + O_2 This reaction takes place in two steps as given below : Step 1. H_2O_2 + I^(-) to H_2O + IO^(-) (slow) Step 2. H_2O_2 +IO^(-) to H_2O + I^(-) + O_2 (fast) Determine the order of reaction.

Answer»

SOLUTION :ORDER = 1 + 1 = 2
Discussion
27.

Consider the decomposition of N_(2)O_(5) as N_(2)O_(5) to 2" NO"_(2)+(1)/(2)O_(2) The rate of reaction is given by -(d[N_(2)O_(5)])/(DT)=(1)/(2)(d[NO_(2)])/(dt)=2(d[O_(2)])/(dt)=k_(1)[N_(2)O_(5)] Therefore, -(d[N_(2)O_(5)])/(dt)=k_(1)[N_(2)O_(5)] +(d[NO_(2)])/(dt)=2k_(1)[N_(2)O_(5)]=k_(1)'[N_(2)O_(5)] +(d[O_(2)])/(dt)=(1)/(2)k_(1)[N_(2)O_(5)]=k_(1)''[N_(2)O_(5)] Choose the correct option

Answer»

`k_(1)=k_(1)'=k_(1)''`
`k_(1)=2" K"_(1)'=k_(1)''`
`2" k"_(1)=k_(1)'=4" k"_(1)''`
`4" k"_(1)=2" k"_(1)'=k_(1)''`

Solution :`-(d[N_(2)O_(5)])/(dt) = (1)/(2)(d[NO_(2)])/(dt) = 2""(d[O_(2)])/(dt)`
`k_(1)[N_(2)O_(5)] = (1)/(2) k_(1)'[N_(2)O_(5)] = 2k_(1)''[N_(2)O_(5)]`
or `k_(1) = (1)/(2)k_(1)' = 2k_(1)'' or 2k_(1) = k_(1)' = 4k_(1)''`
Discussion
28.

Consider the decomposition of N_2O_(5(g)) to form NO_(2(g)) and O_(2(g)) . At a particular instant N_2O_5 disappears at a rate of 2.5xx10^(-2) "mol dm"^(-3)s^(-1) . At what rates are NO_2 and O_2 formed ? What is the rate of the reaction ?

Answer»

Solution :`2N_2O_5(g)rarr4NO_(2(g))+O_2(g))`
from the stoichiometry of the REACTION of the reaction.
`-1/2(d[N_2O_5])/(dt)=1/4(d[NO_2])/(dt)=-(d[NO_2])/(dt)=2(-d[N_2O_5])/(dt)`
RATE of disappearance of `N_2O_(5(g))` is `2.5xx10^(-2) "mol dm"^(-3)s^(-1)`
`:.` The rate of formation of `NO_2` at this TEMPERATURE is `2xx2.5xx10^(-2)=5xx10^(-2)"mol dm"^(3)s^(-1)` .
`-1/2(d[N_2O_5])/(dt)=-(d[O_2])/(dt)`
`:. (d[O_2])/(dt) = 1/2xx2.5xx10^(-2) "mol dm "^(-3)s^(-1)=1.25xx10^(-2)" mol dm "^(-3)s^(-1)`
Discussion
29.

Consider the decomposition of hydrogen peroxide in the alkaline medium which is catalysed by iodide ions. This reaction takes place in two steps as given below step-1 H_(2)O_(2) + I^(-1) to H_(2)O + IO^(-)(Slow) Step II H_(2)O_(2) + IO^(-) to H_(2)O + I^(-) +O_(2) (fast) a) Write the rate law expression and determine the order of reactan w.r.t H_(2)O_(2) b) What is the molecularity of each individual step?

Answer»

SOLUTION :a) Rate law expression:
Rate = `k[H_(2)O_(2)][I^(-)]`
Order of REACTION w.r.t `H_(2)O_(2)=1`
B) Molecularities of both step-I and step-II are two.
Discussion
30.

Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions. 2H_(2)O_(2) overset(OH^(-)) to 2H_(2)O+O_(2) The This reaction takes place in two steps as given below: Step I H_(2)O_(2)+I^(-) to H_(2)O+IO^(-) (slow) Step IIH_(2)O_(2) +IO^(-) to H_(2)O +I^(-) +O_(2) (fast) (a) Writethe rate expression and determine the order of reaction wrt H_(2)O_(2). (b) What is the molecularity of each individual step?

Answer»

Solution :(a)Rate `=k[H_(2)O][IO^(-)]`
As, there is one molecule of `H_(2)O_(2)` involved in the SLOW step, ORDER of the reaction WRT `H_(2)O_(2)`is 1.
(B) Molecularity of step `I=2`.
Molecularity of step `II =2.`
Discussion
31.

Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions : 2" H"_(2)"O"_(2)overset("OH"^(-))to2" H"_(2)"O"+"O"_(2) This reaction takes place in two steps as given below : Step-I. "H"_(2)"O"_(2)+I^(-)to"H"_(2)"O"+" IO"^(-)("slow") Step-II."H"_(2)"O"_(2)+"IO"^(-)to"H"_(2)"O"+I^(-)+"O"_(2)("fast") (a) Write the rate law expression and determine the order of reaction w.r.t. "H"_(2)"O"_(2). (b) What is the molecularity of each individual step ?

Answer»


ANSWER :(a) Rate `=K[H_(2)O_(2)][I^(-)]," ORDER w.r.t. "H_(2)O_(2)=1." (b)MOLECULARITY : Step-1"=2, "Step-II = 2."`
Discussion
32.

Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions. 2H_2O_2 overset(OH^(-))to 2H_2O + O_2 This reaction takes place in two steps as given below : Step I H_2O_2 + I^(-) to H_2O + IO^(-)(slow ) Step II H_2O_2 + IO^(-) to H_2O + I^(-) + O_2 ( fast ) (a) Write the rate law expression and determine the order of reaction w.r.t. H_2O_2. (b)What is the molecularity of each individualstep ?

Answer»

Solution :(a) Rate `=K[H_2O][IO^(-)]`
As , there is ONE moleculeof `H_2O_2` involved in the slow step, order of the reaction w.r.t. `H_2O_2` is 1 .
(B) Molecularityof step I = 2 .
Molecularity of step II =2 .
Discussion
33.

Consider the data given below (AAK_MCP_16_NEET_CHE_E16_019_Q01.png" width="80%"> On the basis of the given data, arrange the gases in their increasing

Answer»

AltBltCltD
BltDltAltC
CltDltAltB
CltAltDltB

Answer :B
Discussion
34.

Consider the coordination compound [Cr(NH_3)_5CO_3]Cl.Write the ionisation isomer of the above compound.

Answer»

SOLUTION :The IONISATION ISOMER is `[CR(NH_3)_5Cl]CO_3`
Discussion
35.

Consider the coordination compound [Cr(NH_3)_5CO_3]Cl.Write the IUPAC name of the above coordination compound.

Answer»

SOLUTION :Pentammine CARBONATO CHROMIUM(III) CHLORIDE
Discussion
36.

Consider the coordination compound [Cr(NH_3)_5CO_3]Cl.whether the ligands present in the above compound are ambidentate ligands?Why?

Answer»

Solution :No both AMMONIA and carbonate IONS are not ambidentate because they have only one DONOR SITE
Discussion
37.

Consider the coordination compound [Cr(NH_3)_5CO_3]Cl.Name the ligands present in the above compound.

Answer»

SOLUTION :AMMONIA CARBONATE ION
Discussion
38.

Consider the coordination compound [Cr(NH_3)_5CO_3]Cl.Name the central metal ion of the above compound.

Answer»

SOLUTION :CHROMIUM(Cr^+3)
Discussion
39.

Consider the coordination compound [Co(NH_3)_5SO_4]Br.Which type of structural isomerisms is exhibited by the above coordination compound?

Answer»

SOLUTION :IONISATION ISOMERISM
Discussion
40.

Consider the coordination compound [Co(NH_3)_5SO_4]Br.Write the IUPAC name of the above coordination compound.

Answer»

SOLUTION :Pentaamminesulphatocobalt(III0 BROMIDE
Discussion
41.

Consider the coordination compound [Co(NH_3)_5SO_4]Br.What is the primary valance and secondary valance of the central metal cobalt in the above coordination compound?

Answer»

Solution :Primary VALENCE of CO^(+3)(i.e,OXIDATION state) secondary valence of Co=6(i.e.coordination NUMBER)
Discussion
42.

Consider the coordination compound, [Co(NH_(3))_(5)SO_(4)]Br (i) Write IUPAC name of the above coordinationcompound (ii) What is the primary valence and secondary valence of the central metal, cobalt, in the above coordination compound? (iii) Which type of structural isomerism is exhibited by theabove coordination compound ?

Answer»

Solution :(i) pentaamminesulphatocobalt (III) BROMIDE (ii) Primary valence = +3, SECONDARY valence =6 (iii) Ionization isomerism]
Discussion
43.

Consider the coordination compound [Co(NH_3)_5Cl]Cl_2 write the name of the isomerism exhibited by the complex [Pt(NH_3)_2Cl_2] represent the possible isomers

Answer»

SOLUTION :GEOMETRICAL ISOMERISM
Discussion
44.

Consider the compounds Pick up the correct statement/s

Answer»

II is more BASIC than I and III
All are aromatic bases
III is more basic than II
I is more basic than II.

Solution :II is more basic than I and III. Also all the compounds given in the qestion are aromatic in NATURE.
Discussion
45.

Consider the coordination compound [Co(NH_3)_5Cl]Cl_2 write the IUPAC name of the coordination compound.

Answer»

SOLUTION :Pentaamminechloridocobalt(III) CHLORIDE
Discussion
46.

Consider the coordination compound [Co(NH_3)_5Cl]Cl_2 what is the primary valence and secondary valence of the central metal ion in the above co ordination compound.

Answer»

SOLUTION :PRIMARY VALENCY:+3,SECONDARY valency:6
Discussion
47.

Consider the compounds given below: (P) BiCl_(3) (Q) SbCl_(3) (R ) VCl_(5) (S) AlCl_(3) Which would form oxo ion of MO^(+) type when subjected to hydrolysis?

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<P>P,Q,R only
P,Q,S only
P,Q only
P,Q,R,S

Answer :C
Discussion
48.

Consider the compound on left column and match with reactions from right column by which they can be prepared:

Answer»


Answer :`"A"rightarrow"r,t";"B"rightarrow"p,Q,s; C"rightarrow"r,t;D"rightarrow"r,t"`
Discussion
49.

Consider the complexes (P) [Fe(Cl)(F)(CN)(H_(2)O)(en)] (Q) [MoCl_(2)F_(2)(gly)]^(2-) these are having number for optically active isomers a and b respectively then a-b is :

Answer»


SOLUTION :i.e. a=12, b=4.
Discussion
50.

Consider the complex [Pt(S_(5))_(3)]^(2-). If x, y and z are the number of chelates, number of optically active isomers and number of geometrical iosmers respectively, find the value of (x+y+z):

Answer»


Solution :`[PT(S_(5))_(3)]^(2-)`

NUMBER of chelates =3
Number of OPTICALLY active =2(Fnantiomers)
Number of geometrical isomers=0
3+2+0=5
Discussion