Explore topic-wise InterviewSolutions in Current Affairs.

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1.	The reaction of white phosphorus with aqueous `NaOH` gives phosphine along with another phosphorus containing compound. The reacation type, the oxidation states of phosphorus in phosphine and the other product are respectvely:A. redox reaction, -3 and -5B. redox reaction, +3 and +5C. disproportion reaction, -4 and +5D. disproportion reaction, -3 and +5
Answer» Correct Answer - D The disproportion reaction is `P_(4)+NaOH to overset(-3)(PH_(3))+Na_(2)HPoverset(+3)(O_(3))`

2.	Explain why `HNO_(3)` acts only as oxidising agent while `HNO_(2)` can act both as a reducing agent and an oxiding agent?
Answer» Nitrogen can have oxidation number from -3 to +5. The oxidation number of nitrogen in `HNO_(3)` is +5. Thus, increase in oxidation number beyond +5 cannot occur. Hence, `HNO_(3)` cannot act as reducing agent. The oxidation number of nitrogen in `HNO_(3)` can only decreasel thus it acts as an oxidising agent. In `HNO_(3)` the oxidation number of nitrogen is +3. Thus, it can increase or decrease within the range -3 to +5. Hence it can act as an oxidising as well as reducing agent.

3.	Which ordering of Compounds is according to the decreasing order of the oxidation state of nitrogen?A. `HNO_(3), NO,NH_(4)Cl,N_(2)`B. `HNO_(3), NO,N_(2),NH_(4)Cl`C. `HNO_(3),NH_(4)Cl,NO,N_(2)`D. `NO,HNO_(3),NH_(4)Cl,N_(2)`
Answer» Correct Answer - B `[Hoverset(+5)(NO_(3)), Noverset(+2)(O), overset(0)(N_(2)),overset(-3)(NH_(3)Cl)]`

4.	The oxidation number of nitrogen atom in `NH_(4)NO_(3)` are:A. `+3,+3`B. `+3,-3`C. `-3,+5`D. `-5,+3`
Answer» Correct Answer - C `NH_(4)NO_(3) Leftrightarrow NH_(4)^(+)+NO_(3)^(-)` `NH_(4)^(+)x+4=+1` x=-3 `NH_(3)^(-)x+6=-1` x=+5

5.	Balance the following equation by oxidation number method `NaIO_(3)+NaHSO_(3)to Na_(2)SO_(4)+NaHSO_(4)+I_(2)+H_(2)O`
Answer» Writing oxidation number of al the atoms. `overset(+1 +5 -2)(Na IO_(3))+overset(+1 +1 +4 -2)(NaHSO_(3))to overset(+1)(Na_(2))overset(+6 -2)(SO_(4))+overset(+1)(Na)overset(+1 +6 -2)(HSO_(4))+overset(0)(I_(2))+overset(+1 -2)(H_(2)O)` The oxidation no. of I has increased while that of S has increased `Naoverset(+5)(IO_(3)) to overset(0)(I)....(i)` `NaHoverset(+4)(SO_(3)) to NaHoverset(+6)(SO_(4)).....(ii)` Decrease in Ox. no. of I=5 units per molecules `NaIO_(3)` Increase in Ox. no. of S=2 unit per molecule `NaHSO_(3))` Eq. (i) is multiplied by 2 and eq(ii) is multiplied by 5 as to make decrease and increase equal. `2NaIO+5NaHSO_(3)toI_(2)+3NaHSO_(4)+2NaSO_(4)` To balance hdyrogen and oxygen, one `H_(2)O` molecule should be added on RHS. Hence, the balanced equation is `2NaIO_(3)+5NaHSO_(3)to I_(2)+3NaHSO_(4)+2Na_(2)SO_(4)+H_(2)O`

6.	The oxidation state of iodine in `IPO_(4)` isA. `+1`B. `+3`C. `+5`D. `+7`
Answer» Correct Answer - B Let oxidation state of iodine be x x-3=0, x=+3, `therefore PO_(4)^(3-)` has combined oxidation number -3.

7.	Arrange the following in the order of (a) Increasing oxidation number of iodine. `I_(2), HI, HIO_(4),ICI` (b) Increasing oxidation number of cholorine `Cl_(2)O_(7),Cl_(2)O,HCl,CIF_(3),Cl_(2)` (c ) increasing oxidation number of nitrogen. `NH_(3),N_(3)H,N_(2)O,NO,N_(2)O_(5)`
Answer» (a) `HI(-I),I_(2)(0),ICI(+1),HIO_(4)(+7)` (b). `HCl(-1),Cl_(2)(0),Cl_(2)(0),Cl_(2)(+1),CIF_(3)(+3),Cl_(2)O_(7)(+7)` (c ).`NH_(3)(-3),N_(3)H(-1//3),N_(2)O(+1),NO(+2),N_(2)O_(2)(+5)`

8.	In alkaline medium `CIO_(2)` oxidises to `H_(2)O_(2)` and `O_(2)` and itself gets reduced to `Cl^(-)`. How many moles of `H_(2)O_(2)` are oxidised by 1 mole of `ClO_(2)`? Hint: The balanced chemical equation is:A. 1B. 1.5C. 2.5D. 3.5
Answer» Correct Answer - C The balanced chemical equation is: `2ClO_(2)+5H_(2)O_(2)+2OH^(-) to 2Cl^(-)+5O_(2)+6H_(2)O` `2"mol"ClO_(2)-=5"mol"H_(2)O_(2)` `1"mol"ClO_(2)-=2.5"mol"H_(2)O_(2)`

9.	Calculate the oxidation state of underlined (a) `Ba_(2)XeO_(2)` (b). `BaCl_(2)` (c ).`C_(12)H_(22)O_(11)` (d) `IF_(7)` (e ) `Na[Fe(CN)_(5)NO]` (f) `RuO_(4)` (g) `K_(2)TaF_(7)` (h) `Na_(2)MO_(4)` `U_(2)O_(7)^(4-)` (j) C is diamond.
Answer» `(a).0, (b).+2,(c ).0, (d).+7,(e ).3 (f).+8,(g)5,(h).+2,(i)+5,(j)0`

10.	Write the following ionic equation in the molecular form if the reactants are chlorides. `2Fe^(3+)+Sn^(2+) to 2Fe^(2+)+Sn^(4+)`
Answer» For writing the reactants in moleuclar forms, the requisite number of chloride ions are added. `ubrace(2Fe^(3+)+6Cl^(-))+ubrace(Sn^(2+)+2Cl^(-))` Similarly `8Cl^(-)` ions are added on RHS ot neutralise the charges. `underset(2FeCl_(2))ubrace(2Fe^(2+)+4Cl^(-))+underset(SnCl_(4))ubrace(Sn^(4+)+4Cl^(-))` Thus, the balanced molecular equation is `2FeCl_(3)+SnCl_(2)=2FeCl_(2)+SnCl_(4)`

11.	In alkaline medium, `H_(2)O_(2)` reacts with `Fe^(3+) and Mn^(2+)` separately to give:A. `Fe^(4+) and Mn^(4+)`B. `Fe^(2+) and Mn^(2+)`C. `Fe^(2+) and Mn^(4+)`D. `Fe^(4+) and Mn^(2+)`
Answer» Correct Answer - C `2K_(3)[overset(3+)(Fe)(CN)_(6)]+2KOH+2H_(2)O_(2)to2K_(4) [overset(2+)(Fe)(CN)_(6)]+2H_(2)O+O_(2)` `overset(2+)(MnSO_(4))+H_(2)O_(2)tooverset(4+)(MnO_(2))+H_(2)SO_(4)`

12.	The pair in which phosphours atoms have a formed oxidation state of `+3` isA. Orthophosphorous and phyrophosphorous acidsB. Pyrophosphorous and hypophosphoric acidsC. Orthophosphorous and hypophosphoric acidsD. Phyrophosphorous and pyrophosphoric acids
Answer» Correct Answer - A Orthophosphorous acid=`H_(3) overset(+3)(PO_(3))` Pyrophosphorous acid=`H_(2)overset(+3)(P_(2)O_(5))`

13.	Statement-1: Spectator ions are the species that are presetnt in the solution but do not take part in the reaction. Because Statement-2: The phenomena of formation of `H_(2)O_(2)` by the oxidation of `H_(2)O` is known as auto-oxidation. Hint: `Zn+2H^(+)+2Cl^(-) to Zn^(2+)+2Cl^(-)+2Cl^(-)+H_(2)`. Here `Cl^(-)` ion is spectator ion.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-1D. Statement-1 is false, statement-2 is false
Answer» Correct Answer - B `Zn+2H^(+)+2Cl^(-) to Zn^(2+)+2Cl^(-)+H_(2)`. Here `Cl^(-)` ion is spectator ion.

14.	Freshly prepared, bright blue coloured solution of sodium in liquid ammonia can be used to reduce the organic functional moieties. In this, the actual reducing species isA. `[Na(NH_(3))_(n)]^(+)`B. `[H_(2)(NH_(3))_(n)]`C. `[NaNH_(2)(NH_(3))_(n)]`D. `[e(NH_(3))_(n)]^(+)`
Answer» Correct Answer - D Solved electron acts as reducing agent `Na+(m+n)NH_(3) to [Na(NH_(3))_(n))]^(+)+[e(NH_(3))_(n)]^(-)"solved electron"`

15.	Number of moles of `MnO_(4)^(-)` required to oxidise one mole of ferrous oxalate completely in acidic medium will beA. 7.5 molesB. 0.2 molesC. 0.6 molesD. 0.4 moles
Answer» Correct Answer - D `2MnO_(4)^(-)+16H^(+)+5C_(2)O_(4)^(2-) to 2Mn^(2+)+10CO_(2)` Number of moles of `MnO_(4)^(-)` required to oxidise `=(2)/(5)=0.4`

16.	In the redox reaction, `x KMnO_(4) + NH_(3) rarr y KNO_(3) + MnO_(2) + MnO_(2) + KOH + H_(2)O`, x and y areA. X=4,Y=6B. X=3,Y=8C. X=8,Y=6D. X=8,Y=3
Answer» Correct Answer - D Balanced equation is `8KMnO_(4)+3NH_(3) to8MnO_(4)+3KNO_(3)+5KOH+2H_(2)O`

17.	(a). Which compound among the following has the lowest oxidation number of Mn? `KMnO_(4), K_(2)MnO_(4), MnO_(2) and Mn_(2)O_(3)` (b). Which compound among the following has the higest oxidation number of P? `PH_(3),H_(3)PO_(2),PCl_(3) and H_(3)PO_(4)` (c ). Which compound among the following has the highest oxidation number of chlorine ? `HClO_(4),HOCl, CIF_(3),HCIO_(3)and HCl`
Answer» (a).`Mn_(2)O_(3) (b) H_(3)PO_(4) (c )CH_(2)Cl_(2) (d) HCl`

18.	The oxidation number of `Cr` in `K_(2)Cr_(2)O_(7)` is
Answer» Let the Ox.no. of Cr in `K_(2)Cr_(2)O_(7)` be x. We know that, Ox. No of K=+1 Ox. No.of O=-2 `2("Ox.no.K")+2("Ox.no.Cr")+7("Ox.no.O")=0` `{:(,2(+1),+,2(x),+,7(-2),=0),(,+2,+,2x,-,14,=0):}` or 2x=+14-2=+12 or `x=+(12)/(2)=+6` Hence oxidation number of Cr in `K_(2)Cr_(2)O_(7)` is +6.

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