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.	Oxidation numbers of sulphur in H_(2)S_(3) molecule are
Answer» `-1, -1 and -1` `+1, +1 and +1` `-1, 0 and -1` `+1, 0 and -1` ANSWER :C

2.	Oxidation numbers of sodium, mercury in sodium amalgam are
Answer» zero, zero ` +1.- 1 ` `-2, +2` ` 0, +1` SOLUTION :NA - Hg it is an alloy. Hg is a MAJOR compoud

3.	Oxidation number of the metal ion in the compound [CO(NH_(3))_(5)Cl]Cl_(2) is +3. Calculate the oxidation number of the complex ion.
Answer» SOLUTION :Chlorine present outside the complex coordination sphere are chloride ions. Since there are two `CL^(-)` ions outside, the OXIDATION number of complex ion is +2. The ion is givne as `[CO(NH)_(3))_(5) Cl]^(2+)`

4.	Oxidation number of sulphur in oleum (H_(2)S_(2)O_(7)) is
Answer» `+4` `+2` `-2` `+6` ANSWER :D

5.	Oxidation number of .S. in (CH_(3))_(2) SO is:
Answer» Solution : Alkyl GROUPS have (+1) oxidation state . `:. (CH_3)_(2)SOimplies +2+x-2=0impliesx=0` Oxidation state of SULPHUR is ZERO

6.	Oxidation number of P of H_(3)PO_(4) is similar with P of which compound ?
Answer» `PH_(3)` `P_(2)O_(3)` `P_(2)O_(7)^(-4)` All of these Solution :`H_(2)PO_(4)=3(+1)+x+4(-2)=0` `therefore3+x-8=0` `thereforex=+5` `P_(2)O_(7)^(-4)=2x+7(-2)=-4` `therefore2x=10` `thereforex=+5`

7.	Oxidation number of S in S_2Cl_2 , is
Answer» `+1` `+6` 0 `-1` SOLUTION :`UL(S_2)Cl_2,2x-2=0impliesx=+1`

8.	Oxidation number of P in P_(2)O_4^(4-)is
Answer» `+3` `+4` `+5` `+6` SOLUTION :`O^(-)-OVERSET(O)overset(\|\|)underset(O^(-))underset(\|)P-O-overset(O^(-))overset(\|\|)underset(O)underset(\|\|)P-O^(-),P=+5`

9.	Oxidation number of oxygen in potassium ozonide is _____
Answer» SOLUTION :`(-1)/3`

10.	Oxidation number of oxygen in super oxide is...
Answer» `-(1)/(2)` `+(1)/(2)` `-1` `+1` ANSWER :A

11.	Oxidation number of oxygen in KO_3, KO_2 and K_2Ois
Answer» `-1/3 , -1/2 , -1` `-1/3 , -1/2 , -1/2` `-1,-1,-1` `-2,-2,-2` SOLUTION :`KO_(3)^(1/3)""KO_(2)^(1/2)""K_2overset(-1)O_(2)`

12.	Oxidation number of oxygen in KO_(2) i
Answer» `-1` `-2` `-1/2` `+1/2` SOLUTION :`KO_(2)1+2x=0,x=1/2`

13.	Oxidation number of nitrogen in NaNO_(2) is
Answer» `+2` `+3` `+4` `-3` ANSWER :B

14.	Oxidation number of Li in Li_(2)O ?
Answer» Solution :OXYGEN have -2 OXIDATION state in ALKALI metal OXIDE elements so Li having +1 oxidation state.

15.	Oxidation number of metal in the complex [Co(NH_(3))_(4)Cl_(2)]^(+) is
Answer» `+1` `+2` `+3` `+6` Answer :C

16.	Oxidation number of iron in Na_(2)[Fe(CN)_(5)NO]
Answer» `+2` `+3` `+1` 0 Solution :`Na_(2)[FE(CN)_(5)OVERSET(+1)(NO)]`

17.	Oxidation number of iodine in IO_(3)^(-), IO_(4)^(-), KI and I_(2) respectively are:
Answer» `-1, -1, 0, +1` `+3, +5, +7, 0` `+5, +7, -1, 0` `-1, -5, -1,0` Answer :C

18.	Oxidation number of iodine in IO_(3)^(-),IO_(4)^(-),KIandI_(2) respectively are ____
Answer» `-1, -1, 0, +1` `+3,+5,+7,0` `+5,+7,-1,0` `-1,-5,-1,0` Solution :

19.	The oxidation number of fluorine in all its compounds is equal to
Answer» ANSWER :A

20.	Calculate the oxidation number of iron in Fe_(3)O_(4)?
Answer» `1//2` `2//6` `8//3` `3//2` ANSWER :C

21.	Oxidation number of 'Co' in Hg[Co(SCN)_(4)]
Answer» `+2` `+1` `+3` `+5` ANSWER :A

22.	Oxidation number of CI in NOCIO_4 is
Answer» `+7` `-7` `+5` `-5` Solution :`NOClO_(4)impliesNO^(+)ClO_(4)^(-),X-8=-1, x =+7`

23.	Oxidation number of Cl in CaOCl_(2) (bleaching powder) is
Answer» zero, since it CONTAINS `Cl_(2)` `-1` since it contains `CL^(-)` `+1`, since it contains `CLO^(-)` `+1` and -1 since it contains `ClO^(-)` and `Cl^(-)` Answer :D

24.	Oxidation number of chlorine in NOClO_4is:
Answer» Solution : In `NOClO_4`there is `NO^(+) and CIO_4^(-)`ions .Let OXIDATION STATE of chlorine is x. `:. x- 8 =-1 impliesx =+7`

25.	Oxidation number of chlorine in HClO_(4) is "____________".
Answer» `+1` `-1` `-7` `+7` ANSWER :D

26.	Oxidation number of Chlorine in chlorine heptaoxide is
Answer» `+1` `+4` `+6` `+7` Solution :`Cl_(2)O_(7), 2x -14 =0 , X =+7`

27.	Oxidation number of carbon is zero in the compound
Answer» methyl chloride chloroform glucose CARBON TETRACHLORIDE Solution :`C_6H_(12)O_6` `IMPLIES` glucose `6 X +12 +6(-2) =0, x = 0`

28.	Oxidation number of carbon in carbon suboxide (C_(3)O_(2)) is :
Answer» `+ 2//3` `+4 //3 ` `+ 4 ` ` - 4//3` ANSWER :B

29.	Oxidation number of carbon in diamond is "_____________".
Answer» `-4` `+4` `0` `+2` ANSWER :C

30.	Oxidation number of .C. in methyl chloride is -3.
Answer» Solution :FALSE `CH_(3)Cl=1(C)+3(H)+1(-1)0` `thereforeC+3-1=0` `thereforeC+2=0` `thereforeC=-2`

31.	Oxidation number of carbon in C_3O_(2) , Mg_(2)O_3 are respectively :
Answer» `-(4)/3 ,+(4)/3` `+(4)/3 ,-(4)/3` `-(2)/3 ,+(2)/3` `-(2)/3 ,+(4)/3` Solution :`C_(3)O_(2):3x -4 =0 IMPLIES x= (+4)/3` `Mg_(2)C_(2) : +4 +3x = 0 implies x = (-4)/3`

32.	Oxidation number of C in CH_3OH, CH_2O,HCOOH and C_2H_2 is respectively :
Answer» `-2,0,+2,-1` `+2,0,+2,-2` `-2,0,+2,0` `-2,-4,+2,-2` ANSWER :A

33.	Oxidation number of C in CH_(2)Ois
Answer» `-2` `+2` 0 4 Answer :C

34.	Oxidation number of C in CH_(2)O is
Answer» `-2` `+2` 0 4 Answer :C

35.	Oxidation number is the charge which an atom of an element has in its ion or appears to have when present in the combined state. It is also called oxidation state. Oxidation number of any atom in the elementary state is zero. Oxidation number of a monoatomic ion is equal to the charge on it. In compounds of metals with nonmetals, metals have positive oxidation number while non metals have negative oxidation numbers. In compounds of two difference elements, the more electronegative element has negative oxidation number whereas the other has positive oxidation number. In complex ions, the sum of the oxidation number of all the atoms is equal to the charge on the ion. If a compound contains two or more atoms of the same element, they may have same or different oxidation states according as their chemical bonding is same or different. A compound of Xe and F is found to have 53.3% Xe (atomic weight =133). Oxidation number of Xe in this compound is
Answer» `+2` 0 `+4` `+6` Solution :`Xe : F =(53.3)/(133): (46.7)/19 =0 . 398 : 2 . 45 = 1 : 6` Thus E.F. of XENON FLUORIDE is `XeF_6` `:.`O.N of Xe in `XeF_(6)` is +6

36.	Oxidation number is the charge which an atom of an element has in its ion or appears to have when present in the combined state. It is also called oxidation state. Oxidation number of any atom in the elementary state is zero. Oxidation number of a monoatomic ion is equal to the charge on it. In compounds of metals with nonmetals, metals have positive oxidation number while non metals have negative oxidation numbers. In compounds of two difference elements, the more electronegative element has negative oxidation number whereas the other has positive oxidation number. In complex ions, the sum of the oxidation number of all the atoms is equal to the charge on the ion. If a compound contains two or more atoms of the same element, they may have same or different oxidation states according as their chemical bonding is same or different. Oxidation number of sodium in sodium amalgam is
Answer» `+1` 0 `-1` `+2` Solution : O.N of metals in amalgams is ZERO.

37.	Oxidation number is the charge which an atom of an element has in its ion or appears to have when present in the combined state. It is also called oxidation state. Oxidation number of any atom in the elementary state is zero. Oxidation number of a monoatomic ion is equal to the charge on it. In compounds of metals with nonmetals, metals have positive oxidation number while non metals have negative oxidation numbers. In compounds of two difference elements, the more electronegative element has negative oxidation number whereas the other has positive oxidation number. In complex ions, the sum of the oxidation number of all the atoms is equal to the charge on the ion. If a compound contains two or more atoms of the same element, they may have same or different oxidation states according as their chemical bonding is same or different. The oxidation state of the most electronegative element in the products of the reaction between BaO_(2) and H_(2)SO_(4) are
Answer» 0 and- 1 `-1 and -2` `-2 and 0` `-2 and -1` SOLUTION :`BaO_(2)+H_(2)SO_(4)rarrBaSO_(4) +H_(2)O_(2)` . The most ELECTRONEGATIVE atoms is O. It appears in `SO_(4)^(2-)`where its O.N, is –2 and in `H_2O_2`where it is –1 . Therefore , option (b) is correct .

38.	Oxidation number and valency of oxygen in OF_(2) are
Answer» `+1,2` `+2,2` `+1,1` `+2,1` ANSWER :B

39.	Oxidation number and Covalency of sulphur in S_(8) molecule are respectively
Answer» 6 and 8 0 and 8 0 and 2 6 and 2 Answer :C

40.	Oxidation is the process which involves ......... of electrons and is accompanied by an ......... in the oxidation number of the substance.
Answer» SOLUTION :LOSS, INCREASE

41.	Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overall chemical change is called redox reaction. Redox reactions are of three types: (i) Intermolecular redox reactions, (ii) Autoredox or disproportionation reaction, and (iii) Intramolecular redox reactions.Which of the following shows auto-redox change: (I) 2HCHỌ+ NaOH rarr HCOONa +H_2O (II) Cl_2 + H_2O rarr HCI+HCIO (III) 2Cu^(+) rarr Cu^(2+) + Cu (IV) Cr+2H_(2)O+3OCl^(-) rarr Cr^(3+) + 3Cl^(-) + 6OH^(-)
Answer» I, II, III I, IV II, IV II, III Solution :It INVOLVES oxidation of ONE MOLECULE on the cost of other.

42.	Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overall chemical change is called redox reaction. Redox reactions are of three types: (i) Intermolecular redox reactions, (ii) Autoredox or disproportionation reaction, and (iii) Intramolecular redox reactions.Which of the following show intramolecular redox change : (I) 2KClO_3 rarr 2KCI + 3O_2 (II) (NH_4)_(2) Cr_(2)O_(7) rarrN_(2) + Cr_(2)O_(3) + 4H_(2)O (III) NH_(4) NO_(2) rarr N_2 + 2H_(2)O (IV) 2Cu^(+) rarrCu^(2+)Cu
Answer» I, II, III I, IV II, III II, IV Solution :It INVOLVES REDOX change of two SPECIES in a MOLECULE .

43.	Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overall chemical change is called redox reaction. Redox reactions are of three types: (i) Intermolecular redox reactions, (ii) Autoredox or disproportionation reaction, and (iii) Intramolecular redox reactions.Select the species which can act as oxidant and reductant both : (1) H_2SO_3 (II) H_2 O_2 (III) O_3 (IV) HNO_3 (V) CI_2
Answer» I,II,III, IV I,II,III, V II,IH,IV, V III,IV,V Solution : `O_3` REDUCES PEROXIDES to OXIDES `BaO_(2)+O_(3) rarrBaO+2O_2`

44.	Oxidant is a substance in which the oxidation number of one of the atoms ……………….and reductant is a substacne in which oxidation number of one of atoms ……………..
Answer» ANSWER :DECREASE , INCREASES

45.	Comment on the statement : oxidation and reduction are complementary processes.
Answer» ANSWER :T

46.	Oxalic acid dihydrate, H_(2)C_(2)O_(4). 2H_(2)O(s) is often used as a primary reagent to standardise sodium hydroxide solution. Which of these facts are reasons to choose this substance as a primary standard ? I. It is diprotic. II. It is a stable compound that can be weighed directly in air. III. It is available in pure form.
Answer» III only I and II only II and III only I, II and III Solution :N//A

47.	overset(HO^(-))underset("(conjugate-addition")to(A) ,product (A) is :
Answer» SOLUTION :Michael reaction + Intra MOLECULAR ALDOL

48.	overset(A)(-) is form by hemolytic fission but overset(B)- is form by heterolytic fission of C-Cl bond of CH_(3)-Cl
Answer» Solution :`A= .CH_(3)+ .CL` `B= overset(+)(C )H_(3) + Cl^(-)`

49.	Outer shell 'octet' configuration is observed for the elements of the group
Answer» 2 8 18 32 Answer :C

50.	Out off N_(2)O,SO_(2),I_(3)^(+),I_(3)^(-),H_(2)O,NO_(2)^(-),N_(3)^(-), the linear species are:
Answer» `NO_(2)^(-),I_(3)^(+),H_(2)O` `N_(2)O,I_(3)^(+),N_(3)^(-)` `N_(2)O,I_(3)^(-),N_(3)^(-)` `N_(3)^(-),I_(3)^(-),NO_(2)^(-)` ANSWER :C