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.

The correct order of solubility is

Answer»

`CaCO_(3)ltKHCO_(3)ltNaHCO_(3)`
`KHCO_(3)ltCaCO_(3)ltNaHCO_(3)`
`NaHCO_(3)ltCaCO_(3)ltKHCO_(3)`
`CaCO_(3)ltNaHCO_(3)ltKHCO_(3)`

Solution :Biacarbonates are more soluble than carbonates and bicarbonates of IIA group ELEMENTS are more soluble than IA group.
So the order of SOLUBLITY `CaCO_(3) lt NaHCO_(3) lt KHCO_(3)`
Discussion
2.

The correct order of S_N2//E2 ratio of the % yield of product of the following halide is : (P)CH_3-CH_2-undersetunderset(I)(|)oversetoverset(Ph)(|)C-CH_3 (Q)CH_3-undersetunderset(Ph)(|)CH-undersetunderset(I)(|)CH-CH_3 (R)CH_3-CH_2-I (S)CH_3-CH-undersetunderset(I)(|)CH-CH_3

Answer»

R gt S gt Q gt P
R gt Q gt S gt P
P gt R gt S gt Q
Q gt P gt R gt S

ANSWER :A
Discussion
3.

The decreasing order of second ionisation potential of K, Ca, Ba is

Answer»

`K gt CA gt Ba`
`Ba gt Ca gt K`
`K gt Ba gtCa`
`K = Ba = Ca`

SOLUTION :As the size of atom INCREASES ionisation potential decreases so the order is `K gt Ca gt Ba`
Discussion
4.

The correct order of reducing character of alkali metals is

Answer»

`Rb LT K lt Na lt Li`
`Li lt Na lt K lt Rb`
`Na lt K lt Rb lt Li `
`Rb lt Na lt K lt Li`

Solution :Li has the most negative `E^(@)` values of other alkali METALS become more and more negative as we , MOVE down the group from Na to Rb. THUS , option (c) is correct .
Discussion
5.

The correct order of reactivity towards the electophilic substitution of the compounds aniline (I), benzene (II) and nitrobenzene (III)is

Answer»

`III GT II gt I`
`II gt III gt I`
`I LT II gt III`
`I gt I gt III`

ANSWER :D
Discussion
6.

The correct order of reactivity towards electrophilic substitution is

Answer»

benzenegtphenolgtbenzoic acidgtchlorobenzen
phenolgtbenzenegtchlorobenzenegtbenzoic acid
chlorobenzenegtbenzoic acidgtphenolgtbenzene
benzoic acidgtchlorobenzenegtbenzenegtphenol.

Solution :In GENERAL, ELECTRON releasing groups actiate and electron withdrawing groups deactivate the BENZENE ring TOWARDS electrophilic substitution. Hence, the correct order is
Discussion
7.

The correct order of reactivity of functional group is…….

Answer»

`= CO gt - OH gt - C -= N gt - OR`
`-COOH gt - COOR gt - CONH_(2) gt CHO`
`-COOH gt - NO_(2) gt C= O gt - NH`
`-C -= C - gt - COOH - X gt - COX`

Answer :B
Discussion
8.

The correct order of reactivity of alkene towards an electrophile is mentioned in :

Answer»

`CH_2=CH-Cl GT CH_2 = CH -OCH_3`
`CH_2=CHCl LT CH_2 =C Cl_2`
`CH_2=CH_2 gt CH_3 -CH=CH_2`
`CH_2=CH-OCH_3 gt CH_2=CH-CH_2-OH`

ANSWER :D
Discussion
9.

the correct order of rate of S_(N^(1)) is:

Answer»




SOLUTION :NA
Discussion
10.

The correct order of radius of given ions is :

Answer»

`O^(2) gt F^(-) , S^(2-) , CL^(-)`
`S^(2-) gt O^(2-) F^(-) gt Cl^(-)`
`S^(2-) gt Cl^(-) gt O^(2-) gt F^(-)`
`S^(2-) gt Cl^(-) gt O^(2-) gt F^(-)`

SOLUTION :
` :. ` Ionic radii of : `(S^(2-) " and " Cl^(-)) gt (O^(2-) , F^-))`…..(i)
As negative CHARGE is more , the ion has LARGER size.
` :. ` Ionic radius : `S^(2-) gt Cl^(-) " and " O^(2-) gt F^(-)` ….(ii)
From (i) and (ii) results, it is easy to write the order of ionic radius as : `S^(2-) gt Cl^(-) gt O^(2-) gt F^(-)`
Discussion
11.

The correct order of radii is.....

Answer»

`N lt Be lt B `
`F^(-) lt O^(2-) lt N^(3-)`
`NA lt Li lt K`
`Fe^(3+) lt Fe^(2+) lt Fe^(4+)`

Answer :b
Discussion
12.

The correct order of radii is

Answer»

`Al^(3+) gt Mg^(2+) gt NA^(+) gt F^(-) gt O^(2-)` 
`Na^(+) gt Mg^(2+) gt Al^(3+) gt O^(2-) gt F^(-)`
`Na^(+) gt F^(-) gt Mg^(2+) gt O^(2-) gt Al^(3+)`
`O^(2-) gt F^(-) gt Na^(+) gt Mg^(2+) gt Al^(3+)`

ANSWER :D
Discussion
13.

The correct order of priority of the functional groups of organic compounds in the IUPAC system of nomenclature is……

Answer»

`-COOH gt -CHO gt -CN gt -OH gt -OR`
`-COOH gt -CN gt -CHO gt - OR gt -OH`
`-CHO gt -COOH gt -OH gt -CN gt -OR`
`-COOH gt -CHO gt -OH gt -OR gt -CN`

Answer :B
Discussion
14.

The correct order of polarising ability of cations of Alkaline earth metals is

Answer»

`CL^(-) gt BR^(-) gt I^(-) gt F^(-)`
`F^(-) gt I^(-) gt Br^(-) gt Cl^(-)` 
`I^(-) gt Br^(-) gt CF^(-) gt F^(-)`
`F^(-) gt CF^(-)gt Br^(-) gt I^(-1)`

Answer :C
Discussion
15.

The correct order of O-O bond length in hyrdrogen peroxide, ozone and oxygen is

Answer»

`H_(2)O_(2)gt O_(3)gt O_(2)`
`O_(2)gt O_(3)gt H_(2)O_(2)`
`O_(2)gt H_(2)O_(2)gt O_(3)`
`O_(3)gt O_(2)gt H_(2)O_(2)`

SOLUTION :The BOND ORDER for `O_(2),O_(3) and H_(2)O_(2)` decreases in the order: ` 2 gt 1.5 gt 1`
Discussion
16.

The correct order of N compounds in its decresing order of oxidaton states is

Answer»

`HNO_(3),NO,NH_(4)CI,N_(2)`
`HNO_(3),NO,N_(2),NH_(4)CI`
`HNO_(3_, NH_(4)CI,O,N_(2)`
`NO,HNO_(3),NH_(4)cI,N_(2)`

SOLUTION :`Hoverset(+5)NO_(3),overset(+3)NO,overset(0)N_(2),NH_(4)CI`
Discussion
17.

The correct order of N -O bond length in NO, NO_(2)^(-), NO_(3)^(-) and N_2O_4will be

Answer»

`NO GT N_2 O_4 gt NO_2gt NO_(3)^(-)`
`NO gt NO_(3)^(-)gt N_2 O_4gt NO_(2)^(-)`
`NO_(3) ^(-)gt NO_(2) ^(-)gtN_2 O_(4)gt NO`
`N_2 O_4gt NO_(2)^(-)gt NO_(3)^(-)gt NO`

ANSWER :C
Discussion
18.

The correct order of N-compounds in its decreasing order of oxidation states is

Answer»

`HNO_(3),NO,N_(2),NH_(4)Cl`
`HNO_(3),NO,NH_(4)Cl_(4),N_(2)`
`HNO_(3),NH_(4)Cl,NO,N_(2)`
`NH_(4)Cl,N_(2),NO,HNO_(3)`

ANSWER :A
Discussion
19.

The correct order of mobility of the alkali metal ions in aqueous solution is

Answer»

`Rb^(+)GTK^(+)gtNa^(+)gtLi^(+)`
`Li^(+)gtNa^(+)gtK^(+)gtRb^(+)`
`Na^(+)gtK^(+)gtRb^(+)gtLi^(+)`
`K^(+)gtRb^(+)gtNa^(+)gtLi^(+)`.

Solution :On the basis of explanation of ionic mobility extent of an ion GIVEN in Q. 24, the mobility of alkali metal ions in aqueous solution follows the order. `Rb^(+) GT K^(+) gt Na^(+) gt Li^(+)`.
Discussion
20.

The correct order of melting points of IIIA group elements is

Answer»

` B GT AL gt Tl gt In gt Ga`
` B gt Al gt GA gt In gt Tl`
` B gt Al gt Tl gt Ga gt In`
` B gt Al gt In gt Tl gt Ga`

Solution :ORDER of melting POINTS : `B gt Al gt Tl gt In gt Ga`
Discussion
21.

The correct order of melting points is

Answer»

ETHANE `LT` PROPANE `lt` BUTANE
butane `lt` propane `lt` ethane
propane `lt` ethane `lt` butane
ethane `lt` butane `lt` propane

Answer :C
Discussion
22.

The correct order of mathching of the following compounds is:

Answer»

A-iii,B-I,C -II,D-v,E-iv
A-I,B-v,C-iii,D-iv,E-ii
A-v,B-ii,C -I’d-iii,E-iv
A-iv,B-I,C-v,D-iii,E-ii

Solution :is the CORRECT ORDER of MATCHING
Discussion
23.

The correct order of magnitude of bond angles among the compounds CH_(4),NH_(3) and H_(2)O is

Answer»

`H_2 O LT NH_3 ltCH_4`
`CH_4 lt H_2 OltNH_3`
`NH_3 lt CH_4 ltH_2 O`
`NH_3 lt H_2 O lt CH_4`

ANSWER :A
Discussion
24.

The correct order of liquefaction of the gases NH_(3),CO_(2),SO_(2) and HCl is

Answer»

`NH_(3) GT CO_(2) gt HCl gt SO_(2)`
`CO_(2) gt NH_(3) gt SO_(2) gt HCl_(2)`
`HCl gt CO_(2) gt NH_(3) gt SO_(2)`
`SO_(2) gt NH_(3) gt HCl gt CO_(2)`

Solution :Higher the critical TEMPERATURE of a gas, more easily it is liquefied. HENCE, the CORRECT order of liquefaction is
`{:(,SO_(2),gt,NH_(3),gt,HCl,gt,CO_(2)),("Critical TEMP".(K),430,,405,,325,,304):}`
Discussion
25.

Thecorrectorder of IstionizationenthalpyamongfollowingelementsBe, B,C, N, O is

Answer»

`B lt Be lt C lt O N`
` B lt Be lt C lt N lt O`
`Be ltB LTC lt N O`
`Be lt B lt C lt O lt N`

SOLUTION :`Delta_(i) H ` of Be `gt ` B andthat N `gt `O
Discussion
26.

The correct order of IP_2 , values among F, Na, Mg and Al is

Answer»

`MG gt Al gt F gt NA`
`Mg lt Al lt F lt Na`
`Al lt Mg lt F lt Na`
`M gt Al gt F lt Na`

ANSWER :B
Discussion
27.

The correct order of ionisation energy of C, N, O and F is

Answer»

`F LT N lt C lt O`
`C lt N lt O lt F`
`C lt O lt N lt F`
`F lt O lt N lt C`

ANSWER :C
Discussion
28.

The correct order of increasing strength of the compounds (a) CH_(3)CO_(2)H (b) MeOCH_(2)CO_(2)H (c) CF_(3)CO_(2)H (d)

Answer»

`b lt d lt a lt c`
`d lt a lt c lt b`
`d lt a lt b lt c`
`a lt d lt c lt b`

Solution :Meo ACTS as an ELECTRON withdrawing group on a saturated carbon and HENCE ACID (B) is stronger than acid (A and D)
Discussion
29.

The correct order of increasing thermal stability of K_(2)CO_(3) , MgCO_(3) , CaCO_(3)and BeCO_(3) is

Answer»

`K_(2)CO_(3) lt MgCO_(3) lt CaCO_(3) lt BeCO_(3)`
`BeCO_(3) lt MgCO_(3) lt K_(2)CO_(3) lt CaCO_(3)`
`BeCO_(3) lt MgCO_(3) lt CaCO_(3) lt K_(2)CO_(3)`
`MgCO_(3) lt BeCO_(3) lt CaCO_(3) lt K_(2)CO_(3)`

Solution :Stability increases as the basic character of the corresponding HYDROXIDE increases , i.e., option (C) is correct .
Discussion
30.

The correct order of increasing solubility of AgCl in (a) water (b) 0.1M NaCl (c) 0.1 BaCl_(2) (d) 0.1M NH_(3) is

Answer»

`d GT a gt B gt C`
`d gt c gt b gt a`
`b gt a gt d gt c`
`a gt d gt b gt c`

SOLUTION :N//A
Discussion
31.

The correct order of increasing rate of hydrolysis of the following compund is

Answer»

`I LT II lt III`
`III lt II lt I`
`I lt III lt II`
`III lt I lt II`

Solution :The rate determining step in the hydrolysis of an ester is bimolecular, the larger the R, greater will be the steric RETARDATION since HYBRIDISATION of the carbonyl carbon changes from `sp^(2)` to `sp^(3)` (due to increased bonding). The ORDER of increasing rate of hydrolysis is `III lt II lt I`
Discussion
32.

The correct order of increasing polarising power of the cations in the following AlCl_(3) , MgCl_(3) , NaCl is

Answer»

`AlCl_(3) lt MgCl_(3) lt NaCl`
`MgCl_(2) lt NaCl lt AlCl_(3)`
`NaCl lt MgCl_(2) lt AlCl_(3)`
`NaCl lt AlCl_(3) lt MgCl_(2)`

Solution :SINCE the polarising power INCREASES with increase in charge on the CATION ,therefore , polarising power increases in the order :
`overset(+1)(Na)Cl lt overset(+2)(Mg) Cl_(2) lt overset(3+)(Al) Cl_(3)`
Thus , option (c) is correct .
Discussion
33.

The correct order of increasing polarising power of the cations in AlCl_(3), MgCl_(2) and NaCl is

Answer»

`AlCl_(3) lt MgCl_(2) lt NACL`
`MgCl_(2) lt NaCl lt AlCl_(3)`
`NaCl lt MgCl_(2) lt AlCl_(3)`
`NaCl lt AlCl_(3) lt MgCl_(2)`

Solution :Polarising POWER of cations increases with the
increasing charge
`{:(""(+1),(+2),(+3) ),(NaCl lt ,MgCl_(2), ltAlCl_(3)):}`
Discussion
34.

The correct order of increasing ionic character is.....

Answer»

`BeCl_(2) LT MgCl_(2) lt CaCl_(2) lt BaCl_(2)`
`BeCl_(2) lt MgCl_(2) lt CaCl_(2)`
`BeCl_(2) lt BaCl_(2) lt MgCl_(2) lt CaCl_(2)`
`BaCl_(2) lt CaCl_(2) lt MgCl_(2) lt BeCl_(2)`

ANSWER :A
Discussion
35.

According to Fajans'rules, the percentage of covalent character in an ionic compound increase if the cation is highly charged or small in size and the anion is large or cation has pseudoinert gas configuration. As a result of the increasedcovalent character, solubility in less polar solvent increases and the melting point decreases. The correct order of increasing ionic character is

Answer»

`Be Cl _(2) lt Mg Cl_(2) lt CaCl_(2) lt BaCl_(2)`
`Be Cl_(2) lt MgCl_(2) lt BaCl_(2)lt CaCl_(2)`
`BeCl_(2) lt BaCl_(2) lt MgCl_(2) lt CaCl_(2)`
`BaCl_(2) lt CaCl_(2) lt MgCl_(2) lt BeCl_(2)`

Solution :Ionic character of clorides increases as the size of the action increase (Frajan's rules) from `Be ^(2+)` to `Ba^(2+).`
Discussion
36.

The correct order of increasing [H_(3)O^(+)] in the following aqueous solutions is

Answer»

`0.01 M H_(2) S lt 0.01 M H_(2) SO_(4) lt 0.01 M NACL lt 0.01 M NaNO_(2)`
`0.01 M NaCllt 0.01 M NaNO_(2) lt 0.01 M H_(2)S lt 0.01 M H_(2)SO_(4)`
`0.01 M NaNO_(2) lt 0.01 M NaCl lt 0.01 M H_(2)S lt 0.01 M H_(2)SO_(4)`
`0.01 M H_(2) Slt 0.01 M NaNO_(2) lt 0.01 MNaCl lt 0.01 M H_(2) SO_(4)`.

SOLUTION :`H_(2)S` = weak acid, `H_(2)SO_(4)` =strong acid , NaCl= neutral, `NaNO_(2)` = basic. HENCE `[H_(3)O^(+)]` will be in the order : `NaNO_(2)ltNaClltH_(2)SltH_(2)SO_(4)`.
Discussion
37.

The correct order of increasing [H_(3) O^(+)] in the aqueous solution is

Answer»

`0.01 M H_(2) S LT 0.01 M H_(2) SO_(4) lt 0.01 M NaCl lt 0.01 M NaNO_(2)`
`0.01 M NaCl lt 0.01 M NaNO_(2) lt 0.01 M H_(2) S lt 0.01 M H_(2) SO_(4)`
`0.01 M NaNO_(2) lt 0.01 M NaCl lt 0.01 M H_(2) S lt 0.01 M H_(2) SO_(4)`
`0.01 M H_(2) S lt 0.01 M NaNO_(2) lt 0.01 M NaCl lt 0.01 M H_(2) SO_(4)`

Answer :C
Discussion
38.

The correct order of increasing energy of atomic orbitals is

Answer»

5P lt 4f lt 6s lt 5d
5p lt 6s lt 4f lt 5d
4f lt 5p lt 5d lt 6s
5p lt 5d lt 4f lt 6s

Solution :According to Aufbau principle electron are filled in VARIOUS atomic ORBITAL in the increasing order of energy
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3D lt 3p lt 5s lt 4d lt 5p lt 6slt 4f lt 5d lt 6p lt 7s.
Discussion
39.

The correct order of increasing C-O bond length of CO, CO_(3)^(2-) , CO_(2) is

Answer»

`CO_(3)^(2-) lt CO_(2)lt CO`
`CO_(2) lt CO_(3)^(2-) lt CO`
`CO lt CO_(3)^(2-) lt CO_(2)`
`CO lt CO_(2) lt CO_(3)^(2-)`

SOLUTION :The structures of the GIVEN species are

Thus, `CO_(2)` CONTAINS bond , `CO_(2)` contains
DOUBLE bond while `CO_(3)^(2-)` contains between singleltbegt and double bond . Hence , C-Obond LENGTHS will
be in the order : `CO lt CO_(2) lt CO_(3)^(2-)`
Discussion
40.

The correct order of increasing bond length of C-H, C-O, C-C and C= C is………

Answer»

`C- H LT C-O lt C-C lt C= C`
`C- H lt C = C lt C- O lt C- C`
`C - C lt C= C lt C - O lt C - H`
`C - O lt C - H lt C- C lt C = C`

ANSWER :B
Discussion
41.

The correct order of increasing bond angles in the following triatomic species is :

Answer»

`NO_(2)^(+) LT NO_(2) lt NO_(2)^(-) `
`NO_(2)^(-) lt NO_(2) lt NO_(2)^(+)`
`NO_(2)^(-) lt NO_(2)^(+) lt NO_(2)`
`NO_(2)^(+) lt NO_(2)^(-) lt NO_(2)`

ANSWER :B
Discussion
42.

The correct order of increasing bond angles in the following species are :

Answer»

`Cl_2O lt ClO_2 lt ClO_2^-`
`ClO_2 lt Cl_2O lt ClO_2^-`
`Cl_2O^(-) lt ClO_2^(-) lt ClO_2`
`ClO_2^(-) lt Cl_2O lt ClO_2`

SOLUTION :The CORRECT order of increasing bond angle is `Cl_2O^(-) lt ClO_2^(-) lt ClO_2`

In `ClO_2O^-` there are 2 lone pairs of ELECTRONS present on the central chlorine atom. Therefore the bond angle in `ClO_2^-` is less than `118^@` which is the bond angle in `ClO_2` which has less NUMBER of electrons on chlorine.
Discussion
43.

The correct order of increasing boiling points of the following compounds is

Answer»

`I LT II lt III lt IV`
`II lt III lt IV lt I`
`II lt IV lt III lt I`
`II lt III lt I lt IV`

Solution :The order of INCREASING boiling point among the ACID derivatives is
`underset(underset(underset("boiling")("hydrogen"))("Absencce of"))(CH_(3) COCl) lt CH_(3) CO_(2)CH_(3) lt underset(underset(underset("and more surface area")("molecular mass"))("Higher"))((CH_(3)CO)_(2)O) lt underset(underset(underset("hydrogen boding")("INTERMOLECULAR"))("EXTENSIVE"))(CH_(3)CONH_(2))`
Discussion
44.

The correct order of increasing basicity of the given conjugate bases (R=CH_(3)) is

Answer»

`CRCOO^(-) lt NH_(2)^(-) lt HC -= C^(-) lt R^(-)`
`RCO O^(-) lt HC -= C^(-) lt NH_(2)^(-) lt R^(-)`
`RCO O^(-) lt HC -= C^(-) lt R^(-) lt NH_(2)^(-)`
`R^(-) lt HC -= C^(-) lt RCO O^(-) lt NH_(2)^(-)`

Solution :The conjugate acids of the given bases are `RCOOHgtCH-=CHgtNH_(3)gtRH`. Their acidic strengths are in the order : `RCOOHgtCH-=CHgtNH_(3)gtRH`.As STRONG acid has weak conjugate BASE, the BASIC strenths of the given bases will be in the order :
`RCOO^(-)ltHC-=C^(-)ltNH_(2)^(-)LTR^(-)` (Remember that `sp^(3)` C is less electronegative than `sp^(3)` N. Hence, alkyl carbanion `(R^(-))` is more basic than `bar(N)H_(2)`. However, sp-hybridized carbon is more electronegative than `sp^(3)` N Hence `bar(N)H_(2)` is more basic `HC-=bar(C)`.
Discussion
45.

The correct order of increasing basicity of the given conjugate bases (R=CH_3) is

Answer»

`RCOO^(-) LT HC -=C^(-) lt NH_2^(-)lt R^(-)`
`RCOO^(-) lt HC -=C^(-) lt R^(-) lt NH_2^-`
`R^(-) lt HC-=C^(-) lt RCOO^(-) lt NH_2^(-)`
`RCOO^(-) lt NH_2^(-) lt HC -=C^(-) lt R^(-)`

ANSWER :A
Discussion
46.

The correct order of increasing basicity of the given conjugate bases (R=CH_(3)) is :

Answer»

`RCObarO LT HC -=BARC lt barNH_(2)`
`barR lt HC -= RCObarO lt barNH_(2)`
`RCObarO lt barNH_(2) lt HC -= barC lt barR`
`RCObarO lt HC-=barC lt barNH_(2) lt barR`

Answer :D
Discussion
47.

The correct order of increasing acidic strength of the following acids is

Answer»

`I LT II lt III`
`I lt III lt II`
`III lt II lt I`
`II lt III lt I`

Solution :The `-I` effect of `CL` increases the ACIDIC strength of carboxylic acid. Nearer the `Cl` ATOM to `-COOH` group, stronger is the acid
Discussion
48.

The correct order of increasing acid strength of the compounds. (a) CH_(3)CO_(2)H""(b) MeOCH_(2)CO_(2)H""(c)CF_(3)CO_(2)H "" (d)

Answer»

`B LT d lt a lt C`
`d lt a lt c lt d`
`d lt a lt b lt c`
`a lt d lt c lt b`

ANSWER :C
Discussion
49.

The correct order of hydration enthalpies of alkali metal ions is:

Answer»

`LI^(+) GT NA^(+) gt K^(+) gt RB^(+) gt Cs^(+)`
`Na^(+) gt Li^(+) gt K^(+) gt Rb^(+) gt Cs^(+)`
`Li^(+) gt Na^(+) gt K^(+) gt Cs^(+) gt Rb^(+)`
`Na^(+) gt Li^(+) gt K^(+) gt Cs^(+) gt Rb^(+)`

Solution :`Li^(+) gt Na^(+) gt K^(+) gt Rb^(+) gt Cs^(+)`
Discussion
50.

The correct order of hybridisation of the central atom in the following species NH_(3), [PtCl_(4)]^(2-), PCl_(5) and BCl_(3) is

Answer»

`DSP^(2), dsp^(3), SP^(2) and sp^(3)`
`sp^(3), dsp^(2), dsp^(3), sp^(2)`
`dsp^(2), sp^(2), sp^(3), dsp^(3) `
`dsp^(2), sp^(3), sp^(2), dsp^(3)`

Answer :B
Discussion