94420 + Interview Questions in Chemistry in Class 12 Page 610 InterviewSolution

30451.	The group satisfying the secondary valencies of a cation in a complex are called:
Answer» Ligands Radicals PRIMARY valencies None Answer :A

Discussion

30452.	The group reagment for the test of alcohol is :
Answer» cerric AMMONIUM nitrate Schiff s REAGENT MOLISCH's ragment bromine water Solution :N//A

Discussion

30453.	The group reagent for the test of alcohols is :
Answer» Cerric AMMONIUM nitrate Schiff's reagent Molisch's reagent Bromine water Solution :N//A

Discussion

30454.	The group reagent for the analysis of group IIIB is:
Answer» NH_4Cl+NH_4OH+H_2S` `HCI+H_2O` `NH_4CI+NH_4OH+(NH_4)_2CO_3+H_2S` `CaCO_3` ANSWER :A

Discussion

30455.	The group reagent for group III is:
Answer» DIL HCL `NH_4OH` in the PRESENCE of `H_2S` `NH_4OH` is the presence of `NH_4Cl` `(NH_4)_2CO_3` in the presence of `NH_4OH` ANSWER :C

Discussion

30456.	The group of molecules having identical shape is:
Answer» `PCl_(5)IF_(5), XeO_(2)F_(2)` `BF_(3),PCl_(3),XeO_(3)` `SF_(4),XeF_(4),C Cl_(4)` `CIF_(3),XeOF_(2),XeF_(3)^(+)` Solution :`CIF_(3) to ` HYBRIDISATION `=3+(1)/(2)[7-3]=5(sp^(3)d)` `XeOF_(2) to` Hybridisation `=3+(1)/(2)[8-4]=5(Sp^(3)d)` `XeF_(3)^(+) to` Hybridation `=3+(1)/(2)[8-2-1]=5(sp^(3)d)` All MOLECULES have `sp^(3)`hybridzation and 2 lone pairs. HENCE all haveidentical (T-shape).

Discussion

30457.	The group of metals which is known as ferrous metals is:
Answer» FE,Co,Ni Ru,RH,Pd Os,Ir,Pt Cr,Mn,Cu Answer :A

Discussion

30458.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)Qwhich will give borax beat Test with bluebeard?
Answer» `Cr^(3+)` `CO^(3+)` `Ni^(2+)` `CD^(2+)` Answer :b

Discussion

30459.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QWhich component of KCIO, used in above reaction with HCl and given Salt
Answer» `KCIO_(3)` KCL `KClO_(3)` NASCENT oxygen Answer :d

Discussion

30460.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QWhich metal salt gives a violet coloured bead in the borax bead test?
Answer» `FE^(2+)` `NI^(2+)` `Co^(2+)` `Mn^(2+)` Answer :d

Discussion

30461.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QColour of cobalt chloride is
Answer» pink black colourless green ANSWER :B

Discussion

30462.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QWhich of the following pairs of ions cannot be separated by HS in dilute hydrochloric acid?
Answer» `BI^(3+) Sn^(4+)` `AL^(3+) HG^(2+)` `ZN^(2+). Ca^(2+)` `Ni^(2+)Cu^(2+)` Answer :a

Discussion

30463.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QIn above paragraph, given compound 'D' is
Answer» `K_4[FE(CN)_6]` `K_4[Mn(CN)_6)` `K_4[CO(CN)_6]` `K_4[Ni(CN)_6]` ANSWER :C

Discussion

30464.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QIn above paragraph, given salt 'A' is
Answer» Coch Zns Cos NİS Answer :a

Discussion

30465.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QIn above paragraph, given salt 'B' is
Answer» ZNS NIS Cos MnS Answer :B

Discussion

30466.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QFormation of a rose red precipitate when a slightly alkaline solution of Inorganic salt is treated with dimethylglyoxime confirms the presence of
Answer» cobalt zinc iron nickel ANSWER :d

Discussion

30467.	The group of four basic radicals, CO^(2+) Nl^(2+) Mn^(2+) and Żn^(2+) have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in NH_(4)OH. Addition of HS increases the ionization of NH_4OH due to reaction of OH and H lon. NH_4OHharrNH_(4) + OHH_2S hArr2H+S^(2-) Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)QTo avoid the precipitation of hydroxides of Ni*, Co. Mn along with those of the third group cations, the solution should be
Answer» heated with few drops of CONC. `HNO_(3)` heated with excess of `NH_4`CI concentrated to SMALL volume NONE of the above Answer :b

Discussion

30468.	The group obtained by the removal of H-atom of the -OH group of ROH is called
Answer» ALKYL group alkene alkoxy group all of these ANSWER :C

Discussion

30469.	The group linkage present in fats is :
Answer» PEPTIDE linkage ESTER linkage glycoside linkage NONE of these Answer :B

Discussion

30470.	The group II precipitate soluble in yellow ammonium sulphide may be
Answer» `As,SB,Sn` `Ca,Hg,BI,Cd` Both (a) and (B) None of these Solution :YELLOW ammonium sulphide is a SOLVENT for As,Sb and Sa salphides

Discussion

30471.	The group in which all the elemets do not have same number of valency electron is :
Answer» Zero Second First Seventh Answer :A

Discussion

30472.	The group I elements form hydroxides, which are :
Answer» STRONGLY acidic strongly basic weakly acidic neutral Answer :B

Discussion

30473.	The 'group displacement law' was given by :
Answer» Bacqueral Rutherford Madam Curie Soddy and Fajan Answer :D

Discussion

30474.	The group displacement law was proposed by :
Answer» SODDY and Thomson Soddy and RUTHERFORD Soddy and FAJAN Rutherford. Answer :C

Discussion

30475.	The group containing oxides that cannot be reduced by carbon to give the respective metal is :
Answer» `Cu_2o, SnO_2` `FeO_3` `Na_2O, AI_2O_3` PBO,`Fe_2O_3` ANSWER :C

Discussion

30476.	The group commonly present in synthetic detergents is
Answer» RNa RONa RCOONA `RC_6H_(4)SO_3Na` ANSWER :D

Discussion

30477.	The group 15 elements are commonly known as:
Answer» Halogens Normal elements Pnicogens None Answer :C

Discussion

30478.	The ground state electronic configuration of the elements, U, V, W, X, and Y (these symbols do not have any chemical significance) are as follows: U""1s^(2)2s^(2)2p^(3) V""1s^(2)2s^(2)2p^(6)3s^(1) W""1s^(2)2s^(2)2p^(6)3s^(2)3p^(2) X""1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(2) Y""1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6) Determine which sequence of elements satisfy the following statement: (i) element forms a carbonate which is not decomposed by heating (ii) Element is most likely to form coloured ionic compound (iii) element has largest atomic radius (iv) element forms only acidic oxide
Answer» V W Y U V X Y W V W Y X V X W U Solution :(i) Alkali metal carbonates do not decompose even at red HOT `to`V (ii) Transition metal ions having unpaired d-electrons are coloured in aq. Sol./compounds`toX` (iii) In case of Kr VAN der Waals. RADIUS is considered, which is largest atomic radius `to`Y (IV) Si atom has only acid `SiO_(2) to W`

Discussion

30479.	The ground state magnetic property of B_(2) and C_(2) molecules will be
Answer» `B_(2)` paramagnetic and `C_(2)` diamagnetic `B_(2)` diamagnetic and `C_(2)` paramagnetic both are diamagnetic both are paramagnetic. Solution :`B_(2)` MOLECULE `sigma1s^(2), sigma1s^(2), sigma2s^(2),sigma2S^(2),pi2p_(x)^(1)~~pi2p_(y)^(1)` (Two UNPAIRED electrons hence paramagnetic) `C_(2)` molecule `sigma1s^(2), sigma1s^(2),sigma2s^(2), sigma2s^(2),pi2p_(x)^(2)~~pi2p_(7)^(2)` No unpaired electrons hence diamagnetic.

Discussion

30480.	The ground state energy of hydrogen atom is -13.6 eV. Consider an electronic state Psiof He^(+)whose energy, azimuthal quantum number and magnetic quantum number are -3.4 eV, 2 and 0, respectively. Which of the following statement(s) is(are) true for the state Psi?
Answer» It has 3 radial nodes It has 2 angular nodes The nuclear CHARGE experienced by the electron in this state is LESS than 2e, where e is the magnitude of the ELECTRONIC charge It is a 4d state Answer :A::C::D

Discussion

30481.	The ground state electronic configuration ofFe^(3+) (g)ion is:
Answer» `[Ar]3D^(3)4S^(2)` `[Ar]3d^(6)4s^(2)` `[Ar]3d^(5)` `[Ar]3d^(6)` Solution :`[Ar]3d^(5)`

Discussion

30482.	The ground state electronic configuration of the element which is isoelectronic with NO^(+) is
Answer» `1S^(2)2s^(2)2p^(6)3S^(2)3p^(3)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(2)` `1s^(2)2s^(2)2p^(6)` ANSWER :C

Discussion

30483.	The ground state electronic configuration of Cu(29) is [Ar]3d^(10) 4s^(1) and not [Ar]3d^(0)4s^(2). This is indicated by the fact that
Answer» Completely filled sub-shell configuration is more stable than one less filled sub-shell configuration. Cu EXHIBITS +1 IONIC state besides `+2` STATES. Copper turnings react with hot conc. `H_(2)SO_(4)` to evolve `SO_(2)(g)`rather than `H_(2)(g)`. `Cu^(2+)` disproportionates in aqueous solution into `Cu^(2+)` and Cu. Answer :B

Discussion

30484.	The ground state electronic configuration ofoxygen in Li_(2)O is
Answer» `1s^(2)2s^(2)2p^(4)` `1s^(2)2s^(2)2p^(6)` `1s^(2)2s^(2)2p^(2)` `1s^(2)2s^(2)2p^(3)` Solution :In`Li_(2)O` OXYGEN eixsts as `O^(2-)`.Its configuration is `1s^(2)2s^(2)2p^(6)`.

Discussion

30485.	the ground state electronic configuration of chromium can be written as
Answer» `[AR] 3d^1 4s^1` `[Ar] 3d^1 4s^2` `[Ar] 3d^5 4s^1` `[NE] 3d^4 4s^1` ANSWER :C

Discussion

30486.	The ground state electronic configuration for chromium atom (z=24) is
Answer» `[Ar]3D^(5)4s^(1)` `[Ar](3d^(4))4s^(2)` `[Ar]3d^(6)` `[Ar]4s^(1)4p^(5)` ANSWER :A

Discussion

30487.	The groun dhaving isoelectronic species is
Answer» `O^(2-),F^(-), NA^(+), MG^(2+)` `O^(-), F^(-), Na, Mg^(2+)` `O^(2-), F^(-), Na, Mg^(2+)` `O^(-), F^(-), Na^(+), Mg^(2+)` ANSWER :A

Discussion

30488.	The ground state configuration of oxygen in OF_(2) is
Answer» `1s^(2)2s^(2)2P^(4)` `1s^(2)2s^(2)2p^(6)` `1s^(2)2s^(2) 2p^(2)` `1s^(2)2s^(2)2p^(3)` Solution :In `OF_(2)`, OXYGEN exists as `O^(2+).(1s^(2) 2s^(2) 2p^(2))`

Discussion

30489.	The Grignard reagent, CH_(3)CH_(2)MgBr, can be used to prepare
Answer» ethane 3-methyl-3-pentanol propanoic acid all of these. Answer :D

Discussion

30490.	The grey cast iron contains:
Answer» IRON carbide Silicon carbide silicon dioxide Graphite Answer :D

Discussion

30491.	The grey cast iron contains :
Answer» Irion CARBIDE SILICON carbide Silicon DIOXIDE Graphite Answer :D

Discussion

30492.	The green pigment chlorophyll contains ................... Ion surrounded by a modified porphyrin ligand called ..................
Answer» ANSWER :`MG^(2+)`, corrinring

Discussion

30493.	The greenhouse effect is caused by
Answer» CO `NO_2` `CO_2` NO SOLUTION :`CO_2`

Discussion

30494.	The green pigment of plants essential for the formation of carbohydrates by photosynthesisis
Answer» Acrophyll Lyphyll Chlorophyll None of the above Answer :C

Discussion

30495.	The green mass formed when copper is exposed to moist air has the same composition as that of
Answer» Azurite Malachite CUPRITE COPPER pyrites Answer :B

Discussion

30496.	The green coloured salt formed on heating (NH_(4))_(2) Cr_(2)O_(7) is "................."
Answer» ANSWER :`Cr_(2)O_(3)`

Discussion

30497.	The greatest strain is involved in cycloalkane when the bond angle is:
Answer» `60^@` `90^@` `120^@` `108^@` Answer :B

Discussion

30498.	The greatest percentage of CO is in :
Answer» COAL gas Producer gas Water gas Oil gas Answer :C

Discussion

30499.	The greater the value of r_(+)//r_(-).
Answer» the LOWER will be the CN the HIGHER the VALUE of CN the higher will be the NUMBER of cations the lower will be the number of anions Answer :B

Discussion

30500.	The greater the ionic character of the carbon metal bond:
Answer» The more reactive is the ORGANOMETALLIC COMPOUND The less reactive is the organometallic compound both are correct None is correct Answer :A

Discussion

Explore topic-wise InterviewSolutions in .

The group satisfying the secondary valencies of a cation in a complex are called:

The group reagment for the test of alcohol is :

The group reagent for the test of alcohols is :

The group reagent for the analysis of group IIIB is:

The group reagent for group III is:

The group of molecules having identical shape is:

The group of metals which is known as ferrous metals is:

The group obtained by the removal of H-atom of the -OH group of ROH is called

The group linkage present in fats is :

The group II precipitate soluble in yellow ammonium sulphide may be

The group in which all the elemets do not have same number of valency electron is :

The group I elements form hydroxides, which are :

The 'group displacement law' was given by :

The group displacement law was proposed by :

The group containing oxides that cannot be reduced by carbon to give the respective metal is :

The group commonly present in synthetic detergents is

The group 15 elements are commonly known as:

The ground state magnetic property of B_(2) and C_(2) molecules will be

The ground state energy of hydrogen atom is -13.6 eV. Consider an electronic state Psiof He^(+)whose energy, azimuthal quantum number and magnetic quantum number are -3.4 eV, 2 and 0, respectively. Which of the following statement(s) is(are) true for the state Psi?

The ground state electronic configuration ofFe^(3+) (g)ion is:

The ground state electronic configuration of the element which is isoelectronic with NO^(+) is

The ground state electronic configuration of Cu(29) is [Ar]3d^(10) 4s^(1) and not [Ar]3d^(0)4s^(2). This is indicated by the fact that

The ground state electronic configuration ofoxygen in Li_(2)O is

the ground state electronic configuration of chromium can be written as

The ground state electronic configuration for chromium atom (z=24) is

The groun dhaving isoelectronic species is

The ground state configuration of oxygen in OF_(2) is

The Grignard reagent, CH_(3)CH_(2)MgBr, can be used to prepare

The grey cast iron contains:

The grey cast iron contains :

The green pigment chlorophyll contains ................... Ion surrounded by a modified porphyrin ligand called ..................

The greenhouse effect is caused by

The green pigment of plants essential for the formation of carbohydrates by photosynthesisis

The green mass formed when copper is exposed to moist air has the same composition as that of

The green coloured salt formed on heating (NH_(4))_(2) Cr_(2)O_(7) is "................."

The greatest strain is involved in cycloalkane when the bond angle is:

The greatest percentage of CO is in :

The greater the value of r_(+)//r_(-).

The greater the ionic character of the carbon metal bond: