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1.

Which of the following lanthanoid ions is diamagnetic? (Atomic number of `Ce=58,Sm=62,Eu=63,Yb=70`]A. `Ce^(2+)`B. `Sm^(2+)`C. `Eu^(2+)`D. `Yb^(2+)`

Answer» Correct Answer - D
Lanthanoid ion with no unpiared electronis diamagnetic in nature.
`Ce_(58) =[Xe]4f^(2)5d^(0)6s^(2)`
`Ce^(2+) =[Xe]4f^(2)` (two unpaired electrons)
`Sm_(62) = [Xe]4f^(6)5d^(0)6s^(2)`
`Sm^(2+) = [Xe]4f^(6)` (six unpaired electrons)
`Eu_(63) =[Xe]4f^(6)5d^(0)6s^(2)`
`Eu^(2+) =[Xe]4f^(7)` (seven unpiared electrons)
`Yb_(70)=[Xe]4f^(14)5d^(0)6s^(2)`
`Yb^(2+) = [Xe]4f^(14)` (no unpaired electrons)
Because of the absence of unpaired electrons, `Yb^(2+)` is diamagnetic.
Discussion
2.

Lanthanides areA. 14 elements in the sixth period (At.no=90 to 103) that are filling 4f sub-levelB. 14 elements in the seventh period (At.no.=90 to 103) that are filling 5f sub-levelC. 14 elements in the sixth period (At.no.=58 to 171) that are filling 4f sub-level).D. 14 elements in the seventh period (At. No.=58 to 71) that are filling 4f sub-level

Answer» Correct Answer - C
Lanthanides are the 14 elements of IIIB group and sixth period (at no. =58 to 71) that are filling sub-shell of antipenultimate shell from 1 to 14. Actually, they are placed below the periodic table horizontal row as lanthanide series.
Discussion
3.

The common oxidation states of Ti areA. `+2, +3`B. `+3, +4`C. `3,-4`D. `+2,+3,+4`

Answer» Correct Answer - B
The most common oxidation state of titanium are `+3` and `+4`.
Discussion
4.

Photographic plates and films have an essential ingredient ofA. silver nitrateB. silver bromideC. sodium chlorideD. oleic acid

Answer» Correct Answer - B
The photographic plate of film consists of a glass plate or thin strip of celluloid which is coated with the thin layer of an emulsion of silver bromide dispersed in getalin.
Discussion
5.

The oxidation state of `Cr` in `K_(2)Cr_(2)O_(7)` is:A. `+5`B. `+3`C. `+6`D. `+7`

Answer» Correct Answer - C
Let the oxidation state of Cr is x
`K_(2)Cr_(2)O_(7)`
`therefore 2(+1)+2x=7(-2)=0`
`2+2x-14=0`
`2x-12=0`
`2x=12`
`x=12/2 = +6`
Discussion
6.

Which of the following elements shows maximum number of different oxidation states in its compounds ?A. EuB. LaC. GdD. Am

Answer» Correct Answer - D
Oxidation states shown by elements are as follows:
La`=+`
Eu and Gd `=+2` and `+3`
Am `=+2+3+4+5` and `+6`
Am shows maximum number of different oxidation state due to its larger size and low ionisation energy.
Discussion
7.

`CuSO_(4)` when reacts with KCN forms `CuCN` which is insoluble in water. It is soluble in excess of KCN due to the formation of the complex.A. `K_(3)[Cu(CN)_(4)]`B. `K_(3)[Cu(CN)_(4))]`C. `Cu(CN)_(2)`D. `Cu[Kcu(CN)_(4)]`

Answer» Correct Answer - B
`CuSO_(4)` reacts with KCN to give a white precipitate of cuprous cyanide and cyanogen gas. The cuprous cyanide dissolves in excess of KCN forming `K_(3)[Cu(CN)_(4)]`
`CuSO_(4) + 2KCN to K_(2)SO_(4) + Cu(CN)_(2)`
`2Cu(CN)_(2) to 2CuCN + CN - CN`
`CuCN + 3KCN to K_(3)[Cu(CN)_(4)]`
Discussion
8.

Which of the following may be considered to be an organometallic compound?A. Nickel tetracarbonylB. ChlorophyllC. `K_(3)[Fe(C_(2)O_(4))_(3)]`D. `[Co(en)_(3)Cl_(3)]`

Answer» Correct Answer - B
Chlorophyll may be considered to be an organometallic compound because it contain metal carbon bond.
Discussion
9.

Stainaless steel contains iron andA. `Cr + Ni`B. `Cr+ Zn`C. `Zn+Pb`D. `Fe+Cr+Ni`

Answer» Correct Answer - A
Stainless steel is resistant to rusting. It contains `73%` iron, `18%` chromium and `8%` nickel. Stainless steel is used in utensils, cycle, cutlery and automobiles parts.
Discussion
10.

Cuprous compounds such as CuCI, CuCN and CuSCN are the only salts stable in water due toA. high hydration energy of `Cu^(+)` ionsB. their inherit tendency not to disproportionateC. diamagnetic natureD. insolubility in water.

Answer» Correct Answer - A
Cuprous compound contains `Cu^(+)` ion which has small size, so the hydration is maximum and hence,the system has slower energy which result in stability of the compounds.
Discussion
11.

In which of the following compounds transition metal has zero oxidation state ?A. `Fe(CO)_(5)`B. `NH_(2).NH_(2)`C. `NOClO_(4)`D. `CrO_(5)`

Answer» Correct Answer - A
Oxidation state of Fe in `Fe(CO)_(5)` is zero because CO is a neutral ligand and it shows zero oxidation state.
Discussion
12.

More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this isA. more energy difference between 5f and 6d orbitals than that between 4f and 5d-orbitalsB. lesser energy difference between 5f and 6d-orbitals than that between 4f and 5d-orbitalsC. greater metallic character of the lanthanides than that of the corresponding actinidesD. more active nature of the actinides.

Answer» Correct Answer - B
more number of oxidation states are exhibited by the actinides than by the corresponding lanthanides due to lesser energy difference between 5f and 6d orbitals than that between 4f and 5d-orbitals.
Discussion
13.

Which one of the following ions has electronic configuration `[Ar]3d^(6)` ? `(At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)`A. `Ni^(3+)`B. `Mn^(3+)`C. `Fe^(3+)`D. `Co^(3+)`

Answer» Correct Answer - D
`Ni^(3+)(28)=[Ar]3d^(7)`
`Mn^(3+)(25)=[Ar]3d^(4)`
`Fe^(3+)(26) = [Ar]3d^(5)`
`Co^(3+)(27)=[Ar]3d^(6)`
Hence, `Co^(3+)` has `3d^(6)` electronic configuration correct answer is (d).
Discussion
14.

For the four successive transition elements (Cr, Mn, Fe, and Co), the stability of `+2` oxidation state will be there in which of the following order ? `(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)`A. `Fe gt Mn gt Co gt Cr`B. `Co gt Mn gt Fe gt Cr`C. `Cr gt Mn gt Co gt Fe`D. `Mn gt Fe gt Cr gt Co`

Answer» Correct Answer - D
This can be understood on the basis of `E^(@)` values for `M^(2+)//M`
`E^(@)//V " " Cr " " Mn " " Fe" " Co`
`M^(2+)//M " " -0.90 " "-1.18 " " -0.44 " " -0.28`
`E^(@)` value for Mn is more negative than expected from general trend due to extra stability of half-filled electronic configuration of `Mn^(2+)` ion.
Thus, the correct order should be
`Mn gt Cr gt Fe gt Co`
An examination of `E^(@)` values for redox couple `M^(3+)//M^(2+)` shows that `Cr^(2+)` is strong reducing agent `(E^(@))_(M^(3+)//M^(2+))=0.41 V` and liberates `H_(2)` from dilute acids.
`2Cr^(2+)(aq) + 2H^(+) (aq) to 2Cr^(3+)(aq) + H_(2) uarr (g)`
`therefore` The correct order is `Mn gt Fe gt Cr gt Co`.
Discussion
15.

When `CuSO_(4)` is electrolysed, using Pt electrodesA. Copper is liberated at cathode, sulphur at anodeB. Copper is liberated at cathode, oxygen at anodeC. Sulphur is liberated at cathode, oxygen at anodeD. Oxygen is liberated at cathode, copper at anode.

Answer» Correct Answer - B
During electrolyses, copper is deposited at the cathode while oxygen is liberated at anode. The following reactions occur at the electrodes.
At anode `2H_(2)O to 4H^(+) + O_(2) + 4e^(-)`(Oxidation)
At cathode `Cu^(2+) + 2e^(-) to Cu(s)` (Reduction)
Discussion
16.

The reaction of aqueus `KMnO_(4)` with `H_(2)O_(2)` in acidic conditions givesA. `Mn^(+)` and `O_(2)`B. `Mn^(2+)` and `O_(2)`C. `Mn^(2+)` and `O_(3)`D. `Mn^(4+)` and `MnO_(2)`

Answer» Correct Answer - B
The reaction of aqueous `KMnO_(4)` with `H_(2)O_(2)` in acidic medium is
`3H_(2)SO_(4) + 2KMnO_(4) + 5H_(2)O to 5O_(2) + 2MnSO_(4) + 8H_(2)O + K_(2)SO_(4)`
In the above reaction, `KMnO_(4)` with `H_(2)O_(2)` to `O_(2)` and itself i.e., `[MnO_(4)^(-)]` gets reduced to `Mn^(2+)` ion as `MnSO_(4)`. Hence, aqueous solution of `KMnO_(4)` with `H_(2)O_(2)` yields `Mn^(2+)` and `O_(2)` in acidic conditions.
Discussion
17.

Reason of lanthanoid contraction isA. negligible screening effect of f-orbitalsB. increasing nuclear changeC. decreasing nuclear chargeD. decreasing screening effect

Answer» Correct Answer - A
Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides. This is due to the imperfect shielding [or poor screening effect] of f-orbitals due to thieir diffused shape which unable to counterbalance the effect to the increased nuclear charge. Hence, the net results is a contraction in size of lanthanoids.
Discussion
18.

Because of lanthnoid contraction, which of the following pairs of elements have nearly same atomic radii ? (Number in the parenthesis are atomic numbers)A. Ti(22) and Zr(40)B. Zr(40) and Nb(41)C. Zr (40)and Hf(72)D. Zr(40) and Ta(73)

Answer» Correct Answer - C
Because of the lanthonid contraction Zr (atomic radi 160 pm) and Hf (atomic radii 158 pm) have nearly same atomic radii.
Lanthanoids include the elements from lanthanim `La(Z=57)` to lutentium Lu=(Z=71). Zirconium Zr(40) belong to the second transition series (5d). Lanthanoid contraction is associated with the intervention of the 4f orbitals which are filled before the 5d-series of elements starts. The filling of 4f orbitals before 5d-orbitals results in regular decrease in atomic radii which compensates the expected increase in atomic size with increasing atomic number. As a result of this lanthnoid contraction, the elements of second and third transition series have almost similar atomic radii.
Discussion
19.

`HgCl_(2)` and `I_(2)` both when dissolved in water containing `I^(-)` ions the pair of species formed is:A. `HgI_(2)mI_(3)^(-)`B. `HgI_(2),I^(-)`C. `Hgl_(4)^(2-), I_(3)^(-)`D. `Hg_(2)I_(2),I^(-)`

Answer» Correct Answer - C
`HgCl_(2)` and `I_(2)` both when dissolved in water containing `I^(-)` ions, the pair of species formed is `HgI_(4)^(2-)` and `I_(3)^(-)`.
In aqueous solution, `I_(2)` reacts with `I^(-)` and maintains the following equilibrium
`I_(2)+I^(-) ltimplies I_(3)^(-)`
`Hg^(2+)` gives ppt. of `HgI_(2)` on reaction with `I^(-)`.
But `HgI_(2) +2I^(-) to underset("Red ppt")(HgI_(2))darr+2Cl^(-)`
`HgI_(2) + 2I^(-) ltimplies [HgI_(4)]^(2-)`
`HgI_(2) + 2I^(-) ltimplies [HgI_(4)]^(2) + 2I^(-) ltimplies [HgI_(4)]^(2-)`
Discussion
20.

Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified `KMnO_(4)` for complete oxidation ?A. `FeSO_(4)`B. `FeSO_(3)`C. `FeC_(2)O_(4)`D. `Fe(NO_(2))_(2)`

Answer» Correct Answer - A
`FeSO_(4)` will require the least amountof acdifified `KMnO_(4)` for complete oxidation.
Discussion
21.

Name the gas that can readily decolourise acidified `KMnO_(4)` solution:A. `CO_(2)`B. `SO_(2)`C. `NO_(2)`D. `P_(2)O_(5)`

Answer» Correct Answer - B
`SO_(2)` gas can readily oxidise acidified `KMnO_(4)` solution because `KMnO_(4)` is an oxidising agent and `SO_(2)` act as reducing agent.
`2MnO_(4)^(-) + 5SO_(2) +2H_(2)O to 2Mn^(2+) + 5SO_(4)^(2-) + 4H^(+)`
While other options such as `NO_(2)` (strong oxidising agent), `CO_(2)` (neither oxidising agent nor reducing agent) cannot decolourise acidified `KMnO_(4)` solution.
Discussion