Q 11 :    

The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is ___________ .

 

Ti2+,Cr2+ and V2+.                                           [2024]

  • 3

     

  • 0

     

  • 1

     

  • 2

     

(A)

           M2+             M3++e-               EM3+/M2+°

           H++e-      12H2                             EH+/H2°

          ________________________________________

          M2++H+     M3++12H2          Ereaction°=EH+/H2°-EM3+/M2+°

         

         If M2+ liberates hydrogen from dilute acid, then:

         Ereaction°=EH+/H2-EM3+/M2+>0

         0-EM3+/M2+>0

         EM3+/M2+<0

          As ETi3+/Ti2+=-0.37 V, EV3+/V2+=-0.26 V, ECr3+/Cr2+=-0.41 V

                 Ti2+, V2+ and Cr2+ can liberate hydrogen from dilute acids.

 



Q 12 :    

Arrange the following elements in the increasing order of the number of unpaired electrons in it.

 

(A)  Sc                  (B)  Cr                  (C)  V                   (D)  Ti                    (E)  Mn  

 

Choose the correct answer from the options given below:                         [2024]

  • (A) < (D) < (C) < (B) < (E)

     

  • (B) < (C) < (D) < (E) < (A)

     

  • (A) < (D) < (C) < (E) < (B)

     

  • (C) < (E) < (B) < (A) < (D)

     

(C)

Element and electronic
configuration
Number of unpaired
electrons
(A)       Sc21=[Ar]183d14s2                         1
(B)       Cr24=[Ar]183d54s1                         6
(C)      V23=[Ar]183d34s2                         3
(D)      Ti22=[Ar]183d24s2                         2
(E)      Mn25=[Ar]183d54s2                         5

 



Q 13 :    

The electronic configuration of Cu(II) is 3d9 whereas that of Cu(I) is 3d10. Which of the following is correct?               [2024]

  • Cu(I) and Cu(II) are equally stable

     

  • Cu(II) is less stable

     

  • Cu(II) is more stable

     

  • Stability of Cu(I) and Cu(II) depends on nature of copper salts.

     

(C)

The stability of Cu2+ (aq) rather than Cu+ (aq) is due to the much more negative Hhyd of Cu2+ (aq) than Cu+ (aq), which more than compensates for the second ionisation enthalpy of Cu.

 



Q 14 :    

Given below are two statements:

 

Statement I: The higher oxidation states are more stable down the group among transition elements unlike p-block elements.

 

Statement II: Copper can not liberate hydrogen from weak acids.

 

In the light of the above statements, choose the correct answer from the options given below:               [2024]

  • Statement I is true but Statement II is false

     

  • Both Statement I and Statement II are true

     

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are false

     

(B)

Statement I: In d block elements stability of higher oxidation state increases down the group. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not. In p block elements, stability of lower oxidation state increases down the group due to inert pair effect.

Statement II: ECu2+/Cu=+0.34V,  EH+/H2=0V

For the reaction: Cu+2H+Cu2++H2

  Ereaction=EH+/H2-ECu2+/Cu=0V-0.34V=-0.34V

Since potential of the reaction is negative copper cannot liberate hydrogen from dilute acids.

 



Q 15 :    

The spin-only magnetic moment value of the ion among Ti2+,V2+,Co3+ and Cr2+, that acts as strong oxidising agent in aqueous solution is _______ B.M. (Near integer).

 

(Given atomic numbers : Ti : 22, V : 23, Cr : 24, Co : 27)                 [2024]



(5)

              ECo3+/Co2+=+1.97V

              Since ECo3+/Co2+>0,Co3+ is a strong oxidizing agent.

             Co3+=18[Ar]3d6

            Number of unpaired electrons in Co3+(n)=4

            Spin only magnetic moment (μ)

                             =n(n+2)BM=4(4+2)BM=4.9BM5BM

 



Q 16 :    

Total number of ions from the following with noble gas configuration is _______ .

 

Sr2+(z=38),Cs+(z=55),La2+(z=57),Pb2+(z=82),Yb2+(z=70) and Fe2+(z=26)                [2024]



(2)

Ion Electronic configuration
Sr2+38 [Kr]365s0 (noble gas)
Cs+55 [Xe]546s0 (noble gas)
La2+57 [Xe]545d1
Pb2+82 [Xe]544f145d106s2
Yb2+70 [Xe]544f14
Fe2+26 [Ar]183d6

 



Q 17 :    

A transition metal ‘M’ among Sc, Ti, V, Cr, Mn and Fe has the highest second ionisation enthalpy. The spin-only magnetic moment value of M+ ion is ______. B.M. (Near integer) (Given atomic number Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26)                   [2024]



(6)

   Second ionization energy increases from left to right in periodic table.

Metal Sc Ti V Cr Mn Fe Co Ni Cu Zn
Second
ionization
energy
(kJ/mol)
1235 1309 1414 1592 1509 1561 1644 1752 1958 1734

    Cr has more second ionization energy than Mn and Fe because the electron is removed from stable d5 configuration.

    Cr24([Ar]183d54s1)IE1Cr+([Ar]183d5)IE2Cr2+([Ar]183d4)

    Thus the metal M is Cr. Number of unpaired electrons in Cr+(n) = 5.

    Spin only magnetic moment

    (μ)=n(n+2) BM=5(5+2) BM=5.92 BM