Q: Identify the incorrect statement.

Oxidation state of Cr in CrO42− and Cr2O72− is +6

Q: The manganate and permanganate ions are tetrahedral, due to :

∙ Manganate (MnO42−) : ⇒ π -bonds are of dπ – pπ type∙ Permanganate (MnO4−) : ⇒ π -bonds are of dπ – pπ type

Q: The electronic configurations of Eu (Atomic no. 63), Gd (Atomic NO. 64) and Tb (Atomic No. 65) are:

Eu − [Xe]4f7,6s2Gd − [Xe]4f7,5d1,6s2 T6 − [Xe]4f9,6s2

Q: Which of the following does not give oxygen on heating?

(NH4)2Cr2O7⟶ΔN2+Cr2O3+2H2O 2KClO3⟶Δ2KCl+3O2 Zn(ClO3)2⟶ΔZnCl2+3O2 K2Cr2O7⟶Δ2 K2CrO4+Cr2O3+23O2

Q: KMnO4 can be prepared from K2MnO4 as per the reaction:3MnO42−+2H2O⇌2MnO4−+MnO2+4OH− The reaction can go to completion by removing OH- ins by adding

MnO4− oxidizes HCl and SO2 to Cl2 and SO3 respectively. MnO4− does not oxidizes H2CO3 (maximum oxidation state of C )

Question 1

The calculated spin only magnetic moment of Cr2+ ion is:

Accepted Answer

Electronic configuration of Cr – [Ar] 3d54s1 Electronic configuration of Cr2+ – [Ar] 3d4  Number of unpaired e−=4 Spin only magnetic moment = n(n+2)​ n = number of unpaired e− Spin only magnetic moment 	=4(4+2)​ =24​ BM=4.9 BM

Question 2

Identify the incorrect statement.

Accepted Answer

Oxidation state of Cr in CrO42−​ and Cr2​O72−​ is +6

Question 3

The manganate and permanganate ions are tetrahedral, due to :

Accepted Answer

∙ Manganate (MnO42−​) 		:	  ⇒ π -bonds are of dπ – pπ  type∙ 	Permanganate (MnO4−​) :	 ⇒ π -bonds are of dπ – pπ  type

Question 4

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :

Accepted Answer

Co3+=[Ar]3 d6 ,Unpaired e−(n)=4 Spin magnetic moment =4(4+2)​=24​ BMCr3+=[Ar]3 d3 ,Unpaired e−(n)=3 Spin magnetic moment =3(3+2)​=15​  BMFe3+=[Ar]3 d5 , Unpaired e−(n)=5 Spin magnetic moment =5(5+2)​=35​ BMNi2+=[Ar]3 d8 , Unpaired e−(n)=2 Spin magnetic moment =2(2+2)​=8​  BM

Question 5

The electronic configurations of Eu (Atomic no. 63), Gd (Atomic NO. 64) and Tb (Atomic No. 65) are:

Accepted Answer

Eu − [Xe]4f7,6s2Gd − [Xe]4f7,5d1,6s2 T6​  − [Xe]4f9,6s2

Question 6

Magnetic moment 2.83BM  is given by which of the following ions?(At. nos. Ti=22,Cr=24,Mn=25,Ni=28  ):

Accepted Answer

Ni+2  has two unpaired electron.

Question 7

Reason of lanthanoid contraction is :

Accepted Answer

Due to poor shielding of f-orbitals nucleus will exert a strong attraction causes lanthanoid contraction.

Question 8

Which of the following does not give oxygen on heating?

Accepted Answer

(NH4​)2​Cr2​O7​⟶Δ​N2​+Cr2​O3​+2H2​O 2KClO3​⟶Δ​2KCl+3O2​ Zn(ClO3​)2​⟶Δ​ZnCl2​+3O2​  K2​Cr2​O7​⟶Δ​2 K2​CrO4​+Cr2​O3​+23​O2​

Question 9

Which of the following lanthanoid ions is diamagnetic?(At nos.Ce=58,Sm=62,Eu=63,Yb=70  )

Accepted Answer

70Yb=[Xe]4f145 d06 s2 Yb+2=[Xe]4f4 n=0∴  Diamagnetic

Question 10

KMnO4​  can be prepared from K2​MnO4​  as per the reaction:3MnO42−​+2H2​O⇌2MnO4−​+MnO2​+4OH− The reaction can go to completion by removing OH- ins by adding

Accepted Answer

MnO4−​  oxidizes HCl  and SO2​  to Cl2​  and SO3​  respectively. MnO4−​  does not oxidizes H2​CO3​ (maximum oxidation state of C )

Question 11

Which of the following statements about the interstitial compounds is incorrect?

Accepted Answer

Interstitial compound are chemically inert.

Question 12

The reason for greater range of oxidation states in actinoids is attributed to:

Accepted Answer

It is a fact.

Question 13

Gadolinium belongs of 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?

Accepted Answer

64​ GD = 		54​ [Xe]6s24f75d1

Question 14

Write down the electronic configuration of:(i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+ (viii) Th4+

Accepted Answer

The electronic configurations of various ions are as shown:Cr3+: [Ar]3d3Pm3+: [Xe]4f4Cu+: [Ar]3d10Ce4+: [Xe]Co2+: [Ar]3d7Lu2+: [Xe]4f145d1Mn2+: [Ar]3d5Th4+:[Rn]

Question 15

Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?

Accepted Answer

Electronic configuration of Mn2+ is 3 d5 . This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is'very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6 . It can lose one electron easily to achieve a stable configuration 3d5 .

Question 16

Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number.

Accepted Answer

+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number as 3d orbitals acquire only one electron in each of five 3d orbitals (remain half-filled) and the electronic repulsion is least and the nuclear charge increases. In the second half of the first row transition series, electrons pair up in 3d orbitals. This increases the electronic repulsion.

Question 17

To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Accepted Answer

When d orbitals are half-filled or completely filled, the oxidation state is more stable. Thus, Mn (Z = 25) has electronic configuration [Ar]3d54s2.  +2 oxidation state is most stable as it has stable electronic configuration (half filled 3d orbital). Also, Zn (Z = 30) has electronic configuration [Ar]3d104s2. +2 oxidation state is most stable as it has stable electronic configuration (completely filled 3d orbital).

Question 18

What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d3,3d5,3d8 and 3d4?

Accepted Answer

Up to Mn, the maximum stable oxidation state corresponds to sum of s and d electrons. After Mn, there is decrease in the stability of higher oxidation states. The stable oxidation states of the transition elements with the following d electron configurations in the ground state of their atoms are as shown. 3d34s2: +5 3d54s1:  +1,+63d54s2:  +2,+73d84s2:  +23d44s2: +1,+6

Question 19

Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Accepted Answer

MnO4−​: Mn has +7 oxidation number and group number is 7. CrO42−​: Cr has +6 oxidation state and group number is 6.VO3−​: V has +5 oxidation state and the group number is 5.

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Solution for Class 12, Chemistry, The d-and f-Block Elements

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