RBSE Solutions for Class 12 Chemistry Chapter 8 d and f Block Elements

Rajasthan Board RBSE Class 12 Chemistry Chapter 8 d and f Block Elements

RBSE Class 12 Chemistry Chapter 8 Text Book Type Questions

RBSE Class 12 Chemistry Chapter 8 Multiple Choice Questions

Question 1.
Which of the following shows highest oxidation state of +7 ?
(a) Co
(b) Cr
(c) Mn
(d) v

Question 2.
The number of unpaired electrons in Fe²⁺ are:
(a) 4
(b) 5
(c) 3
(d) 6

Question 3.
In which of the following compound the oxidation state of Fe is zero ?
(a) FeSO₄
(b) [Fe(CO)₅]
(c) K₄[Fe(CN)₆]
(d) FeCl₃

Question 4.
Which of the following has maximum magnetic moment ?
(a) V³⁺
(b) Cr³⁺
(c) Fe³⁺
(d) C³⁺

Question 5.
What is the common oxidation state of lanthanoid series?
(a) + 1
(b) + 4
(c) + 2
(d) + 3

Question 6.
Lanthanoid contraction is due to increase in:
(a) effective nuclear charge
(b) atomic number
(c) size of 4 f – orbital
(d) none of these

Question 7.
One of the member of lanthanoid series shows +4 oxidation state is :
(a) Ce
(b) Lu
(c) Eu
(d) Pm

Question 8.
Which of the following is diamagnetic ?
(a) Cu²⁺
(b) Zn²⁺
(c) Cr²⁺
(d) Ti²⁺

Question 9.
Which of the following has maximum first ionisation enthalpy ?
(a) Ti
(b) Mn
(c) Fe
(d) Ni

Question 10.
In which of the following ions, all the electrons are paired ?
(a) Cr²⁺
(b) Cu²⁺
(c) Cu⁺
(d) Ni²⁺
Answers:
1. (c) 2. (a) 3. (b) 4. (C)5. (d) 6. (e)7. (a) 8. (b) 9. (c) 10. (c)

RBSE Class 12 Chemistry Chapter 8 Very Short Answer Type Questions

Question 1.
Zn is not considerd as transition element. Explain why ?
Answer:
Zinc (Z=30) has the configuration 3d¹⁰4s². It does not have partially filled d-subshell in its elementry form or in commonly occuring oxidation state (Zn²⁺ = 3d¹⁰). Therefore, it is not regarded as transition element.

Question 2.
Ti⁴⁺ ion is colourless. Give reason.
Answer:
Ti⁴⁺ ion colourless :
Ti⁴⁺ (3s²3p⁶) does not contain Linpaired electrons in d – subshell.

Question 3.
What are transuranic elements ?
Answer:
Transuranic elements – After Uranium (Z = 92), all elements are artificial and unstable. These are not found in nature. Therefore these are known as transuranic elements or super heavy elements.

Question 4.
Why an element shows its highest oxidation state in oxides and fluorides ?
Answer:
Because oxygen and fluorine have small size and high electro negativity, therefore element shows its highest oxidation state in oxides and fluorides.

Question 5.
Arrange the following in decreasing order of acidity :
MnO, Mn₂O₃, MnO₂.
Answer:
MnO₃ > Mn₂O₃ > MnO

Question 6.
Write the general electronic configuration of : inner transition elements.
Answer:
(n − 2) f^0-14 (n-1) d^0,1ns²

Question 7.
Transition elements show variable oxidation states. Why ?
Answer:
The variable oxidation states of transition elemetns are due to participation of inner (n − 1)d and outer ns electrons.

Question 8.
All the compounds are Sc are colourless. Give reason.
Answer:
Sc has generally + 3 oxidation state so there is no unpaired electron, therefore, all the compounds of Sc are colourless.

Question 9.
Write the number of upaired electrons in Gd (Z = 64).
Answer:
Gd (Z = 64) : [Xe]4f⁷5d¹6s². So number of unpaired electrons = 8

Question 10.
The magnetic moment of a compound of transition element is 3.9 B.M. Write the number of upaired electrons in the element.
Answer:
Magnetic moment of any compound is given by
µ = \(\sqrt { n(n+2) }\)
Where n is the no. of unpaired electrons.
So,
3.9 = \(\sqrt { n(n+2) }\)
Squaring both side
15.21 = n² + 2n … (1)
This equation (1) satisfying for n = 3. So no. of unpaired electrons in this elements will be three.

RBSE Class 12 Chemistry Chapter 8 Short Answer Type Questions

Question 1.
What is lanthanoid contraction ? Explain.
Answer:
The regular decrease in the size of lanthanoid ions from La³⁺ to Lu³⁺ is known as lanthanoid contraction. This regular decrease in size due to the greater effect of the increased nuclear charge as compared to screening effect.

Question 2.
What are alloys ? Give one use.
Answer:
An alloy is homogeneous mixture of two or more metals and non-metals. An important alloy of lanthanoid metals is misch metal. It contains 95% lanthanoid metal and about 5% iron and traces of S, C, Ca or Al. It is used in magnesium based alloy to produce bullets. shells, lighter flints etc.

Question 3.
Write electronic configuration of Cu²⁺ Calculate its magnetic moment.
Answer:
Electronic configuration of Cu²⁺ will be [Ar]3d⁹ so
magnetic moment is given by.
µ = \(\sqrt { n(n+2) }\)
Where n=1
(No. of unpaired electron)
µ = \(\sqrt { 1(1+2) }\) = √3 = 1.73 B.M.

Question 4.
Transition elements generally form coloured compounds. Why?
Answer:
Due to the presence of unpaired electrons, transition elements generally form coloured compounds.

Question 5.
Give reasons :
(a) Mn shows highest oxidation state in 3d series of transition elements.
(b) The configuraton of both Cr²⁺ and Mn³⁺ is d<sup.4 but Cr²⁺ is reducing agent and Mn³⁺ is oxidising agent.
Answer:
(i) Manganses (Z = 25) shows maximum number of oxidation states because its electronic configuration is 3d⁵4s². As 3d and 4s are close in energy, it has maximum number of electrons to lose or share (as all the 3d electrons are unpaired). Hence it shows hightest oxidation state from +2 to +7.

(ii) E^o value for Cr³⁺ / Cr²⁺ is negative (-0.41. V) whereas E^o value for Mn³⁺ / Mn²⁺ is positive (+ 1.57V). Hence Cr²⁺ ions can easily undergo oxidation to give Cr³⁺ ions and therefore, acts as strong reducing agent. On the other hand Mn³⁺ can easily undergo reduction to give Mn²⁺ and hence acts as oxidising agent.

Question 6.
Explain the following:
(a) The size of 5d transition elements is almost similar to 4d transition elements.
(b) Transition elements forms coordinate bonds
Answer:
(a) The size of one group of 5d and 4d transition series is almost same. In group 3 elements, there is regular increase in size from Sc to Y and from Y to La. But after this the size of one group in 4d to 5d series is almost same because lanthanoids come in between them: and size decreases due to lanthanoid contraction.

(b) Transition elements have vacant d – orbitals which can accommodate the ligands. Hence they forms coordinate bonds. The tendency to form coordinate compounds increases with decrease in atomic size and increase in positive charge, in general.

Question 7.
Write four differences between lanthanoids and actinoids.
Answer:
Differences between lanthanoids of actinoids Property Lanthanoids Actinoids

Property	Lanthanoids	Actinoids
1. Bonding energies	Bonding energies of 4/ are higher.	Bonding energies of 5/ are lower.
2. Radioactivity	Except promethium, these are non-radioactive.	All actinoids are radioactive.
3. Basic character	Lanthanoid compounds are less basic.	Actinoid compounds are more basic
4. Tendency to form oxo ions	They do not form oxo-ions.	They form oxo ions such as WO⁺₂,NpO⁺₂, PuO⁺₂.

Question 8.
The atomic radius of Zr (57) and Hf (72) is almost same. Give reason.
Answer:
Due to lanthanoid contraction. Hf (72) and Zr (57) have almost same radius.

Question 9.
The ionisation potential of Au (79) and Ag (47) are almost same. Explain.
Answer:
Size of Au (79) and Ag (47) is same but not have same ionisation enthalpy becasue Zeff of Au is more due to poor screening effect of f-orbitals so ionisation enthalpy of Au (79) is more than Ag (47).