Faq Transition Elements JEE

FAQ Transition Elements JEE

1. Copper becomes green when exposed to moist air for a long period. This is due to:-

(1) the formation of a layer of cupric oxide on the surface of copper.

(2) the formation of a basic copper sulphate layer on the surface of the metal

(3) the formation of a layer of cupric hydroxide on the surface of copper.

(4) the formation of a layer of basic carbonate of copper on the surface of copper.

Solution:

copper becomes green when exposed to moist air for a long period of time because of the formation of a layer of basic carbonate of copper on the surface of copper.

2Cu + H₂O + CO₂ + O₂ → Cu(OH)₂ + CuCO₃

Therefore, option (4) is the correct answer.

2. Which one of the following exhibits the largest number of oxidation states?

(1) Mn(25)

(2) V(23)

(3) Cr (24)

(4) Ti (22)

Solution:

Manganese can show 6 oxidation states from +2 to +7. It has 5 unpaired electrons in 3d orbitals and 2 electrons in the 4s orbitals.

Therefore, option (1) is the correct answer.

3. The type of isomerism present in nitro- pentamidine chromium (III) chloride is

(1) optical

(2) linkage

(3) ionization

(4) polymerisation

Solution:

The nitro group can attach to metal through nitrogen as (-N0₂ ) or through oxygen as nitrito (-ONO). So isomerism in nitro-pentamine chromium (III) chloride is linkage isomerism.

Therefore, option (2) is the correct answer.

4. Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect?

(1) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.

(2) Ferrous oxide is more basic in nature than ferric oxide.

(3) Ferrous compounds are relatively more ionic than the corresponding ferric compounds.

(4) Ferrous compounds are less volatile than the corresponding ferric compounds.

Solution:

Ferrous oxide is more basic in nature than ferric oxide. Ferrous compounds are less volatile since ferrous compounds are more ionic. Ferrous compounds are relatively more ionic than the corresponding ferric compounds. Ferric compounds are less volatile than the corresponding ferrous compounds. So statement (1) is incorrect.

Therefore, option (1) is the correct answer.

5. Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown-red vapours of:

(1) CrO₃

(2) Cr₂O₃

(3) CrCl₃

(4) CrO₂Cl₂

Solution:

K₂Cr₂O₇ + 6H₂SO₄ + 4NaCl → 2KHSO₄ + 4NaHSO₄ + 2CrO₂Cl₂ + 3H₂O

Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown-red vapours of CrO₂Cl₂

Therefore, option (4) is the correct answer.

6. The actinoids exhibit more oxidation states in general than the lanthanoids. This is because

(1) the 5f orbitals extend further from the nucleus than the 4f orbitals

(2) the 5f orbitals are more buried than the 4f orbitals

(3) there is a similarity between 4f and 5f orbitals in their angular part of the wave function

(4) the actinoids are more reactive than the lanthanoids.

Solution:

Since the distance between the nucleus and 5f orbitals is more than the distance between the nucleus and 4f orbitals, the hold of the nucleus on valence electron decreases in actinides. So actinoids exhibit more oxidation states in general.

Therefore, option (1) is the correct answer.

7. Which of the following is not formed when H₂S reacts with acidic K₂Cr₂O₇ solution?

(1) K₂SO₄

(2) Cr₂(SO₄)₃

(3) S

(4) CrSO₄

Solution:

When H₂S reacts with acidic K₂Cr₂O₇ solution

3H₂S + K₂Cr₂O₇ + 4H₂SO₄ → 3S + Cr₂(SO₄)₃ + K₂SO₃ + 7H₂O

Therefore, option (4) is the correct answer.

8. The lanthanide contraction is responsible for the fact that

(1) Zr and Y have about the same radius

(2) Zr and Nb have a similar oxidation state

(3) Zr and Hf have about the same radius

(4) Zr and Zn have the same oxidation state.

Solution:

The pairs of elements such as Zr-Hf, Mo-W, Nb-Ta, etc possess almost the same properties.

Therefore, option (3) is the correct answer.

9. Which of the following statements is false?

(1) has a Cr – O – Cr bond

(2) is tetrahedral in shape

(3) Na₂Cr₂O₇ is a primary standard in volumetry

(4) Na₂Cr₂O₇ is less soluble than K₂Cr₂O₇

Solution:

Na₂Cr₂O₇ is a secondary standard in volumetry. Na₂Cr₂O₇ is more soluble than K₂Cr₂O₇

Therefore, option (3) and (4) is the correct answer.

10. In context with the transition elements, which of the following statements is incorrect?

(1) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s

and 3d electrons are used for bonding.

(2) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in

bonding decreases.

(3) In addition to the normal oxidation states, the zero oxidation state is also shown by these

elements in complexes.

(4) In the highest oxidation states, the transition metal show basic character and form cationic

complexes.

Solution:

In the highest oxidation states, transition metals form anionic complexes. So statement (4) is incorrect.

Therefore option (4) is the correct answer.

FAQs

Q: What are a portion of the significant points to cover on the Transition components?

Ans: Change components are essentially the components having a place with the d and f gathering of the advanced occasional table. Assuming we go through the earlier year JEE question papers we observe pretty much consistently something like a few inquiries are posed from this section. The ideas relating to change components are more significant as they establish the groundwork for additional points that are presented in coordination compounds. Primary ideas engaged with this part are the colouration of progress components, numerous oxidation states, patterns in their ionization potential, ionic radii both down the gathering and across the period, attractive properties and so forth As indicated by JEE prospectus for science, the responses of potassium permanganate and potassium dichromate are vital and furthermore their readiness strategies.

Q: Which reference books would one be able to follow for progress components”?

Ans: While covering “progress components” for JEE science, one should go through the NCERT class twelfth science course book section 1. An energetic stroll through this part from NCERT provides you with a sound thought of the different points you want to cover for JEE and furthermore fosters the basics in regards to the section. Aside from this, you can pursue a reference book to see a few directions in the properties not clarified exhaustively in the NCERT course reading. A few real books for this point incorporate books from writers like J.D. Lee, and so forth You can likewise allude books by O.P Tandon for rehearsing inquiries on this theme.

Q: What are the overall properties of the d-block components?

Ans: The properties of the d-block components are as per the following;

• Every one of the components in the third series, aside from copper, are viewed as great diminishing specialists.
• The ionization capability of the change components are said to increment by insignificant sums because of the protecting impact.
• The progress components have less reactivity. This reactivity diminishes because of the outcome in the nuclear number. also, high enthalpies of atomization.
• Because of the safeguarding of the valence shell electrons, given by the electrons present in the d orbitals of the penultimate shell, the nuclear radii of the change components go through a reduction by insignificant sums.