What are lanthanoids? Give their general electronic configurations. What is lanthanoid contraction?

Lanthanoids are the elements with atomic numbers from 57 to 71 following the element lanthanum. These elements are also called as 4 f -series elements, since the newly added electron enters the antepenultimate 4 f -subshell. The general electronic configuration of lanthanoids is Xe 4 f 1 - 14 5 d 0 - 1 6 s 2 . The atomic size or the ionic radii of tripositive lanthanoid ions decrease steadily from La to Lu due to increasing nuclear charge and electrons entering inner n - 2 f orbital. This gradual decrease in the size with an increasing atomic number is called lanthanoid contraction.

Predict which of the following will be coloured in aqueous solution: Sc 3 + , Fe 3 + , Ti 4 + , V 3 + (Atomic nos. of Sc , Fe , Ti and V are 21 , 26 , 22 and 23 respectively)

The colour of the d-block compounds depends on the presence of unpaired electrons. The electronic configurations of the given ions can be given as: Sc 3 + = Ar Fe 3 + = Ar 3 d 5 Ti 4 + = Ar V 3 + = Ar 3 d 2 Therefore, Fe 3 + and V 3 + have five and two unpaired electrons respectively, and these two will be coloured in aqueous solution.

Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25 .

The element with atomic number 25 is Mn . Its outer configuration can be written as 3 d 5 4 s 2 . The divalent ion, i.e., Mn 2 + will have a configuration of 3 d 5 . Therefore, the number of unpaired electrons = 5 The magnetic moment = n n + 2 BM Where, n = number of unpaired electrons Therefore, magnetic moment = 5 5 + 2 = 5 . 92 BM Note:- BM means Bohr magneton.

(a) Why are Zn 2 + salts white while Cu 2 + salts are blue. (b) What is meant by ‘disproportionation’? Write the disproportionation reaction of Cu + in aqueous solution.

The colour of a salt is the result of the unpaired electron(s) present in the metal ion. The unpaired electron absorbs particular wavelengths in the visible region and transmits the complementary colour. Cu 2 + has an unpaired electron as its electronic configuration is Ar 3 d 9 . Whereas Zn 2 + has no unpaired electrons as its electronic configuration is [ Ar ] 3 d 10 . Therefore, Zn 2 + salts are white while Cu 2 + salts are blue. Disproportionation reactions are those in which one element undergoes oxidation and reduction both simultaneously. This is called disproportionation. The disproportionation reaction of Cu + in aqueous solution is: 2 Cu + ( aq ) → Cu 2 + ( aq ) + Cu ( s ) Here, one Cu + is getting reduced to Cu and the other is getting oxidized to Cu 2 + ion.

Write down the steps involved in the preparation of KMnO 4 from MnO 2 .

Potassium permanganate is manufactured from pyrolusite ( MnO 2 ), a mineral. (i) Pyrolusite is powdered and fused with potassium hydroxide in the presence of air or oxidising agent like potassium nitrate. MnO 2 is converted to potassium manganate. 2 MnO 2 + 4 KOH + O 2 → 2 K 2 MnO 4 + 2 H 2 O (ii) When neutral or acidic solution of potassium manganate is allowed to stand, it undergoes disproportionation to potassium permanganate and manganese dioxide. 3 MnO 4 2 - + 4 H + → 2 MnO 4 - + MnO 2 + 2 H 2 O

Why does light green Fe 2 + solution change to brown on exposure to air?

When a salt solution of ferrous ion ( Fe 2 + ) is exposed to air, oxidation takes place to form a salt of ferric ion ( Fe 3 + ), and it will give the brown colour.

Transition metals and their many compounds act as a good catalyst. Why?

Transition metals and their many compounds act as good catalyst because, these metals show variable oxidation states and forms complexes. They form unstable intermediate compounds. They provide a new path with lower activation energy of reaction. They also provide a suitable surface for the reaction to occur.

You are supplied with a concentrated solution Na 2 CrO 4 . How will you obtain K 2 Cr 2 O 7 from this? Write the equation involved.

The concentrated sodium chromate solution on reaction with an acid, gets converted to sodium dichromate as shown in the equation below: 2 Na 2 CrO 4 + 2 H + → Na 2 Cr 2 O 7 + 2 Na + H 2 O Sodium dichromate when fused with KCl forms orange crystals of potassium dichromate. Na 2 Cr 2 O 7 + 2 KCl → K 2 Cr 2 O 7 + 2 NaCl Potassium dichromate being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.

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Explain why ${Ce}^{4 +}$ is a good oxidizing agent whereas ${Sm}^{+ 2}$ is a good reducing agent.

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