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.

Explain why Ce 4 + is a good oxidizing agent whereas Sm + 2 is a good reducing agent.

The most stable oxidation state in Lanthanoids is + 3 . Thus Ce 4 + will take up an electron and undergoes reduction and acts as an oxidizing agent. Whereas Sm + 2 will lose an electron and undergoes oxidation and act as a good reducing agent.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - The d- and f- Block Elements>Meghalaya Board-2020>Q 11

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12th Meghalaya Board

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

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