Calculate the number of unpaired electrons in Cr 3 + and V 3 + ions.

An unpaired electron is an electron that occupies an orbital of an atom singly, rather than as part of an electron pair. Each atomic orbital of an atom has a capacity to contain two electrons with opposite spins. The electronic configuration of various ions are given below: Cr 3 + : 3 d 3 , Number of unpaired electrons is 3 . V 3 + : 3 d 2 , Number of unpaired electrons is 2 .

Zn 2 + salts are colourless while Cu 2 + salts are coloured. Why ?

The colour of ions is due to presence of unpaired electrons, which on excitation and de-excitation phenomenon, produce colour. In Zn 2 + 3d orbitals are completely filled, and these salts appear white. Whereas in Cu 2 + 3d has 9 electrons that means one electron is unpaired. So it appears coloured.

What are alloys? Name the metals used for the formation of bronze.

Alloys are the solid mixture of two or more metals. They are made by first melting the constituent metals and then mixing them thoroughly in fixed proportions. Bronze, alloy traditionally composed of copper and tin.

Cu(I) has d 10 configuration, while Cu II had d 9 configuration. Still, Cu(II) is more stable in an aqueous solution than Cu(I). Why?

Stability depends on the hydration energy(It is the amount of energy released when one mole of ions undergoes hydration) of the ions when they bond to the water molecules. Hence, Cu 2 + ion has a greater charge density than Cu + ion and thus forms much stronger bonds releasing more energy and the extra energy needed for the second ionization of the copper is more than compensated for by the hydration, so much so that the Cu + ion loses an electron to become Cu 2 + which can then release this hydration energy.

Use Hund’s rule to derive the electronic configuration of Ce 3 + ion and calculate its magnetic moment.

According to Hund's rule, the electronic configuration of Ce 3 + ion is Xe 4 f 1 . Here, the number of unpaired electrons is one. The magnetic moment is determined by the number of unpaired electrons and is calculated by using the spin formula. It is given as: µ = n n + 2 = 1 1 + 2 = 3 = 1 . 73 B . M . Where n is the number of unpaired electrons.

In the first transition series of elements, which element shows highest oxidation state?

The first transition series of d block is started from Sc and the last element of the first transition series is Zn . In the transition series, oxidation states of metal increase with the increase in the number of unpaired electrons in d subshell. Since, in first transition series Mn has 5 unpaired electrons in d subshell in addition to 2 electrons in s orbital. Therefore Mn exhibits highest oxidation state + 7 in first transition series. The atomic number of Mn is 25 . Electronic configuration of Mn = Ar 3 d 5 4 s 2

Zn 2 + salts are colourless while Cu 2 + salts are coloured. Why ?

When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed or transition metals and their ions are generally coloured due to the d transition of the unpaired electron. So, in Zn 2 + 3d orbitals are completely filled, and these salts appear white. Whereas in Cu 2 + 3d has 9 electrons. So it appears in blue colour.

Give two uses of potassium permanganate.

The uses of Potassium permanganate KMnO 4 are the following: (i) It is used as an oxidising agent in the laboratory for the volumetric analysis. (ii) It is used for the bleaching of wool, cotton, silk, and other textile fibers.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - The d- and f- Block Elements>Assam Board-2020>Q 15

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12th Assam Board

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What is lanthanide contraction? Why actinide contraction is greater from element to element that lanthanide contraction ?

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