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Why Sulphuric acid is more acidic than selenic acid?

Acidic strength of oxyacids H2SO4 and H2SeO4. Sulphur and Selenium belong to group sixteen in the periodic table. Sulphur is more electronegative the selenium and can polarize and weaken the O-H bond to a greater extent than selenium in their oxyacids, making sulphuric acid a stronger acid.


The Slater's Rules: (Estimating the extent of shielding):

The quantitative prediction of specific energy level is not possible if an atom have more than one electron (multi-electrons system). In multi-electron system each electron acts as a shield for electron further away from the nucleus, reducing the attraction between the nucleus and distant electron. The Slater provides a set of rules which help to predict the extent of shielding.

In the Slater rule determine the actual charge felt by an electron and also allow you to estimate the effective nuclear charge Zeff  from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell"

“Slater Defined Z* or Zeff (effective nuclear charge) as a measure of nuclear attraction for an electron.

Thus Z* =Z-S Where Z is the nuclear charge and S is the shielding constant.

Step 1: Write the electron configuration of the atom in the following form:

              (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . .


Step (2):  Electron in higher group do not shield the electrons in the lower group.

Step (3): For ns and np valence electrons:

(A) The electrons in same ns and np group contribute 0.35, except the 1s which shield 0.30

(B) Electrons within the n-1 group contribute 0.85

(C) Electrons within the n-2 or lower groups contribute (shield) 1.00

Step (4): For nd and nf valence electrons:

(A) Electrons in same nd and nf group contribute 0.35

(B) Electrons in the group to the left contribute 1.00

Shielding constant (S) obtained is subtracted from Z to get Z*

 Solved Questions:

(1) What is the shielding constant (S) experienced by a 3d electron in the bromine atom?

What is the shielding constant (S) experienced by a 3d electron in the bromine atom?

Step (1): Write the electronic configuration of Bromine in the appropriate form.

Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

Br: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p5)

Step (2): Use the Slater Rule to calculate the shielding constant for the electron. And ignore the group of electron to the right of the 3d electrons, these electrons do not contribute to the shielding constant.

(S)3d = (d electrons-1)x 0.35+ (remaining electrons)x1.00= 21.15

(S)3d = (10-1)x 0.35+ (18)x1.00= 21.15

Solved Questions:

(2) A given compound AB whose electronegative difference is 1.9 . Atomic radius of A and B are 4 and 2 Angstroms the distance between A and B means dA-B is ?

(3) Stevenson & Schromaker Equation: Determination covalent radius of Heterogeneous Molecules.

Related Questions:

(1) What are the Amphoteric metals ? gives Examples.

(2) Name of total metalloids present in periodic table ?

(3) Total numbers of elements which are liquid at normal temperature is ?

(4) What is Mendeleev's periodic table ? give important features and draw back of Mendeleev's table.

(5) What is atomic density ? give the periodicity of atomic density in periods and groups.

(6) What is atomic volume ? and what is periodicity of atomic volume in groups and periods ?

(7) Why there are 2, 8 and 8 elements in first, second and third periodic of periods table respectively ? Explain.

(8) In alkali metal group which is the strongest reducing agent in aqueous solution and why?

(9) The electron affinity of sulphur is greater than oxygen. Why?

(10) The first ionization energy of carbon atom is greater than that of boron atom, whereas reverse is true for the second ionization energy. Explain.

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