Covalent radius of 82 Pb is 1 . 53 A o . Calculate its electronegativity at the Alfred-Rochow scale:

Alfred-Rochow scale of electronegativity measures the electrostatic force of attraction, acting between the nucleus and the valence electrons of an atom. The most electronegative atom is fluorine. The Alfred-Rochow scale of electronegativity ( χ ) for an element is given by the formula: χ AR = 0 . 744 + 0 . 359 × Z eff r 2 . where, Z eff is the effective nuclear charge acting on the valence electrons; r is the covalent radius of the atom. The formula for the effective nuclear charge is given as: Z eff = Atomic Number ( Z ) - Shielding Factor ( σ ) . If n is taken as the ultimate shell, and ( n - 1 ) is taken as the penultimate shell, then the rules for calculating the effective nuclear charge according to Slater's rule, go as follows: The shielding factor of ns and np electrons is 0 . 35 each. The shielding factor of ( n - 1 ) s and ( n - 1 ) p electrons is 0 . 85 each. The shielding factor of lower shell s and p electrons is 1 . 00 each. The shielding factor of nd and nf electrons is 0 . 35 each. The shielding factor of all the lower shell electrons is 1 . 00 each. The electronic configuration of Pb 82 is: 1 s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 3 d 10 , 4 s 2 4 p 6 4 d 10 4 f 14 , 5 s 2 5 p 6 5 d 10 , 6 s 2 6 p 2 . Total number of s and p electrons in the 6 th ( nth ) shell = 4 . ⇒ Screening factor = 4 × 0 . 35 = 1 . 4 . Total number of s and p electrons in the 5 th [ ( n - 1 ) th ] shell = 8 . ⇒ Screening factor = 8 × 0 . 85 = 6 . 8 . Total number of lower shell s and p electrons = 26 . ⇒ Screening factor = 1 . 00 × 26 = 26 . Total number of d and f electrons in the 6 th ( nth ) shell = 0 . Total number of d and f electrons in the lower shells = 44 . ⇒ Screening factor = 1 . 00 × 44 = 44 . ∴ Total screening effect of all the electrons in an atom of Pb = 1 . 4 + 6 . 8 + 26 + 44 = 78 . 2 . Z eff = 82 - 78 . 2 = 3 . 8 . The value of r is given as 1 . 53 A o . Putting the values in the Alfred-Rochow scale, we get: ∴ χ Pb AR = 0 . 744 + 0 . 359 × 3 . 8 ( 1 . 53 ) 2 = 1 . 32 . When the difference in the electronegativity between the two atoms increases, then the polarity of the bond also increases.

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Important Questions on Periodic Classification of Elements and General Inorganic Chemistry

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 22

Considering the elements

F, Cl, O

and

S

, the correct order of their electron affinity value is:

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 23

The amount of energy released when a neutral isolated gaseous atom accepts an electron to form gaseous anion is called electron
affinity.

$O_{(g)} + e^{⊖} \overset{Exothermic}{⟶} O_{(g)}^{⊖}$ .... (i)

$O_{(g)}^{⊖} + e^{⊖} \overset{Endothermic}{⟶} O_{(g)}^{2 -}$ .... (ii)

In (2) the energy has to be supplied for addition of second electron due to repulsion between an anion and extra electron. The members of third period have higher electron affinity value than members of second period.

Which process involves the maximum release of energy?

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 1

Calculate the electronegativity of carbon at Pauling's scale?

Given:

$E_{H - H} = 104.2 kcal {mol}^{- 1}$

$E_{C - C} = 83.1 kcal {mol}^{- 1}$

$E_{C - H} = 98.8 kcal {mol}^{- 1}$

Electronegativity of hydrogen $= 2.1$

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 2

Electronegativity of

F

on Pauling scale is

4.0 .

Calculate its value on Mulliken scale.

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 3

The ionisation potential and electron affinity of fluorine are

17.42 eV

and

3.45 eV

respectively. Calculate the electronegativity of fluorine on the Mulliken scale and the Pauling scale:

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 4

Covalent radius of

_{82} Pb

1.53 \overset{o}{A}

. Calculate its electronegativity at the Alfred-Rochow scale:

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 5

$Be$ and $Mg$ have zero value of electron affinity, because:

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Chapterwise/Topicwise Daily Practice Problems (DPP) Inorganic Chemistry Part - 1 JEE Main & Advanced>Chapter 1 - Periodic Classification of Elements and General Inorganic Chemistry>DPP>Q 6

Among the following oxoacids, the correct decreasing order of the acidic strength is:

The EA values of elements depend on which of the following?

Important Questions on Periodic Classification of Elements and General Inorganic Chemistry

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