Answer
Verified
39k+ views
Hint: Slater's rule is used to calculate shielding constant.
Formula used- $ = (0.35 \times n) + (0.85 \times m) + (1.00 \times p)$ where n is number of electrons in n shell, m is number of electrons in n-1 shell, p is number of electrons in the remaining inner shells.
Complete step by step answer:
Write the electronic configuration of Zinc
Atomic number of Zinc is 30
Hence its electronic configuration is$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},3{d^{10}},4{s^2}$
Find the number of electrons in n shell.
In the given case, n=4 and the number of electrons is 2. But we will take only 1 electron.
Find number of electrons in n-1 shell.
There are 18 electrons in the third shell.
Find the number of electrons in the remaining inner shells
The remaining inner shells include the first and second shell. There are 10 electrons in the first and second shell.
Calculate the shielding constant using slater’s rule
Shielding effect =$(0.35 \times 1) + (0.85 \times 18) + (1.00 \times 10) = 25.65$
Option A is correct.
Additional information:
When we move from left to right across a period, in periodic table, the number of electrons increases and hence the strength of shielding increases. Electrons in different orbitals have different tendency to shield. Electrons in s and p orbitals are good shielders whereas electrons in the d and f orbitals are poor shielders. Effective nuclear charge on a particular electron can also be calculated by slater's rule. Effective nuclear charge is the pull of the nucleus on electron or measure of the attraction of an electron towards nucleus. Effective nuclear charge can be calculated by subtracting shielding/ screening constant from atomic number.
Note:
In the atom, each electron is said to experience less than the actual nuclear charge because of shielding or screening constant. It is always caused by the electrons intervening between the nucleus and the valence electron.
Formula used- $ = (0.35 \times n) + (0.85 \times m) + (1.00 \times p)$ where n is number of electrons in n shell, m is number of electrons in n-1 shell, p is number of electrons in the remaining inner shells.
Complete step by step answer:
Write the electronic configuration of Zinc
Atomic number of Zinc is 30
Hence its electronic configuration is$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},3{d^{10}},4{s^2}$
Find the number of electrons in n shell.
In the given case, n=4 and the number of electrons is 2. But we will take only 1 electron.
Find number of electrons in n-1 shell.
There are 18 electrons in the third shell.
Find the number of electrons in the remaining inner shells
The remaining inner shells include the first and second shell. There are 10 electrons in the first and second shell.
Calculate the shielding constant using slater’s rule
Shielding effect =$(0.35 \times 1) + (0.85 \times 18) + (1.00 \times 10) = 25.65$
Option A is correct.
Additional information:
When we move from left to right across a period, in periodic table, the number of electrons increases and hence the strength of shielding increases. Electrons in different orbitals have different tendency to shield. Electrons in s and p orbitals are good shielders whereas electrons in the d and f orbitals are poor shielders. Effective nuclear charge on a particular electron can also be calculated by slater's rule. Effective nuclear charge is the pull of the nucleus on electron or measure of the attraction of an electron towards nucleus. Effective nuclear charge can be calculated by subtracting shielding/ screening constant from atomic number.
Note:
In the atom, each electron is said to experience less than the actual nuclear charge because of shielding or screening constant. It is always caused by the electrons intervening between the nucleus and the valence electron.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
Other Pages
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
Iodoform can be prepared from all except A Acetaldehyde class 12 chemistry JEE_Main
Let the refractive index of a denser medium with respect class 12 physics JEE_Main
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main