Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
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Answer: An electron having a quantum number of one is closer to the nucleus
Explanation:
The Bohr model relies on electrostatic attraction between the nucleus and orbital electron. Hence, the closer an electron is to the nucleus the more closely it is held by the nucleus and the lesser its energy (the more stable the electron is and the more difficult it is to ionize it). The farther an electron is from the nucleus ( in higher shells or energy levels), the less the electrostatic attraction of such electron to the nucleus due to shielding effect. Hence it is less tightly held.
are present in 
<u>Explanation:</u>
It is known that each mole of an element is composed of avagadro's number of molecules. So if we need to determine, we need to divide the number of molecules with the avagadro's number.
So,
As here 
molecules of carbon di oxide is given. So the moles in it will be
No. of moles of carbon dioxide = 
No. of moles = 
moles of carbon dioxide.
Thus,
of carbon dioxide are present in .
