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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Explanation: Plz make brainliest
Answer: 1.9 x 10²⁴ molecules Na
Explanation: To solve for the molecules of Na, we will use the Avogadro's number.
3.2 moles Na x 6.022 x10²³ molecules Na/ 1 mole Nà
= 1.9 x 10²⁴ molecules Na
Answer:
Four
Step-by-step explanation:
The <em>superscripts</em> in an electron configuration tell us how many electrons are in a subshell.
If the electron configuration is 1s¹ 2s¹2p², the total number of electrons is
1 + 1 + 2 = 4
The atom contains four electrons.
<em>Note</em>: this atom is in an <em>excited state</em>, because the 1s and 2s subshells can each hold one more electron.
