So the question ask to calculate the over all equation for a certain element and the following are
- over all equation - FE(OH)3+3NH4CI
- over all ionic equation - e(OH)3 (s) & 3 NH4+ (aq) & 3Cl- (aq) - net ionic equation - Fe+3+3OH-+3NH$++3CI-
Answer:
These properties are basically the inverse of each other.
Explanation:
- Electronegativity is the tendency of an atom to attract an electron and make it a part of its orbital.
Ionization enthalpy, is the energy required to remove an electron from an atom.
- More electronegative atoms have high ionization enthalpies If the energy required to remove an electron is less, i.e. the atom has more tendency to give electron, it would thus have less tendency to take electron.
- Values and tendency of electronegativity in the periodic table: In general, the electronegativity of a non‐metal is larger than that of metal. For the elements of one period the electronegativities increase from left to right across the periodic table. For the elements of one main group the electronegativities decrease from top to bottom across the periodic table. To the subgroup elements, there’s no regular rule.
- Values and tendency of ionization potential in the periodic table: The first ionization energy is the energy which is required when a gaseous atom/ion loses an electron to form a gaseous +1 valence ion. The energy which is required for a gaseous +1 valence ion to loose an electron to form a gaseous +2 valence ion, is called the second ionization energy of an element. In general, the second ionization energy is higher than the first ionization energy of an element.
The first ionization energies of the elements of one period increase from the left to the right across the periodic table. According to the elements of main group, the first ionization energies generally decreases from top to bottom across the periodic table.
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.
