The Ions present in CaCl₂ are,
Ca²⁺ Cl⁻ Cl⁻
Means 1 formula unit contains 1 Ca²⁺ ion and 2 Cl⁻ ions.
Also, 1 mole of CaCl₂ contains 6.022 × 10²³ formula units.
So, 1 mole formula units of CaCl₂ contain,
2 × 6.022 × 10²³ = 1.20 × 10²⁴ Cl⁻ Ions
Now, Calculating number of moles contained by 220 g of CaCl₂,
As,
110.98 g of CaCl₂ = 1 mole
Then,
220 g of CaCl₂ = X moles
Solving for X,
X = (220 g × 1 mol) ÷ 110.98 g
X = 1.98 moles
As,
1 mole contained = 1.20 × 10²⁴ Cl⁻ Ions
Then,
1.98 mole will contain = X Cl⁻ Ions
Solving for X,
X = (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol
X = 2.38 × 10²⁴ Cl⁻ Ions
Answer :
The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.
<u>Percent Ionic Character :</u>
It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :

Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom
Given : Molecule is TiAl₃
Electronegativity of Ti = 2.0
Electronegativity of Al = 1.6 ( From image shared )
Plug the value in above formula :



Value of e⁻¹ = 0.90
Percent ionic character = 1 - 0.90 * 100
Percent Ionic character = 10 %
<u>Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .</u>
<span>08 moles Li3N * 1mole N2/2moles Li3N = 0.04 </span>
Compound X is an isomer of butane with different chain types. It is either straight chain or bend chain.