Answer:
sp³d¹ hybridization
Explanation:
Given Cl as central element with three F substrates ...
The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.
Valence bond theory predicts the following during bonding:
Cl:[Ne]3s²3p²p²p¹3d⁰
=> [Ne]3s²p²p¹p¹d¹
=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹
giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.
Note the following images:
Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.
Answer:
1 has the highest density because it has the most amount of circles in the least amount of space- it is the most densely filled with circles; it is the most dense.
Answer:
= 0.030 M
Explanation:
We can take x to be the concentration in mol/L of Ag2SO4 that dissolves
Therefore; concentration of Ag+ is 2x mol/L and that of SO4^2- x mol/L.
Ksp = 1.4 x 10^-5
Ksp = [Ag+]^2 [SO42-]
= (2x)^2(x)
= 4x^3
Thus;
4x^3 = 1.4 x 10^-5
= 0.015 M
molar solubility = 0.015 M
But;
[Ag+]= 2x
Hence; silver ion concentration is
= 2 x 0.015 M
= 0.030 M