Which does your digestive system use to break down food molecules? Bases or Acids
1 answer:
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Answer:
Z* = 3.55
Explanation:
Slater rule says that:
Z*= Z - S
Z* be the nuclear effective charge
Z is the nuclear charge
S is the shielding constant
First we write the electronic configuration of platinum:
The first Slater rule says that we need to group:
The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.
For the third rule we have two considerations, if is ns or np and if is nd or nf:
For our case, we have an electro that is in ns, so the rule says that
-electrons within same group shield 0.35, except the 1s which shield 0.30
-electrons within the n-1 group shield 0.85
-electrons within the n-2 or lower groups shield 1.00
Now we can proceed with the calculation:
The first consideration in the third rule does not apply as we only have one electron on this level.
The second consideration will be as follow for the level 5, where we have 17 electrons.
Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.
So the result for S=(60*1.00 + 17*0.85) = 74.45
And the equation is: Z* = 78 - 74.45
So Z* = 3.55
Answer:
Option C
Explanation:
The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number
Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6
Equation 1 represents the change in enthalpy for formation of RbCl
Equation 2 represents the sublimation reaction of rubidium
Equation 3 represents the ionization enthalpy of rubidium
Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule
Equation 5 represents the electron affinity of chlorine
To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6
We have to perform operations such as
Equation 1 - equation 2 - equation 3 - equation 4 - equation 5
By performing these operations the intermediate compounds gets cancelled and at last we get equation 6
So Equation 1 ≡ ΔH = -431 kJ/mol
Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8 kJ/mol
Equation 3 ≡ IE1(Rb) = 397.5 kJ/mol
Equation 4 ≡ BE(Cl2) = 226 kJ/mol
Equation 5 ≡ Electron Affinity Cl = -332 kJ/mol
Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol
∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol
