Bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter
nuclear distance between the two atoms when the potential energy of the system reaches its lowest value. given that the atomic radii of h and cl are 25.0 cm and 100. pm , respectively, predict the bond length of the hcl molecule.
2 answers:
Answer:Hence, the bond length in HCl is 125 pm.
Explanation:
Bond length : It is an average distance between the nuclei of two bonded atoms in a molecule.
Also given that bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter-nuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Beyond this if atoms come closer to each other then their will be repulsion between them.
So, the bond length between the Hydrogen and Chlorine atom in HCl molecule is :
Hence, the bond length in HCl is 125 pm.
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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