1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Fantom [35]
3 years ago
11

1. Using the Slater rule, determine the effective nuclear charge of platinum.

Chemistry
1 answer:
AleksandrR [38]3 years ago
3 0

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:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

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

You might be interested in
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,
Gennadij [26K]

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore,  mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

4 0
3 years ago
What is the ionic charge for the calcium ion in CaCO3?
Lunna [17]
<span>The ionic charge of Calcium (Ca) in calcium carbonate (CaCO3) is 2+. CaCO3 has a neutral ionic charge sin CO3 has a 2- charge.</span>
3 0
3 years ago
True or false <br><br> The mass of an electron is about equal to the mass of a proton.
pogonyaev
That is false (please mark me brainliest)
7 0
2 years ago
Q Q 3. (08.02 MC)
AnnyKZ [126]

Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

Explanation:

Given: V_{1} = ?,         M_{1} = 0.55 M

V_{2} = 100.0 mL,        M_{2} = 2.50 M

Formula used to calculate the volume of KBr is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.55 M \times V_{1} = 2.50 M \times 100.0 mL\\V_{1} = 455 mL

Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

7 0
3 years ago
Read 2 more answers
How are properties useful in classifying materials
lukranit [14]
S. Properties can be anything from composition to color to ductility. Without knowing an object's properties, it is impossible to characterize it. For example, looking at some common properties of metals helps to differentiate them from non-metals or metalloids.
5 0
3 years ago
Read 2 more answers
Other questions:
  • Write a balanced formation equation at standard conditions for the compound HI.
    14·2 answers
  • A recipe calls for 1.2 cups of oil. How many liters of oil is this?
    5·2 answers
  • Please answer asap for brainliest!!
    10·1 answer
  • Bromine has an atomic number of 35. How many protons are in an atom of bromine? 17 18 35 52
    5·1 answer
  • Which solution has the lowest freezing point?
    15·2 answers
  • The mass of the quartz crystal shown below is 10 grams, and its volume is 4 cm3. What is the density of quartz?
    13·1 answer
  • Which fossil fuel produces the most carbon dioxide?
    10·1 answer
  • Why do most atoms form chemical bonds?
    10·1 answer
  • Soda pop is carbonated with CO2. Mark puts one bottle of soda pop in the refrigerator and leaves the other out in the hot sunlig
    7·1 answer
  • Can someone explain to me what numbers are acidic, base, or neutral on a ph scale?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!