Answer:
Option (d) is correct
N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺
Explanation:
Total electrons for all the species = 10
So these are <u>iso electronic</u> with each other.
We know
Ionic radii ∝
- Si⁴⁺ has 14 protons and 10 electrons
- Mg²⁺ has 12 protons and 10 electrons
- N³⁻ has 7 protons and 10 electrons
- F⁻ has 9 protons and 10 electrons
- Iso electronic species with greatest number of protons have small size and vice versa.
- So Si⁺⁴ have smallest size and N³⁻ have largest in size
Answer:
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
The atomic orbital and radius increases by gaining electron linearly so even electron numbered atoms are diamagnetic while the odd electron numbered atoms are paramagnetic.
Running through the first 18 elements one can observe that there is an alternative odd number of electrons and an even number proofing that that half of the first 18 elements shows paramagnetism and diamagnetism respectively.
Explanation:
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
A: Na3PO4 + MnCl2 > Mn3(PO4)2 + NaCl
<u>Answer:</u> Copper (I) iodide will precipitate first.
<u>Explanation:</u>
We are given:
of CuCl =
of CuI =
Concentration of
Concentration of
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
Putting values in above equation, we get:
Concentration of copper (I) ion =
Putting values in above equation, we get:
Concentration of copper (I) ion =
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.