Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
Answer:
B. an inner layer is needed because phospholipids always form bilayers.
Explanation:
Lipids are organic compounds which are not soluble in water but they are soluble in organic compounds. They are formed in body by combination of various molecules. The inner layer of phospholipids always form bilayers.
The atomic number of Fluorine is 9
Valence (outer) electron configuration is : 2s²2p⁵
Therefore, it requires 1 electron in the p-orbital to complete its octet of 8 electrons.
Thus, the atom Fluorine generally will become <u>more </u>stable through the formation of an ionic chemical compound by accepting <u>1 </u> electron from another atom. This process will fill its outer energy level.
Ans: A) more, 1
Answer: it is soluble
Explanation: nitrates are soluble.
Answer:
c = λ x ν
v =c/λ=(3*10e8)/(5.00 x 10e-5)=6000000000000
f = v/λ =6000000000000/5.00 x 10e-5=120000000
Explanation:
