Answer:
they will form a compound
Answer:
1. hydrogen - H
2. helium - He
3. sodium - Na
4. magnesium - Mg
5. potassium - K
Explanation:
Hydrogen is the element of group 1 and first period. The atomic number of hydrogen is 1 and the symbol of the element is H.
The electronic configuration of the element hydrogen is:-
Helium is the element of group 18 and first period. The atomic number of helium is 2 and the symbol of the element is He.
The electronic configuration of the element helium is:-
Sodium is the element of group 1 and third period. The atomic number of sodium is 11 and the symbol of the element is Na.
The electronic configuration of the element sodium is:-
Magnesium is the element of group 2 and third period. The atomic number of magnesium is 12 and the symbol of the element is Mg.
The electronic configuration of the element magnesium is:-
Potassium is the element of group 1 and forth period. The atomic number of potassium is 19 and the symbol of the element is K.
The electronic configuration of the element potassium is:-
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
Step (1):
Generation of electrophile: by the action of Lewis acid FeCl₃ on Cl₂ to serve as a source of Cl⁺ (Electrophile)
Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again
What they have in common is that they both have the same number of atoms.