Which element requires the least amount of
energy to remove the most loosely held electron
from a gaseous atom in the ground state?
<h3>Answer-</h3><h3>Na</h3>
Answer:
Bonding Order = number of bonding electrons – number of antibonding electrons/2.
So for CO2, there is a total of 16 electrons, 8 of which are antibonding electrons.
So 16 – 8 = 8; divided by 2 = 4. So, 4 is the bonding order of CO2. The molecular structure of CO2 looks like this:
..~-~~..
O=C=O
..~-~~..
It should be 24 electrons
Answer: X could represent the element of oxidation state (+2) such as (Mg2+, Pb2+, Ba2+, Ca2+, Ba2+, Zn2+, ....etc)
Explanation:
- The formula of the compound XSO4 is a neutral compound that the algebraic summation of the oxidation states of different elements in it must be zero.
- The group SO4 has the oxidation state (2-), that S has (6+) oxidation state and O has (2-) oxidation state, so the oxidation of SO4 = (6+) + (-2*4) = -2.
- It is clear that X must have the oxidation state 2+.
- So, X could be represents by many different elements such as (Mg2+, Pb2+, Ba2+, Ca2+, Ba2+, Zn2+, Fe2+, ....etc)
Answer:
... its called a inogounsess
sExplanation:
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