Answer: C)Anion, it would gain 2 electrons to satisfy the octet rule.
Explanation:
Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.
The electrons are filled according to Afbau's rule in order of increasing energies and thus the electronic configuration of oxygen with 8 electrons is
The cation is formed by loss of electrons and anions are formed by gain of electrons.
In order to complete its octet and get stable, it gains 2 electrons and thus would form an anion.
Answer:
um.. is this a quiz or what girl...
Answer: temperature and salinity.
Explanation:
Answer:
sp3 hybridization
Explanation:
Hybridization means the mixing of atomic orbitals to yield hybrid orbitals with characteristics that are different from that of the isolated atomic orbitals before the combination.
sp3 hybridization occurs when one s orbital is mixed with three p orbitals to yield four sp3 hybrid orbitals which can be used to bond to a central atom.
The central atom is then located at the center of a regular tetrahedron at a bond angle of 109°.
Explanation:
Given -
- An organic compound gives H₂ gas with Na
- On treatment with alkaline iodine it gives yellow ppt.
- On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)
To Find -
- Name the compound and write the reaction involved
Now,
Let A be the organic compound.
Then,
- A + Na → + H₂↑
- A + I₂ → CHI₃ (yellow ppt.)
- A + CrO₃ + H⁺ → C₂H₄O
Now,
Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.
- Functional group of aldehyde = —CHO
And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).
Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.
It means,
We know that 1° alcohol on oxidation gives aldehyde.
Here it gives 2 Carbon aldehyde.
It means,
Here 2 Carbon and 1° alcohol is used.
Now,
Its cleared that Compound A is Ethanol.
Reaction Involved -
- CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
- CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
- CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO
