Answer:
The different magnetic shielding effects of the carbonyl double bond in CH0, the CH₂ (a) and CH₂ (b) protons show different chemical shifts
Because the degree of shielding depends on electron density round the proton, the neighboring carbonyl group will increase this density in geranial form.
Now in CH₂ the electron density is increased around CH₂ (a) protons in neral form due to the carbonyl group, and thus it has a lower z values than the corresponding geranial form.
In the case of CH₃ (b) protons. thus they have lower z values and higher chemical shift than the corresponding nera form.
Explanation:
Solution
Due to the different magnetic shielding effects of the carbonyl double bond in CH0, the CH₂ (a) and CH₂ (b) protons show different chemical shifts.
Since the degree of shielding depends on electron density round the proton, the neighboring carbonyl group will increase this density in geranial form more in case of CH₃ (b) protons. thus they have lower z values and higher chemical shift than the equivalent nera form.
Similarly, the electron density is increased around CH₂ (a) protons in neral form due to the carbonyl group, and thus it has a lower z values than the equivalent geranial form.
Answer:
3
Explanation:
Number of Energy Levels: 3
First Energy Level: 2
Second Energy Level: 8
Third Energy Level: 7
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Answer:
Option "B" is correct.
Explanation:
According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In and , the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.