There are two subshells that are s and p, which are present in the second energy level.
The energy level can be defined as the fixed distances from the nucleus of an atom where electrons may be found. Each energy level is divided into some Subshells. These subshells are known as s-subshell, p-subshell, d-subshell, and f-subshell. This subshell contains some orbitals, these orbitals are the place where there is the maximum probability of getting the electrons. In one orbital, a maximum of two electrons can be present.
Hence, there are two subshells in the second energy level.
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Answer:
The nitrile group
Explanation:
The nitrile group contains the C≡N bond. It should be recalled that triple bond is highly electronegative and withdraws electrons from the C-H bond more effectively than the halogen atom.
The higher effectiveness of the C≡N bond at electron withdrawal greatly reduces the electron density of the C-H bond thereby making the hydrogen atom of the bond highly labile
Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,
When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation, 
= 102 pm
Radius of cation, 
= 196 pm
As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
