Protons : positive charge , about same mass as neutrons , in the nucleus
neutrons : no charge , about the same mass as a proton , in the nucleus
electrons : negative charge , less mass than protons and neutrons , in orbitals outside of the nucleus
I think answer is
C. Something that can be observed or measured while changing the identity of the substance
Explanation:
Formula to calculate hybridization is as follows.
Hybridization = ![\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
So, hybridization of 
is as follows.
Hybridization = ![\frac{1}{2}[V + N - C + A]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5BV%20%2B%20N%20-%20C%20%2B%20A%5D)
= ![\frac{1}{2}[2 + 2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B2%20%2B%202%5D)
= 2
Hybridization of is sp. Therefore, is a linear molecule. There will be only two electron groups through which Be is attached.
Similarly, hybridization of 
is calculated as follows.
Hybridization =
= ![\frac{1}{2}[8 + 2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B8%20%2B%202%5D)
= 5
Therefore, hybridization of is ![sp^{3}d. Therefore, [tex]XeF_{2}](https://tex.z-dn.net/?f=sp%5E%7B3%7Dd.%20Therefore%2C%20%5Btex%5DXeF_%7B2%7D)
is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.
Thus, we can conclude that out of the given options is the correct examples of linear molecules for five electron groups.
The appropriate response is the fourth one. The announcement is valid about this condition beneath is in spite of the fact that it is unequal, it can be adjusted by specifically utilizing observer particles.
I hope the answer will help you.
