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.
.513mol x (102g/1mol)
Essentially, this is .513 x 102
Which equals: 52.326
But because you can only have 3 significant figures, your answer is:
52.3 grams
I hope this Helps!
The reaction will generally form more reactants than products.
Answer:
The answer to your question is: letter D.
Explanation:
Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.
Due to this characteristic, they are stable and do not react with other elements.
a. 1s22s22p4 The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.
b. [Ne]2s22p2 The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.
c. [Ar] 3s1 This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.
d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.
