Answer:
European Union was on grand committee
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.
Answer:
Uh first of all this is algebra but I'll answer this
First distribute the three and 5 (Multiply them by both terms inside parenthesis.
3x-6=5x+20
Then add like terms
8x=14
Divide 8 by 8 and 8 by 14
x = 14/8
Explanation:
The answer is Solid.
This is on account of the substances that develop a solid are packed in a settled, firmly pressed geometric plan.
Answer:
B = (2.953 × 10⁻⁹⁵) N.m⁹
Explanation:
At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.
That is,
Fa + Fr = 0
Fa = - Fr
Fa = (|q₁q₂|)/(4πε₀r²)
Fr = -B/(r^n) but n = 9
Fr = -B/r⁹
(|q₁q₂|)/(4πε₀r²) = (B/r⁹)
|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C
(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²
r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m
(k|q₁q₂|)/(r²) = (B/r⁹)
(k × |q₁q₂|) = (B/r⁷)
B = (k × |q₁q₂| × r⁷)
B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]
B = (2.953 × 10⁻⁹⁵) N.m⁹
