Why is the first layer of many atomic electron clouds 2 Why does the minimum energy have to be 2, say 1

Because the nearest k-layer is a s-sublayer spherical orbit, it can only hold two electrons with opposite spin direction at most, which is calculated according to the principle of minimum energy and nainst equation

Why does the electron cloud tend to be more electronegative when two atoms are bonded
I need an essential explanation. Atoms with high electronegativity tend to be at the top right of the periodic table. From the same period point of view, can high electronegativity attract electron clouds because of the large number of nuclear charges, that is, the large number of positive charges? In other words, electronegativity is the average of electron affinity energy and ionization energy. Can we explain the electron cloud shift from the perspective of affinity energy?
First floor: your answer can be used to explain the size relationship of atomic radius, but I'm talking about molecules. According to the similar theory of molecular orbital, electrons are already molecular. It seems that it doesn't make sense to use the radius size of two atoms to see the deviation degree of electron cloud.
Second floor: you really have a point. Electronegativity is a measure of an atom's ability to attract bonding electrons.

Here you have to look at the definition of electronegativity. Electronegativity is the ability of atoms to attract bonding electron pairs. In molecules, bonding electron pairs tend to the side of atoms with strong electronegativity

How to calculate negative minus negative - 3 - (- 5) =?

-3-（-5）=-3+5＝2

In the motion experiment of a horizontal projectile, if three points a, B and C with horizontal distance of △ s = 0.2m are selected, and the vertical distances between these three points are S1 = 0.1M and S2 = 0.2m respectively, the initial velocity is 0.2m The height of point a from the throw point is______ (g = 10m / S2)

First of all, to determine the idea: given that the horizontal displacement is equal, then the initial velocity only needs to find out the time to complete the horizontal displacement - of course, the vertical displacement is used! The vertical displacement is completed in the same time, so we can use the special law of uniform velocity linear motion △ x = a * T ^ 2 to solve the problem

Given that the arithmetic sequence {an}, Sn represents the sum of the first several terms, A3 + A9 > 0, S9 < 0, then S1, S2 The smallest of Sn is S5
Why?

a3+a9=2a6>0
So A6 > 0
S9=9a5

At the intersection, the car starts from the stop line at an acceleration of 0.5 m / S2 to make a uniform acceleration motion, and there happens to be a bicycle passing the stop line at a speed of 5 m / s in the same direction as the car? What's the longest distance? ② Where does the car catch up with the bike? What was the speed of the car when it caught up?

① It is known from the meaning that the distance between two cars is the farthest when the speed is equal. Let t be the time taken for the car to make the uniform acceleration linear motion with the initial speed of 0, so v = at = V from & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; given a = 0.5m/s, V from = 5m / s, t = 10s can be obtained; The longest distance is x = x from - x steam = V from t-12at2 = 25m. ② when a car overtakes a bicycle, their displacement relative to the parking line is equal. Suppose that the time taken for the car to overtake the bicycle is t ′, then x from = x steam & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; that is: V from t ′ = 12a & nbsp; t / 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Substituting a = 0.5m/s 2V from = 5m / s, we can get the distance from the stop line at this time & nbsp; t '= 20s, the distance from the stop line at this time & nbsp; & nbsp; X = V from t' = 100m, the vehicle speed at this time & nbsp; & nbsp; & nbsp; V automobile = A & nbsp; t '= 10m / s answer: (1) when the vehicle moves for 10s, the farthest distance is 25m; (2) when the vehicle overtakes the bicycle at 100m from the stop line, the vehicle speed is 10m / s

Xiaogang and Xiaolei go to the bookstore along the same road from the school. After Xiaogang walks 400 meters, Xiaolei starts to chase Xiaogang. It is known that Xiaolei's speed is am / s faster than Xiaogang
(1) Try to express how much time Xiaolei needs to catch up with Xiaogang by using algebraic formula
(2) When a = 0.8, find the time for Xiaolei to catch up with Xiaogang

（1）400÷a s
（2）400÷0.8=500s

How to write questions 8, 9, 10 and 12 on 37 sides of math 8

Let's try several ways to refuel by ourselves

If the numbers corresponding to two points a and B on the number axis are X1 and X2, then the line AB =
At 10:00 on October 6

lx1-x2l*(1+k^2)^(1/2)

As shown in the figure, to measure the distance between two points a and B on the opposite bank of the river, select two points c and D 40 meters apart along the river bank, and the measured distance is ()
A. 202B. 203C. 402D. 206

Then BC is the diameter of the circle. Then cab = 90 degree. Then BCD = 45 degree, CDB = 90 degree, BC = cos45 degree, CD = 402, ACB = 60 degree, cab = 90 degree, AC = sin30 degree, BC = 202, ab = tan60 degree, AC = 206, so D is selected

Why is the first layer of many atomic electron clouds 2 Why does the minimum energy have to be 2, say 1