<span>There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.
</span><span>Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons; density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.</span>
The element that will have the lowest electronegativity is an element with a small number of valence electrons and a large atomic radius.
Electronegativity of an element is the ability or power of that element in a molecule to attract electrons to its Valence electrons. The following are the properties of electronegativity:
- It increases across a period from left to right of the periodic table,
- It decreases down the periodic table groups
- Group 1 elements are the least (lowest) electronegative elements. These elements have the lowest valence electrons with a large atomic radius.
- Group 7 elements are the most electronegative elements.
Atomic radius of elements increase down a group because of a progressive increase in the number of shells occupied by electrons which increases the size. But it decreases across a period because electrons are accommodated within the same shell leading to greater attraction by the protons in the nucleus.
Learn more about electronegativity of elements here:
brainly.com/question/20348681
Answer:
1. The electronic configuration of X is: 1s2 2s2 sp6 3s2
2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
3. The formula of the compound form by X and Y is given as: XY
Explanation:
For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:
1s2 2s2 sp6 3s2
To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:
Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below
1s2 2s2 2p4
The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
The formula of the compound form by X and Y is given below :
X^2+ + Y^2- —> XY
Their valency will cancel out thus forming XY
Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.
Solution : Given,
For Accelerator 1 model,
Input energy = 2078.3 J
Wasted energy = 663.1 J
Output energy = 1415.2 J
For Accelerator 2 model,
Input energy = 7690.0 J
Wasted energy = 2337.5 J
Output energy = 5353.5 J
For Accelerator 3 model,
Input energy = 4061.9 J
Wasted energy = 2259.6 J
Output energy = 1802.3 J
Formula used for lowest percentage of energy lost as waste is:
% energy lost as waste = (Total energy wasted / Total input energy ) × 100
For Accelerator 1 model,
% energy lost as waste = 
= 31.90%
For Accelerator 2 model,
% energy lost as waste = 
= 30.39%
For Accelerator 3 model,
% energy lost as waste = 
= 55.62%
So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.
