<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 complete and <span>balanced equation for this single-displacement reaction would be written as follows:
</span><span>Li+NaOH--> LiOH + Na
In this reaction, lithium replace sodium in the compound sodium hydroxide forming lithium hydroxide and sodium metal as products. Hope this answers the question.</span>
Ionic bonding would be the answer because they transfer electrons. This gives them a charge. If it loses electrons, it becomes an cation, with a positive charge. While if they gain an electron, they get a negative charge, and become a anion. Transferring an electron is losing or gaining, therefore your answer would be that since electrons are permanently being transferred, the answer is IONIC BOND.
Covalent bonds is a wrong answer because they share electrons, which gives them no charge (neutral).
Also, metallic bonding is not the correct answer.
So our final answer: A- Ionic bond
