Two alkaline earth metals; Be and Mg exhibit a diagonal relationship. The pair of elements are Be-Al and Mg-Li is Be-Al and Mg-Li.
Six elements make up the majority of alkaline earth metals (Group 2): beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). Compared to the elements in Group 1, this group typically contains smooth, silver metals with a less metallic character. Heavy metals like Ca, Sr, Ba, and Ra are equally as reactive as Alkali Metals Group 1 despite sharing many characteristics with that group. All elements found in alkaline earth metals contain two electrons in their valence shells, giving them an oxidation state of +2. This makes it possible for metals to shed electrons with ease, allowing them to create compounds with more stable ionic connections.
Two alkaline earth metals; Be and Mg exhibit a diagonal relationship. The pair of elements are Be-Al and Mg-Li.
To learn more about alkaline earth metals refer the link:
brainly.com/question/12835232
#SPJ4
valence electron is an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
The correct answer is the first option. The piece of evidence would suggest an unknown substance is an ionic compound is that <span>it conducts electricity when dissolved in water.</span><span> A</span><span>n ionic compound dissociates
into ions when in aqueous solution. These ions move freely in the solution and
can allow the flow of electricity into the solution.</span>
Answer 1:
For compound A:
<span>2.8 g of nitrogen for each 1.6 g of oxygen
</span>Atomic weight of N = 14
Atomic weight of O = 16
Thus, number of moles of N = 2.8/14 = 0.2
and number of moles of O = 1.6/16 = 0.1
Thus, molar ratio of N and O in compound is 2:1.
Therefore, <span>lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen is 2:1. And the compound formed is N2O.
.........................................................................................................................
</span>Answer 2:
For compound B:
5.6 g of nitrogen for each 9.6 g of oxygen
Atomic weight of N = 14
Atomic weight of O = 16
Thus, number of moles of N = 5.6/14 = 0.4
and number of moles of O = 9.6/16 = 0.6
Thus, molar ratio of N and O in compound is 4:6 = 2:3.
Therefore, lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen is 2:3. And the compound formed is N2O3.
