Write the conjugate acids for the following Bronsted bases

[https://pubchem.ncbi.nlm.nih.gov/periodic-table/png/Periodic_Table_of_Elements_w_Chemical_Group_Block_PubChem.png ]
As with literally everything else, the most appropriate way to think about chemical bonding depends heavily on the context. Generally speaking, chemical bonds are when atoms stick together and require a significant energy input to be broken back apart. Lower energy states tend to be more stable, while higher energy states tend to be less stable. Energy is delivered to molecules mostly as heat, which means molecules colliding with each other and exchanging velocity and hence kinetic energy. Photon absorption is another possibility, but the mechanics behind it are more complicated and can only get in the way at the moment. So, chemical bonds can exist at temperatures between 0 kelvin (absolute zero, no atomic movement at all) and the conditions under which all electrons completely dissociate from the nuclei to form plasma. The strength of the bonds in question and the conditions in which they are located will determine the specifics, but obviously some chemical bonds are much more resistant to high temperature than others. Towards one end, you have compounds like hydrogen peroxide that are fully capable of spontaneously dissociating at room temperature and pressure. In about the middle you have substances like wood that will break bonds and combust under ambient conditions if supplied with an ignition source, and towards the other end you have things like concrete or rock that don’t usually burn very well. However and while combustion is a convenient, easily accessible reaction, I should note that many other reactions also exist, most of which are more complicated than applying heat in an oxygen atmosphere.
Before we get there, I should repeat that chemical bonds “glue” positively charged nuclei together with negatively charged electron density. To have a chemical bond, the valence electrons and orbitals of the bonding atoms need to combine. At one extreme, an electron can be completely transferred from one atom to another, resulting in an ionic compound. The most popular example of an ionic compound is table salt, sodium chloride/NaCl/Na+ Cl-. As can be seen, the highly electronegative chlorine atom is able to completely remove one of the valence electrons on the sodium atom, which despite the resulting charges puts both atoms into a noble gas electron configuration and is hence energetically favorable. At the other extreme, we have bonding electron density being split completely equally between the two atoms. This only occurs when two of the same atom are bonded together (H2, O2, N2, some carbon-carbon bonds, etc) and makes intuitive sense because you would not expect either of two identical atoms in identical chemical environments to be more electronegative than the other. In between these two extremes is a spectrum of bond polarization, with electron density skewed to some extent or another to the more electronegative of the two atoms. Please note that the electronegativity values I linked in the previous post do not take into account any other bonds that influence electron distribution and hence the chemical environments around the atoms in the bond of interest, so that table should be used cautiously.
From a bond strength perspective, maximizing the electron density between the two bonding atoms also maximizes the strength and minimizes the length of the bonds. To put it another way, increased electron density shields the positive charges on the nuclei from each other, allowing the nuclei to be closer together. Consequently, ionic bonds are very weak in the sense that the cations and anions can be easily pulled apart, and covalent bonds that distribute electron density evenly between two atoms are much more difficult to pull apart. For the next part, I will neglect the behavior of ionic compounds (also acids and bases, which behave similarly) to focus on covalent bonding. I am also going to neglect the polarization of covalent bonds towards more electronegative atoms because the distribution of the electron probability density inside the molecular bonding orbitals does not affect our understanding of how these orbitals form. With covalent bonds, there are two main bond types that are helpful to think about. In reality, what actually happens is that the atoms and their atomic orbitals combine to whichever state is accessible and lowest in energy, but the process of generating a set of molecular orbitals for each individual molecule is very labor intensive and does not add much to our understanding.
So, to start out with let’s examine the main organic elements: carbon, oxygen, nitrogen, and hydrogen. Hydrogen is easy to deal with because it bonds with the 1s orbital only, and the 1s orbital is a sphere. The remaining three elements have both a spherical 2s orbital and three 2p orbitals that can participate in bonding, which makes things more complicated. In terms of shape, each p-orbital can be thought of as existing in a 3D cartesian coordinate system with the nucleus at the origin. Each orbital then has two lobes parallel to the x, y, or z axes, with a nodal plane (no electron density) oriented in the other two axes. As an example, the p_x orbital will have two lobes parallel to the x axis and no electron density on the yz plane. In practice, the result will look much more like a sphere cut in half than the balloon-shaped lobes usually depicted, but that’s not all that important. I should also mention that opposite lobes have opposite polarizations, and that a + polarized p-orbital lobe on one atom does not have bonding overlap with – polarized p-orbital lobes on other atoms but will have bonding overlap with + polarized p-orbital lobes on other atoms. This becomes important later on when we get into conjugated systems and can explain some oddball bonding behavior much later on.
Anyway, I still haven’t introduced sigma and pi bonding, so let’s do that. Sigma bonds have the bonding orbitals located directly between the bonding atoms are as a result yield the strongest bonds. Pi bonds depend on the bonding overlap of p-orbitals above and below and/or to either side of the axis directly between the two atoms where a sigma bond would form. Pi bonds still put electron density in between the two nuclei and are still bonds, but cannot be as strong as a sigma bond. Since hydrogen has no valence p-orbitals, it cannot participate in pi bonding schemes, but carbon, nitrogen, and oxygen are all fully capable of donating one or two p-orbitals to pi bonds. If three or more atoms in a row all have p-orbitals in the same plane, the potential exists for all of those p-orbitals to combine into one conjugated pi system, which usually offers energy advantages compared to isolated pi bonds. There is quite a bit more complexity along these lines, but this is mostly dealt with in organic chemistry.
Moving back to sigma bonds, I should first note that the number of bonds that an atom can form in most circumstances is equal to the number of unoccupied electron spaces in its valence shell. So, hydrogen can only form one bond before filling the 1s orbital, boron in theory should form five bonds but in practice is only capable of three with a completely empty p-orbital before running out of bonding volume around the small atom, carbon can form four bonds, nitrogen can form three bonds, oxygen can form two bonds, and fluorine can form one bond. During bonding, an atom will usually be thought of as “owning” a number of electrons equal to the number of its valence electrons (hydrogen one, boron three, carbon four, nitrogen five, oxygen six, fluorine seven). Due to orbital overlap, the electrons in the bonds that are in theory “owned” by the other atoms are also thought of as filling out the valence shell of the bonded atom, and in this manner the atoms in organic compounds can achieve electron configurations close to or equaling noble gas configurations despite all of the atoms in the molecule having fewer than the eight valence electrons required to actually be a noble gas or halogen anion. To put it another way, in the absence of bonding all of the atoms in organic chemistry are severely electron-deficient from a valence shell point of view, with bonding the valence shells can (mostly) all be filled without stacking a bunch of extra electrons (that don’t exist in the big picture – the number of protons and electrons is about equal) onto all of the atoms. This would also generate screamingly unstable accumulations of negative charge, so from an energy perspective bonding works out very well for most or all of the atoms involved.
At this point, I have not said anything about how bonds are actually arranged in space around an atom with both s and p valence orbitals. In the 2 shell where most of organic chemistry happens, the 2s and 2p orbitals all occupy roughly the same volume, which brings us to orbital hybridization and lone pairs. With four valence orbitals, we expect to have four bonding/molecular orbitals, each located in a distinct volume. Having a spherical 2s orbital and three p-orbitals at right angles arranged around the same nucleus is not compatible with this, and is not how atoms participate in bonding. Instead, three bonding arrangements are possible: tetrahedral (sp^3), trigonal planar (sp^2), and linear (sp). In the sp^3 case, the 2s and all of the 2p orbitals combine to form four new orbitals, each with one part s-orbital character and three parts p-orbital character. The hybridized orbitals form bonds as far apart as physically possible, resulting in a uniform bond angle of 109.5 degrees and the tetrahedral configuration. Methane (CH4) is an example of a tetrahedral compound – the sp^3 orbitals on the carbon atom bond with the 1s orbitals on the hydrogens, resulting in a perfectly symmetrical arrangement of bonds around the carbon atom. Ammonia (NH3) is also an example of a tetrahedral compound, although you might not expect that on first inspection. More properly, I should write ammonia as :NH3, which is because a nitrogen atom “owns” five valence electrons and can form three bonds. In this case, the open electron spaces are filled by the three bonding hydrogen atoms, with three nitrogen electrons participating in these bonds. The remaining two valence electrons from the nitrogen occupy the nitrogen sp^3 not already bonding with a hydrogen atom, with the absence of another positively charged nucleus meaning that the lone pair will tend to repel the electron density in the N-H bonds, pushing the hydrogens closer together on one end of the molecule and distorting the ideal 109.5 degree bond angles.
In a trigonal planar sp^2 bond scheme, one of the p-orbitals does not participate in hybridization and is free to participate in pi bonds with other atoms. The loss of one p-orbital means that there are only three hybrid orbitals, each with one part s character and two parts p character. The higher fraction of s character means that sp^2 orbitals will be lower in energy than sp^3 orbitals, although whether or not the sp^2 bond scheme is energetically favorable overall also depends on the chemical environment of the remaining non-hybridized p-orbital. Geometrically, the remaining p-orbital will tend to occupy all of the space apart from the nodal plane, pushing the other three bonds into the nodal plane at about 120 degree angles from each other. A linear sp bond scheme is quite similar to the sp^2 bond scheme, but with two p-orbitals not participating in hybridization. The s-orbital and remaining p-orbital generate two hybrid orbitals with one part s character and one part p character, so sp orbitals are the lowest in energy of any of the hybrid orbitals. With two p-orbitals at right angles taking up much of the available volume, the other bonds will default to the volume along the intersection of the nodal planes of both of the p-orbitals. Since the intersection of two planes is a line, linear bonds will tend to be 180 degrees apart.
Once we start getting into larger third row elements or the d and f blocks, things become much more chaotic and complicated. With the organic bonding mostly described here sufficient to form the basis of all biological processes, you can probably imagine the idiosyncrasies exhibited by the heavier atoms, particularly if you view the d and f orbitals as depicted here (https://i.stack.imgur.com/K5EcA.jpg ). If you are wondering why heavy metal poisoning can be so damaging to human bodies, this is much of the reason why.