Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t
1 answer:
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Explanation:
<u>Ligands:</u> In co-ordination chemistry ligands are ion, molecule or any species which donates electron pair to central metal atom.
Depending on the type of interaction Ligands are of three types.
- Sigma donor only
- sigma as well as pi donor
- pi acceptor ligand
let's understand each type of Ligands individually & in more detail.
1 - Sigma donor only: This is a unidirectional interaction, in which filled ligand overlaps (head to head) with central metal atom/ion & donates pair of electron in the LUMO of metal.
generally all the molecules of 2nd period without pi bond comes in this category, below are few example of sigma donor ligands,
2- Pi donor: This in also a unidirectional interaction between ligand & central metal atom but the along with head to head overlap, side overlapping takes place.
generally protonated neutral molecules who have more than one pair to donate show such interaction, for e.g.
NH3 have two lone pair to donate but the energy level of both the lone pairs are different hence when it is neutral it only donates one pair of electron. but when NH3 is protonated to NH2- it have two electron pairs (negative charge+ lone pair) to donate & both the pairs have same energy level. example of such ligands are below,
3- Pi acceptor ligand: This is a bidirectional interaction between ligand & central metal atom/ion, the filled orbital of ligand undergoes head to head to overlap with vacant orbital of central metal atom, & filled D orbital of central metal donates their pair to vacant LUMO of ligand.
depending on the LUMO pi acceptor ligands are further classified into two categories.
The dπ-σ* is seen in molecules of 3rd period onwards without pi bond <em>for e.g.</em>
<em>PH3,</em><em> </em><em>PR</em><em>3</em><em>,</em><em> </em><em>AsR</em><em>3</em><em> </em><em>&</em><em> </em><em>SR</em><em>2</em><em> </em><em>etc</em>
The dπ-π* is seen in molecules of 2nd or3rd period with pi bond <em>for e.g.</em>
CO C N- SC N^- etc
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Answer: The value of for this reaction is 250000.
Explanation:
The given equation is as follows.
... (1)
... (2)
To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.
... (3)
Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.
This equation can also be re-written as follows.
This equation is similar to the equilibrium equation given to us.
Therefore, during this subtraction the equation constants get divided as follows.
Thus, we can conclude that the value of for this reaction is 250000.
Answer:
Helium is created from hydrogen in the sun's core.
Four hydrogen-1 nuclei fuse to produce
- one helium-4 nucleus, two neutrons,
- two positrons, and
- two electron neutrinos.
Explanation:
Step One:
.
Two hydrogen-1 nuclei fuse. One proton will convert to a neutron. The products will be
- one hydrogen-2 nucleus,
- one positron, and
- one electron neutrino.
Step Two:
.
There are plenty of hydrogen-1 nuclei available in the core of the sun. The hydrogen-2 nucleus from step one will fuse with a hydrogen-1 nucleus. The product is
- one helium-3 nucleus.
Step Three
.
Two helium-3 nuclei from step two react with each other. The products are:
- one helium-4 nucleus, and
- two hydrogen-1 nuclei.
The overall reaction will be:
.
In other words, hydrogen nuclei in the core of the sun fuse together to form helium.