Its phosphorus (P)In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
Answer: Step 1, Isomerase.
Explanation:
Form the version of palmitic acid in the step one by changing the double bond within alpha and beta carbon by Isomerase.
B and C are Isomers, the molecule only differ in configuration.
While the number nuclear protons as given is 34, and therefore we deal with the element selenium, there are 2 more electrons than protons, and therefore this species has an overall
2
−
charge.
We represent this selenide ion as
S
e
2
−
. Do I win 5 pounds?
Z= 34, therefore the atom is selenium
Metallic ion because the metallic ions will enter an exited state and release photons energy, in the form of light, as they return to their ground state.
Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.
Explanation:
Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.
--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.
--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.
Note: The Markovnikov rule in organic chemistry states that in alkene addition reactions, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.
