Chemistry, 02.09.2019 17:00, monkemily1

Astraight-chain hydrocarbon that has only one double bond in each molecule has the general formula
(1) cnh2n-6 (3) cnh2n
(2) cnh2n-2 (4) cnh2n+2

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answered: bfell92

- 3) C_nH_2_n

Explanations:- C_nH_2_n_-_6 is not a valid formula for an alkane, alkene or alkyne.

C_nH_2_n_-_2 is the general formula for an alkyne that has a triple bond between carbon atoms. The example for an alkyne is acetylene and it's formula is C_2H_2 .

In these general formulas, n is an integer number and for alkene and alkyne the minimum value of n is 2.

Acetylene has two carbons so the value of n for it is 2. If we plug in the value of n in the formula C_nH_2_n_-_2 then:

C_2H_4_-_2

= C_2H_2

In acetylene, there is a triple bond between the carbon atoms and each hydrogen is bonded to each carbon.

If we plug in n = 2 in the formula, C_nH_2_n then:

C_2H_4

It could be written as:

H_2C=CH_2

There is a has a double bond between the carbon atoms and two H are bonded to each carbon atom so, C_nH_2_n is the right formula of an alkene.

If we plug in n = 2 in the fourth formula then:

C_2H_4_+_2

= C_2H_6

It could be written as H_3C-CH_3 , it has a single bond between carbons and so it is an alkane.

answered: Guest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.