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makvit [3.9K]
2 years ago
8

determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m

etal, or a transition metal. Justify your choices.
Chemistry
1 answer:
mamaluj [8]2 years ago
4 0

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) 1s^22s^22p^63s^23p^5

(b) 1s^22s^22p^63s^23p^63d^74s^2

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(d) 1s^22s^22p^63s^23p^63d^4s^1

Answer :

(a) 1s^22s^22p^63s^23p^5   → Halogen

(b) 1s^22s^22p^63s^23p^63d^74s^2    → Transition metal

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6   → Transition metal

(d) 1s^22s^22p^63s^23p^63d^4s^1   → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: ns^2np^6 where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: ns^2np^5 where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: ns^1 where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: ns^2 where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: (n-1)d^{1-10}ns^{0-2} where n is the outermost shell.

(a) 1s^22s^22p^63s^23p^5

The element having this electronic configuration belongs to the halogen family.

(b) 1s^22s^22p^63s^23p^63d^74s^2

The element having this electronic configuration belongs to the transition family.

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

The element having this electronic configuration belongs to the transition family.

(d) 1s^22s^22p^63s^23p^63d^4s^1

The element having this electronic configuration belongs to the transition family.

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2 years ago
There are three factors in which natural selection occurs. Select one statement from the list below that is a true statement abo
azamat

Answer: There is no list in your question. But the three factors in which natural selection occurs are:

1. A struggle for existence,

2. Variation and

3. Inheritance

Explanation:

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The three major factors through which natural selection occurs are:

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8 0
3 years ago
The pKb of the base cyclohexamine, C6H11NH2, is 3.36. What is the pKa of the conjugate acid, C6H11NH3
bonufazy [111]

Answer:

10.64

Explanation:

Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.

C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻     pKb = 3.36

C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:

C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq)   pKa

We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.

pKa + pKb = 14.00

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6 0
3 years ago
Draw the alkene formed when 1-heptyne is treated with hbr in the presence of peroxide.
Nimfa-mama [501]
<h2>Heptene formed is -</h2><h2>CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr</h2>

Explanation:

The two possibilities when the peroxide is not present

  • CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-C≡CH +HBr → CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_{2}

  • CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_2 + HBr →CH_3-CH_2-CH_2-CH_2-CH_2-CBr_2-CH_3

In presence peroxide,

CH_3-CH_2-CH_2-CH_2-CH_2-C≡CH+ HBr →CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr

  • When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
  • This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
  • One mole of HBr adds to one mole of 1-heptane.
  • The structure of heptene formed is -

CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr

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Answer:

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Explanation:

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2 years ago
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